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Chabay, Matter and Interactions, 3e Chapter 1. Interactions and Motion Reading content Interactions and Motion 1.1. Kinds of Matter 1.2. Detecting Interactions 1.3. Newton's First Law of Motion 1.4. Other Indicators of Interaction 1.5. Describing the 3D World: Vectors 1.6. SI Units 1.7. Velocity 1.8. Momentum 1.9. Change of Momentum 1.10. *The Principle of Relativity 1.11. *Updating Position at High Speed Summary Exercises and Problems

Chapter 2. The Momentum Principle Reading content The Momentum Principle 2.1. System and Surroundings 2.2. The Momentum Principle 2.3. Applying the Momentum Principle 2.4. Momentum Change with Changing Force 2.5. Iterative Prediction of Motion 2.6. Special Case: Constant Force 2.7. Estimating Interaction Times 2.8. Physical Models 2.9. *Derivation: Special-Case Average Velocity 2.10. *Inertial Frames 2.11. *Measurements and Units Summary Exercises and Problems

Chapter 3. The Fundamental Interactions Reading content The Fundamental Interactions 3.1. The Fundamental Interactions 3.2. The Gravitational Force LibraryPirate

3.3. Approximate Gravitational Force Near the Earth's Surface 3.4. Reciprocity 3.5. Predicting Motion of Gravitationally Interacting Objects 3.6. The Electric Force 3.7. The Strong Interaction 3.8. Newton and Einstein 3.9. Predicting the Future of Complex Systems 3.10. Determinism 3.11. Conservation of Momentum 3.12. The Multiparticle Momentum Principle 3.13. Collisions: Negligible External Forces 3.14. Points and Spheres 3.15. Measuring the Gravitational Constant G Summary Exercises and Problems

Chapter 4. Contact Interactions Reading content Contact Interactions 4.1. Tarzan and the Vine 4.2. A Model of A Solid: Balls Connected by Springs 4.3. Tension Forces 4.4. Length of an Interatomic Bond 4.5. The Stiffness of An Interatomic Bond 4.6. Stress, Strain, and Young's Modulus 4.7. Compression (Normal) Forces 4.8. Friction 4.9. Speed of Sound in a Solid and Interatomic Bond Stiffness 4.10. Derivative form of the Momentum Principle 4.11. Analytical Solution: Spring-Mass System 4.12. Analytical Expression for Speed of Sound 4.13. Contact Forces Due to Gases 4.14. *A Vertical Spring-Mass System 4.15. *General Solution for the Mass-Spring System Summary Exercises and Problems

Chapter 5. Rate of Change of Momentum Reading content Rate of Change of Momentum 5.1. Identifying Forces on a System 5.2. Momentum not Changing (Statics) 5.3. Finding The Rate of Change of Momentum 5.4. Curving Motion

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5.5. Rate of Change of Direction 5.6. Why Does the Vine Break? 5.7. Problem Solving Summary Exercises and Problems

Chapter 6. The Energy Principle Reading content The Energy Principle 6.1. The Energy Principle 6.2. The Simplest System: A Single Particle 6.3. Work: Mechanical Energy Transfer 6.4. Update form of the Energy Principle 6.5. Change of Rest Energy 6.6. Proof of the Energy Principl for a Particle 6.7. Work Done by a Nonconstant Force 6.8. Potential Energy in Multiparticle Systems 6.9. Gravitational Potential Energy 6.10. General Properties of Potential Energy 6.11. Plotting Energy vs. Separation 6.12. Applying Gravitational Potential Energy 6.13. Gravitational Potential Energy Near the Earth's Surface 6.14. Electric Potential Energy 6.15. The Mass of a Multiparticle System 6.16. Reflection: Why Energy? 6.17. Identifying Initial and Final States 6.18. *A Puzzle 6.19. *Gradient of Potential Energy 6.20. *Integrals and Antiderivatives 6.21. *Approximation for Kinetic Energy 6.22. *Finding the Formula for Particle Energy Summary Exercises and Problems

Chapter 7. Internal Energy Reading content Internal Energy 7.1. Potential Energy of Macroscopic Springs 7.2. Potential Energy of a Pair of Neutral Atoms 7.3. Path Independence of Potential Energy 7.4. Internal Energy and Thermal Energy 7.5. Energy Transfer due to a Temperature Difference 7.6. Reflection: Forms of Energy 7.7. Power: Energy Per Unit Time

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7.8. Open and Closed Systems 7.9. The Choice of System Affects Energy Accounting 7.10. Energy Dissipation 7.11. Potential Energy and "Conservative" Forces 7.12. * Resonance Summary Exercises and Problems

Chapter 8. Energy Quantization Reading content Energy Quantization 8.1. Photons 8.2. Electronic Energy Levels 8.3. The Effect of Temperature 8.4. Vibrational Energy Levels 8.5. Rotational Energy Levels 8.6. Other Energy Levels 8.7. Comparison of Energy Level Spacings 8.8. *Case Study: How a Laser Works 8.9. *Wavelength of Light Summary Exercises and Problems

Chapter 9. Multiparticle Systems Reading content Multiparticle Systems 9.1. The Motion of the Center of Mass 9.2. Separation of Multiparticle System Energy 9.3. Rotational Kinetic Energy 9.4. The "Point Particle System" 9.5. Analyzing Point Particle and Real Systems 9.6. *Modeling Friction in Detail 9.7. *A Physical Model for Dry Friction 9.8. *Derivation: Kinetic Energy of a Multiparticle System 9.9. *Derivation: The Point Particle Energy Equation Summary Exercises and Problems

Chapter 10. Collisions Reading content Collisions

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10.1. Internal Interactions in Collisions 10.2. Elastic and Inelastic Collisions 10.3. A Head-On Collision of Equal Masses 10.4. Head-On Collisions Between Unequal Masses 10.5. Frame of Reference 10.6. Scattering: Collisions in 2-D and 3-D 10.7. Discovering the Nucleus Inside Atoms 10.8. *Distribution of Scattering Angles 10.9. Relativistic Momentum and Energy 10.10. Inelastic Collisions and Quantized Energy 10.11. Collisions in Other Reference Frames Summary Exercises and Problems

Chapter 11. Angular Momentum Reading content Angular Momentum 11.1. Translational Angular Momentum 11.2. Rotational Angular Momentum 11.3. Translational Plus Rotational Angular Momentum 11.4. The Angular Momentum Principle 11.5. Multiparticle Systems 11.6. Three Fundamental Principles of Mechanics 11.7. Systems with Zero Torque 11.8. Systems with Nonzero Torques 11.9. Predicting Positions When There is Rotation 11.10. Angular Momentum Quantization 11.11. *Gyroscopes 11.12. *More Complex Rotational Situations 11.13. *Rate of Change of A Rotating Vector Summary Exercises and Problems

Chapter 12. Entropy: Limits on the Possible Reading content Entropy: Limits on the Possible 12.1. Statistical Issues 12.2. A Statistical Model of Solids 12.3. Thermal Equilibrium of Blocks in Contact 12.4. The Second Law of Thermodynamics 12.5. What is Temperature? 12.6. Specific Heat Capacity of a Solid 12.7. The Boltzmann Distribution 12.8. The Boltzmann Distribution in a Gas

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Summary Exercises and Problems

Chapter 13. Gases and Engines Reading content Gases and Engines 13.1. Gases, Solids, and Liquids 13.2. Gas Leaks through a Hole 13.3. Mean Free Path 13.4. Pressure and Temperature 13.5. Energy Transfers 13.6. Fundamental Limitations on Efficiency 13.7. A Maximally Efficient Process 13.8. *Why Don't We Attain the Theoretical Efficiency? 13.9. *Application: A Random Walk 13.10. *Derivation: Maximum-Power Efficiency Summary Exercises and Problems

Chapter 14. Electric Field Reading content Electric Field 14.1. New Concepts 14.2. Electric Charge and Force 14.3. The Concept of "Electric Field" 14.4. The Electric Field of a Point Charge 14.5. Superposition of Electric Fields 14.6. The Electric Field of a Dipole 14.7. Choice of System 14.8. Is Electric Field Real? Summary Exercises and Problems

Chapter 15. Electric Fields and Matter Reading content Electric Fields and Matter 15.1. Charged Particles in Matter 15.2. How Insulators become Charged 15.3. Polarization 15.4. Polarization of Insulators 15.5. Polarization of Conductors

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15.6. A Model of a Metal 15.7. Charging and Discharging 15.8. When the Field Concept is less Useful Summary Basic Experiments Additional Experiments Exercises and Problems

Chapter 16. Electric Field of Distributed Charges Reading content Electric Field of Distributed Charges 16.1. Overview 16.2. A Uniformly Charged Thin Rod 16.3. Procedure for Calculating Electric Field 16.4. A Uniformly Charged Thin Ring 16.5. A Uniformly Charged Disk 16.6. Two Uniformly Charged Disks: A Capacitor 16.7. A Spherical Shell of Charge 16.8. A Solid Sphere Charged Throughout Its Volume 16.9. Infinitesimals and Integrals in Science 16.10. *Uniform Thin Rod at an Arbitrary Location 16.11. *Integrating The Spherical Shell Summary Exercises and Problems

Chapter 17. Electric Potential Reading content Electric Potential A Review of Potential Energy 17.2. Systems of Charged Objects 17.3. Potential Difference in a Uniform Field 17.4. Sign of Potential Difference 17.5. Potential Difference in a Nonuniform Field 17.6. Path Independence 17.7. The Potential at one Location 17.8. Potential Difference in an Insulator 17.9. Energy Density and Electric Field 17.10. *Potential of Distributed Charges 17.11. *Integrating the Spherical Shell Summary Exercises and Problems

Chapter 18. Magnetic Field

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Reading content Magnetic Field 18.1. Electron Current 18.2. Detecting Magnetic Fields 18.3. Biot-Savart Law: Single Moving Charge 18.4. Relativistic Effects 18.5. Electron Current and Conventional Current 18.6. The Biot-Savart Law for Currents 18.7. The Magnetic Field of Current Distributions 18.8. A Circular Loop of Wire 18.9. Magnetic Dipole Moment 18.10. The Magnetic Field of a Bar Magnet 18.11. The Atomic Structure of Magnets 18.12. *Estimate of Orbital Angular Momentum of an Electron in an Atom 18.13. *Magnetic Field of a Solenoid Summary Experiments Exercises and Problems

Chapter 19. Electric Field and Circuits Reading content Electric Field and Circuits 19.1. Overview 19.2. Current in Different Parts of a Circuit 19.3. Electric Field and Current 19.4. What Charges make the Electric Field in the Wires? 19.5. Connecting a Circuit: The Initial Transient 19.6. Feedback 19.7. Surface Charge and Resistors 19.8. Energy in a Circuit 19.9. Applications of The Theory 19.10. Detecting Surface Charge Summary Experiments Exercises and Problems

Chapter 20. Circuit Elements Reading content Circuit Elements 20.1. Capacitors 20.2. Resistors 20.3. Work and Power in a Circuit 20.4. Batteries LibraryPirate

20.5. Ammeters, Voltmeters, and Ohmmeters 20.6. Quantitative Analysis of an RC Circuit 20.7. Reflection: The Macro-Micro Connection 20.8. *What are AC and DC? 20.9. *Electrons in Metals 20.10. *A Complicated Resistive Circuit Summary Experiments Exercises and Problems

Chapter 21. Magnetic Force Reading content Magnetic Force 21.1. Magnetic Force on a Moving Charge 21.2. Magnetic Force on a Current-Carrying Wire 21.3. Combining Electric and Magnetic Forces 21.4. The Hall Effect 21.5. Motional Emf 21.6. Magnetic Force in a Moving Reference Frame 21.7. Magnetic Torque 21.8. Potential Energy for a Magnetic Dipole 21.9. Motors and Generators 21.10. *Case Study: Sparks in Air 21.11. *Relativistic Field Transformations Summary Experiments Exercises and Problems

Chapter 22. Patterns of Field in Space Reading content Patterns of Field in Space 22.1. Patterns of Electric Field: Gauss's Law 22.2. Definition of "Electric Flux" 22.3. Gauss's Law 22.4. Reasoning from Gauss's Law 22.5. Gauss's Law for Magnetism 22.6. Patterns of Magnetic Field: Ampere's Law 22.7. Maxwell's Equations 22.8. *The Differential Form of Gauss's Law 22.9. *The Differential Form of Ampere's Law 22.10. *Semiconductor Devices Summary Exercises and Problems

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Chapter 23. Faraday's Law Reading content Faraday's Law 23.1. Curly Electric Fields 23.2. Faraday's Law 23.3. Faraday's Law and Motional emf 23.4. Maxwell's Equations 23.5. Superconductors 23.6. Inductance 23.7. *Some Peculiar Circuits 23.8. *The Differential Form of Faraday's Law 23.9. *Lenz's Rule Summary Exercises and Problems

Chapter 24. Electromagnetic Radiation Reading content Electromagnetic Radiation 24.1. Maxwell's Equations 24.2. Fields Traveling Through Space 24.3. Accelerated Charges Produce Radiation 24.4. Sinusoidal Electromagnetic Radiation 24.5. Energy and Momentum in Radiation 24.6. Effects of Radiation on Matter 24.7. Light Propagation Through A Medium 24.8. Refraction: Bending of Light 24.9. Lenses 24.10. Image Formation 24.11. *The Field of an Accelerated Charge 24.12. *Differential Form of Maxwell's Equations Summary Exercises and Problems

Chapter 25. Waves and Particles Reading content Waves and Particles 25.1. Wave Phenomena 25.2. Multisource Interference: Diffraction 25.3. The Wave Model vs. the Particle Model of Light 25.4. Further Applications of the Wave Model 25.5. Angular Resolution 25.6. Standing Waves LibraryPirate

25.7. *Derivation: Two Slits are Like two Sources Summary Exercises and Problems

Appendices Front matter Matter & Interactions Preface

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Interactions and Motion

This textbook deals with the nature of matter and its interactions. The variety of phenomena that we will be able to explain and understand is very wide, including the orbit of stars around a black hole, nuclear fusion, and the speed of sound in a solid. The main goal of this textbook is to have you engage in a process central to science: the attempt to explain in detail a broad range of phenomena using a small set of powerful fundamental principles. The specific focus is on learning how to model the nature of matter and its interactions in terms of a small set of physical laws that govern all mechanical interactions, and in terms of the atomic structure of matter.

Reading a science textbook efficiently and productively requires “active” reading. To encourage active reading, you will find “Stop and Think” questions, indicated by . Trying to answer a question by using what you already know, as well as material you have just read, helps you learn more than if you just read passively. You might cover the rest of the page while you're thinking.

KEY IDEAS Chapter 1 introduces the notion of interactions between material objects and the changes they produce. Fundamental physics principles apply to all kinds of matter, from galaxies to subatomic particles. Change is an indication of an interaction. Position and motion in 3D space can be described precisely by vectors. The momentum of an object depends on both mass and velocity.

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Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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KINDS OF MATTER We will deal with material objects of many sizes, from subatomic particles to galaxies. All of these objects have certain things in common.

Atoms and Nuclei Ordinary matter is made up of tiny atoms. An atom isn't the smallest type of matter, for it is composed of even smaller objects (electrons, protons, and neutrons), but many of the ordinary everyday properties of ordinary matter can be understood in terms of atomic properties and interactions. As you probably know from studying chemistry, atoms have a very small, very dense core, called the nucleus, around which is found a cloud of electrons. The nucleus contains protons and neutrons, collectively called nucleons. Electrons are kept close to the nucleus by electric attraction to the protons (the neutrons hardly interact with the electrons).

QUESTION Recall your previous studies of chemistry. How many protons and electrons are there in a hydrogen atom? In a helium or carbon atom?

If you don't remember the properties of these atoms, see the periodic table on the inside front cover of this textbook. Hydrogen is the simplest atom, with just one proton and one electron. A helium atom has two protons and two electrons. A carbon atom has six protons and six electrons. Near the other end of the chemical periodic table, a uranium atom has 92 protons and 92 electrons. Figure 1.1 shows the relative sizes of the electron clouds in atoms of several elements but cannot show the nucleus to the same scale; the tiny dot marking the nucleus in the figure is much larger than the actual nucleus.

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Figure 1.1 Atoms of hydrogen, carbon, iron, and uranium. The gray blur represents the electron cloud surrounding the nucleus. The black dot shows the location of the nucleus. On this scale, however, the nucleus would be much too small to see.

The radius of the electron cloud for a typical atom is about 1 × 10−10 meter. The reason for this size can be understood using the principles of quantum mechanics, a major development in physics in the early 20th century. The radius of a proton is about 1 × 10−15 meter, very much smaller than the radius of the electron cloud. Nuclei contain neutrons as well as protons (Figure 1.2). The most common form or “isotope” of hydrogen has no neutrons in the nucleus. However, there exist isotopes of hydrogen with one or two neutrons in the nucleus (in addition to the proton). Hydrogen atoms containing one or two neutrons are called deuterium or tritium. The most common isotope of helium has two neutrons (and two protons) in its nucleus, but a rare isotope has only one neutron; this is called helium-3.

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Figure 1.2 Nuclei of hydrogen, helium, and carbon. Note the very much smaller scale than in Figure 1.1!

The most common isotope of carbon has six neutrons together with the six protons in the nucleus (carbon-12), whereas carbon-14 with eight neutrons is an isotope that plays an important role in dating archaeological objects. Near the other end of the periodic table, uranium-235, which can undergo a fission chain reaction, has 92 protons and 143 neutrons, whereas uranium-238, which does not undergo a fission chain reaction, has 92 protons and 146 neutrons.

Molecules and Solids When atoms come in contact with each other, they may stick to each other (“bond” to each other). Several atoms bonded together can form a molecule—a substance whose physical and chemical properties differ from those of the constituent atoms. For example, water molecules (H2O) have properties quite different from the properties of hydrogen atoms or oxygen atoms. An ordinary-sized rigid object made of bound-together atoms and big enough to see and handle is called a solid, such as a bar of aluminum. A new kind of microscope, the scanning tunneling microscope (STM), is able to map the locations of atoms on the surface of a solid, which has provided new techniques for investigating matter at the atomic level. Two such images appear in Figure 1.3. You can see that atoms in a crystalline solid are arranged in a regular three-dimensional array. The arrangement of atoms on the surface depends on the direction along which the crystal is cut. The irregularities in the bottom image reflect “defects, ” such as missing atoms, in the crystal structure.

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Figure 1.3 Two different surfaces of a crystal of pure silicon. The images were made with a scanning tunneling microscope. Images courtesy of Randall Feenstra, Carnegie Mellon University.

Liquids and Gases When a solid is heated to a higher temperature, the atoms in the solid vibrate more vigorously about their normal positions. If the temperature is raised high enough, this thermal agitation may destroy the rigid structure of the solid. The atoms may become able to slide over each other, in which case the substance is a liquid. At even higher temperatures the thermal motion of the atoms or molecules may be so large as to break the interatomic or intermolecular bonds completely, and the liquid turns into a gas. In a gas the atoms or molecules are quite free to move around, only occasionally colliding with each other or the walls of their container. We will learn how to analyze many aspects of the behavior of solids and gases. We won't have much to say about liquids, because their properties are much harder to analyze. Solids are simpler to analyze than liquids because the atoms stay in one place (though with thermal vibration about their usual positions). Gases are simpler to analyze than liquids because between collisions the gas molecules are approximately unaffected by the other molecules. Liquids are the awkward intermediate state, where the atoms move around rather freely but are always in contact with other atoms. This makes the analysis of liquids very complex.

Planets, Stars, Solar Systems, and Galaxies In our brief survey of the kinds of matter that we will study, we make a giant leap in scale from atoms all the way up to planets and stars, such as our Earth and Sun. We will see that many of the same principles that apply to atoms apply to planets and stars. By making this leap we bypass an important physical science, geology, whose domain of interest includes the formation of mountains and continents. We will study objects that are much bigger than mountains, and we will study objects that are much smaller than

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mountains, but we don't have time to apply the principles of physics to every important kind of matter. Our Sun and its accompanying planets constitute our Solar System. It is located in the Milky Way galaxy, a giant rotating diskshaped system of stars. On a clear dark night you can see a band of light (the Milky Way) coming from the huge number of stars lying in this disk, which you are looking at from a position in the disk, about two-thirds of the way out from the center of the disk. Our galaxy is a member of a cluster of galaxies that move around each other much as the planets of our Solar System move around the Sun (Figure 1.4). The Universe contains many such clusters of galaxies.

Figure 1.4 Our Solar System exists inside a galaxy, which itself is a member of a cluster of galaxies.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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DETECTING INTERACTIONS Objects made of different kinds of matter interact with each other in various ways: gravitationally, electrically, magnetically, and through nuclear interactions. How can we detect that an interaction has occurred? In this section we consider various kinds of observations that indicate the presence of interactions.

QUESTION Before you read further, take a moment to think about your own ideas of interactions. How can you tell that two objects are interacting with each other?

Change of Direction of Motion Suppose that you observe a proton moving through a region of outer space, far from almost all other objects. The proton moves along a path like the one shown in Figure 1.5. The arrow indicates the initial direction of the proton's motion, and the “×'s” in the diagram indicate the position of the proton at equal time intervals.

QUESTION Do you see evidence in Figure 1.5 that the proton is interacting with another object?

Figure 1.5 A proton moves through space, far from almost all other objects. The initial direction of the proton's motion is upward, as indicated by the arrow. The ×'s represent the position of the proton at equal time intervals.

Evidently a change in direction is a vivid indicator of interactions. If you observe a change in direction of the motion of a proton, you will find another object somewhere that has interacted with this proton.

QUESTION Suppose that the only other object nearby was another proton. What was the approximate initial location of this second proton?

Since two protons repel each other electrically, the second proton must have been located to the right of the bend in the first proton's path.

Change of Speed Suppose that you observe an electron traveling in a straight line through outer space far from almost all other objects (Figure 1.6). The path of the electron is shown as though a camera had taken multiple exposures at equal time intervals.

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Figure 1.6 An electron moves through space, far from almost all other objects. The initial direction of the electron's motion is upward and to the left, as indicated by the arrow. The ×'s represent the position of the electron at equal time intervals.

QUESTION Where is the electron's speed largest? Where is the electron's speed smallest?

The speed is largest at the upper left, where the ×'s are farther apart, which means that the electron has moved farthest during the time interval between exposures. The speed is smallest at the bottom right, where the ×'s are closer together, which means that the electron has moved the least distance during the time interval between exposures.

QUESTION Suppose that the only other object nearby was another electron. What was the approximate initial location of this other electron?

The other electron must have been located directly just below and to the right of the starting location, since electrons repel each other electrically. Evidently a change in speed is an indicator of interactions. If you observe a change in speed of an electron, you will find another object somewhere that has interacted with the electron.

Change of Velocity: Change of Speed or Direction In physics, the word “velocity” has a special technical meaning that is different from its meaning in everyday speech. In physics, the quantity called “velocity” denotes a combination of speed and direction. (In contrast, in everyday speech, “speed” and “velocity” are often used as synonyms. In physics and other sciences, however, words have rather precise meanings and there are few synonyms.) For example, consider an airplane that is flying with a speed of 1000 kilometers/hour in a direction that is due east. We say the velocity is 1000 km/hr, east, where we specify both speed and direction. An airplane flying west with a speed of 1000 km/hr would have the same speed but a different velocity. We have seen that a change in an object's speed, or a change in the direction of its motion, indicates that the object has interacted with at least one other object. The two indicators of interaction, change of speed and change of direction, can be combined into one compact statement: A change of velocity (speed or direction or both) indicates the existence of an interaction.

Pushing forward or back, parallel to the direction of the motion, will make an object speed up or slow down but cannot make it turn. To change the direction of the motion you have to push sideways, perpendicular to the motion.

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Diagrams Showing Changes in Velocity In physics diagrams, the velocity of an object is represented by an arrow: a line with an arrowhead. The tail of the arrow is placed at the location of the object, and the arrow points in the direction of the motion of the object. The length of the arrow is proportional to the speed of the object. Figure 1.7 shows two successive positions of a particle at two different times, with velocity arrows indicating a change in speed of the particle (it's slowing down). Figure 1.8 shows three successive positions of a different particle at three different times, with velocity arrows indicating a change in direction but no change in speed.

Figure 1.7 Two successive positions of a particle (indicated by a dot), with arrows indicating the velocity of the particle at each location. The shorter arrow indicates that the speed of the particle at location 2 is less than its speed at location 1.

Figure 1.8 Three successive positions of a particle (indicated by a dot), with arrows indicating the velocity of the particle at each location. The arrows are the same length, indicating the same speed, but they point in different directions, indicating a change in direction and therefore a change in velocity.

We will see a little later in the chapter that velocity is only one example of a physical quantity that has a “magnitude” (an amount or a size) and a direction. Other examples of such quantities are position relative to an origin in 3D space, force, and magnetic field. Quantities having magnitude and direction can be usefully described as “vectors.” Vectors are mathematical quantities that have their own special rules of algebra, similar (but not identical) to the rules of ordinary algebra. Arrows are commonly used in diagrams to denote vector quantities. We will use vectors extensively.

Uniform Motion Suppose that you observe a rock moving along in outer space far from all other objects. We don't know what made it start moving in the first place; presumably a long time ago an interaction gave it some velocity and it has been coasting through the vacuum of space ever since. It is an observational fact that such an isolated object moves at constant, unchanging speed, in a straight line. Its velocity does not change (neither its direction nor its speed changes). We call such motion with unchanging velocity “uniform motion” (Figure 1.9).

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Figure 1.9 “Uniform motion”—no change in speed or direction.

An Object at Rest A special case of uniform motion is the case in which an object's speed is zero and remains zero—the object remains at rest. In this case the object's speed is constant (zero) and the direction of motion, while undefined, is not changing.

Uniform Motion Implies No Net Interaction When we observe an object in uniform motion, we conclude that since its velocity is not changing, either it is not interacting significantly with any other object, or else it is undergoing multiple interactions that cancel each other out. In either case, we can say that there is no “net” (total) interaction. Exercises At the end of a section you will usually find short exercises. It is important to work through exercises as you come to them, to make sure that you can apply what you have just read. Simply reading about concepts in physics is not enough—you must be able to use the concepts in answering questions and solving problems. Make a serious attempt to do an exercise before checking the answer at the end of the chapter. This will help you assess your own understanding.

1.X.1 Which of the following do you see moving with constant velocity?

Answer

(a) A ship sailing northeast at a speed of 5 meters per second (b) The Moon orbiting the Earth (c) A tennis ball traveling across the court after having been hit by a tennis racket (d) A can of soda sitting on a table (e) A person riding on a Ferris wheel that is turning at a constant rate 1.X.2 In which of the following situations is there observational evidence for significant interaction between two

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Answer

objects? How can you tell? (a) A ball bounces off a wall with no change in speed. (b) A baseball that was hit by a batter flies toward the outfield. (c) A communications satellite orbits the Earth. (d) A space probe travels at constant speed toward a distant star. (e) A charged particle leaves a curving track in a particle detector.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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NEWTON'S FIRST LAW OF MOTION The basic relationship between change of velocity and interaction is summarized qualitatively by Newton's “first law of motion”:

NEWTON'S FIRST LAW OF MOTION An object moves in a straight line and at constant speed except to the extent that it interacts with other objects.

The words “to the extent” imply that the stronger the interaction, the more change there will be in direction and/or speed. The weaker the interaction, the less change. If there is no net interaction at all, the direction doesn't change and the speed doesn't change (uniform motion). This case can also be called “uniform velocity” or “constant velocity, ” since velocity refers to both speed and direction. It is important to remember that if an object is not moving at all, its velocity is not changing, so it too may be considered to be in uniform motion. Newton's first law of motion is only qualitative, because it doesn't give us a way to calculate quantitatively how much change in speed or direction will be produced by a certain amount of interaction, a subject we will take up in the next chapter. Nevertheless, Newton's first law of motion is important in providing a conceptual framework for thinking about the relationship between interaction and motion. The English physicist Isaac Newton was the first person to state this law clearly. Newton's first law of motion represented a major break with ancient tradition, which assumed that constant pushing was required to keep something moving. This law says something radically different: no interactions at all are needed to keep something moving!

Does Newton's First Law Apply in Everyday Life? Superficially, Newton's first law of motion may seem at first not to apply to many everyday situations. To push a chair across the floor at constant speed, you have to keep pushing all the time.

QUESTION Doesn't Newton's first law of motion say that the chair should keep moving at constant speed without anyone pushing it? In fact, shouldn't the speed or direction of motion of the chair change due to the interaction with your hands? Does this everyday situation violate Newton's first law of motion? Try to answer these questions before reading farther.

The complicating factor here is that your hands aren't the only objects that are interacting with the chair. The floor also interacts with the chair, in a way that we call friction. If you push just hard enough to compensate exactly for the floor friction, the sum of all the interactions is zero, and the chair moves at constant speed as predicted by Newton's first law. (If you push harder than the floor does, the chair's speed does increase.)

Motion without Friction It is difficult to observe motion without friction in everyday life, because objects almost always interact with many other objects, including air, flat surfaces, and so on. This explains why it took people such a long time to understand clearly the relationship between interaction and change (Newton was born in 1642). You may be able to think of situations in which you have seen an object keep moving at constant (or nearly constant) velocity, without being pushed or pulled. One example of a nearly friction-free situation is a hockey puck sliding on ice. The puck slides a long way at nearly constant speed in a straight line (constant velocity) because there is little friction with the ice. An even better

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example is the uniform motion of an object in outer space, far from all other objects. 1.X.3 Apply Newton's first law to each of the following situations. In which situations can you conclude that the object is undergoing a net interaction with one or more other objects?

Answer

(a) A book slides across the table and comes to a stop. (b) A proton in a particle accelerator moves faster and faster. (c) A car travels at constant speed around a circular race track. (d) A spacecraft travels at a constant speed toward a distant star. (e) A hydrogen atom remains at rest in outer space. 1.X.4 A spaceship far from all other objects uses its rockets to attain a speed of 1 × 104 m/s. The crew then shuts off the power. According to Newton's first law, which of the following statements about the motion of the spaceship after the power is shut off are correct? (Choose all statements that are correct.) (a) The spaceship will move in a straight line. (b) The spaceship will travel on a curving path. (c) The spaceship will enter a circular orbit. (d) The speed of the spaceship will not change. (e) The spaceship will gradually slow down. (f) The spaceship will stop suddenly.

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Answer

OTHER INDICATORS OF INTERACTION Change of Identity Change of velocity (change of speed and/or direction) is not the only indicator of interactions. Another is change of identity, such as the formation of water (H2O) from the burning of hydrogen in oxygen. A water molecule behaves very differently from the hydrogen and oxygen atoms of which it is made.

Change of Shape or Configuration Another indicator of interaction is change of shape or configuration (arrangement of the parts). For example, slowly bend a pen or pencil, then hold it in the bent position. The speed hasn't changed, nor is there a change in the direction of motion (it's not moving!). The pencil has not changed identity. Apparently a change of shape can be evidence for interactions, in this case with your hand. Other changes in configuration include “phase changes” such as the freezing or boiling of a liquid, brought about by interactions with the surroundings. In different phases (solid, liquid, gas), atoms or molecules are arranged differently. Changes in configuration at the atomic level are another indication of interactions.

Change of Temperature Another indication of interaction is change of temperature. Place a pot of cold water on a hot stove. As time goes by, a thermometer will indicate a change in the water due to interaction with the hot stove.

Other Indications of Interactions Is a change of position an indicator of an interaction? That depends. If the change of position occurs simply because a particle is moving at constant speed and direction, then a mere change of position is not an indicator of an interaction, since uniform motion is an indicator of no net interaction.

QUESTION However, if you observe an object at rest in one location, and later you observe it again at rest but in a different location, did an interaction take place?

Yes. You can infer that there must have been an interaction to give the object some velocity to move the object toward the new position, and another interaction to slow the object to a stop in its new position. In later chapters we will consider interactions involving change of identity, change of shape, and change of temperature, but for now we'll concentrate on interactions that cause a change of velocity (speed and/or direction).

Indirect Evidence for an Additional Interaction Sometimes there is indirect evidence for an additional interaction. When something doesn't change although you would normally expect a change due to a known interaction, this indicates that another interaction is present. Consider a balloon that hovers motionless in the air despite the downward gravitational pull of the Earth. Evidently there is some other kind of interaction that

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opposes the gravitational interaction. In this case, interactions with air molecules have the net effect of pushing up on the balloon (“buoyancy”). The lack of change implies that the effect of the air molecules exactly compensates for the gravitational interaction with the Earth. When you push a chair across the floor and it moves with constant velocity despite your pushing on it (which ought to change its speed), that means that something else must also be interacting with it (the floor). The stability of the nucleus of an atom is another example of indirect evidence for an additional interaction. The nucleus contains positively charged protons that repel each other electrically, yet the nucleus remains intact. We conclude that there must be some other kind of interaction present, a nonelectric attractive interaction that overcomes the electric repulsion. This is evidence for an interaction called the “strong interaction” that acts between protons and neutrons in the nucleus.

Summary: Changes as Indicators of Interactions Here then are the most common indicators of interactions: Change of velocity (change of direction and/or change of speed) Change of identity Change of shape Change of temperature Lack of change when change is expected (indirect evidence) The important point is this: Interactions cause change. In the absence of interactions, there is no change, which is usually uninteresting. The exception is the surprise when nothing changes despite our expectations that something should change. This is indirect evidence for some interaction that we hadn't recognized was present, that more than one interaction is present and the interactions cancel each others' effects. For the next few chapters we'll concentrate on change of velocity as evidence for an interaction (or lack of change of velocity, which can give indirect evidence for additional interactions). 1.X.5 You slide a coin across the floor, and observe that it travels in a straight line, slowing down and eventually stopping. A sensitive thermometer shows that the coin's temperature increased. What can we conclude? (Choose all statements that are correct.)

Answer

(a) Because the coin traveled in a straight line, we conclude that it did not interact with anything. (b) Because the coin did not change shape, we conclude that it did not interact with anything. (c) Because the coin slowed down, we conclude that Newton's first law does not apply to objects in everyday life, such as coins. (d) Because the coin's speed changed, we conclude that it interacted with one or more other objects. (e) Because the coin got hot, we conclude that it interacted with one or more other objects.

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DESCRIBING THE 3D WORLD: VECTORS Physical phenomena take place in the 3D world around us. In order to be able to make quantitative predictions and give detailed, quantitative explanations, we need tools for describing precisely the positions and velocities of objects in 3D, and the changes in position and velocity due to interactions. These tools are mathematical entities called 3D “vectors.”

3D Coordinates We will use a 3D coordinate system to specify positions in space and other vector quantities. Usually we will orient the axes of the coordinate system as shown in Figure 1.10: +x axis to the right, +y axis upward, and +z axis coming out of the page, toward you. This is a “right-handed” coordinate system: if you hold the thumb, first, and second fingers of your right hand perpendicular to each other, and align your thumb with the x axis and your first finger with the y axis, your second finger points along the z axis. (In some math and physics textbook discussions of 3D coordinate systems, the x axis points out, the y axis points to the right, and the z axis points up, but we will also use a 2D coordinate system with y up, so it makes sense always to have the y axis point up.)

Figure 1.10 Right-handed 3D coordinate system.

Basic Properties of Vectors: Magnitude and Direction A vector is a quantity that has a magnitude and a direction. For example, the velocity of a baseball is a vector quantity. The magnitude of the baseball's velocity is the speed of the baseball—for example, 20 meters/second. The direction of the baseball's velocity is the direction of its motion at a particular instant—for example, “up” or “to the right” or “west” or “in the +y direction.” A symbol denoting a vector is written with an arrow over it:

Position A position in space can also be considered to be a vector, called a position vector, pointing from an origin to that location. Figure 1.11 shows a position vector that might represent your final position if you started at the origin and walked 4 meters along the x axis, then 2 meters parallel to the z axis, then climbed a ladder so you were 3 meters above the ground. Your new position relative to the origin is a vector that can be written like this:

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In three dimensions a vector is a triple of numbers can be represented as vectors:

. Quantities like the position of an object and the velocity of an object

Figure 1.11 A position vector

and its x, y, and z components.

Components of a Vector Each of the numbers in the triple is referred to as a component of the vector. The x component of the vector is the number v x . is . A component such as v x is not a vector, since it The z component of the vector is only one number. It is important to note that the x component of a vector specifies the difference between the x coordinate of the tail of the vector and the x coordinate of the tip of the vector. It does not give any information about the location of the tail of the vector (compare Figure 1.11 and Figure 1.12).

Figure 1.12 The arrow represents the vector

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, drawn with its tail at location

.

In Figure 1.11 we represented your position vector relative to the origin graphically by an arrow whose tail is at the origin and whose arrowhead is at your position. The length of the arrow represents the distance from the origin, and the direction of the arrow represents the direction of the vector, which is the direction of a direct path from the initial position to the final position (the “displacement”; by walking and climbing you “displaced” yourself from the origin to your final position). Since it is difficult to draw a 3D diagram on paper, when working on paper you will usually be asked to draw vectors that all lie in a single plane. Figure 1.13 shows an arrow in the xy plane representing the vector .

Figure 1.13 The position vector

, drawn at the origin, in the xy plane. The components of the vector specify the displacement from the tail to the tip. The z axis, which is not shown, comes out of the page, toward you.

Vectors and Scalars A quantity that is represented by a single number is called a scalar. A scalar quantity does not have a direction. Examples include the mass of an object, such as 5 kg, or the temperature, such as −20°C. Vectors and scalars are very different entities; a vector can never be equal to a scalar, and a scalar cannot be added to a vector. Scalars can be positive or negative:

Although a component of a vector such as v x is not a vector, it's not a scalar either, despite being only one number. An important property of a true scalar is that its value doesn't change if we orient the xyz coordinate axes differently. Rotating the axes doesn't change an object's mass, or the temperature, but it does change what we mean by the x component of the velocity since the x axis now points in a different direction. 1.X.6 How many numbers are needed to specify a 3D position vector?

Answer

1.X.7 How many numbers are needed to specify a scalar?

Answer

Magnitude of a Vector In Figure 1.14 we again show the vector from Figure 1.11, showing your displacement from the origin. Using a 3D extension of the Pythagorean theorem for right triangles (Figure 1.15), the net distance you have moved from the starting point is

We say that the magnitude

of the position vector

is

The magnitude of a vector is written either with absolute-value bars around the vector as the vector without the little arrow above it, r.

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, or simply by writing the symbol for

Figure 1.14 A vector representing a displacement from the origin.

Figure 1.15 The magnitude of a vector is the square root of the sum of the squares of its components (3D version of the Pythagorean theorem).

The magnitude of a vector can be calculated by taking the square root of the sum of the squares of its components (see Figure 1.15).

MAGNITUDE OF A VECTOR

The magnitude of a vector is always a positive number. The magnitude of a vector is a single number, not a triple of numbers, and it is a scalar, not a vector.

The magnitude of a vector is a true scalar, because its value doesn't change if you rotate the coordinate axes. Rotating the axes changes the individual components, but the length of the arrow representing the vector doesn't change.

Can a Vector Be Positive or Negative? QUESTION Consider the vector

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. Is this vector positive? Negative? Zero?

None of these descriptions is appropriate. The x component of this vector is positive, the y component is zero, and the z component is negative. Vectors aren't positive, or negative, or zero. Their components can be positive or negative or zero, but these words just don't mean anything when used with the vector as a whole. On the other hand, the magnitude of a vector such as

1.X.8 If

, find

is always positive.

.

Answer

1.X.9 Can the magnitude of a vector be a negative number?

Answer

Mathematical Operations Involving Vectors Although the algebra of vectors is similar to the scalar algebra with which you are very familiar, it is not identical. There are some algebraic operations that cannot be performed on vectors. Algebraic operations that are legal for vectors include the following operations, which we will discuss in this chapter: Adding one vector to another vector: Subtracting one vector from another vector: Finding the magnitude of a vector: Finding a unit vector (a vector of magnitude 1): Multiplying (or dividing) a vector by a scalar: Finding the rate of change of a vector:

or or

.

In later chapters we will also see that there are two more ways of combining two vectors: The vector dot product, whose result is a scalar The vector cross product, whose result is a vector

Operations That Are Not Legal for Vectors Although vector algebra is similar to the ordinary scalar algebra you have used up to now, there are certain operations that are not legal (and not meaningful) for vectors: A vector cannot be set equal to a scalar. A vector cannot be added to or subtracted from a scalar. A vector cannot occur in the denominator of an expression. (Although you can't divide by a vector, note that you can legally divide by the magnitude of a vector, which is a scalar.)

1.X.10 The vector (a)

0, −9, 1

(b)

4, −5, 5

(c)

4, 9, 5

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and the scalar h = −2. What is

?

Answer

(d) This is a meaningless expression. 1.X.11 Is 4/ 6, −7, 4 a meaningful expression? If so, what is its value?

Answer

Multiplying a Vector by a Scalar A vector can be multiplied (or divided) by a scalar. If a vector is multiplied by a scalar, each of the components of the vector is multiplied by the scalar:

Multiplication by a scalar “scales” a vector, keeping its direction the same but making its magnitude larger or smaller (Figure 1.16). Multiplying by a negative scalar reverses the direction of a vector.

Figure 1.16 Multiplying a vector by a scalar changes the magnitude of the vector. Multiplying by a negative scalar reverses the direction of the vector.

and the scalar f = −3.0. What is

1.X.12 The vector 1.X.13 If

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, what is

?

?

Answer

Answer

Magnitude of a Scalar You may wonder how to find the magnitude of a quantity like expression can be factored:

, which involves the product of a scalar and a vector. This

The magnitude of a scalar is its absolute value, so:

1.X.14 If

m/s, what is the magnitude of

1.X.15 How does the magnitude of the vector

?

compare to the magnitude of the vector

Answer

?

Answer

Direction of a Vector: Unit Vectors One way to describe the direction of a vector is by specifying a unit vector. A unit vector is a vector of magnitude 1, pointing in some direction. A unit vector is written with a “hat” (caret) over it instead of an arrow. The unit vector â is called “a-hat.”

QUESTION Is the vector

The magnitude of The vector

a unit vector?

is

, so this is not a unit vector. is a unit vector, since its magnitude is 1:

Note that every component of a unit vector must be less than or equal to 1. In our 3D Cartesian coordinate system, there are three special unit vectors, oriented along the three axes. They are called i-hat, j-hat, and k-hat, and they point along the x, y, and z axes, respectively (Figure 1.17):

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Figure 1.17 The unit vectors î, ĵ, . One way to express a vector is in terms of these special unit vectors: We will usually use the form rather than the îĵ form in this book, because the familiar calculus textbooks, emphasizes that a vector is a single entity.

notation, used in many

Not all unit vectors point along an axis, as shown in Figure 1.18. For example, the vectors

are both unit vectors, since the magnitude of each is equal to 1. Note that every component of a unit vector is less than or equal to 1.

Figure 1.18 The unit vector

has the same direction as the vector

, but its magnitude is 1, and it has no physical units.

Calculating Unit Vectors Any vector may be factored into the product of a unit vector in the direction of the vector, multiplied by a scalar equal to the magnitude of the vector.

For example, a vector of magnitude 5, aligned with the y axis, could be written as: Therefore, to find a unit vector in the direction of a particular vector, we just divide the vector by its magnitude:

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CALCULATING A UNIT VECTOR

EXAMPLE Unit Vector If

m/s, then

Remember that to divide a vector by a scalar, you divide each component of the vector by the scalar. The result is a new vector. Note also that a unit vector has no physical units (such as meters per second), because the units in the numerator and denominator cancel. 1.X.16 What is the unit vector in the direction of 0, 6, 0 ? 1.X.17 What is the unit vector â in the direction of

, where

Equality of Vectors

EQUALITY OF VECTORS A vector is equal to another vector if and only if all the components of the vectors are equal.

The magnitudes and directions of two equal vectors are the same:

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Answer

?

Answer

Equal Vectors

If

, then

1.X.18 If 1.X.19 Consider the vectors

, and and

, what is the y component of

?

represented by arrows in Figure 1.19. Are these two vectors equal?

Figure 1.19 Are these two vectors equal? (Exercise 1.X.19.)

Vector Addition

ADDING VECTORS The sum of two vectors is another vector, obtained by adding the components of the vectors.

EXAMPLE Adding Vectors

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Answer

Answer

Warning: Don't Add Magnitudes! The magnitude of a vector is not in general equal to the sum of the magnitudes of the two original vectors! For example, the magnitude of the vector 3, 0, 0 is 3, and the magnitude of the vector −2, 0, 0 is 2, but the magnitude of the vector ( 3, 0, 0 + −2, 0, 0 ) is 1, not 5!

Adding Vectors Graphically: Tip to Tail The sum of two vectors has a geometric interpretation. In Figure 1.20 you first walk along displacement vector , followed by ? The x component C x of your net walking along displacement vector . What is your net displacement vector displacement is the sum of A x and B x . Similarly, the y component C y of your net displacement is the sum of Ay and B y .

GRAPHICAL ADDITION OF VECTORS To add two vectors

and

Draw the first vector Move the second vector

graphically (Figure 1.20): . (without rotating it) so its tail is located at the tip of the first vector.

Draw a new vector from the tail of vector

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to the tip of vector

.

Figure 1.20 The procedure for adding two vectors graphically: draw vectors tip to tail. To add so the tail of

1.X.20 If

is at the tip of

, then draw a new arrow starting at the tail of

and

magnitude of Is

, what is the magnitude of

? What is the magnitude of

? What is the magnitude of

graphically, move and ending at the tip of

? What is the

.

Answer

plus the magnitude of

?

?

1.X.21

and following: (a) (b) (c) (d) (e)

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. Calculate the

Answer

Vector Subtraction The difference of two vectors will be very important in this and subsequent chapters. To subtract one vector from another, we subtract the components of the second vector from the components of the first:

1.X.22 If

and

difference

? What is

, what is the sum

? What is the

Answer

?

Subtracting Vectors Graphically: Tail to Tail To subtract one vector

from another vector

Draw the first vector Move the second vector

graphically:

. (without rotating it) so its tail is located at the tail of the first vector.

Draw a new vector from the tip of vector

to the tip of vector

.

Note that you can check this algebraically and graphically. As shown in Figure 1.21, since the tail of , then the vector should be the sum of and , as indeed it is:

is located at the tip of

Figure 1.21 The procedure for subtracting vectors graphically: draw vectors tail to tail; draw a new vector from the tip of the second vector to the tip of the first vector.

1.X.23 Which of the following statements about the three vectors in Figure 1.22 are correct? (a) (b) (c) (d) (e)

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Answer

Figure 1.22 Exercise 1.X.23.

Commutativity and Associativity Vector addition is commutative:

Vector subtraction is not commutative:

The associative property holds for vector addition and subtraction:

The Zero Vector It is convenient to have a compact notation for a vector whose components are all zero. We will use the symbol vector, in order to distinguish it from a scalar quantity that has the value 0.

For example, the sum of two vectors

to denote a zero

.

Change in a Quantity: The Greek Letter Δ Frequently we will want to calculate the change in a quantity. For example, we may want to know the change in a moving object's position or the change in its velocity during some time interval. The Greek letter Δ (capital delta suggesting “d for difference”) is used to denote the change in a quantity (either a scalar or a vector). We use the subscript i to denote an initial value of a quantity, and the subscript f to denote the final value of a quantity. If a vector denotes the initial position of an object relative to the origin (its position at the beginning of a time interval), and denotes the final position of the object, then

means “change of

” or

(displacement)

means “change of t” or t f − t i (time interval) The symbol Δ (delta) always means “final minus initial, ” not “initial minus final.” For example, when a child's height changes from

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1.1 m to 1.2 m, the change is Δy = +0.1 m, a positive number. If your bank account dropped from $150 to $130, what was the change in your balance? Δ (bank account) = −20 dollars.

Relative Position Vectors Vector subtraction is used to calculate relative position vectors, vectors that represent the position of an object relative to another object. In Figure 1.23, object 1 is at location and object 2 is at location . We want the components of a vector that points from object 1 to object 2. This is the vector obtained by subtraction: . Note that the form is always “final” minus “initial” in these calculations.

Figure 1.23 Relative position vector.

1.X.24 At 10:00 AM you are at location −3, 2, 5 m. By 10:02 AM you have walked to location 6, 4, 25 m. (a) What is

Answer

, the change in your position?

(b) What is Δt, the time interval during which your position changed? 1.X.25 A snail is initially at location . What is

. At a later time the snail has crawled to location , the change in the snail's position?

Answer

Unit Vectors and Angles Suppose that a taut string is at an angle to the +x axis, and we need a unit vector in the direction of the string. Figure 1.24 shows a unit vector pointing along the string. What is the x component of this unit vector? Consider the triangle whose base is Ax and whose hypotenuse is

In Figure 1.24 the angle

. From the definition of the cosine of an angle we have this:

is shown in the first quadrant (θx less than 90°), but this works for larger angles as well. For example, in

Figure 1.25 the angle from the +x axis to a unit vector corresponds to B x being negative.

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is in the second quadrant (θx greater than 90°) and cosθx is negative, which

Figure 1.24 A unit vector whose direction is at a known angle from the +x axis.

Figure 1.25 A unit vector in the second quadrant from the +x axis.

What is true for x is also true for y and z. Figure 1.26 shows a 3D unit vector

and indicates the angles between the unit vector and

the x, y, and z axes. Evidently we can write

These three cosines of the angles between a vector (or unit vector) and the coordinate axes are called the “direction cosines” of the vector. The cosine function is never greater than 1, just as no component of a unit vector can be greater than 1.

Figure 1.26 A 3D unit vector and its angles to the x, y, and z axes.

A common special case is that of a unit vector lying in the xy plane, with zero z component (Figure 1.27). In this case , so that , therefore you can express the cosine of θy as the sine of θx , which is often convenient. However, in the general 3D case shown in Figure 1.26 there is no such simple relationship among the direction angles or among their cosines.

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Figure 1.27 If a vector lies in the xy plane,

.

FINDING A UNIT VECTOR FROM ANGLES To find a unit vector if angles are given: Redraw the vector of interest with its tail at the origin, and determine the angles between this vector and the axes. Imagine the vector 1, 0, 0 , which lies on the +x axis. θx is the angle through which you would rotate the vector 1, . 0, 0 until its direction matched that of your vector. θx is positive, and θy is the angle through which you would rotate the vector 0, 1, 0 until its direction matched that of your vector. θy is positive, and . θz is the angle through which you would rotate the vector 0, 0, 1 until its direction matched that of your vector. θz is positive, and .

EXAMPLE From Angle to Unit Vector A rope lying in the xy plane, pointing up and to the right, supports a climber at an angle of 20° to the vertical (Figure 1.28). What is the unit vector pointing up along the rope?

Figure 1.28 A climber supported by a rope.

Solution Follow the procedure given above for finding a unit vector from angles. In Figure 1.29 we redraw the vector with its tail at the origin, and we determine the angles between the vector and the axes. If we rotate the unit vector 1, 0, 0 from along the +x axis to the vector of interest, we see that we have to rotate through an angle θx = 70°. To rotate the unit

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vector 0, 1, 0 from along the +y axis to the vector of interest, we have to rotate through an angle of angle from the +z axis to our vector is . Therefore the unit vector that points along the rope is this:

. The

Figure 1.29 Redraw the vector with its tail at the origin. Identify the angles between the positive axes and the vector. In this example the vector lies in the xy plane.

FURTHER DISCUSSION You may have noticed that the y component of the unit vector can also be calculated as sin 70° = 0.940, and it is often useful to recognize that a vector component can be obtained using sine instead of cosine. There is, however, some advantage always to calculate in terms of direction cosines. This is a method that always works, including in 3D, and that avoids having to decide whether to use a sine or a cosine. Just use the cosine of the angle from the relevant positive axis to the vector.

EXAMPLE From Unit Vector to Angles A vector points from the origin to the location −600, 0, 300 m. What is the angle that this vector makes to the x axis? To the y axis? To the z axis?

Solution

But we also know that Similarly,

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, and the arccosine gives θx = 153.4°.

FURTHER DISCUSSION Looking down on the xz plane in Figure 1.30, you can see that the difference between θx = 153.4° and θz = 63.4° is 90°, as it should be.

Figure 1.30 Look down on the xz plane. The difference in the two angles is 90°, as it should be.

1.X.26 A unit vector lies in the xy plane, at an angle of 160° from the +x axis, with a positive y component. What is the unit vector? (It helps to draw a diagram.) 1.X.27 A string runs up and to the left in the xy plane, making an angle of 40° to the vertical. Determine the unit vector that points along the string.

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Answer

Answer

SI UNITS In this book we use the SI (Système Internationale) unit system, as is customary in technical work. The SI unit of mass is the kilogram (kg), the unit of distance is the meter (m), and the unit of time is the second (s). In later chapters we will encounter other SI units, such as the newton (N), which is a unit of force. It is essential to use SI units in physics equations; this may require that you convert from some other unit system to SI units. If mass is known in grams, you need to divide by 1000 and use the mass in kilograms. If a distance is given in centimeters, you need to divide by 100 to convert the distance to meters. If the time is measured in minutes, you need to multiply by 60 to use a time in seconds. A convenient way to do such conversions is to multiply by factors that are equal to 1, such as (1 min)/(60 s) or (100 cm)/(1 m). As an example, consider converting 60 miles per hour to SI units, meters per second. Start with the 60 mi/hr and multiply by factors of 1:

Observe how most of the units cancel, leaving final units of m/s. 1.X.28 A snail moved 80 cm (80 centimeters) in 5 minutes. What was its average speed in SI units? Write out the factors as was done above.

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Answer

VELOCITY We use vectors not only to describe the position of an object but also to describe velocity (speed and direction). If we know an object's present speed in meters per second and the object's direction of motion, we can predict where it will be a short time into the future. As we have seen, change of velocity is an indication of interaction. We need to be able to work with velocities of objects in 3D, so we need to learn how to use 3D vectors to represent velocities. After learning how to describe velocity in 3D, we will also learn how to describe change of velocity, which is related to interactions.

Average Speed The concept of speed is a familiar one. Speed is a single number, so it is a scalar quantity (speed is the magnitude of velocity). A world-class sprinter can run 100 meters in 10 seconds. We say the sprinter's average speed is (100 m)/(10 s) = 10 m/s. In SI units speed is measured in meters per second, abbreviated “m/s.” A car that travels 100 miles in 2 hours has an average speed of (100 mi)/(2 hr) = 50 miles per hour (about 22 m/s). In symbols,

where v avg is the “average speed, ” d is the distance the car has traveled, and t is the elapsed time. There are other useful versions of the basic relationship among average speed, distance, and time. For example, expresses the fact that if you run 5 m/s for 7 seconds you go 35 meters. Or you can use

to calculate that to go 3000 miles in an airplane that flies at 600 miles per hour will take 5 hours.

Units While it is easy to make a mistake in one of the formulas relating speed, time interval, and change in position, it is also easy to catch such a mistake by looking at the units. If you had written , you would discover that the right-hand side has units of (m/s)/m, or 1/s, not s. Always check units!

Instantaneous Speed Compared to Average Speed If a car went 70 miles per hour for the first hour and 30 miles an hour for the second hour, it would still go 100 miles in 2 hours, with an average speed of 50 miles per hour. Note that during this 2-hour interval, the car was almost never actually traveling at its average speed of 50 miles per hour. To find the “instantaneous” speed—the speed of the car at a particular instant—we should observe the short distance the car goes in a very short time, such as a hundredth of a second: If the car moves 0.3 meters in 0.01 s, its instantaneous speed is 30 meters per second.

Vector Velocity Earlier we calculated vector differences between two different objects. The vector difference represented a relative position vector—the position of object 2 relative to object 1 at a particular time. Now we will be concerned with the change of

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position of one object during a time interval, and will represent the “displacement” of this single object during the time is the initial 3D position and is the final 3D position (note that as with relative position vectors, we always interval, where calculate “final minus initial”). Dividing the (vector) displacement by the (scalar) time interval tf − ti (final time minus initial time) gives the average (vector) velocity of the object:

DEFINITION: AVERAGE VELOCITY

Another way of writing this expression, using the “Δ” symbol (Greek capital delta, defined in Section 1.5) to represent a change in a quantity, is

Remember that this is a compact notation for

Determining Average Velocity from Change in Position Consider a bee in flight (Figure 1.31). At time ti = 15 s after 9:00 AM, the bee's position vector was . At time tf = . On the diagram, we draw and label the vectors and 15.1 s after 9:00 AM, the bee's position vector was .

Figure 1.31 The displacement vector points from initial position to final position.

Next, on the diagram, we draw and label the vector , with the tail of the vector at the bee's initial position. One useful to make the final vector way to think about this graphically is to ask yourself what vector needs to be added to the initial vector , since

can be written in the form

.

The vector we just drew, the change in position , is called the “displacement” of the bee during this time interval. This displacement vector points from the initial position to the final position, and we always calculate displacement as “final minus

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initial.”

Note that the displacement refers to the positions of one object (the bee) at two different times, not the position of one object relative to a second object at one particular time. However, the vector subtraction is the same kind of operation for either kind of situation. We calculate the bee's displacement vector numerically by taking the difference of the two vectors, final minus initial:

This numerical result should be consistent with our graphical construction. Look at the components of in Figure 1.31. Do is you see that this vector has an x component of +1 and a y component of −0.5 m? Note that the (vector) displacement in the direction of the bee's motion. The average velocity of the bee, a vector quantity, is the (vector) displacement i . Calculate the bee's average velocity:

Since we divided flew in a straight line.

by a scalar (t f − t i ), the average velocity

divided by the (scalar) time interval t f − t

points in the direction of the bee's motion, if the bee

What is the speed of the bee?

What is the direction of the bee's motion, expressed as a unit vector?

Note that the “m/s” units cancel; the result is dimensionless. We can check that this really is a unit vector:

This is not quite 1.0 due to rounding the velocity coordinates and speed to three significant figures. Put the pieces back together and see what we get. The original vector factors into the product of the magnitude times the unit vector:

This is the same as the original vector

.

1.X.29 At a time 0.2 seconds after it has been hit by a tennis racket, a tennis ball is located at 5, 7, 2 m, relative to an origin in one corner of a tennis court. At a time 0.7 seconds after being hit, the ball is located at 9, 2, 8 m.

Answer

(a) What is the average velocity of the tennis ball? (b) What is the average speed of the tennis ball? (c) What is the unit vector in the direction of the ball's velocity? 1.X.30 A spacecraft is observed to be at a location 200, 300, −400 m relative to an origin located on a nearby asteroid, and 5 seconds later is observed at location 325, 25, −550 m. (a) What is the average velocity of the spacecraft?

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Answer

(b) What is the average speed of the spacecraft? (c) What is the unit vector in the direction of the spacecraft's velocity?

Scaling a Vector to Fit on a Graph We can plot the average velocity vector on the same graph that we use for showing the vector positions of the bee (Figure 1.32). However, note that velocity has units of meters per second whereas positions have units of meters, so we're mixing apples and oranges.

Figure 1.32 Average velocity vector: displacement divided by time interval.

Moreover, the magnitude of the vector, 11.18 m/s, doesn't fit on a graph that is only 5 units wide (in meters). It is standard practice in such situations to scale down the arrow representing the vector to fit on the graph, preserving the correct direction. In Figure 1.32 we've scaled down the velocity vector by about a factor of 3 to make the arrow fit on the graph. Of course if there is more than one velocity vector we use the same scale factor for all the velocity vectors. The same kind of scaling is used with other physical quantities that are vectors, such as force and momentum, which we will encounter later.

Predicting a New Position We can rewrite the velocity relationship in the form

That is, the (vector) displacement of an object is its average (vector) velocity times the time interval. This is just the vector version of the simple notion that if you run at a speed of 7 m/s for 5 s you move a distance of (7 m/s)(5 s) = 35 m, or that a car going 50 miles per hour for 2 hours goes (50 mi/hr)(2 hr) = 100 miles. a valid vector relation? Yes, multiplying a vector times a scalar (t f − t i ) yields a vector. We make a further rearrangement to obtain a relation for updating the position when we know the velocity:

Is

THE POSITION UPDATE FORMULA

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This equation says that if we know the starting position, the average velocity, and the time interval, we can predict the final position. This equation will be important throughout our work.

Using the Position Update Formula The position update formula

is a vector equation, so we can write out its full component form:

Because the x component on the left of the equation must equal the x component on the right (and similarly for the y and z components), this compact vector equation represents three separate component equations:

EXAMPLE Updating the Position of a Ball At time ti = 12.18 s after 1:30 PM a ball's position vector is . The ball's velocity at that moment . At time tf = 12.21 s after 1:30 PM, where is the ball, assuming that its velocity hardly is changes during this short time interval?

Solution

Note that if the velocity changes significantly during the time interval, in either magnitude or direction, our prediction for the new position may not be very accurate. In this case the velocity at the initial time could differ significantly from the average velocity during the time interval.

1.X.31 A proton traveling with a velocity of 3 × 105, 2 × 105, −4 × 105 m/s passes the origin at a time 9.0 seconds after a proton detector is turned on. Assuming that the velocity of the proton does not change, what

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Answer

will its position be at time 9.7 seconds? 1.X.32 How long does it take a baseball with velocity 30, 20, 25 m/s to travel from location to location ?

Answer

Instantaneous Velocity The curved colored line in Figure 1.33 shows the path of a ball through the air. The colored dots mark the ball's position at time intervals of one second. While the ball is in the air, its velocity is constantly changing, due to interactions with the Earth (gravity) and with the air (air resistance).

Figure 1.33 The trajectory of a ball through air. The axes represent the x and y distance from the ball's initial location; each square on the grid corresponds to 10 meters. Three different displacements, corresponding to three different time intervals, are indicated by arrows on the diagram.

Suppose we ask: What is the velocity of the ball at the precise instant that it reaches location B? This quantity would be called the “instantaneous velocity” of the ball. We can start by approximating the instantaneous velocity of the ball by finding its average velocity over some larger time interval. The table in Figure 1.34 shows the time and the position of the ball for each location marked by a colored dot in Figure 1.33. We can use these data to calculate the average velocity of the ball over three different intervals, by finding the ball's displacement during each interval, and dividing by the appropriate Δ t for that interval:

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Figure 1.34 Table showing elapsed time and position of the ball at each location marked by a dot in Figure 1.33.

Not surprisingly, the average velocities over these different time intervals are not the same, because both the direction of the ball's motion and the speed of the ball were changing continuously during its flight. The three average velocity vectors that we calculated are shown in Figure 1.35.

QUESTION Which of the three average velocity vectors depicted in Figure 1.35 best approximates the instantaneous velocity of the ball at location B?

Figure 1.35 The three different average velocity vectors calculated above are shown by three arrows, each with its tail at location B. Note that since the units of velocity are m/s, these arrows use a different scale from the distance scale used for the path of the ball. The three arrows representing average velocities are drawn with their tails at the location of interest. The dashed arrow represents the actual instantaneous velocity of the ball at location B.

Simply by looking at the diagram, we can tell that is closest to the actual instantaneous velocity of the ball at location B, because its direction is closest to the direction in which the ball is actually traveling. Because the direction of the instantaneous velocity is the direction in which the ball is moving at a particular instant, the instantaneous velocity is tangent to the ball's path. Of the three average velocity vectors we calculated, best approximates a tangent to the path of the ball. Evidently , the velocity calculated with the shortest time interval, t C − t B , is the best approximation to the instantaneous velocity at location B. If we used even smaller values of Δ t in our calculation of average velocity, such as 0.1 second, or 0.01 second, or 0.001 second, we would presumably have better and better estimates of the actual instantaneous velocity of the object at the instant when it passes location B.

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Two important ideas have emerged from this discussion: The direction of the instantaneous velocity of an object is tangent to the path of the object's motion. Smaller time intervals yield more accurate estimates of instantaneous velocity.

1.X.33 How does average velocity differ from instantaneous velocity? 1.X.34 A comet travels in an elliptical path around a star, in the direction shown in Figure 1.36. Which arrow best indicates the direction of the comet's velocity vector at each of the numbered locations in the orbit?

Answer

Answer

Figure 1.36 A comet goes around a star.

Connection to Calculus You may already have learned about derivatives in calculus. The instantaneous velocity is a derivative, the limit of time interval Δ t used in the calculation gets closer and closer to zero:

as the

In Figure 1.35, the process of taking the limit is illustrated graphically. As smaller values of Δ t are used in the calculation, the average velocity vectors approach the limiting value: the actual instantaneous velocity.

The rate of change of a vector (the derivative) is itself a vector. A useful way to see the meaning of the derivative of a vector is to consider the components:

The derivative of the position vector

gives components that are the components of the velocity, as we should expect.

Informally, you can think of as a very small (“infinitesimal”) displacement, and dt as a very small (“infinitesimal”) time interval. It is as though we had continued the process illustrated in Figure 1.35 to smaller and smaller time intervals, down to an extremely tiny time interval dt with a correspondingly tiny displacement . The ratio of these tiny quantities is the instantaneous velocity.

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The ratio of these two tiny quantities need not be small. For example, suppose that an object moves in the x direction a tiny distance of 1 × 10−15 m, the radius of a proton, in a very short time interval of 1 × 10−23 s:

which is one-third the speed of light (3 × 108 m/s)!

Acceleration Velocity is the time rate of change of position: . Similarly, we define “acceleration” as the time rate of change of . Acceleration, which is itself a vector quantity, has units of meters per second per second, written as m/s/s or velocity: m/s 2.

DEFINITION: ACCELERATION Instantaneous acceleration is the time rate of change of velocity:

Average acceleration can be calculated from a change in velocity:

The units of acceleration are m/s 2.

If a car traveling in a straight line speeds up from 20 m/s to 26 m/s in 3 seconds, we say that the magnitude of the acceleration is (26 − 20)/3 = 2 m/s/s. If you drop a rock, its speed increases 9.8 m/s every second, so its acceleration is 9.8 m/s/s, as long as air resistance is negligible. 1.X.35 Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 5 seconds. What is the magnitude of the acceleration? How does this compare with the acceleration of a falling rock? 1.X.36 Suppose the position of an object at time t is 3 + 5t, 4t2, 2t − 6t3 . What is the instantaneous velocity at time t? What is the acceleration at time t? What is the instantaneous velocity at time t = 0? What is the acceleration at time t = 0?

Answer

Answer

Change of Magnitude and/or Change in Direction There are two parts to the acceleration, the time rate of change of the velocity

:

As we'll see in later chapters, these two parts of the acceleration are associated with pushing or pulling parallel to the motion (changing the speed) or perpendicular to the motion (changing the direction).

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MOMENTUM Newton's first law of motion, An object moves in a straight line and at constant speed except to the extent that it interacts with other objects gives us a conceptual connection between interactions and their effects on the motion of objects. However, this law does not allow us to make quantitative (numerical) predictions or explanations—we could not have used this law to predict the exact trajectory of the ball shown in Figure 1.33, and we could not use this law alone to figure out how to send a rocket to the Moon. In order to make quantitative predictions or explanations of physical phenomena, we need a quantitative measure of interactions and a quantitative measure of effects of those interactions. Newton's first law of motion does contain the important idea that if there is no interaction, a moving object will continue to move in a straight line, with no change of direction or speed, and an object that is not moving will remain at rest. A quantitative version of this law would provide a means of predicting the motion of an object, or of deducing how it must have moved in the past, if we could list all of its interactions with other objects.

Changes in Velocity QUESTION What factors make it difficult or easy to change an object's velocity?

You have probably noticed that if two objects have the same velocity but one is much more massive than the other, it is more difficult to change the heavy object's speed or direction. It is easier to stop a baseball traveling at a hundred miles per hour than to stop a car traveling at a hundred miles per hour! It is easier to change the direction of a canoe than to change the direction of a large, massive ship such as the Titanic (which couldn't change course quickly enough after the iceberg was spotted).

Momentum Involves Both Mass and Velocity To take into account both an object's mass and its velocity, we define a vector quantity called “momentum” that involves the product of mass (a scalar) and velocity (a vector). Instead of saying “the stronger the interaction, the bigger the change in the velocity, ” we now say “the stronger the interaction, the bigger the change in the momentum.” Momentum, a vector quantity, is usually represented by the symbol . We might expect that the mathematical expression for , and indeed this is almost, but not quite, correct. momentum would be simply Experiments on particles moving at very high speeds, close to the speed of light c = 3 × 108 m/s, show that changes in are not really proportional to the strength of the interactions. As you keep applying a force to a particle near the speed of light, the speed of the particle barely increases, and it is not possible to increase a particle's speed beyond the speed of light. Through experiments it has been found that changes in the following quantity are proportional to the amount of interaction:

DEFINITION OF MOMENTUM Momentum is defined as the product of mass times velocity, multiplied by a proportionality factor gamma:

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The proportionality factor γ (lower-case Greek gamma) is defined as

In these equations

represents momentum,

is the velocity of the object, m is the mass of the object,

is the

magnitude of the object's velocity (the speed), and c is the speed of light (3 × 108 m/s). Momentum has units of kg · m/s. To calculate momentum in these units, you must specify mass in kg and velocity in meters per second.

This is the “relativistic” definition of momentum. Albert Einstein in 1905 in his Special Theory of Relativity predicted that this would be the appropriate definition for momentum at high speeds, a prediction that has been abundantly verified in a wide range of experiments.

EXAMPLE Momentum of a Fast-Moving Proton Suppose that a proton (mass 1.7 × 10−27 kg) is traveling with a velocity of 2 × 107 m/s. (a) What is the momentum of the proton? (b) What is the magnitude of the momentum of the proton?

Solution (a)

(b)

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Approximate Expression for Momentum In the example above, we found that γ = 1.007. Since in that calculation we used only two significant figures, we could have used the approximation that γ ≈ 1.0 without affecting our answer. Let's examine the expression for γ to see whether we can come up with a guideline for when it is reasonable to use the approximate expression Looking at the expression for γ

we see that it depends only on the ratio of the speed of the object to the speed of light (the object's mass doesn't appear in this expression). If

is a very small number, then

, so γ ≈ 1.

APPROXIMATION: MOMENTUM AT LOW SPEEDS

Some values of

and γ are displayed in Figure 1.37. From this table you can see that even at the very high speed where

, which means that

, the relativistic factor γ is only slightly different from 1. For large-scale

objects such as a space rocket, whose speed is typically only about 1 × 104 m/s, we can ignore the factor γ, and momentum is to a good approximation . It is only for high-speed cosmic rays or particles produced in high-speed particle accelerators that we . need to use the full relativistic definition for momentum,

Figure 1.37 Values of γ calculated for some speeds. γ is shown to four decimal places, which is more accuracy than we will usually need.

From this table you can also get a sense of why it is not possible to exceed the speed of light. As you make a particle go faster and

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faster, approaching the speed of light, additional increases in the speed become increasingly difficult, because a tiny increase in speed means a huge increase in momentum, requiring huge amounts of interaction. In fact, for the speed to equal the speed of light, the momentum would have to increase to be infinite! There is a cosmic speed limit in the Universe, 3 × 108 m/s. We will repeatedly emphasize the role of momentum throughout this textbook because of its fundamental importance not only in classical (prequantum) mechanics but also in relativity and quantum mechanics. The use of momentum clarifies the physics analysis of certain complex processes such as collisions, including collisions at speeds approaching the speed of light. 1.X.37 A good sprinter can run 100 meters in 10 seconds. What is the magnitude of the momentum of a sprinter whose mass is 65 kg and who is running at a speed of 10 m/s? 1.X.38 What is the momentum of an electron traveling at a velocity of 0, 0, −2 × 108 m/s? (Masses of particles are given on the inside back cover of this textbook.) What is the magnitude of the momentum of the electron?

Answer

Answer

Direction of Momentum Like velocity, momentum is a vector quantity, so it has a magnitude and a direction. The direction is the same as the direction of the velocity.

QUESTION A leaf is blown by a gust of wind, and at a particular instant is traveling straight upward, in the +y direction. What is the direction of the leaf's momentum?

The mathematical expression for momentum can be looked at as the product of a scalar part times a vector part. Since the mass m must be a positive number, and the factor gamma (γ) must be a positive number, this scalar factor cannot change the direction of the vector (Figure 1.38). Therefore the direction of the leaf's momentum is the same as the direction of its velocity: straight up (the +y direction).

Figure 1.38 The expression for momentum is the product of a scalar times a vector. The scalar factor is positive, so the direction of an object's momentum is the same as the direction of its velocity.

1.X.39 A comet's momentum is in the direction of the arrow labeled b in Figure 1.39. What arrow best describes the direction of the comet's velocity?

Figure 1.39 A comet's momentum is in the direction of the arrow labeled b.

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Answer

Using Momentum to Update Position If you know the momentum of an object, you can calculate the change in position of the object over a given time interval. This is straightforward if the object is traveling at a speed low enough that the approximate expression for momentum can be used, since in this case . For speeds near the speed of light, we must use the exact expression

This expression was derived simply by rearranging the equation chapter.

. The detailed derivation is given at the end of the

EXAMPLE Displacement of a Fast Proton A proton with constant momentum 0, 0, 2.72 × 10−19 kg · m/s leaves the origin 10.0 seconds after an accelerator experiment is started. What is the location of the proton 2 ns later? (ns = nanosecond = 1 × 10−19 s)

Solution

The proton traveled 28 cm in 2 ns.

EXAMPLE Displacement of an Ice Skater An ice skater whose mass is 50 kg moves with constant momentum 400, 0, 300 kg · m/s. At a particular instant in her skating program she passes location 0, 0, 3 m. What was her location at a time 3 seconds earlier?

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Solution

1.X.40 A 4.5 kg bowling ball rolls down an alley with nearly constant momentum of 0.9, 0, 22.5 kg · m/s, starting from the origin. What will be the location of the ball after 2 seconds? 1.X.41 At time t1 = 12 s, a car with mass 1300 kg is located at 94, 0, 30 m and has momentum 4500, 0, −3000 kg · m/s. The car's momentum is not changing. At time t2 = 17 s, what is the position of the car?

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Answer

Answer

CHANGE OF MOMENTUM In the next chapter we will introduce “the Momentum Principle, ” which quantitatively relates change in momentum to the strength and duration of an interaction. In order to be able to use the Momentum Principle we need to know how to calculate changes in momentum. Momentum is a vector quantity and proportional to the velocity, so just as was the case with velocity, there are two aspects of momentum that can change: magnitude and direction. A mathematical description of change of momentum must include either a change in the magnitude of the momentum, or a change in the direction of the momentum, or both. As we will see in later chapters, pushing parallel to the motion changes the magnitude of the momentum, but to change the direction of the momentum requires pushing perpendicular (sideways) to the motion.

Change of the Vector Momentum The change in the momentum during a time interval is a vector: . This vector expression captures both to a changes in magnitude and changes in direction. Figure 1.40 is a graphical illustration of a change from an initial momentum . Place the initial and final momentum vectors tail to tail, then draw a vector from initial to final. This is the final momentum . This is the same procedure you used to calculate relative position vectors by subtraction, or vector representing displacement vectors by subtraction. The rule for subtracting vectors is always the same: Place the vectors tail to tail, then draw from the tip of the initial vector to the tip of the final vector. This resultant vector is “final minus initial.”

Figure 1.40 Calculation of

.

Change in Magnitude of Momentum If an object's speed changes (that is, the magnitude of its velocity changes), the magnitude of the object's momentum also changes. Pushing in the direction of the momentum (and velocity) makes the magnitude of the momentum increase (the speed increases); pushing backwards makes the magnitude decrease (the object slows down).

Change of Direction of Momentum There are various ways to specify a change in the direction of motion. For example, if you used compass directions, you could say that an airplane changed its direction from 30° east of north to 45° east of north: a 15° clockwise change. One can imagine various other schemes, involving other kinds of coordinate systems. The standard way to deal with this is to use vectors, as in Figure 1.41, which shows the changing momentum of an object traveling at constant speed along a circular path.

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Figure 1.41 The momentum of an object traveling in a circle changes, even if the magnitude of momentum does not change.

EXAMPLE Change in Momentum of a Ball Figure 1.42 shows a portion of the trajectory of a ball in air, subject to gravity and air resistance. At location B, the ball's momentum is . At location C, the ball's momentum is . (a) Find the change in the ball's momentum between these locations, and show it on the diagram. (b) Find the change in the magnitude of the ball's momentum.

Figure 1.42 A portion of the trajectory of a ball moving through air, subject to gravity and air resistance. The arrows represent the momentum of the ball at the locations indicated by letters.

Solution (a)

Both the x and y components of the ball's momentum decreased, so is consistent with the graphical subtraction shown in Figure 1.43.

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has negative x and y components. This

Figure 1.43 Graphical calculation of

. The result is also shown superimposed on the ball's path, midway between the initial and final locations.

(b)

FURTHER DISCUSSION The magnitude of the ball's momentum decreased, which means that there must have been a component of gravitational and air resistance forces acting on the ball in a direction opposite to the ball's motion, which slowed it down. The direction of the ball's momentum also changed, as is seen in Figure 1.42, which means that there must have been components of the forces acting perpendicular to the ball's motion.

EXAMPLE Change in Momentum and in Magnitude of Momentum Suppose you are driving a 1000 kilogram car at 20 m/s in the +x direction. After making a 180 degree turn, you drive the car at 20 m/s in the −x direction. (20 m/s is about 45 miles per hour or 72 km per hour.) (a) What is the change of magnitude of the momentum of the car

?

(b) What is the magnitude of the change of momentum of the car

?

Solution (a)

(b)

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So

.

FURTHER DISCUSSION This calculation shows something important. Change of magnitude, even more dramatic example:

, isn't the same as magnitude of change,

. Here's an

QUESTION A ball with momentum 5, 0, 0 kg · m/s rebounds from a wall with nearly the same speed, so its final momentum is ? What is ? approximately −5, 0, 0 kg · m/s. What is

The change of the magnitude of the momentum, is

, is zero (5−5 = 0), but the magnitude of the change of the momentum, .

1.X.42 A tennis ball of mass 57 g travels with velocity 50, 0, 0 m/s toward a wall. After bouncing off the wall, the tennis ball is observed to be traveling with velocity −48, 0, 0 m/s.

Answer

(a) Draw a diagram showing the initial and final momentum of the tennis ball. (b) What is the change in the momentum of the tennis ball? (c) What is the change in the magnitude of the tennis ball's momentum? 1.X.43 The planet Mars has a mass of 6.4 × 1023 kg, and travels in a nearly circular orbit around the Sun, as shown in Figure 1.44. When it is at location A, the velocity of Mars is 0, 0, −2.5 × 104 m/s. When it reaches

Answer

location B, the planet's velocity is −2.5 × 104, 0, 0 m/s. We're looking down on the orbit from above the north poles of the Sun and Mars, with +x to the right and +z down the page. (a) What is

, the change in the momentum of Mars between locations A and B?

(b) On a copy of the diagram in Figure 1.44, draw two arrows representing the momentum of Mars at locations C and D, paying attention to both the length and direction of each arrow. (c) What is the direction of the change in the momentum of Mars between locations C and D? Draw the vector on your diagram.

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,

Figure 1.44 The nearly circular orbit of Mars around the Sun, viewed from above the orbital plane (+x to the right, +z down the page). Not to scale: the sizes of the Sun and Mars are exaggerated.

Average Rate of Change of Momentum The rate of change of the vector position is such an important quantity that it has a special name: “velocity.” In Section 1.7 we discussed how to find both the average rate of change of position (average velocity) and the instantaneous rate of change of position (instantaneous velocity) of an object. The average rate of change of momentum and the instantaneous rate of change of momentum are also extremely important quantities. In some situations, we will only be able to find an average rate of change of momentum:

AVERAGE RATE OF CHANGE OF MOMENTUM

This quantity is a vector, and points in the direction of

.

EXAMPLE Average Rate of Change of Momentum If the momentum of a ball changes from 1, 2, 0 kg · m/s to 0.5, 0, 0.5 kg · m/s in half a second, the average rate of change of momentum of the ball is

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Parallel and Perpendicular Change In Figure 1.45 we see the parallel and perpendicular components of a change of momentum, . To make a nonzero , involving a change in direction, requires pushing perpendicular (sideways) to the initial momentum perpendicular component .

Figure 1.45 The parallel and perpendicular components of the momentum change,

.

1.X.44 A hockey puck was sliding on ice with momentum 0, 0, 10 kg · m/s when it was kicked, and the new momentum was 7, 0, 12 kg · m/s. (a) What was (b) What was

? ? It helps to draw a diagram.

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Answer

*THE PRINCIPLE OF RELATIVITY Sections marked with a “*” are optional. They provide additional information and context, but later sections of the textbook don't depend critically on them. This optional section deals with some deep issues about the “reference frame” from which you observe motion. Newton's first law of motion only applies in an “inertial reference frame, ” which we will discuss here in the context of the principle of relativity. A great variety of experimental observations has led to the establishment of the following principle:

THE PRINCIPLE OF RELATIVITY Physical laws work in the same way for observers in uniform motion as for observers at rest.

This principle is called “the principle of relativity.” (Einstein's extensions of this principle are known as “special relativity” and “general relativity.”) Phenomena observed in a room in uniform motion (for example, on a train moving with constant speed on a smooth straight track) obey the same physical laws in the same way as experiments done in a room that is not moving. According to this principle, Newton's first law of motion should be true both for an observer moving at constant velocity and for an observer at rest. For example, suppose that you're riding in a car moving with constant velocity, and you're looking at a map lying on the dashboard. As far as you're concerned, the map isn't moving, and no interactions are required to hold it still on the dashboard. Someone standing at the side of the road sees the car go by, sees the map moving at a high speed in a straight line, and can see that no interactions are required to hold the map still on the dashboard. Both you and the bystander agree that Newton's first law of motion is obeyed: the bystander sees the map moving with constant velocity in the absence of interactions, and you see the map not moving at all (a zero constant velocity) in the absence of interactions. On the other hand, if the car suddenly speeds up, it moves out from under the map, which ends up in your lap. To you it looks like “the map sped up in the backwards direction” without any interactions to cause this to happen, which looks like a violation of Newton's first law of motion. The problem is that you're strapped to the car, which is an accelerated reference frame, and Newton's first law of motion applies only to nonaccelerated reference frames, called “inertial” reference frames. Similarly, if the car suddenly turns to the right, moving out from under the map, the map tends to keep going in its original direction, and to you it looks like “the map moved to the left” without any interactions. So a change of speed or a change of direction of the car (your reference frame) leads you to see the map behave in a strange way. The bystander, who is in an inertial (nonaccelerating) reference frame, doesn't see any violation of Newton's first law of motion. The bystander's reference frame is an inertial frame, and the map behaves in an understandable way, tending to keep moving with unchanged speed and direction when the car changes speed or direction.

The Cosmic Microwave Background The principle of relativity, and Newton's first law of motion, apply only to observers who have a constant speed and direction (or zero speed) relative to the “cosmic microwave background, ” which provides the only backdrop and frame of reference with an absolute, universal character. It used to be that the basic reference frame was loosely called “the fixed stars, ” but stars and galaxies have their own individual motions within the Universe and do not constitute an adequate reference frame with respect to which to measure motion. The cosmic microwave background is low-intensity electromagnetic radiation with wavelengths in the microwave region, which pervades the Universe, radiating in all directions. Measurements show that our galaxy is moving through this microwave radiation with a large, essentially constant velocity, toward a cluster of a large number of other galaxies. The way we detect our motion relative to the microwave background is through the “Doppler shift” of the frequencies of the microwave radiation, toward higher frequencies in front of us and lower frequencies behind. This is essentially the same phenomenon as that responsible for a fire

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engine siren sounding at a higher frequency when it is approaching us and a lower frequency when it is moving away from us. The discovery of the cosmic microwave background provided major support for the “Big Bang” theory of the formation of the Universe. According to the Big Bang theory, the early Universe must have been an extremely hot mixture of charged particles and high-energy, short-wavelength electromagnetic radiation (visible light, x-rays, gamma rays, etc.). Electromagnetic radiation interacts strongly with charged particles, so light could not travel very far without interacting, making the Universe essentially opaque. Also, the Universe was so hot that electrically neutral atoms could not form without the electrons immediately being stripped away again by collisions with other fast-moving particles. As the Universe expanded, the temperature dropped. Eventually the temperature was low enough for neutral atoms to form. The interaction of electromagnetic radiation with neutral atoms is much weaker than with individual charged particles, so the radiation was now essentially free, dissociated from the matter, and the Universe became transparent. As the Universe continued to expand (the actual space between clumps of matter got bigger), the wavelengths of the electromagnetic radiation got longer, until today this fossil radiation has wavelengths in the relatively low-energy, long-wavelength microwave portion of the electromagnetic spectrum.

Inertial Frames of Reference It is an observational fact that in reference frames that are in uniform motion with respect to the cosmic microwave background, far from other objects (so that interactions are negligible), an object maintains uniform motion. Such frames are called “inertial frames” and are reference frames in which Newton's first law of motion is valid. All of these reference frames are equally valid; the cosmic microwave background simply provides a concrete example of such a reference frame.

QUESTION Is the surface of the Earth an inertial frame?

No! The Earth is rotating on its axis, so the velocity of an object sitting on the surface of the Earth is constantly changing direction, as is a coordinate frame tied to the Earth (Figure 1.46). Moreover, the Earth is orbiting the Sun, and the Solar System itself is orbiting the center of our Milky Way galaxy, and our galaxy is moving toward other galaxies. So the motion of an object sitting on the Earth is actually quite complicated and definitely not uniform with respect to the cosmic microwave background.

Figure 1.46 Axes tied to the Earth rotate through 90° in a quarter of a day (6 hours).

However, for many purposes the surface of the Earth can be considered to be (approximately) an inertial frame. For example, it takes 6 hours for the rotation of the Earth on its axis to make a 90° change in the direction of the velocity of a “fixed” point. If a

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process of interest takes only a few minutes, during these few minutes a “fixed” point moves in nearly a straight line at constant speed due to the Earth's rotation, and velocity changes in the process of interest are typically much larger than the very small velocity change of the approximate inertial frame of the Earth's surface. Similarly, although the Earth is in orbit around the Sun, it takes 365 days to go around once, so for a period of a few days or even weeks the Earth's orbital motion is nearly in a straight line at constant speed. Hence for many purposes the Earth represents an approximately inertial frame despite its motion around the Sun.

The Special Theory of Relativity Einstein's Special Theory of Relativity (published in 1905) built on the basic principle of relativity but added the conjecture that the speed of a beam of light must be the same as measured by observers in different frames of reference in uniform motion with respect to each other. In Figure 1.47, observers on each spaceship measure the speed of the light c emitted by the ship at the top to be the same (c = 3 × 108) m/s, despite the fact that they are moving at different velocities.

Figure 1.47 Light emitted by the top spaceship is measured to have the same speed by observers in all three ships.

This additional condition seems peculiar and has far-reaching consequences. After all, the map on the dashboard of your car has different speeds relative to different observers, depending on the motion of the observer. Yet a wide range of experiments has confirmed Einstein's conjecture: all observers measure the same speed for the same beam of light, c = 3 × 108 m/s. (The color of the light is different for the different observers, but the speed is the same.)

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On the other hand, if someone on the ship at the top throws a ball or a proton or some other piece of matter, the speed of the object will be different for observers on the three ships; it is only light whose speed is independent of the observer. Einstein's theory has interesting consequences. For example, it predicts that time will run at different rates in different frames of reference. These predictions have been confirmed by many experiments. These unusual effects are large only at very high speeds (a sizable fraction of the speed of light), which is why we don't normally observe these effects in everyday life, and why we can use nonrelativistic calculations for low-speed phenomena. However, for the Global Positioning System (GPS) to give adequate accuracy it is necessary to take Einstein's Special Theory of Relativity into account. The atomic clocks on the satellites run slower than our clocks, due to the speed of the satellites, and the difference in clock rate depends on γ. Although γ for the GPS satellites orbiting the Earth is nearly 1, it differs just enough that making the approximation γ ≈ 1 would make the GPS hopelessly inaccurate and useless. It is even necessary to apply corrections based on Einstein's General Theory of Relativity, which correctly predicts an additional change in clock rate due to the gravity of the Earth being weaker at the high altitude of the GPS satellites. 1.X.45 A spaceship at rest with respect to the cosmic microwave background emits a beam of red light. A different spaceship, moving at a speed of 2.5 × 108 m/s toward the first ship, detects the light. Which of the following statements are true for observers on the second ship? (More than one statement may be correct.) (a) They observe that the light travels at 3 × 108 m/s. (b) They see light that is not red. (c) They observe that the light travels at 5.5 × 108 m/s. (d) They observe that the light travels at 2.5 × 108 m/s.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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Answer

*UPDATING POSITION AT HIGH SPEED If and . But at high speed it is more complicated to determine the velocity from the (relativistic) in terms of : momentum. Here is a way to solve for

The expression above gives the magnitude of , in terms of the magnitude of . To get an expression for the vector that any vector can be factored into its magnitude times a unit vector in the direction of the vector, so

But since

and

are in the same direction,

, so

THE RELATIVISTIC POSITION UPDATE FORMULA

Note that at low speeds

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, and the denominator is

, recall

so the formula becomes the familiar

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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SUMMARY Interactions (Section 1.2 and Section 1.4) Interactions are indicated by Change of velocity (change of direction and/or change of speed) Change of identity Change of shape of multiparticle system Change of temperature of multiparticle system Lack of change when change is expected

Newton's first law of motion (Section 1.3) An object moves in a straight line and at constant speed, except to the extent that it interacts with other objects. Vectors (Section 1.5) A 3D vector is a quantity with magnitude and a direction, which can be expressed as a triple x, y, z . A vector is indicated by an arrow: . A scalar is a single number. Legal mathematical operations involving vectors include: Adding one vector to another vector Subtracting one vector from another vector Multiplying or dividing a vector by a scalar Finding the magnitude of a vector Taking the derivative of a vector

Operations that are not legal with vectors include: A vector cannot be added to a scalar. A vector cannot be set equal to a scalar. A vector cannot appear in the denominator (you can't divide by a vector). A unit vector

has magnitude 1.

A vector can be factored using a unit vector:

.

Direction cosines: The symbol Δ The symbol Δ (delta) means “change of”: Δt = tf − ti , Δ always means “final minus initial.”

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.

Velocity and change of position (Section 1.7) Definition of average velocity

Velocity is a vector. is the position of an object (a vector). t is the time. Average velocity is equal to the change in position divided by the time elapsed. SI units of velocity are meters per second (m/s). The position update formula

The final position (vector) is the vector sum of the initial position plus the product of the average velocity and the elapsed time. Definition of instantaneous velocity

The instantaneous velocity is the limiting value of the average velocity as the time elapsed becomes very small. Velocity in terms of momentum

Momentum (Section 1.8) Definition of momentum

where

(lower-case Greek gamma)

Momentum (a vector) is the product of the relativistic factor “gamma” (a scalar), mass, and velocity.

Combined into one equation:

.

Approximation for momentum at low speeds

Useful numbers: Radius of a typical atom: about 1 × 10−10 meter Radius of a proton or neutron: about 1 × 10−15 meter Speed of light: 3 × 108 m/s These and other useful data and conversion factors are given on the inside back cover of the textbook.

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Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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EXERCISES AND PROBLEMS Exercises (“X”) are small-scale questions that deal with basic definitions and quantities and require few steps in reasoning. Their purpose is to give you practice in dealing with the basic concepts. Problems (“P”) are larger applications that require some thought and may involve several steps in reasoning. The problems give you practice in applying what you are learning.

Section 1.2 1.X.46 Give two examples (other than those discussed in the text) of interactions that may be detected by observing: (a) Change in velocity (b) Change in temperature (c) Change in shape (d) Change in identity (e) Lack of change when change is expected 1.X.47 Which of the following observations give conclusive evidence of an interaction? (Choose all that apply.) (a) Change of velocity, either change of direction or change of speed (b) Change of shape or configuration without change of velocity (c) Change of position without change of velocity (d) Change of identity without change of velocity (e) Change of temperature without change of velocity. Explain your choice. 1.X.48 Moving objects left the traces labeled A–F in Figure 1.48. and Figure 1.49. The dots were deposited at equal time intervals (for example, one dot each second). In each case the object starts from the square. Which trajectories show evidence that the moving object was interacting with another object somewhere? If there is evidence of an interaction, what is the evidence?

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Figure 1.48 Exercise 1.X.48. Also see Figure 1.49.

Figure 1.49 Exercise 1.X.48. Also see Figure 1.48.

Section 1.3 1.X.49 Which of the following observers might observe something that appears to violate Newton's first law of motion? Explain why. (a) A person standing still on a street corner (b) A person riding on a roller coaster

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(c) A passenger on a starship traveling at 0.75c toward the nearby star Alpha Centauri (d) An airplane pilot doing aerobatic loops (e) A hockey player coasting across the ice 1.X.50 Why do we use a spaceship in outer space, far from other objects, to illustrate Newton's first law? Why not a car or a train? (More than one of the following statements may be correct.) (a) A car or train touches other objects, and interacts with them. (b) A car or train can't travel fast enough. (c) The spaceship has negligible interactions with other objects. (d) A car or train interacts gravitationally with the Earth. (e) A spaceship can never experience a gravitational force. 1.X.51 A spaceship far from all other objects uses its thrusters to attain a speed of 1 × 104 m/s. The crew then shuts off the power. According to Newton's first law, what will happen to the motion of the spaceship from then on? 1.X.52 Some science museums have an exhibit called a Bernoulli blower, in which a volleyball hangs suspended in a column of air blown upward by a strong fan. If you saw a ball suspended in the air but didn't know the blower was there, why would Newton's first law suggest that something must be holding the ball up? 1.X.53 Place a ball on a book and walk with the book in uniform motion. Note that you don't really have to do anything to the ball to keep the ball moving with constant velocity (relative to the ground) or to keep the ball at rest (relative to you). Then stop suddenly, or abruptly change your direction or speed. What does Newton's first law of motion predict for the motion of the ball (assuming that the interaction between the ball and the book is small)? Does the ball behave as predicted? It may help to take the point of view of a friend who is standing still, watching you.

Section 1.5 Vector equality; components 1.X.54 Consider a vector , and another vector statements must be true? Some, all, or none of the following may be true:

. If

, then which of the following

(a) ux = px (b) uy = py (c) uz = pz (d) The direction of

is the same as the direction of

.

Vectors and scalars 1.X.55 Does the symbol

represent a vector or a scalar?

1.X.56 Which of the following are vectors? (a) 5 m/s (b)

−11, 5.4, −33 m

(c) (d) v z 1.X.57 Does the symbol

represent a vector or a scalar?

1.X.58 Which of the following are vectors? (a) 3.5 (b) 0 (c) 0.7, 0.7, 0.7 (d) 0, 2.3, −1 (e) −3 × 106 (f) 3 · 14, 0, −22 1.X.59 Which of the following are vectors? (a)

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(b)

(c)

(d) 5 ·

Magnitude of a vector 1.X.60 What is the magnitude of the vector

, where

?

1.X.61 Figure 1.50 shows several arrows representing vectors in the xy plane. (a) Which vectors have magnitudes equal to the magnitude of (b) Which vectors are equal to

?

?

Figure 1.50 Exercise 1.X.61. 1.X.62 In Figure 1.51 three vectors are represented by arrows in the xy plane. Each square in the grid represents one meter. For each vector, write out the components of the vector, and calculate the magnitude of the vector.

Figure 1.51 Exercise 1.X.62.

Illegal Operations with Vectors 1.X.63 Is 3 + 2, −3, 5 a meaningful expression? If so, what is its value?

Multiplying a Vector by a Scalar 1.X.64 On a piece of graph paper, draw arrows representing the following vectors. Make sure the tip and tail of each arrow you draw are clearly distinguishable. (a) Placing the tail of the vector at 5, 2, 0 , draw an arrow representing the vector (b) Placing the tail of the vector at −5, 8, 0 , draw an arrow representing the vector 1.X.65 A proton is located at statements are true? (a)

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. An electron is located at

. Label it . Label it

.

.

. Which of the following

(b) (c)

1.X.66 The following questions refer to the vectors depicted by arrows in Figure 1.52. (a) What are the components of the vector

? (Note that since the vector lies in the xy plane, its z component is zero.)

(b) What are the components of the vector

?

(c) Is this statement true or false? (d) What are the components of the vector

?

(e) Is this statement true or false? (f) What are the components of the vector

?

(g) Is this statement true or false?

Figure 1.52 Exercise 1.X.66. 1.X.67

(a) In Figure 1.53, what are the components of the vector (b) If

, what are the components of

(c) If the tail of vector (d) If the tail of vector

were moved to location

?

? , where would the tip of the vector be located?

were placed at location −1, −1, −1 m, where would the tip of the vector be located?

Figure 1.53 Exercise 1.X.67.

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by the scalar f, where 0.02, −1.7, 30.0 and f = 2.0?

1.X.68 What is the result of multiplying the vector 1.X.69

, putting the tail of the vector at −3, 0, 0 . Label the

(a) On a piece of graph paper, draw the vector vector

.

(b) Calculate the vector

, and draw this vector on the graph, putting its tail at −3, −3, 0 , so you can compare it to

the original vector. Label the vector (c) How does the magnitude of

compare to the magnitude of

(d) How does the direction of (e) Calculate the vector

.

compare to the direction of

? ?

, and draw this vector on the graph, putting its tail at −3, −6, 0 , so you can compare it

to the other vectors. Label the vector (f) How does the magnitude of

.

compare to the magnitude of

(g) How does the direction of

compare to the direction of

? ?

(h) Does multiplying a vector by a scalar change the magnitude of the vector? (i)

The vector

has a magnitude three times as great as that of

, and its direction is opposite to the direction of

. What is the value of the scalar factor a?

Unit vectors 1.X.70 What is the unit vector in the direction of −300, 0, 0 ? 1.X.71 What is the unit vector in the direction of 2, 2, 2 ? What is the unit vector in the direction of 3, 3, 3 ? 1.X.72 Write the vector 1.X.73

as the product

.

(a) On a piece of graph paper, draw the vector (b) Calculate the magnitude of

. Put the tail of the vector at the origin.

.

(c) Calculate , the unit vector pointing in the direction of

.

(d) On the graph, draw . Put the tail of the vector at 1, 0, 0 m so you can compare (e) Calculate the product of the magnitude

times the unit vector ,

1.X.74 A proton is located at (a) What is

and

.

.

.

, the vector from the origin to the location of the proton?

(b) What is

?

(c) What is , the unit vector in the direction of

?

1.X.75 Write each of these vectors as the product of the magnitude of the vector and the appropriate unit vector: (a)

0, 0, 9.5

(b)

0, −679, 0

(c)

3.5 × 10−3, 0, −3.5 × 10−3

(d)

4 × 106, −6 × 106, 3 × 106

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1.X.76 A cube is 3 cm on a side, with one corner at the origin. What is the unit vector pointing from the origin to the diagonally opposite corner at location 3, 3, 3 cm? What is the angle from this diagonal to one of the adjacent edges of the cube? 1.X.77 Which of the following are unit vectors? (Numerical values are given to only 3 significant figures.) (a)

0, 0, −1

(b)

0.5, 0.5, 0

(c)

0.333, 0.333, 0.333

(d)

0.9, 0, 0.1

(e)

0, 3, 0

(f)

1, −1, 1

(g)

0.577, 0.577, 0.577

(h)

0.949, 0, −0.316

1.X.78 Two vectors,

and

, are equal:

. Which of the following statements are true?

(a) (b) gx = fx (c) fz = gz (d) The directions of (e) The magnitudes of

and and

may be different. may be different.

Vector addition and subtraction 1.X.79 Imagine that you have a baseball and a tennis ball at different locations. The center of the baseball is at 3, 5, 0 m, and the center of the tennis ball is at −3, −1, 0 m. On a piece of graph paper, do the following: (a) Draw dots at the locations of the center of the baseball and the center of the tennis ball. (b) Draw the position vector of the baseball, which is an arrow whose tail is at the origin and whose tip is at the location of the baseball. Label this position vector . Clearly show the tip and tail of each arrow. (c) Complete this equation:

= < _________ , _________ , _________ > m.

(d) Draw the position vector of the tennis ball. Label it (e) Complete this equation: (f)

.

= < _________ , _________ , _________ > m.

Draw the relative position vector for the tennis ball relative to the baseball. The tail of this vector is at the center of the baseball, and the tip of the vector is at the center of the tennis ball. Label this relative position vector .

(g) Complete the following equation by reading the coordinates of from your graph: , _________ > m. (h) Calculate this difference: = < _________ , _________ , _________ > m. (i)

Is the following statement true?

_________ .

(j)

Write two other equations relating the vectors

,

, and

, and

.

(k) Calculate the magnitudes of the vectors (l)

Calculate the difference of the magnitudes

(m) Does Change of a vector; relative position

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?

,

.

.

= < _________ , _________

1.X.80 In Figure 1.54, and . Calculate the position of object 2 relative to object 1, as a relative position vector. Before checking the answer, see whether your answer is consistent with the appearance of the vector shown in the diagram. What is the position of object 1 relative to object 2, as a vector?

Figure 1.54 Exercise 1.X.80. 1.X.81

(a) What is the vector whose tail is at 9.5, 7, 0 m and whose head is at 4, −13, 0 m? (b) What is the magnitude of this vector?

1.X.82 A man is standing on the roof of a building with his head at the position 12, 30, 13 m. He sees the top of a tree, which is at the position −25, 35, 43 m. (a) What is the relative position vector that points from the man's head to the top of the tree? (b) What is the distance from the man's head to the top of the tree? 1.X.83 A star is located at 6 × 1010, 8 × 1010, 6 × 1010 m. A planet is located at −4 × 1010, −9 × 1010, 6 × 1010 m. (a) What is the vector

pointing from the star to the planet?

(b) What is the vector

pointing from the planet to the star?

1.X.84 A planet is located at −1 × 1010, 8 × 1010, −3 × 1010 . A star is located at 6 × 1010, −5 × 1010, 1 × 1010 . (a) What is

, the vector from the star to the planet?

(b) What is the magnitude of (c) What is

?

, the unit vector (vector with magnitude 1) in the direction of

?

1.X.85 A proton is located at . An electron is located at . What is the vector pointing from the electron to the proton? What is the vector pointing from the proton to the electron?

Section 1.7 1.X.86 In the expression

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, what is the meaning of

? What is the meaning of Δ t?

1.X.87 A “slow” neutron produced in a nuclear reactor travels from location 0.2, −0.05, 0.1 m to location −0.202, 0.054, 0.098 m in 2 microseconds (1 μs = 1 × 10−6 s). (a) What is the average velocity of the neutron? (b) What is the average speed of the neutron? 1.X.88 The position of a baseball relative to home plate changes from 15, 8, −3 m to 20, 6, −1 m in 0.1 second. As a vector, write the average velocity of the baseball during this time interval. 1.P.89 A spacecraft traveling at a velocity of −20, −90, 40 m/s is observed to be at a location 200, 300, −500 m relative to an origin located on a nearby asteroid. At a later time the spacecraft is at location −380, −2310, 660 m. (a) How long did it take the spacecraft to travel between these locations? (b) How far did the spacecraft travel? (c) What is the speed of the spacecraft? (d) What is the unit vector in the direction of the spacecraft's velocity? 1.X.90 Start with the definition of average velocity and derive the position update formula from it. Show all steps in the derivation. 1.X.91 At time t1 = 12 s, a car is located at 84, 78, 24 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of the car? (The velocity is constant in magnitude and direction during this time interval.) 1.X.92 At a certain instant a ball passes location 7, 21, −17 m. In the next 3 seconds, the ball's average velocity is −11, 42, 11 m/s. At the end of this 3 second time interval, what is the height y of the ball? 1.X.93 Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time Position

6.3 s −3.5, 9.4, 0 m

6.8 s −1.3, 6.2, 0 m

7.3 s 0.5, 1.7, 0 m

(a) Between 6.3 s and 6.8 s, what was the bee's average velocity? Be careful with signs. (b) Between 6.3 s and 7.3 s, what was the bee's average velocity? Be careful with signs. (c) Of the two average velocities you calculated, which is the best estimate of the bee's instantaneous velocity at time 6.3 s? (d) Using the best information available, what was the displacement of the bee during the time interval from 6.3 s to 6.33 s? 1.X.94 The position of a golf ball relative to the tee changes from 50, 20, 30 m to 53, 18, 31 m in 0.1 second. As a vector, write the velocity of the golf ball during this short time interval. 1.X.95 The crew of a stationary spacecraft observe an asteroid whose mass is 4 × 1017 kg. Taking the location of the spacecraft as the origin, the asteroid is observed to be at location −3 × 103, −4 × 103, 8 × 103 m at a time 18.4 seconds after lunchtime. At a time 21.4 seconds after lunchtime, the asteroid is observed to be at location −1.4 × 103, −6.2 × 103, 9.7 × 103 m. Assuming that the velocity of the asteroid does not change during this time interval, calculate the vector velocity of the asteroid. 1.X.96 An electron passes location 0.02, 0.04, −0.06 m, and 2 μs later is detected at location 0.02, 1.84, −0.86 m (1 microsecond is 1 × 10−6 s). (a) What is the average velocity of the electron? (b) If the electron continues to travel at this average velocity, where will it be in another 5 μs? 1.P.97 Figure 1.55 shows the trajectory of a ball traveling through the air, affected by both gravity and air resistance. Here are the positions of the ball at several successive times:

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location

t(s)

position (m)

A

0.0

0, 0, 0

B

1.0

22.3, 26.1, 0

C

2.0

40.1, 38.1, 0

(a) What is the average velocity of the ball as it travels between location A and location B? (b) If the ball continued to travel at the same average velocity during the next second, where would it be at the end of that second? (That is, where would it be at time t = 2 seconds?) (c) How does your prediction from part (b) compare to the actual position of the ball at t = 2 seconds (location C)? If the predicted and observed locations of the ball are different, explain why.

Figure 1.55 Problem 1.P.97. 1.P.98 At 6 seconds after 3:00, a butterfly is observed leaving a flower whose location is 6, −3, 10 m relative to an origin on top of a nearby tree. The butterfly flies until 10 seconds after 3:00, when it alights on a different flower whose location is 6.8, −4.2, 11.2 m relative to the same origin. What was the location of the butterfly at a time 8.5 seconds after 3:00? What assumption did you have to make in calculating this location?

Section 1.8 1.X.99 Which of the following statements about the velocity and momentum of an object are correct? (a) The momentum of an object is always in the same direction as its velocity. (b) The momentum of an object can be either in the same direction as its velocity or in the opposite direction. (c) The momentum of an object is perpendicular to its velocity. (d) The direction of an object's momentum is not related to the direction of its velocity. (e) The direction of an object's momentum is tangent to its path. 1.X.100 In which of these situations is it reasonable to use the approximate formula for the momentum of an object, instead of the full relativistically correct formula? (a) A car traveling on an interstate highway (b) A commercial jet airliner flying between New York and Seattle (c) A neutron traveling at 2700 meters per second (d) A proton in outer space traveling at 2 × 108 m/s (e) An electron in a television tube traveling 3 × 106 m/s 1.X.101 Answer the following questions about the factor γ (gamma) in the full relativistic formula for momentum: (a) Is γ a scalar or a vector quantity?

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(b) What is the minimum possible value of γ? (c) Does γ reach its minimum value when an object's speed is high or low? (d) Is there a maximum possible value for γ? (e) Does γ become large when an object's speed is high or low? (f) Does the approximation γ ≈ 1 apply when an object's speed is low or when it is high? 1.X.102 A baseball has a mass of 0.155 kg. A professional pitcher throws a baseball 90 miles per hour, which is 40 m/s. What is the magnitude of the momentum of the pitched baseball? 1.X.103 A proton in an accelerator attains a speed of 0.88c. What is the magnitude of the momentum of the proton? 1.X.104 A hockey puck with a mass of 0.4 kg has a velocity of 38, 0, −27 m/s. What is the magnitude of its momentum,

?

1.X.105 A baseball has a mass of about 155 g. What is the magnitude of the momentum of a baseball thrown at a speed of 100 miles per hour? (Note that you need to convert mass to kilograms and speed to meters/second. See the inside back cover of the textbook for conversion factors.) 1.X.106 What is the magnitude (in kg · m/s) of the momentum of a 1000 kg airplane traveling at a speed of 500 miles per hour? (Note that you need to convert speed to meters per second.) 1.X.107 If a particle has momentum

, what is the magnitude

of its momentum?

1.X.108 An electron with a speed of 0.95c is emitted by a supernova, where c is the speed of light. What is the magnitude of the momentum of this electron? 1.X.109 A “cosmic-ray” proton hits the upper atmosphere with a speed 0.9999c, where c is the speed of light. What is the magnitude of the momentum of this proton? Note that ; you don't actually need to calculate the speed . 1.X.110 A proton in an accelerator is traveling at a speed of 0.99c. (Masses of particles are given on the inside back cover of this textbook.) (a) If you use the approximate nonrelativistic formula for the magnitude of momentum of the proton, what answer do you get? (b) What is the magnitude of the correct relativistic momentum of the proton? (c) The approximate value (the answer to part a) is significantly too low. What is the ratio of magnitudes you calculated (correct/approximate)? Such speeds are attained in particle accelerators. 1.X.111 An object with mass 1.6 kg has momentum 0, 0, 4 kg · m/s. (a) What is the magnitude of the momentum? (b) What is the unit vector corresponding to the momentum? (c) What is the speed of the object? 1.X.112 When the speed of a particle is close to the speed of light, the factor γ, the ratio of the correct relativistic momentum to the approximate nonrelativistic momentum , is quite large. Such speeds are attained in particle accelerators, and at these speeds the approximate nonrelativistic formula for momentum is a very poor approximation. Calculate γ for the case . where 1.X.113 An electron travels at speed

, where c = 3 × 108 m/s is the speed of light. The electron travels in the

direction given by the unit vector

= 0.655, −0.492, −0.573 . The mass of an electron is 9 × 10−31 kg.

(a) What is the value of

? You can simplify the calculation if you notice that .

(b) What is the speed of the electron?

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(c) What is the magnitude of the electron's momentum? (d) What is the vector momentum of the electron? Remember that any vector can be “factored” into its magnitude times its unit vector, so that

.

1.X.114 If p/m is 0.85c, what is v in terms of c?

Section 1.9 1.X.115 A tennis ball of mass m traveling with velocity v x , 0, 0 hits a wall and rebounds with velocity −v x , 0, 0 . (a) What was the change in momentum of the tennis ball? (b) What was the change in the magnitude of the momentum of the tennis ball? 1.X.116 A 50 kg child is riding on a carousel (merry-go-round) at a constant speed of 5 m/s. What is the magnitude of the change in the child's momentum in going all the way around (360°)? In going halfway around (180°)? Draw a diagram showing the initial vector momentum and the final vector momentum, then subtract, then find the magnitude. 1.P.117 Figure 1.56 shows a portion of the trajectory of a ball traveling through the air. Arrows indicate its momentum at several locations. At various locations, the ball's momentum is:

(a) Calculate the change in the ball's momentum between each pair of adjacent locations. (b) On a copy of the diagram, draw arrows representing each

you calculated in part (a).

(c) Between which two locations is the magnitude of the change in momentum greatest?

Figure 1.56 Problem 1.P.117.

Computational Problems These problems are intended to introduce you to using a computer to model matter, interactions, and motion. You will build on these small calculations to build models of physical systems in later chapters. Some parts of these problems can be done with almost any tool (spreadsheet, math package, etc.). Other parts are most easily done with a programming language. We recommend the free 3D programming environment VPython, obtainable at http://vpython.org. Your instructor will introduce you to an available computational tool and assign problems, or parts of problems, that can be addressed using the chosen tool. 1.P.118

(a) Write a program that makes an object move from left to right across the screen with velocity v x , 0, 0 . Make v x a

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variable, so you can change it later. Let the time interval for each step of the computation be a variable dt, so that the position x increases by an amount v x dt each time. (b) Modify a copy of your program to make the object run into a wall and reverse its direction. (c) Make a modification so that v x is no longer a constant but changes smoothly with time. Is the change of the speed of the object clearly visible to an observer? Try to make one version in which the speed change is clearly noticeable, and another in which it is not noticeable. (d) Corresponding to part (c), make a computer graph of x vs. t, where t is the time. (e) Corresponding to part (c), make a computer graph of v x vs. t, where t is the time. 1.P.119

(a) Write a program that makes an object move at an angle. (b) Change the component of velocity of the object in the x direction but not in the y direction, or vice versa. What do you observe? (c) Start the object moving at an angle and make it bounce off at an appropriate angle when it hits a wall.

1.P.120 Write a program that makes an object move smoothly from left to right across the screen with speed v, leaving a trail of dots on the screen at equal time intervals. If the dots are too close together, leave a dot every N steps, and adjust N to give a nice display. 1.P.121 This problem requires a 3D programming environment. Assume SI units, and give the objects reasonable and distinguishable attributes (such as sizes and colors). Run the program after each part, to make sure your display is correct. (a) Create a sphere representing a basketball, at location −5, 2, −3 m. (b) Create an arrow to represent the position of the basketball, relative to the origin. (c) Create a second sphere to represent a volleyball, at location −3, −1, 3.5 m. (d) Create an arrow to represent the position of the volleyball relative to the origin. (e) Using symbolic names (no numbers) for the positions and attributes of these balls, create an arrow to represent the position of the volleyball relative to the basketball. (f) Change the positions of the basketball and the volleyball to −4, 2, 5 m and 3, 1, −2 m, respectively. If you correctly used symbolic names in the previous part, the arrow representing the relative position vector should automatically adjust. Does it? (g) Add to your program instructions to print the position of each ball, and the position of the tip and the tail of the arrow (all 3D vectors). 1.P.122 Write a program to display the motion of a cart traveling on a track at constant velocity. (a) Create a box to represent a track 2 meters long, 10 cm wide, and 5 cm high. Orient the long axis of the track along the x axis. (b) Create a second box to represent a low-friction cart that is 10 cm long, 4 cm high, and 6 cm wide. Give this box a symbolic name so you can refer to it later in the program. (c) Position the cart so it is 1 cm above the track, and its left edge lines up with the left end of the track. (d) Create a vector variable to represent the velocity of the cart. Give the cart an initial velocity of 0.2, 0, 0 m/s. (e) Using a time step of 0.01 seconds, write a loop to use the position update formula to animate the motion of the cart as it travels at constant velocity from one end of the track to the other. (f) Change your program to start the cart at the right end of the track and move to the left. (g) What would happen if the cart had a nonzero y velocity component? Try it. (h) Have your cart leave a trail behind it as it moves. 1.P.123 Write a program to display the motion of a hockey puck sliding at constant velocity on ice. (a) Represent the ice surface by a box that is large in the x and z dimensions and thin in the y dimension. Represent the puck by a small cylinder, positioned just above the ice surface. (If you do not have a 3D programming environment,

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represent the puck by a circle rather than a cylinder.) Give the puck a symbolic name so you can refer to it later in the program. (b) Position the puck at one corner of the ice surface. (c) Create a vector variable to represent the velocity of the puck. Assign velocity components so the puck will slide diagonally across the ice. (d) Using a time step of 0.001 second, write a loop to use the position update formula to animate the motion of the puck as it slides from one corner of the ice surface to another.

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The Momentum Principle

A fundamental principle relates the change in a quantity to the interaction causing the change, and it is powerful because it applies in absolutely every situation. The Momentum Principle is the first of three fundamental principles of mechanics that together make it possible to predict and explain a very broad range of real-world phenomena (the other two are the Energy Principle and the Angular Momentum Principle). The Momentum Principle makes a quantitative connection between amount of interaction and change of momentum.

KEY IDEAS One or more objects can be considered to be a “system.”

▪ Everything not in the system is part of the “surroundings.” The momentum of a system can be changed only by interactions with the surroundings.

▪ Force is a quantitative measure of interaction. ▪ Force is a vector (it has magnitude and direction). The change in momentum of a system is equal to the net impulse applied, which involves:

▪ The total force exerted on it by all objects in the surroundings, and ▪ The time interval over which the interaction occurs. The motion of an object can be predicted by iteratively updating its momentum and position.

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Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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SYSTEM AND SURROUNDINGS One or more objects can be considered to be a “system.” Everything that is not included in the system is part of the “surroundings.” The Momentum Principle relates the change in momentum of a system to the amount of interaction with its surroundings. No matter what system we choose, the Momentum Principle will correctly predict the behavior of the system. In this chapter we will always consider a single object to be the system, and everything else in the Universe to be the surroundings. However, in later chapters we will see that it is possible, and often useful, to choose a system comprising several objects. A fundamental principle such as the Momentum Principle applies to any system, no matter how complex.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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THE MOMENTUM PRINCIPLE The Momentum Principle is a fundamental principle that is also known as Newton's second law. It restates and extends Newton's first law of motion in a quantitative, causal form that can be used to predict the behavior of objects. The validity of the Momentum Principle has been verified through a very wide variety of observations and experiments, involving large and small objects, moving slowly or at speeds near the speed of light. It is a summary of the way interactions affect motion in the real world.

The Momentum Principle is a fundamental principle because: It applies to every possible system, no matter how large or small (from clusters of galaxies to subatomic particles), and no matter how fast it is moving. It is true for every kind of interaction (electric, gravitational, etc.) It relates an effect (change in momentum) to a cause (an interaction).

THE MOMENTUM PRINCIPLE

The change of momentum of a system (the effect) is equal to the net force acting on the system times the duration of the interaction (the cause). The time interval Δt must be small enough that the net force is nearly constant during this time interval. As usual, the capital Greek letter delta (Δ) means “change of” (something), or “final minus initial.” We will see that the time interval Δt must be small enough that the net force on the system does not change significantly during this time interval. To begin to understand this equation, we will consider each quantity involved.

Change of Momentum

As we saw in the previous chapter, the change in momentum of a system can involve a change in the magnitude of momentum a change in the direction of momentum change in both magnitude and direction

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2.X.1 A ball's momentum changes from

to

. What is

?

Answer

Force

Scientists and engineers employ the concept of “force” to quantify interactions between two objects. Force is a vector quantity because a force has a magnitude and is exerted in a particular direction. Examples of forces include the following: The repulsive electric force a proton exerts on another proton The attractive gravitational force the Earth exerts on you The force that a compressed spring exerts on your hand The force on a spacecraft of expanding gases in a rocket engine The force of the surrounding air on the propeller of an airplane or airboat

Measuring a Force Measuring the magnitude of the velocity of an object (in other words, measuring its speed) is a familiar task, but how do we measure the magnitude of a force? A simple way to measure force is to use the stretch or compression of a spring. In Figure 2.1 we hang a block from a spring, and note that the spring is stretched a distance s. Then we hang two such blocks from the spring, and we see that the spring is stretched twice as much. By experimentation, we find that any spring made of the same material and produced to the same specifications behaves in the same way.

Figure 2.1 Stretching of a spring is a measure of force.

Similarly, we can observe how much the spring compresses when the same blocks are supported by it (Figure 2.2). We find that one

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block compresses the spring by the same distance , and two blocks compress it by stretch, because the length of the spring decreases.)

. (Compression can be considered negative

Figure 2.2 Compression of a spring is also a measure of force.

Unit of Force We can use a spring to make a scale for measuring forces, calibrating it in terms of how much force is required to produce a given stretch. The SI unit of force is the “newton,” abbreviated as “N.” One newton is a rather small force. A newton is approximately the downward gravitational force of the Earth on a small apple, or about a quarter of a pound (Figure 2.3). If you hold a small apple at rest in your hand, you apply an upward force of about one newton, compensating for the downward pull of the Earth.

Figure 2.3 The magnitude of the gravitational force by the Earth on a small apple is about 1 newton.

The Meaning of net In physics the word net has a precise meaning: it is the sum of all contributions to a quantity, both pluses and minuses. Because a force has a magnitude and a direction, it is a vector, so the net force on a system is the vector sum of all the forces on a system, acting for some time Δt, that exerted on the system by all the objects in the surroundings. It is the net force causes a change of momentum.

DEFINITION OF NET FORCE

The “net” force acting on a system at an instant is the vector sum of all the forces exerted on the system by all the objects in

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the surroundings at that instant.

Since the Momentum Principle relates force to change in momentum over a time interval, we see that a newton can be expressed in terms of kilograms, meters, and seconds. The net force acting on a system at an instant is the vector sum of all of the forces exerted on the system by all the objects in the surroundings, which are called “external” forces. There may be forces internal to the system, exerted by one object in the system on another object in the system, but such internal forces cannot change the momentum of the system. We will see in detail why this is so in Chapter 3, but the basic reason is that internal forces cancel each other's effects. A force that object 1 in a system exerts on object 2 in the system does change the momentum of object 2, but object 2 exerts a force in the opposite direction on object 1 that changes the momentum of object 1 in the opposite way, so that the changes in momenta of the two objects add up to zero. 2.X.2 Two external forces, the system?

and

, act on a system. What is the net force acting on

Answer

The Earth and the air are part of the surroundings, so the force exerted by the Earth and the force exerted by the air are both external forces. They are forces exerted on the chosen system (the ball) by objects in the surroundings (the Earth and the air). Forces that atoms in the ball exert on neighboring atoms in the ball are internal forces. The internal forces don't contribute to the net force that changes the ball's momentum as predicted by the Momentum Principle,

.

QUESTION A ball falling toward the Earth consists of many atoms. Each atom in the ball exerts forces on neighboring atoms in the ball, the Earth exerts forces on each atom in the ball, and the air exerts forces on atoms at the surface of the ball. Taking the ball as the system, and the Earth and air as the surroundings, which of these forces are external forces and which are internal forces?

We find experimentally that the magnitude of the net force acting on an object affects the magnitude of the change in its momentum. Many introductory physics laboratories have air tracks like the one illustrated in Figure 2.4. The long triangular base has many small holes in it, and air under pressure is blown out through these holes. The air forms a cushion under the glider, allowing it to coast smoothly with very little friction.

Figure 2.4 A block mounted on a nearly frictionless air track, pulled by a spring.

Suppose that we place a block on a glider on a long air track, and attach a calibrated spring to it (Figure 2.4). We pull on the spring so that it is stretched a distance s, so that it exerts a force F on the block, and we pull for a short time. Choose the block and glider as the system, so that the net force on the system is just the spring, since friction with the track is very small due to the air cushion. We observe that the momentum of the system of block and glider increases from zero to an amount mv (since v d1, as indicated in Figure 6.25. Therefore more work is done on the small star.

Figure 6.25 Two stars move toward each other during a short time interval. The internal work for the two-star system is F times the change in the separation distance of this pair of objects. The displacements are shown overly large for clarity.

The impulse (and momentum change) have the same magnitude for both stars because both forces act for the same time interval Δt. However, the work done is different because the two forces act through different distances. This is an important difference between impulse and work, and therefore between the Momentum Principle and the Energy Principle. Consider the system consisting of both stars. The internal work Wint is the work done by forces internal to the system (not exerted by objects in the surroundings): Change of potential energy is defined as the negative of the internal work: Since ΔU for the two stars is negative, the potential energy decreased, which would imply that the kinetic energy increased, which is indeed what happened.

Change in Separation of Two Objects The stars got closer to each other; their separation decreased by an amount (d1+d2). We can calculate internal work by multiplying the force one object exerts on another by the change in the separation between the two objects. The individual positions of the objects don't matter. The way the algebra works out, the internal work is calculated as though a single force F acted through a distance equal to the change in the separation of the pair, as in Figure 6.26.

Figure 6.26 We can calculate internal work as though one of the stars were stationary and the other moved.

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Next we'll prove this more generally. Again consider a system of two interacting stars. We define the relative position vector to be the difference between the positions of two stars, as shown in Figure 6.27. When the stars move a little closer together, star 1 does work on star 2, and star 2 does work on star 1. Since both stars are inside the system, the change in gravitational potential energy of the system is the negative of the sum of these work terms. Assume that the changes in position are small enough that we can consider the forces to be constant:

involves the change of position of each star (Figure 6.27). We can rewrite this in terms of Δ

The term

, the

change in relative position of the two stars:

Figure 6.27 Two stars whose relative positions are changing (in this case, the stars move closer together).

According to our definition of

,

Using this result we can rewrite the change in potential energy:

is the position of star 2 relative to star 1. Therefore we can calculate the amount of change of potential energy, ΔU, in terms of the relative displacements or separation between pairs of particles, and any change of potential energy is associated with a change of “shape,” with pairs of particles getting closer together or farther apart. The algebra works out in such a way that the force we use in calculating ΔU is just one of the pairs of forces, not both. The result is the same if the interaction is an electric interaction instead of a gravitational interaction, since the reasoning above depended only on reciprocity of forces, and not on any other details of the force.

For every pair of interacting particles in a system there is a potential energy term (U 12, U 23 etc.). Change of potential energy is associated with change of distance between pairs of particles. We write U 12 for the potential energy of the pair consisting of particles 1 and 2, U 23 for pair 2 and 3, and so on. If a system consists of two particles, there are three energy terms: two particle energies E1 and E 2, and one potential energy term, U 12, since there is only one pair of particles. (Of course the particle energies can be further subdivided into rest energy plus kinetic energy.)

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If a system consists of three particles, there are six energy terms: three particle energies E1, E 2, and E 3, and three potential energy terms, U 12, U 23, and U 31, since there are three pairs of particles.

Force Is the Negative Gradient of U Since it is the relative displacement that matters, even if both objects move we can pretend that object 1 remains stationary, and just calculate the work done on object 2. The force is in a line with , so we can evaluate the dot product and write the following, , and r is the distance from object 1 to object where F r is the component of the gravitational force on object 2 in the direction of 2 (Figure 6.28):

Figure 6.28 The r component of the gravitational force, Fr .

Therefore if we already knew the formula for potential energy, we could calculate the associated force like this: Fr = −dU/dr. A rate of change of a quantity with respect to position such as this is called a “gradient.” For example, if a hill is said to have “a 7% grade,” this means that it rises 7 m vertically for every 100 meters horizontally—the tangent of the angle of the hill (the slope) is 0.07 (Figure 6.29).

Figure 6.29 The gradient is the rate of change of y with respect to x.

FORCE IS THE NEGATIVE GRADIENT OF POTENTIAL ENERGY

All that remains to be done to find a formula for gravitational energy is to think of an expression whose (negative) derivative is the component of the gravitational force acting on m 2 in the direction of . From Figure 6.28 we see that this component of the gravitational force is negative: F r = −G(m 1m 2/r 2), where r is the center-to-center distance. The minus sign reflects the fact that the attractive gravitational force on object 2 points back toward object 1.

QUESTION Can you think of a function of r whose (negative) derivative with respect to r has this value of F r? (If necessary, review the appendix on basic calculus.)

The derivative of r n with respect to r is nr n−1. We have a r −2 factor in the gravitational force, so n = −1, and the “antiderivative” of r −2 (the function whose derivative gives r −2) must be −r −1.

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QUESTION Therefore, what is the formula for gravitational energy?

Being careful about the signs, we find that U = −Gm 1m 2/r.

QUESTION Check that the negative gradient of this gravitational energy is indeed the force, by differentiating with respect to r.

What about the minus sign on U? Is it correct? Yes, because the potential energy decreases as the particles get closer together. When stars fall toward each other, their kinetic energies increase and the pair-wise potential energies must decrease to more and more negative values. In summary,

GRAVITATIONAL POTENTIAL ENERGY

r is the center-to-center separation of m 1 and m 2.

QUESTION Could there be an additional constant in this equation?

It is true that the gradient of U = (−Gm 1m 2/r + constant) would also give us the correct gravitational force, because the derivative of a constant is zero. However, as we will soon discuss in more detail, when the objects are very far apart the total energy of the system must be equal just to the sum of the particle energies, and the potential energy must be zero.

QUESTION When the particle separation r is very large, what is the value of −Gm 1m 2/r?

When r is very large, −Gm 1m 2/r is nearly zero. Therefore, for the potential energy to be zero when the separation is very large, the value of the constant in U = (−Gm 1m 2/r + constant) must be zero, in order for U to be zero at large separations. The pair-wise gravitational potential energy must be simply U = −Gm 1m 2/r, with no added constant. 6.X.18 Calculate the gravitational potential energy of the interacting pair of the Earth and a 1 kg block sitting on the Answer surface of the Earth. You would need to supply the absolute value of this result to move the block to a location very far from the Earth (actually, you would need to use even more energy than this due to the gravitational potential energy associated with the Sun–block interacting pair).

The Minus Sign

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It may seem odd that the expression for gravitational potential energy has a minus sign in it. Figure 6.30 shows that although U is negative, it does increase with increasing separation r, which means that you have to do work to move two gravitationally attracting objects farther apart.

Figure 6.30 Gravitational potential energy is negative. It increases with increasing separation r.

Here are two little exercises that will help you see why the minus sign is appropriate: 6.X.19 If a meteor falls toward the Earth, the kinetic energy increases and the gravitational potential energy (which is proportional to −1/r) must decrease. Suppose that the separation r decreases from 1 × 108 m to 1 × 107 m. What is (−1/r f ) − (−1/r i )? Is this change negative, as expected? 6.X.20 If a spacecraft coasts away from the Earth, the kinetic energy decreases and the gravitational potential energy (which is proportional to −1/r) must increase. Suppose that the separation r increases from 1 × 107 m to 1 × 108 m. What is (−1/r f ) − (−1/r i )? Is this change positive, as expected?

Answer

Answer

EXAMPLE A Robot Spacecraft Leaves an Asteroid A robot spacecraft lands on an asteroid, picks up a sample, and blasts off to return to Earth; its total mass is 1500 kg. When it is 200 km (2 × 105 m) from the center of the asteroid, its speed is 5.0 m/s, and the rockets are turned off. At the moment when it has coasted to a distance 500 km (5 × 105 m) from the center of the asteroid, its speed has decreased to 4.1 m/s. You can use these data to determine the mass of the asteroid. Before starting your analysis, draw a diagram of the initial and final states you will consider in your analysis. Label your diagram to show the distances involved. This will help avoid mistakes in your calculation. (a) Draw and label the diagram. (b) Calculate the mass of the asteroid. Make an accurate calculation, not a rough, approximate calculation.

Solution (a) Figure 6.31 shows the situation. It is important to keep in mind that r is measured center to center in U = −Gm 1m 2/r.

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Figure 6.31 Initial and final states. System: Asteroid and spacecraft Surroundings: Nothing significant Initial state: Rockets off, 5.0 m/s, 200 km from center of asteroid Final state: Speed 4.1 m/s, 500 km from center of asteroid Energy Principle: E f = E i + W

By choosing both the asteroid and the spacecraft to be in the system, we won't have to calculate work, since there are no significant external forces. No change in rest energies. No work by surroundings (no interacting objects in surroundings). As discussed below, the kinetic energy of the massive asteroid changes very little: .

In many cases it is useful to solve the problem algebraically before putting in numbers, as is done here,

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because it reduces the number of calculations and makes it easier to track down algebra errors. Note that m canceled; we didn't need to know or use the value of 1500 kg for the mass.

FURTHER DISCUSSION We could not have used the Momentum Principle to carry out an accurate analysis without writing an iterative computer program because we didn't know how much time the process took. The velocity was changing at a nonconstant rate, so knowing the distance wouldn't give us the time, since we didn't know the average velocity. We were able to use the Energy Principle because we knew something about distances. 6.X.21 What was the advantage of choosing an “inclusive” system (including both the asteroid and the spacecraft)? 6.X.22 Concerning the separation r in U = −Gm 1m 2/r : Is this the distance from the spacecraft to the surface of the asteroid, or from the spacecraft to the center of the asteroid?

Answer

Answer

Multiparticle Systems: Same Time Interval, Different Displacements In the previous example the spacecraft pulls just as hard on the asteroid as the asteroid pulls on the spacecraft (reciprocity of gravitational forces). Yet we claimed that the kinetic energy of the asteroid hardly changed compared to the change in the spacecraft's kinetic energy. We can show why by using the Momentum Principle and the Energy Principle together.

Apply the Momentum Principle First consider the Momentum Principle. The two different forces act on the asteroid and the spacecraft for the same amount of time Δt. Since the magnitude of the two forces is the same, the magnitude of the impulse is the same, and the magnitude of the change of momentum is the same. Since v = p/M at low speeds, the change in speed of the asteroid is extremely small compared to the change in speed of the spacecraft (ratio of m/M), and if the asteroid was initially at rest it will move a very small distance compared to the displacement of the spacecraft.

Apply the Energy Principle Next consider the Energy Principle. The work done on the asteroid by the spacecraft is extremely small, because the displacement of the asteroid is extremely small. The two objects exert the same magnitude of force on each other, but through enormously different distances. In the example above, the spacecraft moved 300 km, whereas the asteroid moved roughly (m/M) of that distance, or about (1500/2 × 1016)(3 × 105 m) ≈ 2 × 10−8 m, or about 100 atomic diameters! This illustrates yet another important difference in the role played by forces acting in and on a multiparticle system in the Momentum Principle (all forces act for the same time interval, and the net force is all that matters) and in the Energy Principle (different forces may act through very different distances, and change of shape matters). Another way of seeing why the asteroid gets very little kinetic energy is to recall that kinetic energy can be expressed in terms of momentum:

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The asteroid and spacecraft experience the same change in the magnitude of momentum, but the p2 term is divided by a huge mass in the case of the asteroid. When a ball falls toward the Earth, the magnitude mg of the force that the Earth exerts on the ball is the same as that which the ball exerts on the Earth, and both experience the same change in the magnitude of momentum in the same time interval. However, the massive Earth acquires an extremely small velocity, so the Earth moves an extremely small distance while the ball moves a large distance. Same force mg, very different distances, very different amounts of work, and the Earth gets very little kinetic energy.

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GENERAL PROPERTIES OF POTENTIAL ENERGY There are some general properties of potential energy that are true not only for gravitational interactions but also for other kinds of interactions as well, including electric interactions. Here is a list of important properties: Potential energy depends on the separation between pairs of particles, not on their individual positions. (As we saw with gravitational interactions, this depends on the reciprocity of gravitational and electric forces.) Potential energy must approach zero as the separation between particles becomes very large, as we will soon show. If an interaction is attractive, potential energy becomes negative as the distance between particles decreases. (We saw this in the case of gravitational interactions.) If an interaction is repulsive, potential energy becomes positive as the distance between particles decreases.

Let's look in a bit more detail at the consequences of the fact that potential energy depends on the separation between the interacting objects.

QUESTION Suppose that two particles move together, with the same velocity (Figure 6.32). In that case, what is What is ΔU 12?

for this two-particle system?

Figure 6.32 The two particles move together, with the same velocity.

In this case the relative positions don't change,

is zero, and there is no change in the pair-wise potential energy.

QUESTION Suppose instead that the two particles rotate around each other (Figure 6.33). What can you say about ΔU 12?

Figure 6.33 The two particles rotate around each other.

This is a bit trickier, because if there is rotation, isn't zero, because perpendicular to and to the force , so the dot product

is now a rotating vector. In rotation, however, is is zero. So again there is no change in the pair-wise potential

energy.

QUESTION What can you conclude about potential energy for a rigid system?

Evidently U is constant for a rigid system, one whose shape doesn't change. For the potential energy to change, there must be a change of shape or size. Conversely, a change of shape or size is an indicator that U may have changed.

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U → 0 When the Particles Are Very Far Apart Consider a system consisting of two identical stars, which attract each other gravitationally. If the stars are at rest, very, very far apart—almost infinitely far apart, so that the force one exerts on the other is nearly zero—then the total energy of the two-star system must be simply the sum of their rest energies:

This implies that the potential energy associated with the interaction of these two stars must be zero if the stars are “infinitely” far apart. More generally, the total energy of a multiparticle system is given by the expression Esystem = (E 1 + E 2 + ···) + U, where E 1 = γm 1c 2 is the particle energy of particle 1, and so on. When the particles are very far apart, we must define U to be zero, so that the total energy of these farapart particles is simply the sum of the particle energies:

POTENTIAL ENERGY OF A PAIR OF PARTICLES THAT ARE VERY FAR APART

Attractive Interaction: U Becomes Negative as r Decreases If the particles start to come closer together, we find that U starts to become negative if the interactions are attractive, while U starts to become positive if the interactions are repulsive. To see this, first consider an isolated system consisting of two stars that are very far apart, initially at rest, so the initial energy of the system is just the sum of the rest energies of the two stars. At first slowly, then more and more rapidly, their kinetic energies increase due to their mutual gravitational attractions (Figure 6.34).

Figure 6.34 Initially two stars are very far apart, at rest, but they attract each other.

QUESTION U was initially zero, so what must be the sign of U when the stars have acquired significant kineticenergies?

If the system is isolated, there are no external forces and hence no external work, so Esys does not change. If the individual particle energies increase (so the total kinetic energy increases), the pair-wise potential energy must decrease. Because U was defined to be zero initially, U must become negative, as we found for gravitational potential energy. A graph of the energies is shown in Figure 6.35, omitting the very large rest energies of the stars. Since the rest energies don't change, K + U is a constant.

Figure 6.35 The energies of two stars (with rest energies omitted) as a function of separation r. U must be zero at large r. As r decreases, K increases and U decreases. The total energy of the two-star system is constant, and K + U is constant.

Does this negative sign make sense in terms of the definition of potential energy change, ΔU = −Wint? The attractive gravitational forces that

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the stars exert on each other certainly do a positive amount of work on each star, so the potential energy change must indeed be negative.

Repulsive Interaction: U Becomes Positive as r Decreases Next consider an isolated system consisting of two protons that are very far apart, and suppose that initially they have high speeds, heading inward (Figure 6.36). Their mutual electric repulsions make the protons slow down (their electric repulsion is enormously larger than their gravitational attraction).

Figure 6.36 Initially two protons are very far apart, heading inward. The mutual repulsions slow the protons down.

QUESTION U was initially zero, so what must be the sign of U when the protons have slowed down somewhat?

Again, there is no external work done on this isolated multiparticle system, so Esys does not change. The particle energies decrease, so the potential energy must increase; U becomes positive (Figure 6.37). Also, the repulsive electric forces do negative work on each other, so ΔU = −Wint is positive. The graph in Figure 6.37 omits the rest energies of the protons. Since the rest energies don't change, K + U is a constant.

Figure 6.37 The energies of two protons as a function of separation (with rest energies omitted). U must be zero at large r. As r decreases, K decreases and U increases. The total energy of the two-proton system is constant, and K + U is constant.

Note that the change in potential energy for the two stars or the two protons was associated with a change in shape (change in configuration) of the two-particle system. If there is no change in shape or size, there is no change in potential energy.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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PLOTTING ENERGY vs. SEPARATION A graphical representation of energy can be exceptionally helpful in reasoning about a process. In Figure 6.38 we show an energy graph for the spacecraft leaving the asteroid, the process analyzed in Section 6.9 (the graph continues beyond the separation r f = 500 km). We plot the pair-wise gravitational potential energy U, the spacecraft's kinetic energy K, and the sum K + U, as a function of the separation distance r between the center of the asteroid and the spacecraft. We omit the rest energies, which are not changing during the motion.

Figure 6.38 Energy vs. the separation distance between the asteroid and the spacecraft. (We omit the rest energies, which aren't changing.)

Important features of the graph are these: The gravitational potential energy U is negative because the interaction is attractive, not repulsive, and the potential energy increases with increasing separation r. As the potential energy increases, the kinetic energy K of the spacecraft decreases (the gravitational force exerted on the spacecraft by the asteroid is slowing it down). Because there is negligible work done on the combined system of asteroid plus spacecraft, K f + U f = K i + U i , so K + U has a constant value at all times, and the graph of K + U is a straight horizontal line on the graph, representing a constant positive value. As the spacecraft gets farther and farther from the asteroid, you can see from the graph that its kinetic energy K will continue to fall, and the graph of K will approach the K + U line at large separation. When the two objects are very far apart, U is nearly zero, and K + U is just K. This is an example of an “unbound” system, because the separation between spacecraft and asteroid will increase without bound. An unbound system has K + U greater than zero. As an example of a “bound” system, again consider the spacecraft and the asteroid but this time give the spacecraft a smaller initial velocity, headed away from the asteroid. As a function of the center-to-center asteroid-to-spacecraft separation r, we again plot a graph of the potential energy −GMm/r (Figure 6.39), and a graph of kinetic energy K. As the spacecraft moves away from the asteroid, its kinetic energy K falls as the pair-wise potential energy U rises. You can see on the graph that K falls to zero, and it cannot go negative, because kinetic energy is always a positive quantity. What happens next? The spacecraft momentarily comes to a stop (a turning point), then turns around and speeds up toward the asteroid; eventually it will crash onto the asteroid's surface.

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Figure 6.39 Energy vs. the separation distance between the asteroid and the spacecraft. (We omit the rest energies, which aren't changing.) At r 1 the potential energy is U 1 and the kinetic energy is K1 .

We add to the graph a horizontal line whose height above or below the axis represents the total kinetic plus potential energy of the two-object system. The line on the graph for K+U is horizontal because no external work is done on this isolated system, and there is no change of the rest masses (no change of identity), so K + U is constant during the motion.

QUESTION Is K + U positive or negative for the system shown in Figure 6.39?

The horizontal line representing the constant value of K + U is below the axis representing U = 0, so K + U is negative. This corresponds to a “bound” state of the spacecraft and asteroid; the spacecraft and asteroid cannot get completely away from each other. The vertical distance from the U curve up to the K + U line is equal to K, the total kinetic energy of the spacecraft and asteroid. At all separations between the spacecraft and the asteroid, the height of the K + U line above the U curve tells you the total kinetic energy. Notice that when K falls to zero, K + U = U and the right end of the K + U line touches the U graph. The shaded area in Figure 6.39 represents a forbidden region, where the kinetic energy K would be negative, no matter what the value of K +U (more about this in the following discussion). 6.X.23 The asteroid–spacecraft separation increases from r 1 to r 2 (Figure 6.39). Does the kinetic energy of the asteroid–spacecraft system increase, decrease, or remain constant?

Answer

Limits on Possible Motion The energy graph shows at a glance the limits on the motion for a given energy. In Figure 6.40, the asteroid and spacecraft with K + U < 0 cannot get farther away from each other than r 3, a bound that is set by the potential-energy curve. At a larger separation r 4, the kinetic energy would have to be negative, which is “classically forbidden” (that is, impossible according to the laws of classical, , which is always positive. At high speeds, K prequantum mechanics). Kinetic energy cannot be negative. At low speeds, 2

2

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= γ mc − mc , which is always positive, because γ is always greater than 1.

Figure 6.40 The asteroid and spacecraft cannot get farther away from each other than r 3 ; at r 4 the kinetic energy would be negative. (We omit the rest energies, which aren't changing.)

If the system has the total K + U shown (K + U < 0), the separation of the asteroid and spacecraft can never be larger than r 3, and we say that this is a “bound” system. An energy graph shows at a glance the range of possible positions achievable during the motion, for a given total K + U. (We should mention, however, that in the world of atoms where a full analysis requires quantum mechanics, in some cases a system can “tunnel through” a region that is classically forbidden!) An energy graph can show at a glance whether a system is “bound” or “unbound.” An example of a bound system is a planet in circular or elliptical orbit around a star: the planet cannot escape. Another example is an electron bound to an atom. An example of an unbound system is the Earth plus a spacecraft whose initial speed is great enough that the spacecraft will get away from the Earth and never come back. For another example, if an amount of energy greater than or equal to the ionization energy is supplied to an atom, an electron can become unbound and escape from the atom. An unbound system has total K + U ≥ 0, which is easy to see on an energy graph, where it is clear that arbitrarily large separations are possible. Only when K + U ≥ 0 can the system become separated by large distances, because K is never negative, and U goes to zero at large separations. 6.X.24 In this energy graph for some system (Figure 6.41), consider the various energy states indicated. Which of Answer these values for the energy K + U (A, B, or C) represent a bound state? Which represent an unbound state (the particle can escape)? Which represents a trapped state with enough energy to be unbound but with a barrier that (classically) prevents escape?

Figure 6.41 An energy graph for some system. Which states are bound? (Exercise 6.X.24.)

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Drawing Energy Graphs It is important to be able to draw (create) energy graphs as well as to be able to read them. Here is a practical scheme: Draw U vs. r for the particular interaction (gravitational, electric). At some r where you happen to know K, plot the point (r, K). Add that value of K to the value U has at the same separation r. Plot K + U at that r, then draw a horizontal line through the point. At some other r, find a K that when added to U at that r gives K + U. Now you have two points on the K graph; sketch the behavior of K vs. r.

EXAMPLE Making an Energy Graph Let's follow through these steps for the energy graph for the spacecraft leaving the asteroid, which is shown again in Figure 6.42.

Figure 6.42 Energy vs. the separation distance between the asteroid and the spacecraft. (We omit the rest energies, which aren't changing.)

Solution We know the shape of U for gravity; it is −GMm/r. If r is small, U is a large negative number. If r is large, U is nearly zero. Plot two such points, then sketch the behavior of U vs. r. We know that K is nonzero at large r, because the spacecraft escapes, so at large r, plot some positive K. At that distant location, U is nearly zero, so the K you just plotted is nearly the same as K + U. Draw a horizontal line representing K + U for all separations. At the initial separation r i , determine graphically a value of K that when added to the value of U at r i yields the known value of K + U. The value of K plus the value of U must be the same as K + U. Sketch the behavior of K vs. r. Note that its shape is the mirror image of U vs. r.

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Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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APPLYING GRAVITATIONAL POTENTIAL ENERGY In the next sections we apply the Energy Principle to the analysis of a variety of situations involving gravitational potential energy. We will frequently use energy diagrams as a powerful tool.

Application: Energy Graphs and Types of Orbits In Figure 6.43 are plotted graphs of potential energy, and potential plus kinetic energy, for three different orbits of a planet and star. The following exercise asks you to interpret the meaning of these graphs.

Figure 6.43 What kind of motion is represented by each of these situations? (Exercise 6.X.25).

6.X.25 In Figure 6.43, what kind of motion is represented by the situation with K + U = A? B? C? Think about the range of r in each situation. For example, C represents a circular orbit (constant r).

Answer

Application: Escape Speed An analysis based on the Energy Principle will allow us to answer the following question: With what minimum speed v must a spacecraft leave the surface of an airless planet (no air resistance) of mass M and radius R in order that it can coast away without ever coming back (Figure 6.44)? Assume that there are no other objects nearby.

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Figure 6.44 A spacecraft is launched with momentum p i from near an airless planet's surface; it has momentum p f when it has traveled far away.

If you did the Moon voyage problem in Chapter 3 you found that there was a minimum initial speed required to reach the Moon. With an initial speed smaller than this, the spacecraft fell back to Earth without reaching the Moon. Your determination of the minimum initial speed was done by trial and error. However, we can now use energy relationships to calculate such minimum speeds directly. As is often the case, an energy diagram can be very helpful. Consider Figure 6.45. Motions with K + U = A, B, C, or D all start from the surface of an airless planet, with a separation R between the center of the planet and the spacecraft. (Remember that the gravitational force outside a uniform sphere is exactly the same as it would be if all the mass collapsed to the center of the sphere.)

Figure 6.45 What is the minimum K required to escape?

QUESTION Which of these motions starts with the largest kinetic energy K? Which starts with the least K?

Each of the horizontal lines represents motion with K +U constant. Evidently A is the highest value of K + U, and therefore involves the largest initial kinetic energy. D is the lowest value of K + U, and starts with the least kinetic energy.

QUESTION For which of these motions does the spacecraft “escape” and never come back?

For motions with K + U = A, B, or C, no matter how far from the planet the spacecraft gets, there is still some kinetic energy, so the spacecraft will never return. This is another example of an unbound state being associated with a positive value of K + U. In contrast, with motion K + U = D the spacecraft will reach a maximum separation from the planet, at which point its kinetic energy has fallen to zero. The spacecraft will fall back to the planet.

QUESTION How would you describe the motion that requires the least possible initial kinetic energy to achieve escape?

Evidently the least costly escape is to have K = 0 when the spacecraft has reached a distance very far from the planet. At a large separation, U = 0, so K + U = 0. This corresponds on the diagram to a horizontal line lying on the axis. Because K + U doesn't change (no external work on the planet–spacecraft system), it must also be true that K + U = 0 at the start of the motion:

MINIMAL CONDITION FOR ESCAPE

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K+U=0

The minimal initial kinetic plus potential energy of the system composed of planet plus spacecraft is this, assuming that the kinetic energy of the planet is negligible:

This lets us calculate the minimum speed, which is called the “escape speed” v esc , for a planet with mass M and radius R. Though we speak of “escape speed,” it is wrong to say that the spacecraft has “escaped from gravity.” As long as the spacecraft is a finite distance from the planet, it feels a finite gravitational attraction to the planet. If you give the spacecraft more than the minimum kinetic energy, it not only escapes but also has nonzero kinetic energy when it is far away (for example, see motions A and B in Figure 6.45). It is interesting to note that the initial direction of motion doesn't matter. An energy analysis is insensitive to direction.

Bound vs. Unbound States If the initial velocity points directly away from the planet, the path is a straight line. If the initial velocity points in some other direction, the path can be shown to be a parabola if the initial speed is exactly equal to escape speed and a hyperbola for higher speeds. If the speed is below escape speed, there is no escape. The various possible orbits (Figure 6.46) can be classified on the basis of their total kinetic plus potential energy, with initial speed v i :

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Figure 6.46 Possible orbital trajectories.

This is a specific example of an important general principle. If the kinetic plus potential energy is negative, the system is “bound”: the objects cannot become widely separated, because that would require that the kinetic energy become negative, which is impossible (classically). If the kinetic plus potential energy is positive, the system is “unbound” or “free”: the objects can become widely separated, with net kinetic energy (and zero potential energy).

BOUND AND UNBOUND STATES If K + U < 0, the system is in a bound state. If K + U ≥ 0, the system is unbound (free).

EXAMPLE Escape Speed What is escape speed from Earth?

Solution System: Earth + spacecraft Surroundings: Nothing significant (ignore Sun and Moon) Initial state: At surface of Earth, with speed Final state: Spacecraft extremely far away from Earth, nearly at rest Energy Principle:

When we know there is no change in rest energies it is okay to omit these terms from the Energy Principle equation. Negligible change in kinetic energy of Earth. No work by surroundings.

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6.X.26 This escape speed is slightly larger than the launch speed that was found necessary to reach the Moon in the problem in Chapter 3 on the Ranger 7 mission to the Moon. Explain why.

Answer

6.X.27 Turn the argument around. If an object falls to Earth starting from rest a great distance away, what is the Answer speed with which it will hit the upper atmosphere? (Actually, a comet or asteroid coming from a long distance away might well have an even larger speed, due to its interaction with the Sun.) Small objects vaporize as they plunge through the atmosphere, but a very large object can penetrate and hit the ground at very high speed. Such massive impact is thought to have killed off the dinosaurs (see T. Rex and the Crater of Doom, Walter Alvarez, Princeton University Press, 1997).

Application: The Moon Voyage We have been applying energy considerations to two-particle systems. A homework problem in Chapter 3 involves predicting the motion of the Ranger 7 spacecraft as it travels from Earth to the Moon (where it crash lands). In our simple model for this voyage there are three “particles”: the spacecraft with mass m, the Earth with mass MEarth, and the Moon with mass MMoon.

QUESTION How many interaction pairs are there in this three-particle system?

In a three-particle system there are three interaction pairs: U 12, U 13, and U 23. We can write the gravitational potential energy like this:

In our simple model, the Earth and Moon were fixed in space, so the Earth–Moon term of this expression doesn't change. Also, there is no change of identity of the three particles, so their rest masses do not change. Therefore the energy equation for the system can be written like this, where v is the speed of the spacecraft, the only moving particle in the model system (and v > R). Note that the potential energy curve must be continuous, with no breaks. 6.X.76 One often hears the statement, “Nuclear energy production is fundamentally different from chemical energy production (such as burning of coal), because the nuclear case involves a change of mass.” Critique this statement. Discuss the similarities and differences of the two kinds of energy production. 6.X.77 Figure 6.71 shows the path of a comet orbiting a star.

Figure 6.71

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(a) Rank-order the locations on the path in terms of the magnitude of the comet's momentum at each location, starting with the location at which the magnitude of the momentum is the largest. (b) Rank-order the locations on the path in terms of the comet's kinetic energy at each location, starting with the location at which the kinetic energy is the largest. (c) Consider the system of the comet plus the star. Which of the following statements are correct? (A) External work must be done on the system to speed up the comet. (B) As the comet slows down, energy is lost from the system. (C) As the comet's kinetic energy increases, the gravitational potential energy of the system also increases. (D) As the comet slows down, the kinetic energy of the system decreases. (E) As the kinetic energy of the system increases, the gravitational potential energy of the system decreases. (d) Still considering the system of the comet plus the star, which of the following statements are correct? a. The sum of the kinetic energy of the system plus the gravitational potential energy of the system is a positive number. b. At every location along the comet's path the gravitational potential energy of the system is negative. c. The gravitational potential energy of the system is inversely proportional to the square of the distance between the comet and star. d. The sum of the kinetic energy of the system plus the gravitational potential energy of the system is the same at every location along this path. e. Along this path the gravitational potential energy of the system is never zero. (e) Rank-order the locations on the path in terms of the potential energy of the system at each location, largest first. (Remember that −3 > −5). 6.X.78 Figure 6.72 is a graph of the energy of a system of a planet interacting with a star. The gravitational potential energy U g is shown as the thick curve, and plotted along the vertical axis are various values of K + U g.

Figure 6.72

Suppose that K + U g of the system is A. Which of the following statements are true? (A) The potential energy of the system decreases as the planet moves from r to r .

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1

2

(B) When the separation between the two bodies is r 2, the kinetic energy of the system is (A − B). (C) The system is a bound system; the planet can never escape. (D) The planet will escape. (E) When the separation between the two bodies is r 2, the kinetic energy of the system is (B − C). (F) The kinetic energy of the system is greater when the distance between the star and planet is r 1 than when the distance between the two bodies is r 2.

Suppose instead that K + U g of the system is B. Which of the following statements are true? (A) When the separation between the planet and star is r 2, the kinetic energy of the system is zero. (B) The planet and star cannot get farther apart than r 2. (C) This is not a bound system; the planet can escape. (D) When the separation between the planet and star is r 2, the potential energy of the system is zero. 6.X.79 Under certain conditions the interaction between a “polar” molecule such as HCl located at the origin and an ion located along the x axis can be described by a potential energy U = −b/x 2, where b is a constant. What is F x , the x component of the force on the ion? What is Fy , the y component of the force on the ion?

Sections 6.10, 6.11 6.X.80 The radius of the Moon is 1750 km, and its mass is 7 × 1022 kg. What would be escape speed from an isolated Moon? Why was a small rocket adequate to lift the lunar astronauts back up from the surface of the Moon? 6.P.81 The radius of Mars (from the center to just above the atmosphere) is 3400 km (3400 × 103 m), and its mass is 0.6 × 1024 kg. An object is launched straight up from just above the atmosphere of Mars. (a) What initial speed is needed so that when the object is far from Mars its final speed is 1000 m/s? (b) What initial speed is needed so that when the object is far from Mars its final speed is 0 m/s? (This is called the “escape speed.”) 6.P.82 The radius of Mars (from the center to just above the atmosphere) is 3400 km (3.4 × 106 m), and itsmass is 6 × 1023 kg. An object is launched straight up from just above the atmosphere of Mars. (a) What initial speed is needed so that when the object is far from Mars its final speed is 2000 m/s? (b) What initial speed is needed so that when the object is far from Mars its final speed is 0 m/s? (This is called the “escape speed.”) 6.P.83 The escape speed from an asteroid whose radius is about 10 km is only 10 m/s. If you throw a rock away from the asteroid at a speed of 20 m/s, what will be its final speed? 6.P.84 The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 35 m/s, what will be its final speed? 6.P.85 Use energy conservation to calculate analytically (that is, without doing a numerical integration) the final speed of the spacecraft just before it hits the Moon. Include the gravitational effect of the Moon. Use a launch speed of 1.3 × 104 m/s. Modify your program for the Chapter 3 problem on the Ranger 7 mission to the Moon to print out the speed of the spacecraft when it hits the surface of the Moon, and compare this value to your analytical result. What questions that could be addressed in the numerical integration are you not able to answer by doing this energy calculation? 6.P.86 A spacecraft is coasting toward Mars. The mass of Mars is 6.4 × 1023 kg and its radius is 3400 km (3.4 × 106 m). When the spacecraft is 7000 km (7 × 106 m) from the center of Mars, the spacecraft's speed is 3000 m/s. Later, when the spacecraft is 4000 km (4 × 106 m) from the center of Mars, what is its speed? Assume that the effects of Mars's two tiny moons, the other

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planets, and the Sun are negligible. Precision is required to land on Mars, so make an accurate calculation, not a rough, approximate calculation. 6.P.87 Calculate the speed of a satellite in a circular orbit near the Earth (just above the atmosphere). If the mass of the satellite is 200 kg, what is the minimum energy required to move the satellite from this near-Earth orbit to very far away from the Earth? 6.P.88 A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4 × 1010 m (inside the orbit of Mercury), at which point its speed is 8.17 × 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 × 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.) 6.P.89 An electron is traveling at a speed of 0.95c in an electron accelerator. An electric force of 1.6 × 10−13 N is applied in the direction of motion while the electron travels a distance of 2 m. What is the new speed of the electron? 6.P.90 You stand on a spherical asteroid of uniform density whose mass is 2 × 1016 kg and whose radius is 10 km (104 m). These are typical values for small asteroids, although some asteroids have been found to have much lower average density and are thought to be loose agglomerations of shattered rocks. (a) How fast do you have to throw the rock so that it never comes back to the asteroid and ends up traveling at a speed of 3 m/s when it is very far away? (b) Sketch graphs of the kinetic energy of the rock, the gravitational potential energy of the rock plus asteroid, and their sum, as a function of separation (distance from center of asteroid to rock). Label the graphs clearly. 6.P.91 This problem is closely related to the spectacular impact of the comet Shoemaker-Levy with Jupiter in July 1994: http://www.jpl.nasa.gov/sl9/sl9.html

A rock far outside our Solar System is initially moving very slowly relative to the Sun, in the plane of Jupiter's orbit around the Sun. The rock falls toward the Sun, but on its way to the Sun it collides with Jupiter. Calculate the rock's speed just before colliding with Jupiter. Explain your calculation and any approximations that you make. MSun = 2 × 1030 kg, MJupiter = 2 × 1027 kg Distance, Sun to Jupiter = 8 × 1011 m Radius of Jupiter = 1.4 × 108 m

Section 6.12 6.X.92 Use energy conservation to find the approximate final speed of a basketball dropped from a height of 2 meters (roughly the height of a professional basketball player). Why don't you need to know the mass of the basketball? 6.X.93 (a) A 0.5 kg teddy bear is nudged off a window sill and falls 2 m to the ground. What is its kinetic energy at the instant it hits the ground? What is its speed? What assumptions or approximations did you make in this calculation? (b) A 1.0 kg flowerpot is nudged off a window sill and falls 2 m to the ground. What is its kinetic energy at the instant it hits the ground? What is its speed? How do the speed and kinetic energy compare to that of the teddy bear in part (a)? 6.X.94 You throw a ball of mass 1.2 kg straight up. You observe that it takes 3.1 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.55 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward. (a) Use the Momentum Principle to determine the speed that the ball had just after it left your hand. (b) Use the Energy Principle to determine the maximum height above your hand reached by the ball. 6.P.95 In the rough approximation that the density of the Earth is uniform throughout its interior, the gravitational field strength (force per unit mass) inside the Earth at a distance r from the center is gr/R, where R is the radius of the Earth. (In actual fact, the outer layers of rock have lower density than the inner core of molten iron.) Using the uniform-density approximation, calculate the amount of energy required to move a mass m from the center of the Earth to the surface. Compare with the amount of energy required to move the mass from the surface of the Earth to a great distance away.

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Section 6.13 6.X.96 Two protons are a distance 4 × 10−9 m apart. What is the electric potential energy of the system consisting of the two protons? If the two protons move closer together, will the electric potential energy of the system increase, decrease, or remain the same? Aproton and an electron are a distance 4 × 10−9 m apart. What is the electric potential energy of the system consisting of the proton and the electron? If the proton and the electron move closer together, will the electric potential energy of the system increase, decrease, or remain the same? Which of the following statements are true? (A) In some situations charged particles released from rest would move in a direction that increases electric potential energy, but not in other situations. (B) If released from rest two protons would move closer together, increasing the potential energy of the system. (C) If any two charged particles are released from rest, they will spontaneously move in the direction in which the potential energy of the system will be decreased. 6.X.97 Which of the diagrams in Figure 6.73 corresponds to a system of two electrons that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)?

Figure 6.73

Which of the diagrams corresponds to a system of a proton and an electron that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)? 6.P.98 Four protons, each with mass M and charge +e, are initially held at the corners of a square that is d on a side. They are then released from rest. What is the speed of each proton when the protons are very far apart?

Section 6.14 6.P.99

(a) A particle with mass M and charge +e and its antiparticle (same mass M, charge −e) are initially at rest, far from each other. They attract each other and move toward each other. Make a graph of energy terms vs. separation distance r between the two particles. Label the various energies involved in this process. Include the rest energy of the particles, assuming that the other energy terms are comparable to the rest energy. (b) When the particle and antiparticle collide, they annihilate and produce a different particle with rest mass m (much smaller than M) and charge +e and its antiparticle (same rest mass m, charge −e). When these two particles have

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moved far away from each other, how fast are they going? Is this speed large or small compared to c? (c) Now take the specific case of a proton and antiproton colliding to form a positive and negative pion. Each pion has a rest mass of 2.5 × 10−28 kg. When the pions have moved far away, how fast are they going? (d) How far apart must the two pions be (in meters) for their electric potential energy to be negligible compared to their kinetic energy? Be explicit and quantitative about your criterion and your result. 6.P.100 For some isotopes of some very heavy nuclei, including nuclei of thorium, uranium, and plutonium, the nucleus will fission (split apart) when it absorbs a slow-moving neutron. For example, plutonium-241, with 94 protons and 147 neutrons, can fission when it absorbs a neutron and becomes plutonium-242. The two fission fragments, called “daughter” nuclei, can be almost any two nuclei whose charges Q 1 and Q 2 add up to 94e (where e is the charge on a proton), and whose nucleons add up to 242 protons and neutrons (Pu-242, formed from Pu-241 plus a neutron). One of the possible fission modes involves nearly equal fragments, silver nuclei (Ag-121) each with electric charge Q 1 = Q 2 = 47e. The rest masses of the two daughter nuclei add up to less than the rest mass of the original parent nucleus. (In addition to the two main fission fragments there are typically one or more free neutrons in the final state; in your analysis make the simplifying assumption that there are no free neutrons, just two daughter nuclei.) Objects involved in the reaction, where the rest mass is given in atomic mass units, and 1 amu = 1.66054 × 10−27 kg: Nucleus

Protons

Neutrons

Charge

Rest mass

Pu-242

94

148

+94e

242.007

Ag-121

47

74

+47e

120.894

Although in most problems you solve in this course it is adequate to use values of constants rounded to 2 or 3 significant figures, in this problem you must keep 6 significant figures throughout your calculation. Problems involving mass changes require many significant figures because the changes in mass are small compared to the total mass. In this problem you must use the following values of constants, accurate to 6 significant figures: Constant

Value to 6 significant figures

c (speed of light)

2.99792 × 108 m/s

e (charge of a proton)

1.60218 × 10−19 coulomb 8.98755 × 1009N · m2/C 2

There are three states you should consider in your analysis: (1) The initial state of the Pu-242 nucleus, before it fissions. (2) The state just after fission, when the two silver nuclei are close together, and momentarily at rest. (3) The state when the silver nuclei are very far away from each other, traveling at high speed. A: The final speed of the fission products Your first task is to determine the final speed of each of the daughter nuclei in state (3), when they are far from each other. Figure 6.74 shows three important states in the process:

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Figure 6.74

Which diagram (a, b, or c) depicts the state of the original plutonium nucleus? We'll take this state to be the initial state. Which diagram depicts the state when the two daughter nuclei have reached their final speeds? We'll take this state to be the final state. Compare the initial and final states of the system. Which quantities have changed: potential energy, rest energy, kinetic energy? What will be the total kinetic energy of the two daughter nuclei when they are very far apart? Assume for the moment that when the two silver nuclei are very far apart that the speed is small compared to the speed of light. In that case, what is the final speed of one of the nuclei? (Although this is a very high speed, it is still a rather small fraction of the speed of light, so the low-speed approximation for the kinetic energy is good enough for our purposes.) B: Fission as an energy source The kinetic energy of the fast-moving daughter nuclei is eventually absorbed in the surrounding material and raises the temperature of that material. In a fission power plant, this thermal energy is used to boil water to drive a steam turbine and generate electricity. If a mole of plutonium undergoes this fission reaction, how much kinetic energy is generated? For comparison, only around 1 × 106 joules are obtained from burning a mole of gasoline, which is why energy from fission is of great interest, if some way can be found to deal with the waste products. These silver fission nuclei are not the same as the silver nuclei found in nature; the fission-product isotopes have excess neutrons and are “radioactive,” which means that they emit electrons, alpha particles (helium nuclei), and gamma rays (highenergy photons). These radiations can be easily blocked by burying the wastes, but if radioactive nuclei leak into the ground water and are consumed by humans, the emissions can damage tissues inside the body and cause cancer and genetic damage. C: Testing the simple model of fission A simple model of fission is that when the original plutonium nucleus splits, the two daughter nuclei are “born” at rest, right next to each other. We can test this model by determining what the Energy Principle tells us about the center-to-center distance of these two nuclei just after the plutonium nucleus splits. In Figure 6.74, which two states do NOT involve any change in rest energy? It is convenient to take these two states to be the initial and final states. Compare these initial and final states of the system. Which quantities have changed: potential energy, rest energy, kinetic energy? Using energy considerations, calculate the distance between centers of the silver nuclei just after fission, when they are

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momentarily at rest. Keep at least 6 significant figures in your calculations. A proton or neutron has a radius r of roughly 1 × 10−15 m, and a nucleus is a tightly packed collection of nucleons of radius R. Therefore the volume of the nucleus, , is approximately equal to the volume of one nucleon, , times . So the radius R of a nucleus is about N 1/3 times the radius r of

the number N of nucleons in the nucleus:

one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.3 × 10−15 m)N 1/3 . What is the radius of a silver-121 nucleus? On paper, make a careful scale drawing of the two silver nuclei just after fission, and label the drawing with the distances that you just calculated. If the two silver nuclei are nearly touching, this would be consistent with our model of fission, in which the Pu-242 nucleus fissions into two pieces that are initially nearly at rest. How big is the gap between the surfaces of the two nuclei? (If you have done the calculations correctly, you will indeed find that the gap is a rather small fraction of the center-to-center distance, which means that our simple model for the fission process is a pretty good model.) Note: One way to check all the calculations in this problem is to apply the Energy Principle to a different choice of initial and final states than those that you considered above. You should of course get the same results. 6.P.101 In a fusion reaction, the nuclei of two atoms join to form a single atom of a different element. In such a reaction, a fraction of the rest energy of the original atoms is converted to kinetic energy of the reaction products. A fusion reaction that occurs in the Sun converts hydrogen to helium. Since electrons are not involved in the reaction, we focus on the nuclei. Hydrogen and deuterium (heavy hydrogen) can react to form helium plus a high-energy photon called a gamma ray:

Objects involved in the reaction, where the rest mass is given in atomic mass units, and 1 amu = 1.66054 × 10−27 kg are: Particle

Protons

Neutrons

Charge

Rest mass

1H (proton)

1

0

+e

1.0073

2H (deuterium)

1

1

+e

2.0136

3He (helium)

2

1

+2e

3.0155

gamma ray

0

0

0

0

Although in most problems you solve in this course you should use values of constants rounded to 2 or 3 significant figures, in this problem you must keep at least 6 significant figures throughout your calculation. Problems involving mass changes require many significant figures because the changes in mass are small compared to the total mass. In this problem you must use the following values of constants, accurate to 6 significant figures: Constant

Value to 6 significant figures

c (speed of light)

2.99792 × 108 m/s

e (charge of a proton)

1.60218 × 10−19 coulomb

atomic mass unit

1.66054 × 10−27 kg 8.98755 × 1009N · m2/C 2

A proton (1H nucleus) and a deuteron (2H nucleus) start out far apart. An experimental apparatus shoots them toward each other (with equal and opposite momenta). If they get close enough to make actual contact with each other, they can react to form a helium-3 nucleus and a gamma ray (a high energy photon, which has kinetic energy but zero rest energy). Consider

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the system containing all particles. Work out the answers to the following questions on paper, using symbols (algebra), before plugging numbers into your calculator. A: Bringing the particles together Which diagram in Figure 6.75 depicts the initial state in the process of bringing the particles together?

Figure 6.75

Which diagram depicts the final state in the process of bringing the particles together? Compare the initial state and final states of the system. Which quantities have changed: potential energy, rest energy, kinetic energy? The deuterium nucleus starts out with a kinetic energy of 2.07 × 10−13 joules, and the proton starts out with a kinetic energy of 4.14 × 10−13 joules. The radius of a proton is 0.9 × 10−15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch? B: Reaction to make helium Now that the proton and the deuterium nucleus are touching, the reaction can occur. Take the final state from the previous process to be the initial state of the system for this new process. Which diagram depicts the final state? Compare the initial state and final states of the system. Which quantities have changed: potential energy, rest energy, kinetic energy? What is the kinetic energy of the reaction products (helium nucleus plus photon)? C: Gain of kinetic energy What was the gain of kinetic energy in this reaction? (The products have more kinetic energy than the original particles did when they were far apart. How much more?) D: Fusion as energy source Kinetic energy can be used to drive motors and do other useful things. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated? (For comparison, around 1 × 106 joules are obtained from burning a mole of gasoline.) 6.P.102 Many heavy nuclei undergo spontaneous “alpha decay,” in which the original nucleus emits an alpha particle (a helium nucleus containing two protons and two neutrons), leaving behind a “daughter” nucleus that has two fewer protons and two fewer neutrons than the original nucleus. Consider a radium-220 nucleus that is at rest before it decays to radon-216 by alpha decay. The mass of the radium-220 nucleus is 219.96274 u (unified atomic mass units) where 1 u = 1.6603 × 10−27 kg (approximately the mass of one nucleon).

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The mass of a radon-216 nucleus is 215.95308 u, and the mass of an alpha particle is 4.00151 u. Radium has 88 protons, radon 86, and an alpha particle 2. (a) Make a diagram of the final state of the radon-216 nucleus and the alpha particle when they are far apart, showing the momenta of each particle to the same relative scale. Explain why you drew the lengths of the momentum vectors the way you did. (b) Calculate the final kinetic energy of the alpha particle. For the moment, assume that its speed is small compared to the speed of light. (c) Calculate the final kinetic energy of the radon-216 nucleus. (d) Show that the nonrelativistic approximation was reasonable. 6.P.103 A proton (1.6726 × 10−27 kg) and a neutron (1.6749 × 10−27 kg) at rest combine to form a deuteron, the nucleus of deuterium or “heavy hydrogen.” In this process, a gamma ray (highenergy photon) is emitted, and its energy is measured to be 2.2 MeV (2.2 × 106 eV). (a) Keeping all five significant figures, what is the mass of the deuteron? Assume that you can neglect the small kinetic energy of the recoiling deuteron. (b) Momentum must be conserved, so the deuteron must recoil with momentum equal and opposite to the momentum of the gamma ray. Calculate approximately the kinetic energy of the recoiling deuteron and show that it is indeed small compared to the energy of the gamma ray. 6.P.104 Uranium-235 fissions when it absorbs a slow-moving neutron. The two fission fragments can be almost any two nuclei whose charges Q 1 and Q 2 add up to 92e (where e is the charge on a proton, e = 1.6 × 10−19 coulomb), and whose nucleons add up to 236 protons and neutrons (U-236; U-235 plus a neutron). One of the possible fission modes involves nearly equal fragments, palladium nuclei with Q 1 = Q 2 = 46e. The rest masses of the two palladium nuclei add up to less than the rest mass of the original nucleus. (In addition to the two main fission fragments there are typically one or more free neutrons in the final state; in your analysis make the simplifying assumption that there are no free neutrons, just two palladium nuclei.) The rest mass of the U-236 nucleus is 235.996 u (unified atomic mass units), and the rest mass of each Pd-118 nucleus is 117.894 u, where 1 u = 1.7 × 10−27 kg (approximately the mass of one nucleon). (a) Calculate the final speed v, when the palladium nuclei have moved far apart (due to their mutual electric repulsion). Is this speed small enough that p2/(2m) is an adequate approximation for the kinetic energy of one of the palladium nuclei? (It is all right to go ahead and make the nonrelativistic assumption first, but you then must check that the calculated v is indeed small compared to c.) (b) Using energy considerations, calculate the distance between centers of the palladium nuclei just after fission, when they are starting from rest. (c) A proton or neutron has a radius of roughly 1 × 10−15 m, and a nucleus is a tightly packed collection of nucleons. Experiments show that the radius of a nucleus containing N nucleons is approximately (1.3 × 10−15 m) × N 1/3 . What is the approximate radius of a palladium nucleus? Draw a sketch of the two palladium nuclei in part (b), and label the distances you calculated in parts (b) and (c). 6.P.105 One of the thermonuclear or fusion reactions that takes place inside a star such as our Sun is the production of helium-3 (3He, with two protons and one neutron) and a gamma ray (high-energy photon) in a collision between a proton (1H) and a deuteron (2H, the nucleus of “heavy” hydrogen, consisting of a proton and a neutron): The rest mass of the proton is 1.0073 u (unified atomic mass unit, 1.7 × 10−27 kg), the rest mass of the deuteron is 2.0136 u, the rest mass of the helium-3 nucleus is 3.0155 u, and the gamma ray is massless. (a) The strong interaction has a very short range and is essentially a contact interaction. For this fusion reaction to take place, the proton and deuteron have to come close enough together to touch. The approximate radius of a proton or neutron is about 1 × 10−15 m. What is the approximate initial total kinetic energy of the proton and deuteron

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required for the fusion reaction to proceed, in joules and electron volts (1 eV = 1.6 × 10−19 joule)? (b) Given the initial conditions found in part (a), what is the kinetic energy of the 3He plus the energy of the gamma ray, in joules and in electron-volts? (c) The net energy released is the kinetic energy of the 3He plus the energy of the gamma ray found in part (b), minus the energy input that you calculated in part (a). What is the net energy release, in joules and in electron volts? Note that you do get back the energy investment made in part (a). (d) Which of the following potential energy curves (1–4) in Figure 6.76 is a reasonable representation of the interaction in this fusion reaction? Why?

Figure 6.76 Potential energy curves.

As we will study later, the average kinetic energy of a gas molecule is

, where k is the “Boltzmann constant,” 1.4 ×

10−23 joule/Kelvin (J/K), and T is the absolute or Kelvin temperature, measured from absolute zero (so that the freezing point of water is 273 K). The approximate temperature required for the fusion reaction to proceed is very high. This high temperature, required because of the electric repulsion barrier to the reaction, is the main reason why it has been so difficult to make progress toward thermonuclear power generation. Sufficiently high temperatures are found in the interior of the Sun, where fusion reactions take place. 6.P.106 A nucleus whose mass is 3.917268 × 10−25 kg undergoes spontaneous “alpha” decay. The original nucleus disappears and there appear two new particles: a He-4 nucleus of mass 6.640678 × 10−27 kg (an “alpha particle” consisting of two protons and two neutrons) and a new nucleus of mass 3.850768 × 10−25 kg. (Note that the new nucleus has less mass than the original nucleus, and it has two fewer protons and two fewer neutrons.) When the alpha particle has moved far away from the new nucleus (so that the electric potential energy is negligible), what is the combined kinetic energy of the alpha particle and new nucleus? 6.P.107 Apendulum (see Figure 6.77) consists of a very light but stiff rod of length L hanging from a nearly frictionless axle, with a mass m at the end of the rod.

Figure 6.77

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(a) Calculate the gravitational potential energy as a function of the angle θ, measured from the vertical. (b) Sketch the potential energy as a function of the angle θ, for angles from −210° to +210°. (c) Let s = Lθ = the arc length away from the bottom of the arc. Calculate the tangential component of the force on the mass by taking the (negative) gradient of the energy with respect to s. Does your result make sense? (d) Suppose that you hit the stationary hanging mass so it has an initial speed v i . What is the minimum initial speed needed for the pendulum to go over the top (θ = 180°)? On your sketch of the potential energy (part b), draw and label energy levels for the case in which the initial speed is less than, equal to, or greater than this critical initial speed. 6.P.108

Show the validity of the relation

when m ≠ 0, by making these substitutions:

Computational Problems 6.P.109 If you have already done the problem in Chapter 3 on the Ranger 7 mission to the Moon, you can use the program you wrote for that problem, and go directly to part (b). Otherwise: (a) Write a program to model the journey of a spacecraft coasting from the Earth to the Moon. Start the spacecraft at a height of 50 km above the Earth's surface with a speed of 1.3 × 104 m/s. This is approximately the speed it would have after all the rockets have fired, and is high enough to be above most of the Earth's atmosphere. Include the spacecraft's interactions with both the Earth and the Moon. Use a dt of 5 seconds. Make sure you stop the program when the spacecraft reaches the Moon's surface (not its center!). The data you need may be found on the inside back cover of this book; the mass of the spacecraft was 173 kg. If you are interested in more of the details of this trip, see the problem in Chapter 3 on the Ranger 7 mission to the Moon. (b) Add a calculation of the work done by the gravitational forces of the Earth and the Moon to your analysis of sending a spacecraft to the Moon. You need to approximate the work by adding up the amount of work done by gravitational forces along each step of the path:

Compare the numerical value of the work with the change in the kinetic energy (final kinetic energy just before crashing on the Moon, minus initial kinetic energy when released above the Earth's atmosphere). Try shorter time steps to make sure that the errors introduced by a finite time step are not significant. Report your results. 6.P.110 Energy conservation is a powerful check on the accuracy of a numerical integration. Modify the program for the Chapter 3 problem on the Ranger 7 mission to the Moon to plot graphs of kinetic energy, of gravitational potential energy, and of the sum of the kinetic energy and the gravitational potential energy, vs. position. Does the kinetic plus potential energy remain constant? What if you vary the step size (which varies the accuracy of the numerical integration)? Vary the launch speed, and explain the effect that this has on your graphs.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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Internal Energy

KEY IDEAS Macroscopic systems are composed of many interacting particles. The potential energy associated with stretching or compressing interatomic bonds is similar to the potential energy of macroscopic springs. The potential energy difference between two states does not depend on how a system gets from the initial to the final state (path independence). Internal energy includes deformation, rotation, vibration, and thermal energy. Thermal energy is random kinetic and potential energy of atoms and interatomic bonds within an object.

▪ Temperature is a measure of thermal energy. ▪ A temperature difference between system and surroundings leads to energy transfer (Q). ▪ Specific heat capacity (C) is the amount of energy required to raise the temperature of a given amount of a substance by 1 kelvin. The choice of system affects the form of the Energy Principle.

▪ If there are energy transfers between the system and the surroundings, the system is an “open” system; otherwise it is a “closed” system.

▪ A system is in a “steady state” if there are energy inputs and outputs but no net change in the energy of the system. Energy can be dissipated throughout a system and the surroundings in a form that cannot be fully retrieved.

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POTENTIAL ENERGY OF MACROSCOPIC SPRINGS In Chapter 4 we used the behavior of idealized, macroscopic springs as a model for the dynamic behavior of the springlike interatomic bonds in solid objects. This was a productive model, because it allowed us to understand the nature of contact forces (tension and compression forces) and because using this model we were able to predict the speed of sound in different materials. This model continues to be useful in understanding the flow of energy in macroscopic systems (objects composed of many atoms, such as wires, tables, vines, and humans). To apply it, we need to find an equation for the potential energy of a mass–spring system.

An Ideal Spring A real spring has a nonzero mass and can be damaged in various ways: it can be deformed or broken, the metal can fatigue, and so on. There is a limit to how much a real spring can be stretched (it will eventually break) or compressed (the coils will eventually touch each other, turning the spring into a stiff hollow rod). An idealized spring has none of these problems. There are no limits on the stretch s in the ideal spring force equation (here we're calling the axis of the spring the “s” axis): Despite its limitations, an ideal spring proves to be a useful model for both real macroscopic springs and for springlike interatomic bonds. To determine the potential energy associated with the spring force, recall that the (negative) gradient of potential energy is equal to the associated force. So to find the spring potential energy U s we need a solution to this equation:

which simplifies to

QUESTION Can you think of a function of s whose derivative with respect to s is k s s?

The derivative with respect to s of

is k ss, and the negative gradient of

is the force −k ss.

IDEAL SPRING POTENTIAL ENERGY

s is stretch, measured from the equilibrium point.

The function

is an upturned parabola (Figure 7.1). Note that the stretch s appears squared in the expression for spring energy.

This reflects the fact that either a lengthening (positive stretch) or a compression (negative stretch) of the spring involves increased potential energy.

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Figure 7.1 Potential energy as a function of stretch s for an ideal spring.

Peculiar Features of the Ideal Spring Potential Energy This ideal spring potential energy curve has some peculiar features: As the absolute value of stretch becomes large, U becomes infinite, instead of approaching zero. (This curve is sometimes called an “infinite potential energy well.”) For a system of a mass and ideal spring, K + U is always greater than zero, so apparently there can't be any bound states of the system.

As a result of these peculiarities, this ideal potential energy function doesn't fully represent the nature of a macroscopic spring you can hold in your hand, or of an interatomic bond. One possible modification to the equation would be to subtract a constant ES. Such a function would still satisfy the gradient equation above, but it would allow for bound states of the system, as in Figure 7.2.

For the ideal spring, the constant is conventionally set to zero, but the negative offset may be needed when analyzing real systems.

Figure 7.2 Subtracting a constant ES shifts the ideal spring potential energy curve downward.

Real Springs We are primarily interested in helical (spiral) springs. A modest force applied to a helical spring gives a substantial change in the length of the spring, even though the total length of the coiled wire hardly changes. A large stretch of the spring with a small force implies a much smaller spring stiffness than the wire would have when straightened out. If you stretch a helical spring far enough, it straightens out into a long straight wire, and it suddenly becomes very difficult to lengthen the wire any further. If you continue to stretch the straight wire, eventually it yields and then breaks. If you compress a helical spring so far that its coils run into each other, it suddenly gets extremely hard to compress the spring any further. Therefore,

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the linear relation between stretch and force, F s = −k ss, is valid only over a limited region of stretch, although this linear range is much wider than for the microscopic “spring.” In many situations the entire phenomenon of interest takes place within the valid range of the linear force approximation. One can speak of an “infinite” potential energy well as shown as the dashed curve in Figure 7.3, which is an idealization of the real situation. , we approximate the real curve by a parabola. The actual Within the linear range where Fs = −k ss is the negative gradient of bottom of the parabola is at a negative value of the potential energy. Work would be required to stretch the wire out straight. However, in our calculations with macroscopic springs we're almost always interested only in changes in the potential energy, so we shift the origin to the bottom of the well and measure U s from there.

Figure 7.3 The dashed curve represents the spring potential energy

within the linear range where Fs = −ks s. The colored solid curve represents a real macroscopic spring.

In Figure 7.3 (which is still an idealization of the real situation) there are very steep sides of the potential energy well, corresponding to the large forces required to compress a spring whose coils are in contact and the large forces required to stretch the spring after it has straightened out into a straight wire. Since force is the negative gradient of the potential energy, a steep slope corresponds to a large force.

Energy in an Oscillating Spring–Mass System In an oscillating spring–mass system such as the one discussed in Chapter 4, the energy of the system is continually changing from kinetic energy to potential energy, and back again. Consider a block on a low friction surface, connected by a helical spring to a wall. If you compress the spring and then let go, the block acquires kinetic energy (Figure 7.4), with a loss of spring potential energy. If you stretch the spring and then let go, the block also acquires kinetic energy, again with a loss of spring potential energy (Figure 7.5).

Figure 7.4 As the compressed spring expands the kinetic energy of the block increases, and spring potential energy decreases.

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Figure 7.5 As the stretched spring contracts, the kinetic energy of the block increases and spring potential energy decreases.

QUESTION At the moment that the spring is released, what is the kinetic energy of the spring–mass system? What is the total energy of the system?

Since the block is released from rest, its kinetic energy is zero. The energy of the system is equal to the energy stored in the spring, (omitting the constant rest energy and constant −E S).

QUESTION At what point in the oscillation is the kinetic energy of the system highest? At that moment, what is the potential energy of the system?

When the contracting spring reaches its equilibrium length, the energy stored in the spring is a minimum,

. The maximum

possible amount of spring potential energy has now been converted to kinetic energy, so this is the instant when the kinetic energy of the system is highest.

QUESTION At what point in the oscillation will the kinetic energy of the system be lowest?

At either turning point of the oscillation—spring fully compressed or spring fully extended—the instantaneous speed of the mass is zero. The kinetic energy of the system is lowest here (zero in fact), and all the energy has been momentarily converted back into spring potential energy. Since the absolute value of the stretch is highest at these locations, the potential energy is also highest.

Does the Wall Do Work on the System? For the system of spring + block, the surroundings include the Earth, the low friction table, and the wall. Both the table and the wall touch the system and therefore exert contact forces on it. The table and Earth exert forces perpendicular to the block's motion, so they do zero work on the system (assuming negligible friction).

QUESTION The force by the wall on the spring is parallel to the block's motion. Does it do nonzero work on the system?

This force does zero work, because the point at which the force acts (the left end of the spring) undergoes zero displacement. So

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This force does have an effect on the momentum of the system, however, because the time interval over which it acts is not zero. The momentum of the mass–spring system is changed by the force on the spring by the wall. (Think about what would happen if the wall were not there.) This is another example of the difference between the Momentum Principle (all forces act for the same time interval) and the Energy Principle (different forces may act over different distances).

Flow of Energy in the Mass–Spring System Since the surroundings do no work on the system, the total energy of the system must be constant (in the absence of friction), and energy flows from potential to kinetic and back again, as shown in Figure 7.6.

Figure 7.6 Kinetic energy (red), potential energy (purple), and the sum of kinetic + potential energy (green) for an oscillating spring–mass system.

EXAMPLE Energy in a Spring–Mass System A mass of 0.2 kg is attached to a horizontal spring whose stiffness is 12 N/m. Friction is negligible. At t = 0 the spring has a stretch of 3 cm and the mass has a speed of 0.5 m/s. (a) What is the amplitude (maximum stretch) of the oscillation? (b) What is the maximum speed of the block?

Solution System: Mass and spring Surroundings: Earth, table, wall (neglect friction, air) Initial state: s i = 3 cm, v i = 0.5 m/s Final state: v f = 0 (maximum stretch)

From the discussion in the preceding section, we know that no work is done by the surroundings, because the point where the wall force is applied does not move. We also know that when the stretch is maximum, the block is momentarily at rest (a turning point).

We could use any known state as the initial state in part (b), because the energy of the system is

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constant. (a) Find the maximum stretch: Energy Principle:

(b) Find the maximum speed: Initial state: s i = 3 cm, v i = 0.5 m/s Final state: Maximum speed (s f = 0)

FURTHER DISCUSSION Assuming negligible energy dissipation by friction or air resistance, the energy of the system never changes. If we can find K +U for one state, we can always use this as the initial state in any calculation. 7.X.1 How many joules of energy can you store in a spring whose stiffness is 0.6 N/m, by starting from a relaxed spring and stretching it 20 cm? 7.X.2 A horizontal spring with stiffness 0.5 N/m has a relaxed length of 15 cm. A mass of 20 grams is attached and you stretch the spring to a total length of 25 cm. The mass is then released from rest and moves with little friction. What is the speed of the mass at the moment when the spring returns to its relaxed length of 15 cm?

Answer

Answer

Nonhelical Macroscopic Springs We saw in Chapter 4 that a bar of metal can be treated as a spring, since stretching or compressing the bar stretches or compresses the interatomic bonds (Figure 7.7). Double the force produces double the change in length, as long as we're in the range where the spring approximation to the interatomic force is adequate. This is usually described for a bar in terms of Young's modulus Y, which we studied in Chapter 4, where the tension force FT (per unit area) is proportional to the stretch ΔL (per unit length):

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Since ΔL, like s, represents a change in length, and FT is a springlike force (tension force), this equation can be rewritten to parallel the spring force equation:

The effective “spring stiffness” of the metal bar is YA/L. If you stretch the bar too much, two things can make the bar stop behaving in a springlike manner. You might exceed the interatomic stretch for which k ss is a good approximation to the magnitude of the interatomic force. Also, the regular array of atoms may be disrupted by dislocations of the crystal structure, leading to large-scale slippage of crystal planes. Suddenly the bar “yields” and grows very much longer with little applied force.

Figure 7.7 A bar of metal can be treated as a spring. For a limited range, double the force gives double the stretch.

A straight object is not usually used as a longitudinal spring but may be used in a bending mode. For example, when a diving board bends under a load, atomic bonds in the upper part of the board are stretched, and atomic bonds in the lower part are compressed (Figure 7.8). The net effect is a springlike behavior of the diving board, with the amount of bend proportional to the applied force.

Figure 7.8 The bending of a diving board stretches and compresses interatomic bonds.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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POTENTIAL ENERGY OF A PAIR OF NEUTRAL ATOMS In Chapter 6 we found an equation for electric potential energy for pairs of charged particles. In Chapter 4 we studied the springlike forces between electrically neutral atoms (a neutral object has equal amounts of positive and negative charge, so it has zero net charge). These interatomic forces are the superposition of electric attractions and repulsions among the positively charged protons and negatively charged electrons of which the atoms are made. When two electrically neutral atoms are far from each other (Figure 7.9), they exert almost no force on each other because the attractions between unlike charges (electrons in atom one with protons in atom two, and vice versa) are nearly equal to the repulsions between like charges. However, when the atoms come quite near each other, the electron clouds distort in such a way that the two atoms attract each other. Molecules and solid objects made of two or more atoms are bound together by these electric forces.

Figure 7.9 Two neutral atoms interact very little when far apart, attract each other at intermediate distances, and repel each other at very short distances.

However, if you try to push two atoms even closer together, you eventually encounter a rapidly increasing repulsion. The attraction at a small distance and the repulsion when very close are due ultimately to the superposition of the electric forces of the various protons and electrons in the atoms, with the probable locations of these particles governed by the laws of quantum mechanics.

Potential Energy for Springlike Interatomic Bonds Since an interatomic bond behaves like a spring as long as it is not stretched or compressed excessively, we can guess that the equation for interatomic potential energy has the same form as the potential energy for an ideal spring (Figure 7.10):

APPROXIMATE INTERATOMIC POTENTIAL ENERGY

Figure 7.10 The interatomic potential energy is approximately springlike if the bond is not stretched or compressed too much.

What About Large Separations? Of course, the potential energy of a real interatomic bond can't become infinitely large at large separation—it needs to approach zero. A more realistic potential energy function for the interatomic bond is called the Morse potential energy and is shown in Figure 7.11.

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The Morse potential energy is described by this equation:

The Morse approximation is designed to model the basic behavior of the interatomic force and is in approximate agreement with experimental measurements of the interatomic force.

Figure 7.11 The Morse potential energy function U M , representing the potential energy of two interacting neutral atoms as a function of separation.

The parameter E M gives the depth of the potential energy at its minimum, and the parameter r eq (on the order of 1 × 10−10 m) is the distance at which the potential energy curve is a minimum. The parameter α is adjusted to give the curve the right width. When the interatomic separation r is very large, U M = 0, as it should. The depth of the potential energy well, EM, is typically a few electron volts (recall that 1 eV is equal to 1.6 × 10−19 joule). For example, the potential energy well for oxygen, O 2, is about 5 eV deep. (As you saw in Chapter 6, a typical ionization energy is also in the range of a few electron volts.) However, for the noble gases such as helium and argon the well is so shallow that these atoms don't form stable diatomic molecules at room temperature, because collisions easily provide enough energy to break the molecules apart.

QUESTION What is the physical significance of the minimum of the curve?

At the minimum of the potential energy curve, the slope is zero, so the force between the two atoms is zero (F = −dU/dr). This is the position of stable equilibrium.

QUESTION What is the direction of the force on an atom at a distance smaller than the equilibrium distance?

The left side of the curve has a large negative slope, so the force F = −dU/dr will be large and to the right (in the +r direction), away from the other atom. This makes sense; when the atoms are too close together they push each other apart. Likewise, if the atom moves to the right (farther away) there should be an attractive restoring force to the left. The slope of this region of the curve is positive, so the sign of the force F = −dU/dr is negative, meaning its direction is to the left, as expected. As indicated by the dashed curve in Figure 7.12, the bottom of the interatomic potential energy curve can be approximated by the springlike potential energy function:

Because we can make this approximate fit, we know that for ordinary small oscillations around the equilibrium point the two atoms

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will act as though they are connected by a spring with a −k ss force. This is the rationale for modeling a solid as a collection of balls connected by springs.

Figure 7.12 The bottom of the interatomic potential energy curve may be approximated by a parabola.

Bound and Unbound States with Interatomic Potential Energy In Figure 7.13, a horizontal line represents motion in which the sum of the kinetic energies and the potential energy of a two-atom system does not change. A horizontal line corresponds to a possible state of the system when it is isolated from external forces.

Figure 7.13 A horizontal line represents constant kinetic plus potential energy.

7.X.3 At separation r 1 in Figure 7.13, what is the physical significance of the quantity A? Of the quantity B? Of the quantity C? 7.X.4 In Figure 7.14, which of the states are bound states of a two-atom system? Which are unbound states?

Figure 7.14 Which are bound states? Which are unbound states?

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Answer

Answer

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PATH INDEPENDENCE OF POTENTIAL ENERGY In a system of several interacting particles, potential energy may be associated with an interaction that depends solely on position. In contrast, in a later section we will consider interactions such as air resistance that depend on speed and do not have an associated potential energy. For example, spring energy depends on the stretch s, and gravitational and electric energy depend on the separation r. It is easy to show that the following must be true:

PATH INDEPENDENCE OF POTENTIAL ENERGY Change in potential energy ΔU doesn't depend on the path taken. For a round trip, the change in ΔU is zero.

Here is the reasoning: In Figure 7.15, change in potential energy U of the system when one of its particles moves along some path from location A to location B depends solely on the initial and final values of r or s, so it doesn't matter what path the particle follows to get from A to B. Similarly, in Figure 7.16 if one of the particles in a multiparticle system starts at point A, moves around, and comes back to the starting point, the change in potential energy U of the multiparticle system must be zero, since the initial and final potential energy depends on pair-wise distances r or s, and these return to their original values in a round trip.

Figure 7.15 Potential energy difference is independent of the path between the initial and final locations.

Figure 7.16 A particle moves from A along a path and returns to A. There is a connection between the fact that change of potential energy is independent of path, and the fact that ΔU is zero over a round trip.

QUESTION In Figure 7.17, suppose that along path 1 from A to B the change in potential energy is +5 J. What will be the change in potential energy along the path 1 in the opposite direction, from B to A?

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Figure 7.17 Forward and reversed paths, and a round trip.

Going backwards from B to A along this same path the change would be −5 J. Since the change is independent of the path, you could take any path to return from B to A, such as path 2, with a change of −5 J. Any round trip from A to B and back to A will yield a change of zero. A simple example is projectile motion with negligible air resistance (Figure 7.18). You throw a ball up in the air with speed v, and when it returns to your hand it has regained the same speed v that you gave it (the gravitational potential energy change of the ball + Earth system is zero, so the change in the kinetic energy is zero). Another example is the motion of a mass on a spring in the absence of friction: every time the mass passes a particular position, it has the same speed (same s implies same v). Note that the direction of the velocity may change, but neither the kinetic energy nor the potential energy depend on direction.

Figure 7.18 A ball goes up and down with negligible air resistance.

7.X.5 During one complete oscillation of a mass on a spring (one period), what is the change in potential energy of the mass + spring system, in the absence of friction?

Answer

EXAMPLE A Rebounding Block A metal block of mass 3 kg is moving downward with speed 2 m/s when the bottom of the block is 0.8 m above the floor (Figure 7.19). When the bottom of the block is 0.4 m above the floor, it strikes the top of a relaxed vertical spring 0.4 m in length. The stiffness of the spring is 2000 N/m. (a) The block continues downward, compressing the spring. When the bottom of the block is 0.3 m above the floor, what is its speed? (b) The block eventually heads back upward, loses contact with the spring, and continues upward. What is the maximum height reached by the bottom of the block above the floor? (c) What approximations did you make?

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Figure 7.19 A block falls onto a vertical spring, compresses the spring, then rebounds upward.

Both Earth and spring are part of the system, so there are two potential energy terms, one for (block + spring) and one for (block + Earth). Since the mass of the spring is negligible, there is not a (spring + Earth) potential energy term.

We could equally well have used the final state from part (a) as the initial state in part (b).

Solution (a) Find the speed of block when it is 0.3 m above the floor. System: Earth, block, spring Surroundings: Nothing significant Initial state: Block 0.8 m above floor, moving downward, v i = 2 m/s, spring relaxed Final state: Block 0.3 m above floor, spring compressed Energy Principle:

(b) Find the maximum height. Initial state: Same as in part (a) Final state: Block at highest point, v f = 0, spring relaxed Energy Principle:

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(c) Approximations: Air resistance and dissipation in the spring are negligible. U g ≈ mgy near Earth's surface. ΔK Earth is negligible.

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INTERNAL ENERGY AND THERMAL ENERGY In Figure 7.20 are two identical objects each consisting of two balls connected by identical springs, one compressed (and kept compressed by a string with negligible mass tied around it) and the other relaxed. If these two objects move at the same speed v, they have the same kinetic energy , where M is the total mass, but the object with the compressed spring has more “internal energy.”

Figure 7.20 The object with the compressed spring has more “internal energy.” It is kept compressed by a string of negligible mass.

It is often useful to split the total internal energy into various categories of interest to us in a particular situation. For example, if an object rotates about its center of mass, this internal energy is called “rotational energy” (Figure 7.21). If two balls connected by a spring oscillate relative to the center of mass of the system, this internal energy is called “vibrational energy” (Figure 7.22). If an object is stretched or compressed, there is an associated internal energy change, as in Figure 7.20. When you eat, you increase your internal energy in the form of more “chemical energy.” If the temperature of an object increases, there is an increase in the internal energy that is called “thermal energy.”

Figure 7.21 More rotation means more internal energy in the form of “rotational energy.”

Figure 7.22 More vibration means more internal energy in the form of “vibrational energy.”

From a fundamental perspective, all of these internal energy increases can be thought of as contributing to an increase in the mass of the system: Msys = E sys /c 2. However, for many practical macroscopic calculations it makes sense to focus just on the changes in the thermal energy, or vibrational energy, or rotational energy, or deformational energy, or chemical energy, because these changes are extremely small compared to the huge total rest energy Mc2.

INTERNAL ENERGY

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In this section we will focus specifically on thermal energy, which is an important category of internal energy. In our idealized models of spring–mass systems, the sum of kinetic plus spring potential energy is constant. In real spring–mass systems, however, the oscillations die away (“damp down”) with time, which means that the kinetic plus spring potential energy must be decreasing. The decrease in the amplitude of the oscillations is mainly due to air resistance or to friction with a solid, and some energy is transferred to the surroundings, to the air or to a solid along which the mass slides. On the microscopic level, if we could look inside the solid, we would see increased atomic motion—the atoms in the solid have gained energy, and we call this form of internal energy thermal energy. A macroscopic indication of this increased atomic motion is an increase in the temperature of the object. The most important difference between thermal energy and vibrational or rotational internal energy is that thermal energy is random in character, not organized into macroscopically recognizable motion or deformation. When two objects collide and there is a loss of some of the original kinetic energy, the increased internal energy can be seen in the form of organized rotation or vibration, or visible deformation, but there may also be random motions at the atomic level, and it is this random component of the internal energy that we call thermal energy and whose change is associated with a change in temperature. Merely increasing the rotational or vibrational internal energy of a system doesn't affect its temperature. We can measure macroscopic spring potential energy and kinetic energy with simple measurements that don't require a microscope. However, when the temperature of a solid object increases, the increased energy inside the solid is not visible to the naked eye, and we face the problem of how to evaluate the increase in the thermal energy.

Microscopic Kinetic and Spring Potential Energy The increased microscopic energy in a solid is in two forms. First, the atoms on average are moving around faster, with increased kinetic energy . Second, there is increased spring potential energy in the interatomic bonds (“springs” in our model of solids), where k s is the stiffness of an interatomic bond modeled as a spring (Figure 7.23).

Figure 7.23 In our model solid we must account for the kinetic energy of every ball and the spring energy of every spring.

In principle the microscopic, thermal energy in a solid could be evaluated by simultaneously measuring at some instant the momenta of all of the atoms, and the stretches or compressions of all of the springs (that is, changes in atomic positions away from their equilibrium positions, together with knowledge of the spring stiffness). Since there are about 1 × 1023 atoms (and even more “springs”) in a macroscopic object, it is in practice impossible to evaluate the energy this way.

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We might try to make microscopic measurements of just one atom (and its attached “springs”), assume that this is the average energy of each atom, and then multiply by the number of atoms there are in the object to get the total energy. That doesn't work either, because at a given instant we may choose an atom that is momentarily sitting motionless at its equilibrium position (zero energy), but an instant later will be displaced away from its equilibrium position and moving rapidly. Energy keeps getting passed back and forth among the many atoms and springs, and measuring the energy of just one atom and its attached springs at one instant doesn't give us the correct average energy we would need to determine the total energy of the object.

Temperature In the 1800s it was realized that for many systems the temperature is essentially a measure of the average energy of the atoms in the system. Therefore we can use a thermometer to measure thermal energy. A familiar example is the mercury thermometer (silver colored) or the alcohol thermometer (tinted red for visibility). In these thermometers, a very thin column of liquid expands more than the glass when heated. By tradition, marks on the thermometer are placed to represent “degrees of temperature,” but we could just as well place marks to indicate the average energy of the molecules in the thermometer, in joules (Figure 7.24). One kelvin (or one Celsius degree of temperature) is equivalent to an average molecular energy of about 1 × 10−23 joules.

Figure 7.24 A thermometer might be calibrated in joules instead of in degrees.

This is not the whole story. In a later chapter we will make more precise the relationship between temperature and thermal energy, and we will see that temperature is more directly related to a quantity called “entropy” than it is to energy. Nevertheless, for most ordinary systems at room temperature it is true that temperature is a good measure of the average thermal energy of the atoms. How does a thermometer work? When two objects are in good contact with each other, the average kinetic energies of the molecules in the two objects slowly come to be equal. At a location where two objects touch each other, molecules collide with each other, and the faster molecules on average lose kinetic energy to the slower molecules in a collision. Once the average kinetic energies of molecules in both objects have come to be equal, additional collisions on average make no further change. To measure the thermal energy of an object, we place a thermometer in contact with the object. We wait for the temperatures (average molecular kinetic energies) of the object and thermometer to equilibrate, then we read the thermometer. A useful thermometer must have a relatively small mass compared to the object of interest, so that attaching the thermometer doesn't add or subtract much energy to or from the object. Another advantage of a small thermometer is that it will reach thermal equilibrium quicker than a large thermometer will. How was the calibration between one kelvin (or one Celsius degree) and energy established? In one of a series of classic

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experiments performed by Joule in the 1840s, a paddle wheel in water was turned by a falling weight. The work done by the Earth turned the paddle wheel, which stirred the water. When the paddle wheel stopped, the increase in the internal energy of the water was equal to the work done by the Earth, assuming little rise in internal energy of the paddle wheel and container, and little energy transfer to the surroundings. It was found for water that the energy required to raise the temperature of one gram of water by one kelvin (1 K) is 4.2 joules. The “heat capacity” of an object is the amount of energy required to raise its temperature one kelvin. The “specific heat capacity” is the heat capacity on a per-gram or per-mole or per-atom basis, and is a property of the material. The specific heat capacity of water is 4.2 joules per kelvin per gram (Figure 7.25). Other materials have different specific heat capacities. We will study an atomic theory of specific heat capacity in a later chapter.

Figure 7.25 Energy input of 4.2 joules into a gram of water raises the temperature by 1 K. We say that the specific heat capacity of water is 4.2 J/K/gram.

Volume and Temperature A liquid thermometer works because an ordinary liquid has a larger volume at higher temperature. Why does this happen? For large oscillations of atoms whose interaction is described by a potential energy such as that shown in Figure 7.26, the average stretch increases. Because the actual potential energy curve is not really symmetric around the equilibrium point, with increasing energy the average interatomic bond length gets slightly longer. This is why an object typically expands at higher temperature. Higher temperature implies larger amplitude of oscillations and higher energy, and there is a slight shift in the center of the oscillations at higher energy.

Figure 7.26 For larger oscillations of two bound atoms the average separation increases.

In the interior of a solid the potential energy curve for an atom is symmetric, not asymmetric, because the potential energy is associated with forces exerted by atoms to the left and to the right. However, since the bond lengthening can start at the surface and propagate into the interior, a solid ultimately expands at higher temperatures.

Other Kinds of Thermometers There are many other kinds of thermometers. All materials change in some way when they get hotter or colder, and some of these effects are the basis for useful temperature indicators. Liquid crystals are increasingly used in thermometers. In a liquid crystal there is some molecular order over long distances in the liquid, unlike a true liquid in which there is hardly any long-range order. In one form of liquid crystal thermometer, helical molecules remain roughly parallel to each other, and the distance between coils in the helix changes with temperature. Because the distance between coils is comparable to the wavelength of visible light, there is an effect on the way light is reflected by or transmitted through the liquid crystal. The result is that a change in temperature affects the

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appearance of the material.

Calculations Using Specific Heat Capacity Joule found that it takes 4.2 J of energy input to a gram of water to raise its temperature 1 K. If such experiments are done on other materials, the temperature rise is found to be different. For example, the specific heat capacity of ethanol is found to be 2.4 J/K/gram, and the specific heat capacity of copper is only 0.4 J/K/gram. One of the many unusual and useful properties of water is that it has a very large specific heat capacity on a per-gram basis, which means that it is difficult to change its temperature. Here is the meaning of specific heat capacity on a per-gram basis, where ΔE thermal is the rise in the internal energy of the system, in the form of increased atomic kinetic and potential energy:

SPECIFIC HEAT CAPACITY C ON A PER-GRAM BASIS

In principle, it should be possible to predict the specific heat capacity of a material if something is known about its atomic structure, since a temperature rise is a measure of increased energy at the atomic level. Comparisons of calculations with experimental values of the specific heat capacity data are a good test of our understanding of atomic models of solids, liquids, and gases. In later chapters you will learn how to predict the specific heat capacity of a solid (based on the ball-and-spring model of a solid) and of a gas. If the specific heat capacity of a material is known, the amount of energy transfer into an object can be determined by measuring the temperature rise.

EXAMPLE Energy Input Raises the Temperature You stir 12 kg of water vigorously, doing 36,000 joules of work. If the container is well insulated (so that all of your energy input goes into increasing the energy of the water), what temperature rise would you expect?

Solution

Including units helps avoid making calculational mistakes. Note that a lot of work is required to achieve a small temperature change.

EXAMPLE Thermal Equilibrium

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A 300 gram block of aluminum at temperature 500 K is placed on a 650 gram block of iron at temperature 350 K in an insulated enclosure. At these temperatures the specific heat capacity of aluminum is approximately 1.0 J/K/gram, and the specific heat capacity of iron is approximately 0.42 J/K/gram. Within a few minutes the two metal blocks reach the same common temperature Tf . Calculate Tf .

Solution System: The two blocks Surroundings: Due to the insulation, no objects exchange energy with the chosen system Initial state: Different temperatures, not in contact Final state: Blocks have come to thermal equilibrium, same Tf Energy Principle: ΔE Al + ΔE Fe = 0 (The total energy of the two blocks does not change, because there is no energy transfer from or to the surroundings.)

Solving for the final temperature, we find Tf = 429 K. In words, what happens in this process is that the aluminum temperature falls from 500 K to 429 K, and the iron temperature rises from 350 K to 429 K. The thermal energy decrease in the aluminum is equal to the thermal energy increase in the iron.

7.X.6 300 grams of water whose temperature is 25° Celsius are added to a thin glass containing 800 grams of water at 20° Celsius (about room temperature). What is the final temperature of the water? What simplifying assumptions did you have to make in order to determine your approximate result?

Answer

7.X.7 Niagara Falls is about 50 meters high. What is the temperature rise in kelvins of the water from just before to Answer just after it hits the rocks at the bottom of the falls, assuming negligible air resistance during the fall and that the water doesn't rebound but just splats onto the rock? It is helpful (but not essential) to consider a 1 gram drop of water.

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ENERGY TRANSFER DUE TO A TEMPERATURE DIFFERENCE Work is mechanical energy transfer into or out of a system involving forces acting through macroscopic displacements. When a hot object is placed in contact with a cold object, energy is transferred from the hot object to the cold one, but there are no macroscopic forces or displacements, so we don't refer to this kind of energy transfer as work. Instead, we speak of “energy transfer Q due to a temperature difference.” At the microscopic level there is actual work; when a hot block is placed in contact with a cold block (Figure 7.27), at the interface the atoms in the two blocks collide with each other, and do work on each other. The atoms in the hot block have greater average kinetic energy than the atoms in the cold block, so in an individual collision it is likely that a fast-moving atom in the hot block loses energy to a slow-moving atom in the cold block.

Figure 7.27 A hot block in contact with a cold block.

It can happen that the atom in the hot block happens to be moving slowly and gains energy when it is hit by an atom in the cold object that happens to be moving fast, but this is less likely. On average there is energy flow (“microscopic work”) from the hot block to the cold block. These energy changes will diffuse throughout the blocks. If left to themselves, the two blocks will eventually come to the same temperature, intermediate between their two initial temperatures.

Q = ENERGY TRANSFER DUE TO A TEMPERATURE DIFFERENCE The symbol Q represents an amount of energy that flows from the surroundings into a system, due to a temperature difference between the system and the surroundings. One can also call this process “microscopic work.” Q can be positive or negative.

If we choose the cold block as our system of interest, this system gains energy (at the expense of the surroundings, the hot block). At the atomic level, this energy increase is the result of work done on the atoms at the surface of the cold block. Usually we are unable to observe these atomic-level interactions directly. We can infer the amount of energy transfer into the system by observing the temperature rise of that system, if we know the specific heat capacity of the material. It is common practice to denote energy transfer due to a temperature difference (microscopic work) by the letter Q and work (macroscopic, mechanical energy transfer) by the letter W. The energy equation for an “open” (not isolated) system is then That is, energy transfer due to a temperature difference and mechanical energy transfers into or out of the system produce a change in the energy of the system. The energy of the system includes particle energy and potential energy, at the microscopic or macroscopic level.

Sign of Q

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Like W, Q can be negative. If there is energy transfer out of a system, we give Q a negative sign, because the change in the energy of the system is a decrease. See Figure 7.28 and Figure 7.29. Typically such transfer would be due to the system having a higher temperature than the surroundings. Similarly, if the system does work on its surroundings rather than the other way round, the sign of W is negative. (Some older textbooks on thermal physics reverse the sign of W, counting energy outputs from the system as positive.)

Figure 7.28 If energy flows from the surroundings into the system, Q is positive.

Figure 7.29 If energy flows from the system into the surroundings, Q is negative.

It is thought that for the Universe as a whole Q and W are zero, and the energy of the Universe does not change. However, even a small system may be effectively closed, with its energy unchanging, if it is thermally well insulated (to prevent the flow of energy due to a temperature difference) and if there are no other energy transfers from the surroundings.

Terminology: The Word Heat The technical meaning of the word “heat” in science is not the same as its everyday meaning, just as the scientific meaning of the word “work” is different from its everyday meaning. In science the word “heat” is reserved to refer to energy transfer across a system boundary due to a temperature difference (microscopic work); this quantity is represented by the symbol Q. One cannot say “there is heat in the object;” instead, one says there is thermal energy Ethermal in the object. The amount of energy in the object might increase due to “heat” Q(microscopic work). However, even professional scientists sometimes slip in their usage of the word “heat”! To avoid confusion, we will avoid the use of the word “heat” as a noun. Instead, we will speak of “energy transfer Q due to a temperature difference” between system and surroundings, and change of “thermal energy” inside the system ΔE thermal (in the form of increased atomic kinetic energy and potential energy). 7.X.8 Suppose that you warm up 500 grams of water (half a liter, or about a pint) on a stove while doing 5 × 104 J Answer of work on the water with an electric beater. The temperature of the water is observed to rise from 20°C to 80°C. What was the change in the thermal energy of the water? Taking the water as the system, how much transfer of energy Q due to a temperature difference was there across the system boundary? What was the energy change of the surroundings?

Other Kinds of Energy Transfers Work W and energy transfer Q due to a temperature difference are very common kinds of energy transfers into or out of an open

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system, but there are several other kinds of energy transfer. Here is a list, together with examples of each type. W, mechanical work ; a force acts through a macroscopic distance: Throw a ball, increasing the ball's kinetic energy; compress a macroscopic spring. Q, energy transfer due to a temperature difference between system and surroundings: Boil water over a hot flame; a warm house loses energy to the cold outside air. Matter transfer: Put gasoline into a car; natural gas is piped into the home. Mechanical waves: Sound enters a microphone and moves a plate to generate electric signals; ocean waves drive an electric generator. Electricity: Electrical current enters the house and is used to run appliances; electrical current stores energy in the batteries of an electric car. Electromagnetic radiation, including visible light, infrared light, gamma rays, x-rays, and ultraviolet light: Sunlight absorbed by the Earth raises the temperature (and the thermal energy) of the daylight side of the Earth. We will usually account for this kind of energy transfer in terms of photons (packets of electromagnetic energy) crossing the system boundary, to be discussed in a later chapter.

A general form of the Energy Principle is this:

THE ENERGY PRINCIPLE

Other energy transfers into a system may include matter transfer, mechanical waves, electric current, and electromagnetic radiation.

7.X.9 An electric hot plate raises its own internal energy and the internal energy of a cup of water by 8000 J, and there is at the same time 1000 J transferred to the cooler air (that is, Q = −1000 J). How much energy was transferred to the hot plate in the form of electricity?

Answer

The Steady State A system is in a steady state if its energy does not change despite the fact that there are energy transfers between the system and the surroundings. In the steady state, the rates of energy flow into and out of the system must be equal.

EXAMPLE An Electric Heater An electric heater receives an energy input of 5000 J of electric energy. During this time the heating element is maintained at a constant high temperature. What is ΔE thermal for the heater, and what is Q, the energy transfer between the heater's hot heating element and the cooler air?

Solution System: Heater

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Surroundings: Electric wires, air Energy Principle: ΔE sys = W + Q + other energy transfers Since there is no internal energy change, ΔE thermal = ΔE sys = 0. This is a steady-state situation.

FURTHER DISCUSSION In this situation the sign of Q is negative, reflecting the fact that energy is transferred from the heating element to the air. We describe this situation as a steady state, not equilibrium, because there are energy transfers into and out of the system. However, these energy transfers, plus and minus, add up to zero, so there is no change internally. If the heater were in an insulating enclosure and no energy could flow into the air, as electric energy continually flowed into the heater the temperature of the heating element would keep increasing until the element melted.

Q Is Not the Same as ΔE thermal It is very important to see the difference between an energy transfer Q between system and surroundings due to a temperature difference, and a change in the thermal energy inside a system, associated with a change in the temperature of the system. In the previous example of the electric heater, the thermal or internal energy of the heating element did not change (ΔE sys = 0, steady state), but Q = −5000 J. 7.X.10 In a certain time interval, natural gas with energy content of 10000 J was piped into a house during a winter Answer day. In the same time interval sunshine coming through the windows delivered 1000 J of energy into the house. The temperature of the house didn't change. What was ΔE thermal of the house, and what was Q, the energy transfer between the house and the outside air?

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REFLECTION: FORMS OF ENERGY There are three fundamental kinds of energy in a multiparticle system: rest energy kinetic energy potential energy

All other forms of energy are examples of these basic forms. For example, chemicals in a system can react with each other and raise the temperature (and thermal energy) of the system, so we speak of “chemical energy,” of which food is an example, with the molecule ATP providing energy storage in the body. The “chemical energy” is the kinetic energy and electric potential energy of molecules. Change of shape (configuration) is associated with change of potential energy, and there may also be changes of the kinetic energy of the molecules. Nuclear energy is similar. Nucleons can be more or less tightly bound to particular nuclei, associated with different amounts of nuclear potential energy (change of shape) and different amounts of kinetic energy of the nucleons. Also, a change of identity in a nuclear reaction can be associated with a (huge) kinetic energy change associated with a change of rest mass.

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POWER: ENERGY PER UNIT TIME In technical usage, the word “power” is defined to mean “energy per unit time.” From the point of view of energy usage, it makes no difference whether you take a minute or a month to climb a flight of stairs, yet from a practical point of view you certainly do notice the rate of energy usage, called power. The units of power are joules per second, or watts (honoring James Watt, the developer of the first efficient steam engine).

POWER Energy per unit time (J/s or watts)

We can construct an equation for the instantaneous power associated with the work done by a force, through the following reasoning. If a force does an amount of work in a time Δt, then:

so instantaneously

INSTANTANEOUS POWER

EXAMPLE Light Bulb Power How much energy is required to run a 100 watt light bulb for an hour?

Solution

7.X.11 In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. Answer The height of the falls is about 50 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (1 × 106 watts)? 7.X.12 A vehicle with a mass of 1000 kg has an engine whose maximum power output is 50 kilowatts (about 67

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Answer

horsepower; one horsepower is 746 watts). At a speed of 20 m/s (about 45 miles/hour) and maximum power, determine the maximum acceleration by calculating the force that is acting.

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OPEN AND CLOSED SYSTEMS We have already had some experience with the difference between a closed system and an open system. Because of the importance of this distinction, in this section we will go more deeply into the issues. Figure 7.30 shows transfers of money into and out of a bank during a particular time period, and the corresponding change in accounts inside the bank.

Figure 7.30 A bank is an open system.

QUESTION We have shown a situation in which more money is transferred into the bank than is transferred out. During these transactions, does the total amount of money inside the bank remain unchanged?

In Figure 7.30 $5000 came in and $3000 went out, for a net gain of $2000. We say that the bank is an “open system”—a portion of the world open to transfers in and out, and therefore subject to changes in its internal amount of money. During times when the bank does not permit transfers, the bank is temporarily a “closed system” and its total internal amount of money is unchanged, although there may be changes in the form of money inside the bank such as transfers between checking accounts and savings accounts. Now consider Figure 7.31, which shows energy transfers during a particular time period in the winter, into and out of a house that is heated by gas, and the corresponding change in the energy inside the house. In this situation more energy comes into the house than goes out, and as a result the temperature inside the house rises.

Figure 7.31 A house is an open system.

QUESTION During these energy transactions, the total amount of energy in the whole Universe is unchanged, but is the amount of energy inside the house unchanged?

The amount of energy inside the house increased by 2000 joules, and the energy of the surroundings decreased by 2000 joules. We

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say that the house is an “open system”—a portion of the Universe open to energy transfers, and therefore subject to changes in its internal amount of energy. If the house could be perfectly insulated against heat leakage (and solar radiation), and the gas (and electricity) turned off, the house would be a “closed system” with respect to energy and its total internal amount of energy would be unchanged, although there may be changes in the form of energy inside the house, such as the family dog converting chemical energy into kinetic energy when chasing its tail. These considerations lead to the following scheme for keeping track of energy. Choose some portion of the Universe and mentally surround it by a dashed line marking the boundary (Figure 7.32). Then the energy transfer into the system across the system boundary minus the energy transfer out of the system across the system boundary is equal to the change of the energy inside the system.

Figure 7.32 The change of energy of the system is equal to energy transferred into the system across the system boundary minus energy transferred out of the system across the system boundary.

The change in energy inside the system, ΔE sys , can be either a positive or a negative amount of energy. For example, take an electric car as the system of interest: Charge the battery: ΔE sys is positive (more energy stored in the battery). Do work on the car by pushing it: ΔE sys is positive (more kinetic energy). Run the headlights (radiate electromagnetic energy): ΔE sys is negative (less energy stored in the battery).

The car is an open system whose total amount of energy changes due to energy transfers into or out of the system: the energy of an open system is not constant. In contrast, the total energy of the Universe remains constant, because there are compensating energy changes in the surroundings of the car: the electric company lost a store of chemical energy by running its generators to charge the car, you used up some chemical energy to push the car, and the light from the headlights is absorbed by the surroundings and leads to a rise in temperature of the surroundings. The measurements that we are able to make confirm the premise that the Universe is a closed system whose total amount of energy never changes, although the form of the energy may change. For any closed system, inflow = outflow = 0, so ΔE sys = 0. The energy of a closed system does not change.

The Universe as a whole is the most important example of a closed system, but it is often the case that a portion of the Universe can be considered to be a closed system, at least approximately. For example, put hot water and ice cubes into a very well-insulated container. During the short time that the ice melts and the water gets somewhat cooler, we can neglect the small amount of energy leakage through the walls of the insulated container. There is an increase in the energy of the ice cubes (which changed from solid to liquid), and a decrease in the energy of the hot water, but negligible net change in the energy inside the container, which is approximately a closed system whose total energy is (approximately) unchanged.

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THE CHOICE OF SYSTEM AFFECTS ENERGY ACCOUNTING What you choose to include in your chosen system affects the form of the energy equation for that system. For example, when we analyze a system consisting of an airless Earth plus a ball moving upward, the energy of the system does not change, because there are no energy transfers into or out of our chosen system (Figure 7.33).

Figure 7.33 System: Earth + ball.

System: Earth + Ball System: Earth + ball Surroundings: Nothing significant Initial state: Ball moving upward Final state: Ball has risen a distance h, and is momentarily at rest Energy Principle: Since only the difference in U matters, we can set U i = 0.

System: Ball On the other hand, when we analyze a system consisting of just the ball (Figure 7.34), the Earth (which is now outside our chosen system) does work on the ball, and the energy of the ball increases:

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Figure 7.34 System: ball only.

System: Ball Surroundings: Earth Energy Principle: There are no potential energy terms for a single-particle system. The Earth does negative work because the force is opposite to the displacement.

Energy terms move back and forth across the equal sign in the energy equation, depending on our choice of the system of interest. It is extremely important to be clear about the choice of system, because this affects the form of the energy equation and helps avoid serious mistakes.

Warning: Avoid Double Counting! QUESTION A student said, “The Earth exerts a downward force mg as the ball rises a distance h and does an amount of work −mgh, and there is also an increase in gravitational potential energy +mgh, so the kinetic energy decreases by an amount −2mgh.” What is wrong with the student's statement?

The problem here is double-counting due to a lack of clarity about the choice of system. If the Earth is part of the system, the force mg is internal to the system and does no external work on the system; rather, there is work done by internal forces on both the Earth and the ball, and the negative of this work by internal forces is the change in potential energy. On the other hand, if the ball alone is the system, there is no shape change of the ball and no change of gravitational potential energy, but there is an external force mg that does an amount of work −mgh. In the next section we will illustrate the issue of choice of system by analyzing the same problem three times, using three different choices of system. We will find that the form of the Energy Principle will be different in each case, but we obtain the same physical results each time.

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The Woman, the Barbell, and the Earth To help avoid double-counting mistakes, it is useful practice to write the energy equation for various choices of system in a more complicated situation, that of a woman applying a constant force F to lift a barbell of mass m from rest through a distance h, at which point the barbell is not only higher above the Earth but also has acquired some speed v (Figure 7.35). We're considering a time when the barbell is still headed upward, before the woman has brought the barbell to a stop above her.

Figure 7.35 A woman lifts a barbell a distance h, exerting a force F. In the final state the barbell is moving upward with speed v.

For convenience in the following discussion, let Ew represent the following energy terms:

System: Woman + Barbell + Earth System: Woman + barbell + Earth (Figure 7.36)

Figure 7.36 System: woman + barbell + Earth.

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Surroundings: Nothing significant Initial state: Barbell at rest Final state: Barbell has moved upward a distance h, and has speed v Energy Principle:

If we choose the system to consist of the woman, the barbell, and the Earth, there are no energy transfers into or out of our system, so energy does not change, and the energy equation looks like this as the woman lifts the barbell a distance h, starting from rest (we neglect the tiny kinetic energy of the recoiling Earth). This allows us to calculate the change in the woman's energy from our final equation:

As can be seen from this result, the net effect is a decrease in her energy, because the barbell and the Earth are farther apart, and the barbell goes faster, both of which represent energy increases in other parts of the system. The energy term ΔE w represents changes in energy associated with the woman, mainly a decrease in her stored chemical energy, because she must perform chemical reactions with food (or with chemicals stored in body tissues after eating food) in order to be able to move her arm. The woman has to burn more energy than the rest of the system gets, since she not only has to supply energy to the rest of the system but also has to supply her own energy increases: increased kinetic energy and gravitational potential energy associated with her moving, raised arm, and increased thermal energy, because her temperature rises a bit when she exerts herself. Warning! It is tempting, but wrong, to include a term for the change in the gravitational energy of the barbell and to write the energy equation for the barbell alone like this:

The error here is in assigning some gravitational energy to the barbell alone. Gravitational potential energy is associated with a change in the shape of a multi-object system; in particular, to changes in the separations of pairs of interacting objects. The barbell alone is not a pair of objects and does not change shape.

System: Barbell System: Barbell (Figure 7.37)

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Figure 7.37 System: barbell only.

Surroundings: Woman and Earth Initial state: Barbell at rest Final state: Barbell has moved upward a distance h, and has speed v Energy Principle:

If we choose the system to consist just of the barbell, the woman does (positive) work +Fh on the system, and the Earth does (negative) work −mgh on the system (because mg is down and the displacement h is up). Compare this equation to the equation we obtained when considering the system of woman + barbell + Earth. We see that energy terms have moved to the other side of the equation as a result of our different choice of system.

QUESTION Which is bigger, F or mg? Why?

Since the barbell's upward momentum increases, F must be larger than mg, since the net force in the upward direction is F − mg.

QUESTION Compare the equations for system 1 (woman + barbell + Earth) and system 2 (barbell only). In terms of F, calculate ΔE w .

By comparing the two equations, we see that the change in the energy associated with the woman must be −Fh. The sign makes sense: we had concluded earlier that her energy change must be negative.

System: Barbell + Earth System: Barbell + Earth Surroundings: Woman Initial state: Barbell at rest

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Final state: Barbell has moved upward a distance h, and has speed v Energy Principle:

If we choose the system to consist of the barbell plus the Earth (Figure 7.38), this system changes shape as the woman pushes the two pieces of the system apart. It is almost as though she is stretching an invisible spring connecting the two objects.

Figure 7.38 System: barbell + Earth.

QUESTION Does the woman do any work on the Earth? Why or why not?

The woman's feet push down on the Earth but do negligible work, because there is essentially no displacement there (the Earth hardly moves). Work done by a force is calculated by taking into account the displacement of the point where the force is applied. In this case, the point of application hardly moves, so the force does hardly any work.

Further Discussion Looking back over the equations obtained for different choices of system, you can see how energy terms that represent transfers across the system boundary for one choice of system become changes of energy inside the system for a different choice of system. Comparing equations for different choices of system can be useful in determining an unknown quantity (such as the energy change in the woman, −Fh, in the example you just worked through). Also, analyzing a process for more than one choice of system is a good check on your calculations and your understanding. 7.X.13 Consider a harmonic oscillator (mass on a spring without friction). Taking the mass alone to be the system, Answer how much work is done on the system as the spring of stiffness k s contracts from its maximum stretch A to its relaxed length? What is the change in kinetic energy of the system during this motion? For what choice of system does energy remain constant during this motion?

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ENERGY DISSIPATION The total energy of the Universe does not change, but useful energy is often “dissipated” into forms less useful to us. Push a chair across the floor, and some of the work that you do goes into raising the temperature of the floor and the chair, rather than into increasing the macroscopic kinetic energy of the chair. Throw a ball up into the air, and some of the initial energy is dissipated into increased microscopic energy of the air. Sliding friction, air resistance, viscous friction—all of these phenomena are examples of energy dissipation discussed in the rest of the chapter.

Air Resistance Air resistance isn't a major factor when you drop a metal ball a short distance. A video sequence of a falling metal ball shows that the ball moves faster and faster as it falls (an effect that is hard to observe by eye alone). The time between adjacent frames in Figure 7.39 is 1/15 second, and the increasing distances between heights of the ball in adjacent frames show that the speed is getting faster and faster. (Also note the increasing blur due to faster motion of the ball.) The gravitational attraction of the Earth acting on the ball makes the momentum of the ball continually increase. A curve is drawn along the tops of the ball images. The major visible marks on the vertical meter sticks are 10 cm apart. In 7/15 second the ball falls about 1 meter.

Figure 7.39 A sequence of video frames of a falling metal ball, which travels farther in each frame, reflecting the increase in its speed.

In contrast, if you drop a bowl-shaped paper coffee filter, a video sequence of the falling filter shows that the filter's speed does increase at first, but instead of continuing to gain speed it quickly reaches a constant speed despite the gravitational force acting on it (Figure 7.40):

Figure 7.40 A sequence of video frames of a falling coffee filter, which speeds up briefly, then falls at constant speed.

In the video sequence shown above, the time between adjacent frames is again 1/15 second. In 16/15 second the filter falls about 1 meter (the dense ball took only 7/15 second to fall that far). A curve is drawn through the centers of the coffee filter images. The nearly straight line in the later part of the motion indicates motion at constant speed.

Dependence of Air Resistance on Speed The restraining effect of the air is called “air resistance” or “drag.” You have probably observed that it is harder to move something quickly through a fluid such as water or air than to move it slowly. This suggests that air resistance acting on a falling object might depend on the speed of the object.

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QUESTION Consider a falling coffee filter at two different times: (1) while the filter's speed is still increasing, and (2) after a constant speed has been reached. How do you know the air resistance is larger at the second instant than at the first instant?

While the filter's speed is increasing there must be a nonzero net force, which means that the air resistance is smaller than the gravitational force (Figure 7.41). Later, however, the net force is zero, because the speed is no longer increasing, which means that the force of the air has grown to be as large as the gravitational force (which is nearly constant at mg and in fact increases very slightly as the coffee filter gets closer to the Earth). Since the change in the density of the air is small over the distance the filter falls, and the change in the gravitational force is very small, we can conclude that the change in the air resistance is due to a speed dependence of that force.

Figure 7.41 Forces acting on a falling coffee filter at Time 1 when the filter's speed is increasing and at Time 2 when a constant speed has been reached.

Experiments show that the magnitude of the air resistance force is approximately proportional to the square of an object's speed. You can do a simple experiment with coffee filters to determine the form of the speed dependence, at least for low speeds (see Problem 7.P.60).

Terminal Speed A sky diver initially speeds up due to the gravitational force acting downward, but the sky diver's speed doesn't keep getting bigger and bigger, because the upward air resistance force increases with speed. Eventually the sky diver reaches a “terminal speed” and falls thereafter at constant speed despite the gravitational force. This terminal speed for falling humans is so high (about 60 m/s, or about 135 miles per hour) that hitting the ground without a parachute is normally fatal, although there have been cases of people hitting deep snow at terminal speed and surviving. 7.X.14 A sky diver whose mass is 90 kg is falling at a terminal speed of 60 m/s. What is the magnitude of the force of the air on the sky diver?

Answer

Dependence of Air Resistance on Cross-Sectional Area It is easy to observe that the effect of air resistance increases as the crosssectional area of an object increases—a sky diver falls much more slowly with an open parachute than with a closed parachute. A low-density paper coffee filter has a large cross-sectional area, so air resistance has a large effect on a falling coffee filter. In contrast, a high-density metal ball has a small crosssectional area,

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but a large gravitational force acts on it, and air resistance may be negligible in comparison with gravity.

Dependence of Air Resistance on Shape QUESTION From your own experience, would you expect larger air resistance (at a particular speed) for a pointed object or a blunt object?

As you might expect, the air resistance force depends not only on density, cross-sectional area, and speed, but also on the shape of the object. There is a larger air resistance force on a blunt object such as a coffee filter than on a ball with the same cross-sectional area. Streamlined objects such as arrows have reduced air resistance.

Dependence of Air Resistance on Air Density QUESTION If you could double the density of the air through which a coffee filter falls, what change in the air resistance force would you expect?

You would probably expect a larger air resistance force in denser air. In fact, measurements of a variety of ordinary-sized objects moving through air or other fluids show that air resistance is proportional to the density ρ of the air. For example, there is less air resistance at higher altitudes, where the air is less dense.

An Empirical Equation for Air Resistance An equation describing the air resistance force on a moving object must incorporate all the effects we have listed above: dependence on speed, crosssectional area, shape of the object, and density of the air. The shape effect is usually taken into account by measuring the air resistance force (for example, by measuring the terminal speed) and determining an experimental “drag coefficient” C such that the following equation expresses the experimental results for the friction force:

APPROXIMATE AIR RESISTANCE FORCE (EMPIRICAL) For blunt objects at ordinary speeds:

where velocity:

; blunter objects have higher values of C (ρ is the density of the air). The direction is opposite to the

How important is the effect of air resistance in the everyday world? In Chapter 2 we analyzed the motion of a ball thrown through the air. However, we neglected air resistance, which can have a sizable effect. In Problem 7.P.68 you are asked to include the air resistance force given above. You find that a baseball thrown at high speed by a professional baseball pitcher goes only about half as far in air as it would go in a vacuum.

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Even when we include air resistance in our calculations, we still have ignored a force that can have a major effect on the trajectory. If a ball has spin, there is an effect of fluid flow around the ball that raises the air pressure on the side where the rotational motion is in the same direction as the ball's velocity, and lowers the air pressure on the other side, where the rotational motion is in the opposite direction to the velocity. If the force points to the side, the ball curves; this is an important effect in baseball pitching. If the force points upward (due to “backspin,” as in Figure 7.42), it lifts the ball and extends the range. If the force points downward (due to “topspin”) it decreases the range. The calculations involved in modeling this effect are quite complex, and involve “fluid dynamics” (the study of fluid flow).

Figure 7.42 A spinning ball experiences additional forces.

It is important to keep in mind that air resistance is not a fundamental force like the gravitational or electric force. Air resistance is the average result of a huge number of momentary contact electric interactions of air molecules hitting atoms on the surfaces of the falling object, and the average net air resistance is very difficult to calculate from fundamental principles at the molecular level. 7.X.15 As a ball falls, it has to push aside the air underneath it, and the air molecules acquire additional kinetic energy K air. Therefore, will the final speed of the ball be greater or less than you would predict from the constancy of kinetic energy plus gravitational energy? Why? 7.X.16 A coffee filter of mass 1.4 grams dropped from a height of 2 m reaches the ground with a speed of 0.8 m/s. How much kinetic energy K air did the air molecules gain from the falling coffee filter?

Answer

Answer

Mechanism of Air Resistance We would like to understand the details of the interaction of the air with a falling object. In particular, we'd like to find a reason that the air resistance would depend on the speed of the object, the cross-sectional area of the object, and the density of the air. Let's consider the interactions of the air and object from a microscopic viewpoint. When you hold a coffee filter stationary, air molecules hit it nearly equally from above and below, as seen in Figure 7.43. (Actually there are slightly more collisions per second on the lower surface than on the upper surface, leading to a small upward buoyant force that is very small compared to air resistance.) However, when the filter is moving downward due to the gravitational attraction of the Earth, the bottom side of the filter is running into the air molecules, while the top side is moving away from the air molecules.

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Figure 7.43 Air molecules collide with a falling coffee filter.

On average, the bottom side will have a significantly increased number of collisions per second with air molecules, and with greater impact. On the top side there will be a reduced number of collisions per second, and with less impact. There is a net upward push on the filter. This air resistance force increases with increasing downward speed of the filter, because higher speed increases the rate and impact of collisions on the bottom, and decreases the rate and impact of collisions on the top. Eventually, when the downward speed of the coffee filter has become big enough, the net upward push by the air molecules becomes as large as the downward gravitational pull of the Earth. From then on the filter falls at a constant speed (terminal speed). While this model captures the qualitative aspects of the situation, a simple quantitative calculation along these lines (using techniques discussed in a later chapter on gases) predicts a terminal speed that is a thousand times smaller than is observed! This is a serious failure of our simple model. The difficulty turns out to be that the falling coffee filter establishes macroscopic “wind”-like motions in the air, and the random motions of the air molecules at any particular location are relative to the bulk wind motion at that location. The techniques of “fluid dynamics” are capable of analyzing such phenomena, but these techniques are mathematically quite difficult and well beyond the scope of this introductory textbook. (Our later study of gases will address simpler situations, where there is no large-scale wind motion.) 7.X.17 When a falling object reaches terminal speed, its kinetic energy reaches a constant value. However, the gravitational energy of the system consisting of object plus Earth continues to decrease. Does this violate the principle of conservation of energy? Explain why or why not. 7.X.18 If you let a mass at the end of a string start swinging, at first the maximum swing decreases rather quickly, but once the swing has become small it takes a long time for further significant decrease to occur. Try it! Explain this simple observation.

Answer

Answer

Viscous Friction Very small particles such as dust or droplets of fog falling slowly through air, or small objects moving through a thick liquid such as honey, experience a friction force that is proportional to v rather than v 2. This is called “viscous” friction. The dependence of the friction force on v (for viscous friction) or v 2 (for air resistance) is approximately valid only in specific situations. The expressions for friction forces do not have the wide applicability of those for the gravitational and electric forces. However, even approximate expressions for friction forces allow us to make much better predictions than we could make if we ignored friction completely.

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Viscous Friction in a Mass–Spring System One of the most obvious things about the motion of a real macroscopic spring–mass system is that the amplitude gets smaller and smaller with time. Our model of a spring–mass system that we developed in Chapter 4 is too simple: we should include the effects of friction. See Problem 7.P.67 on numerical integration of a spring–mass system with friction. In the case of “viscous” friction, where the friction force is

, it is possible with elaborate math (or by guessing the

solution and plugging it into the Momentum Principle for the system) to find an analytical solution for x position as a function of time.

The amplitude dies away exponentially with time (Figure 7.44). The angular frequency ω of this “damped” oscillator is

which is approximately

for low friction.

Figure 7.44 Position vs. time for an oscillator with viscous friction. There is no such simple analytical solution if the friction force is independent of the speed (sliding friction) or proportional to v 2 (air resistance for large blunt objects), but such frictional forces are easily handled in a computational model.

Dissipation with Sliding Friction In Chapter 4 we saw that when a block slides on a table, the table exerts a force on the block associated with compression of the interatomic bonds in the table and block. This force has two components: normal to the surface, and parallel to the surface, called the frictional force. Atoms of the block run into atoms in the table, and the resulting interactions stretch and compress interatomic bonds in the objects. The frictional force is nearly independent of the speed of the block, and f ≈ μ k F N , where μ k is the kinetic (motional) coefficient of friction. If the block does not slide across the table, the frictional force is just big enough to make the net force be zero. In this case, f ≤ μ s F N , where μ s is called the coefficient of static friction.

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Figure 7.45 provides an atomic-level picture of friction. This image is a single frame from a molecular dynamics computer simulation of sliding friction between two objects, created by Judith A. Harrison and coworkers at the United States Naval Academy in Annapolis, Maryland. The computations are similar to the numerical integrations you have done, but they involve a large number of objects (atoms). The forces between the atoms are modeled by springlike forces.

Image courtesy of Judith A. Harrison, U.S. Naval Academy. Figure 7.45 A computational model of sliding friction between a diamond tip and diamond surface. The vertical forces are not shown.

In this image, a diamond tip is dragged to the right across a diamond surface. The large spheres represent carbon atoms, and the small ones are hydrogen atoms on the carbon surfaces. Note that the projecting tip has caught on the lower surface, causing interatomic bonds in the two objects to deform. The stretching and compression of interatomic bonds is responsible for the horizontal component of the force exerted on the moving object by the lower object.

A Paradox Involving Friction If we are careful in our energy bookkeeping for a system involving motion with friction, we encounter a paradox that cannot easily be resolved. Consider the situation shown in Figure 7.46.

Figure 7.46 Forces in the x direction on a block pulled with constant speed across a table. (Forces in the y direction are not shown.)

You exert a constant force F, and pull a block a distance d across a table. Because there is friction between the block and the table, the block travels at constant speed. Since the momentum of the block does not change, we conclude correctly that the net force on the block is zero, and that therefore the magnitude of the friction force must also be F. In terms of energy inputs, you have done an amount of work Fd on the block. The friction force, acting in the opposite direction, has apparently done an amount of work −Fd. The net work done on the block therefore would seem to be zero. This is consistent with the fact that the kinetic energy of the block did not seem to change. It is not, however, consistent with the fact that both the block and the table get hotter as they rub together! If the net work done on the block was zero, where did that increased internal energy come from? This is not a “trick question,” but a genuine puzzle. We will return to this issue in a later chapter, when we will deal with it in enough detail to resolve the paradox.

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7.X.19 You move a block slowly to the right a distance of 0.2 m on a table. You apply a constant force of 20 N. How much work do you do? You then move slowly back to the starting point; how much work do you do in this second move? What change has there been in the thermal energy of the block and table?

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Answer

POTENTIAL ENERGY AND “CONSERVATIVE” FORCES In Section 7.3 we saw that, in the absence of air resistance, a ball thrown upward will return to your hand with exactly the same kinetic energy it had when it left your hand. Taking the ball as the system, the work done on the ball by the Earth is negative as the ball ascends and positive as the ball descends; the total work done by the Earth during the round trip is zero, and the energy of the system is unchanged (Figure 7.47).

Figure 7.47 In the absence of air resistance, work done on the ball by the Earth is negative going up and positive coming down, and the total work done on the ball is zero. When the ball has returned to its initial position, its kinetic energy has not changed.

However, if we include air resistance, we find that the total work done by the air resistance force is not zero, and therefore the final kinetic energy of the system (the ball) is less than its initial energy. The work done by the air resistance force is negative as the ball ascends, and it is also negative as the ball descends, because the direction of the air resistance force has changed (Figure 7.48). The kinetic energy of the ball decreases during both parts of the trip.

Figure 7.48 Taking into account air resistance, work done on the ball by the air is negative going up and also negative coming down, and the total work done on the ball is negative (the total work done by the Earth is zero). When the ball has returned to its initial position, its kinetic energy has decreased.

The decrease in kinetic energy of the ball is accompanied by an increase in the internal energy of the air and the ball, and the temperatures of the air and ball rise. The kinetic energy lost by the ball has been randomly and irreversibly distributed among the molecules of the air and the ball. If a force depends only on the position of two interacting objects, it is possible to define a potential energy associated with this force. Gravitational and electric forces are examples of such forces, as is the force exerted by a spring. Such forces are called “conservative forces.” A conservative force is equal to the negative gradient of the associated potential energy. Dissipative forces such as friction forces are called “nonconservative” forces. It isn't possible to define a potential energy for a dissipative force. In defining potential energy, we consider two interacting objects to be pointlike, in the sense that their internal

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states do not change. This is not the case for dissipative forces; the air is a complicated multiparticle system whose internal energy does change. The flow of energy from the ball to the air is irreversible; the energy distributed among the air molecules cannot be recovered and converted back to kinetic energy of the ball. Irreversibility has far-reaching consequences for the statistical or probabilistic behavior of macroscopic objects consisting of large numbers of atoms. We will consider issues of reversibility and irreversibility in more detail in a later chapter.

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* RESONANCE Imagine pushing somebody in a swing. If you time your pushes to match the natural frequency of the swing, you can build up an increasingly large amplitude of the swing. However, larger amplitude means higher speed and a higher rate of energy dissipation.

QUESTION You keep pushing with constant (small) amplitude, and the swing eventually reaches a “steady state” with constant (large) amplitude. What determines the size of this steadystate amplitude?

The steady state is established when the energy dissipated through air resistance on each cycle has grown to be equal to the energy input you make on each cycle. The large steady-state oscillation needs only little pushes at the right times from you to make up for energy dissipation, and the amplitude of the oscillation can be much larger than the amplitude of the pushes (Figure 7.49).

Figure 7.49 Small pushes at the right times lead to a big response.

On the other hand, if you deliberately push at the wrong times, not matching the natural frequency of the swing, the amplitude of the swing won't continually build up (Figure 7.50). To get a large response from an oscillating system, inputs should be made at the natural, free-oscillation frequency of that system. This important feature of driven oscillating systems is called “resonance.”

Figure 7.50 Pushes at the wrong times lead to a small response.

*Analytical Treatment of Resonance Problem 7.P.71 is a numerical integration of a sinusoidally driven spring–mass oscillator, as shown in Figure 7.51.

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Figure 7.51 A sinusoidally driven oscillator.

This problem lets you study the buildup of the steady state as well as the steady state itself. For the steady-state portion of the motion, it is possible to obtain an analytical solution, if the energy dissipation is due to viscous friction. Here is the Momentum Principle for the driven oscillator in the diagram, with viscous friction force −cv:

where x is the position of the mass, D is the amplitude of the sinusoidal motion, and ω D is the variable angular frequency of the driven end of the spring. For more details on how this equation was obtained, see the discussion in Problem 7.P.71. In the steady state x must be a sinusoidal function, with amplitude A and phase shift φ depending strongly on the driving angular frequency ω D: By taking derivatives of this expression to obtain v = dx/dt and dv/dt as a function of time t and plugging these derivatives into the Momentum Principle, it can be shown that this expression for x as a function of time is a solution in the steady state if the amplitude A and phase shift φ have values determined by the following expressions:

The detailed proof is rather complicated, so we omit it. Instead, in the exercises that follow we give you the opportunity to see that these expressions for amplitude and phase shift are in agreement with your experimental observations and the computer modeling of Problem 7.P.71. Figure 7.52 is a graph of the amplitude A as a function of the driving angular frequency ω D for two different values of the viscous friction parameter c.

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Figure 7.52 Amplitude as a function of driving frequency. The damping in the lower curve is five times greater than in the upper curve.

7.X.20 Using the equation for the amplitude A, show that if the viscous friction is small, the amplitude is large when ω D is approximately equal to ω F . Using the equation involving the phase shift φ, show that the phase shift is approximately 0 for very low driving frequency ω D, approximately 180 for very high driving frequency ω D, and 90 at resonance, consistent with your experiment. Answer

7.X.21 Show that with small viscous friction, the amplitude A drops to

of the peak amplitude when the

Answer

driving angular frequency differs from resonance by this amount:

(Hint: Note that near resonance

, so

.)

Given these results, how does the width of the resonance peak depend on the amount of friction? What would the resonance curve look like if there were very little friction?

Resonance in Other Systems We have discussed resonance in the context of a particular system—a mass on a spring driven by a motor. We study resonance because it is an important phenomenon in a great variety of situations. When you choose a radio or television station, you adjust the parameters of an electrically oscillating circuit so that the circuit has a narrow resonance at the chosen station's main frequency, and other stations with significantly different frequencies have little effect on the circuit. Magnetic resonance imaging (MRI) is based on the phenomenon of nuclear magnetic resonance (NMR) in which nuclei acting like toy tops precess in a strong magnetic field. The precessing nuclei are significantly affected by radio waves only at the precession frequency. Because the precession frequency depends slightly on what kind of atoms are nearby, resonance occurs for different radio frequencies, which makes it possible to identify different kinds of tissue and produce remarkably detailed images.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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SUMMARY Potential energy of an ideal spring–mass system:

Approximate interatomic potential energy:

A better approximation to interatomic potential energy:

(More potential energy, Figure 7.53)

Figure 7.53

Thermal energy: Related to the temperature of the object ; also calculable as Energy transfer due to a temperature difference, Q Other energy transfers: Work W, matter transfer, mechanical waves, electricity, electromagnetic radiation Specific heat capacity = energy per gram to raise temperature 1 K

Open system: energy flows into and/or out of system. Closed system: no energy flows into or out of system. The choice of system (open or closed) affects energy accounting. Dissipation of energy Terminal speed Air resistance: collisions between air molecules and moving object

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Sliding friction force due to forming and breaking of bonds between two solid objects Path independence of potential energy Impossibility of describing friction in terms of potential energy Resonance: sensitivity of driven oscillator to driving frequency Air resistance is approximately proportional to speed squared, and in a direction opposite to the velocity:

For small particles (“viscous” friction), fluid friction is approximately proportional to v. *Analytical solution for driven spring–mass system:

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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, where

; viscous friction cv, driven at ω D:

EXERCISES AND PROBLEMS Sections 7.1 7.X.22 A spring has a relaxed length of 6 cm and a stiffness of 100 N/m. How much work must you do to change its length from 5 cm to 9 cm? 7.X.23 A constant force 23, −12, 32 N acts through a displacement 0.12, 0.31, −0.24 m. How much work does this force do? 7.X.24 An object with mass 7 kg moves from a location 22, 43, −41 m near the Earth's surface to location −27, 11, 46 m. What is the change in the potential energy of the system consisting of the object plus the Earth? 7.X.25 A spring whose stiffness is 800 N/m has a relaxed length of 0.66. If the length of the spring changes from 0.55 m to 0.96 m, what is the change in the potential energy of the spring? 7.X.26 A particle moves inside a circular glass tube under the influence of a tangential force of constant magnitude F (Figure 7.54). Explain why we cannot associate a potential energy with this force (which is therefore a “nonconservative” force). How is this situation different from the case of a block on the end of a string, which is swung in a circle?

Figure 7.54 A particle moves in a circular glass tube. 7.X.27 There are three different ways to get from location A to location E in Figure 7.55. Along path 1, you take an elevator directly from location A straight up to location E. Along path 2, you walk from A to B, climb a rope from B to C, then walk from C to E. Along path 3, you walk from location A to D, then climb a ramp up to location E. The following questions focus on the work done on a 80 kg person by the Earth while following each of these paths.

Figure 7.55 Following different paths from A to E. Taking the origin at location A, the coordinates of the other locations are: B 0,−3, 0 ; C −3, 13, 0 ; D 33, 0, 0 ; E 0, 13, 0 . (a) Path 1 (A to E): What is the displacement vector (b)

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for the path from A to E on the elevator?

What is the gravitational force acting on the person along path 1, as a vector? (c) What is the work done by the gravitational force of the Earth on the 77 kg person during the elevator ride from A to E? (d) Path 2 (A to B to C to E): What is the work done by the gravitational force of the Earth on the person as he walks from A to B? (e) What is the work done by the gravitational force of the Earth on the person as he climbs from B to C? (f) What is the work done by the gravitational force of the Earth on the person as he walks from C to E? (g) What is the total work done by the gravitational force of the Earth on the person as the person goes from A to E along path 2? (h) Path 3 (A to D to E): What is the work done by the gravitational force of the Earth on the person as he walks from A to D? (i) What is the work done by the gravitational force of the Earth on the person as he climbs from D to E? (j) What is the total work done by the gravitational force of the Earth on the person as the person goes from A to E along path 3? (k) You calculated the work done by the gravitational force of the Earth on the person along three different paths. How do these quantities compare with each other? 7.X.28 Refer to Figure 7.56. Calculate the change in electric energy along two different paths in moving charge q away from charge Q from A to B along a radial path, then to C along a circle centered on Q, then to D along a radial path. Also calculate the change in energy in going directly from A to D along a circle centered on Q. Specifically, what are U B − U A , U C − U B , U D − U C , and their sum? What is U D − U A ? Also, calculate the round-trip difference in the electric energy when moving charge q along the path from A to B to C to D to A.

Figure 7.56 Change in electric energy along two different paths. 7.P.29 A horizontal spring–mass system has low friction, spring stiffness 200 N/m, and mass 0.4 kg. The system is released with an initial compression of the spring of 10 cm and an initial speed of the mass of 3 m/s. (a) What is the maximum stretch during the motion? (b) What is the maximum speed during the motion? (c) Now suppose that there is energy dissipation of 0.01 J per cycle of the spring–mass system. What is the average power input in watts required to maintain a steady oscillation? 7.P.30 Write an equation for the total energy of a system consisting of a mass suspended vertically from a spring, and include the Earth in the system. Place the origin for gravitational energy at the equilibrium position of the mass, and show that the changes in energy of a vertical spring–mass system are the same as the changes in energy of a horizontal spring–mass system. 7.P.31 A spring with stiffness k s and relaxed length L stands vertically on a table. You hold a mass M just barely touching the top of the spring. (a) You very slowly let the mass down onto the spring a certain distance, and when you let go, the mass doesn't move. How much did the spring compress? How much work did you do? (b) Now you again hold the mass just barely touching the top of the spring, and then let go. What is the maximum

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compression of the spring? State what approximations and simplifying assumptions you made. (c) Next you push the mass down on the spring so that the spring is compressed an amount s, then let go, and the mass starts moving upward and goes quite high. When the mass is a height of 2L above the table, what is its speed? 7.P.32 Figure 7.57 is a potential energy curve for the interaction of two neutral atoms. The two-atom system is in a vibrational state indicated by the heavy black horizontal line. (a) At r = r 1, what are the approximate values of the kinetic energy K, the potential energy U, and the quantity K + U? (b) What minimum energy must be supplied to cause these two atoms to separate? (c) In some cases, when r is large, the interatomic potential energy can be expressed approximately as U = −a/r 6. For large r, what is the algebraic form of the magnitude of the force the two atoms exert on each other in this case?

Figure 7.57 Potential energy diagram. 7.P.33 Design a “bungee jump” apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 10 m long, and that the cords stretch in the jump an additional 24 m for a jumper whose mass is 80 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground). (a) It will help you a great deal in your analysis to make a series of 5 simple diagrams, like a comic strip, showing the platform, the jumper, and the two cords at the following times in the fall and the rebound:

▪ ▪ ▪ ▪ ▪

while cords are slack (shown here as an example to get you started) when the two cords are just starting to stretch when the two cords are half stretched when the two cords are fully stretched when the two cords are again half stretched, on the way up

On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper's velocity. Make the relative lengths of the vectors reflect their relative magnitudes. (b) At what instant is there the greatest tension in the cords? (How do you know?) (c) What is the jumper's speed at this instant, when the tension is greatest in the cords? (d) Is the jumper's momentum changing at this instant or not? (That is, is dpy /dt nonzero or zero?) Give a valid physics reason for your answer. Check to make sure that the magnitudes of the velocity and force vectors shown in your diagram number 4 are consistent with your analysis of parts (c) and (d). (e) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness k s for each of the two cords. (f) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what

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kind of cords you need to buy.) (g) What is the maximum acceleration |ay | = |dv y /dt| (in “g's”; acceleration divided by 9.8 m/s 2) that the jumper experiences? (Note that |dpy /dt| = m|dv y /dt| if v is small compared to c.) (h) What is the direction of this maximum acceleration? (i) What approximations or simplifying assumptions did you have to make in your analysis that might not be adequately valid? 7.P.34 A package of mass m sits on an airless asteroid of mass M and radius R. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed v. We have a powerful spring whose stiffness is k s. How much must we compress the spring? 7.P.35 A package of mass 9 kg sits on an airless asteroid of mass 8.0 × 1020 kg and radius 8.7 × 105 m. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 226 m/s. We have a large and powerful spring whose stiffness is 2.8 × 105 N/m. How much must we compress the spring? 7.P.36 A mass of 0.3 kg hangs motionless from a vertical spring whose length is 0.8 m and whose unstretched length is 0.65 m. Next the mass is pulled down so the spring has a length of 0.9 m and is given an initial speed upward of 1.2 m/s. What is the maximum length of the spring during the following motion? What approximations or simplifying assumptions did you make?

Sections 7.4 7.X.37 Substance A has a large specific heat capacity (on a per gram basis), while substance B has a smaller specific heat capacity. If the same amount of energy is put into a 100 gram block of each substance, and if both blocks were initially at the same temperature, which one will now have the higher temperature? 7.X.38 An oil company included in its advertising the following phrase: “Energy—not just a force, it's power!” In technical usage, what are the differences among the terms energy, force, and power? 7.X.39 Electricity is billed in kilowatt-hours. Is this energy or power? How much is one kilowatt-hour in standard physics units? (The typical cost of one kilowatt-hour is 5 to 10 cents.) 7.X.40 State which of the following are open systems with respect to energy, and which are closed: a car; a person; an insulated picnic chest; the Universe; the Earth. Explain why. 7.X.41 A certain motor is capable of doing 3000 joules of work in 11 seconds. What is the power output of this motor? 7.X.42 Starting from rest, a woman lifts a barbell with a constant force F through a distance h, at which point she is still lifting, and the barbell has acquired a speed v. Let E woman stand for the following energy terms associated with the woman:

The change in the kinetic energy of the barbell is

.

The Energy Principle here is ΔE sys = W (no other energy transfers). We'll consider terms on the left side of the equation (the ΔE sys side, changes in the energy inside the system) and terms on the right side (the W side, energy inputs from the surroundings). I.

System: Woman + barbell + Earth For the system consisting of the woman, the barbell, and the Earth, which terms belong on the left side of the energy equation (the ΔE sys side)? For the system consisting of the woman, the barbell, and the Earth, which terms belong on the right side of the energy equation (the W side)?

II. System: Barbell only For the system consisting of the barbell only, which terms belong on the left side of the energy equation (the ΔE sys side)? For the system consisting of the barbell only, which terms belong on the right side of the energy equation (the W side)?

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III. System: Barbell + Earth For the system consisting of the barbell and the Earth, which terms belong on the left side of the energy equation (the ΔE sys side)? For the system consisting of the barbell and the Earth, which terms belong on the right side of the energy equation (the W side)? 7.X.43 A block of mass m is projected straight upward by a strong spring whose stiffness is k s. When the block is a height y 1 above the floor, it is traveling upward at speed v 1, and the spring is compressed an amount s 1. A short time later the block is at height y 2, traveling upward at speed v 2, and the spring is compressed an amount s 2. In this discussion we'll use the Energy Principle in the form Ef = E i + W, where we assume that thermal transfer of energy (microscopic work) Q between the block and the air is negligible. We can ignore the rest energies, which don't change, and the kinetic energy of the Earth, which hardly changes. I.

System: Universe (block + spring + Earth) For the system consisting of the block, the spring, and the Earth, write the Energy Principle in the update form Ef = E i + W.

II. System: Block + spring For the system consisting of the block plus spring, write the Energy Principle in the update form Ef = E i + W. III. System: Block alone For the system consisting of the block alone, write the Energy Principle in the update form Ef = E i + W. 7.X.44 A horse whose mass is M gallops at constant speed v up a long hill whose vertical height is h, taking an amount of time t to reach the top. The horse's hooves do not slip on the rocky ground, so the work done by the force of the ground on the hooves is zero (no displacement of the force). When the horse started running, its temperature rose quickly to a point at which from then on, heat transferred from the horse to the air keeps the horse's temperature constant. (a) First consider the horse as the system of interest. In the initial state the horse is already moving at speed v. In the final state the horse is at the top of the hill, still moving at speed v. Write out the Energy Principle ΔE sys = W + Q for the system of the horse alone. The terms on the left-hand side should include only energy changes for the system, while the terms on the right-hand side should relate to the surroundings (everything else). Which of the terms are equal to 0? Which of the terms should go on the system (left) side (ΔE sys )? Which of the terms should go on the surroundings (right) side? (b) Next consider the Universe as the system of interest. Write out the energy principle for this system. Remember that terms on the left-hand side should include all energy changes for the system, while the terms on the righthand side should relate to the surroundings. Which of the terms are equal to 0? Which of the terms should go on the system (left) side (ΔE sys )? Which of the terms should go on the surroundings (right) side? 7.P.45 400 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass is 600 grams and initial temperature 20°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water? Explain. What simplifying assumptions did you have to make? 7.P.46 180 grams of boiling water (temperature 100°C, heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass is 1050 grams and initial temperature 26 C (the heat capacity of aluminum is 0.9 J/K/gram). (a) After a short time, what is the temperature of the water? (b) What simplifying assumptions did you have to make? (A) The thermal energy of the water doesn't change. (B) The thermal energy of the aluminum doesn't change. (C) Energy transfer between the system (water plus pan) and the surroundings was negligible during this time. (D) The heat capacities for both water and aluminum hardly change with temperature in this temperature range.

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(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 29,541 J of work, and the temperature of the water and pan increases to 86.9°C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan? 7.P.47 Here are questions dealing with human power. (a) If you follow a diet of 2000 food calories per day (2000 kilocalories), what is your average power consumption in watts? (A food or “large” calorie is a unit of energy equal to 2.4 × 103 J; a regular or “small” calorie is equal to 4.2 J.) Compare with the power consumption in a table lamp. (b) You can produce much higher power for short periods. Make appropriate measurements as you run up some stairs, and report your measurements. Use these measurements to estimate your power output (this is in addition to your basal metabolism—the power needed when resting). Compare with a horsepower (which is about 750 watts) or a toaster (which is about 1000 watts). 7.P.48 Here are questions about human diet. (a) A typical candy bar provides 280 calories (one “food” or “large” calorie is equal to 4.2 × 103 joules). How many candy bars would you have to eat to replace the chemical energy you expend doing 100 sit-ups? Explain your work, including any approximations or assumptions you make. (In a sit-up, you go from lying on your back to sitting up.) (b) How many days of a diet of 2000 large calories are equivalent to the gravitational energy difference for you between sea level and the top of Mount Everest, 8848 m above sea level? (However, the body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.) 7.P.49 Humans have about 60 milliliters (60 cm 3) of blood per kilogram of body mass, and blood makes a complete circuit in about 20 seconds, to keep tissues supplied with oxygen. Make a crude estimate of the additional power output of your heart (in watts) when you are standing compared with when you are lying down. Note that we're not asking you to estimate the power output of your heart when you are lying down, just the change in power when you are standing up. You will have to estimate the values of some of the relevant parameters. Because we're only looking for a crude estimate, try to construct a model that is as simple as possible. Describe the approximations and estimates you made. 7.P.50 During three hours one winter afternoon, when the outside temperature was 0° C (32° F), a house heated by electricity was kept at 20° C (68° F) with the expenditure of 45 kwh (kilowatt · hours) of electric energy. What was the average energy leakage in joules per second through the walls of the house to the environment (the outside air and ground)? The rate at which energy is transferred between two systems is often proportional to their temperature difference. Assuming this to hold in this case, if the house temperature had been kept at 25° C (77° F), how many kwh of electricity would have been consumed? 7.P.51 You observe someone pulling a block of mass 43 kg across a low-friction surface. While they pull a distance of 3 m in the direction of motion, the speed of the block changes from 5 m/s to 7 m/s. Calculate the magnitude of the force exerted by the person on the block. What was the change in internal energy (chemical energy plus thermal energy) of the person pulling the block?

Sections 7.9 7.X.52 Describe a situation in which it would be appropriate to neglect the effects of air resistance. 7.X.53 Describe a situation in which neglecting the effects of air resistance would lead to significantly wrong predictions. 7.X.54 List all the approximations that are made in order to find a simple analytical solution for the motion of a projectile near the Earth's surface. 7.X.55 You are standing at the top of a 50 m cliff. You throw a rock in the horizontal direction with speed 10 m/s. If you neglect air resistance, where would you predict it would hit on the flat plain below? Is your prediction too large or too small as a result of neglecting air resistance? 7.X.56 A coffee filter of mass 1.8 grams dropped from a height of 4 m reaches the ground with a speed of 0.8 m/s. How much kinetic energy K air did the air molecules gain from the falling coffee filter? Start from the Energy Principle, and choose as the system the coffee filter, the Earth, and the air.

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7.X.57 A box with its contents has a total mass of 36 kg. It is dropped from a very high building. (a) After reaching terminal speed, what is the magnitude of the air resistance force acting upward on the falling box? (b) The box survived the fall and is returned to the top of the building. More objects are put into the box, and the box with its contents now has a total mass of 71 kg. The box is dropped, and it reaches a higher terminal speed than before. After reaching terminal speed, what is the magnitude of the air resistance force acting upward on the falling box? (The fact that the heavier object reaches a higher terminal speed shows that the air resistance force increases with increasing speed.) 7.X.58 You drag a block across a table while a friend presses down on the block. The coefficient of friction between the table and the block is 0.6. The vertical component of the force exerted by the table on the block is 190 N. How big is the horizontal component of the force exerted by the table on the block? 7.X.59 You drag a block with constant speed v across a table with friction. Explain in detail what you have to do in order to change to a constant speed of 2v on the same surface. (That is, the puzzle is to explain how it is possible to drag a block with sliding friction at different constant speeds.) 7.P.60 A paper coffee filter in the shape of a truncated cone falls in a stable way and quickly reaches terminal speed because it has a large area (big air resistance) but low mass (small gravitational force). Use a stopwatch to time the drop of coffee filters from as high a starting position as you can conveniently manage. If a stairwell is available, drop the coffee filters there to time a longer fall. This experiment is easier to do with a partner. (a) By stacking coffee filters you can change the mass of a falling object without changing the shape. With this scheme you can explore how the terminal speed depends on the mass. Time the fall for different numbers of stacked filters, taking some care that the shape of the bottom filter is always the same (the filters tend to flatten out when removed from a stack). Start the measurement after the filters have fallen somewhat, to allow them to reach terminal speed. Average the results of several repeated measurements of time and height. Can you think of a simple experiment you can do to verify that the coffee filters do in fact reach terminal speed before you start timing? (b) We want to know quantitatively how air resistance depends on speed. Plot your data for the air resistance force vs. the terminal speed. (The air resistance force is equal to the gravitational force when terminal speed has been reached, so it is proportional to the number of filters in a stack). How does the air resistance force depend on v? (c) There is an important constraint on the graph of speed dependence: Should the curve of the air resistance force vs. terminal speed go through the origin (force and terminal speed both very small), or not? (d) Analysis hint: If you suspect that the force is proportional to v 3, you might plot force vs. v 3 and see whether a straight line fits the data. 7.P.61 Figure 7.58 is a portion of a graph of energy terms vs. time for a mass on a spring, subject to air resistance. Identify and label the three curves as to what kind of energy each represents. Explain briefly how you determined which curve represented which kind of energy.

Figure 7.58 Which curve represents which kind of energy? 7.P.62 You drop a single coffee filter of mass 1.7 grams from a very tall building, and it takes 52 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed. (a) What was the upward force of the air resistance while the coffee filter was falling at terminal speed? (b) Next you drop a stack of five of these coffee filters. What was the upward force of the air resistance while this stack

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of coffee filters was falling at terminal speed? (c) Again assuming that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground? (Hint: Consider the relation between speed and the force of air resistance.) 7.P.63 Design and carry out an experiment to determine the coefficient of sliding friction for a block on a flat plate. 7.P.64 For this problem you will need measurements of the position vs. time of a block sliding on a table, starting with some initial velocity, slowing down, and coming to rest. If you do not have an appropriate laboratory setup for making these measurements, your instructor will provide you with such data. Analyze these data to see how well they support the assertion that the force of sliding friction is essentially independent of the speed of sliding. (Recall that a reason was given for this assertion—the friction force is determined by the strength of temporary interatomic bonds that must be broken.) 7.P.65 You can observe the main effects of resonance with very simple experiments. Hold a spring vertically with a mass suspended at the other end, and observe the frequency of “free” oscillations with your hand kept still. Then stop the oscillations, and move your hand extremely slowly up and down in a kind of slow sinusoidal motion. You will see that the mass moves up and down with the same very low frequency. (a) How does the amplitude (plus or minus displacement from the center location) of the mass compare with the amplitude of your hand? (Notice that the phase shift of the oscillation is 0°; the mass moves up when your hand moves up.) (b) Next move your hand up and down at a significantly higher frequency than the free-oscillation frequency. How does the amplitude of the mass compare to the amplitude of your hand? (Notice that the phase shift of the oscillation is 180°; the mass moves down when your hand moves up.) (c) Finally, move your hand up and down at the freeoscillation frequency. How does the amplitude of the mass compare with the amplitude of your hand? (It is hard to observe, but the phase shift of the oscillation is 90°; the mass is at the midpoint of its travel when your hand is at its maximum height.) (d) Change the system in some way so as to increase the air resistance significantly. For example, attach a piece of paper to increase drag. At the free-oscillation frequency, how does this affect the size of the response?

A strong dependence of the amplitude and phase shift of the system to the driving frequency is called resonance.

Computational Problems 7.P.66 Energy calculations provide a powerful check on the accuracy of a numerical integration. Modify your numerical integration of the motion of a mass on a spring that you did earlier. In addition to plotting graphs of position versus time, also plot graphs of kinetic energy, of spring energy, and of the sum of the kinetic energy and the spring energy, versus time during the motion. It is up to you to choose some value for the potential energy U 0 at the bottom of the well. Label the energy graphs with numerical values and units. Make sure that the energy values are reasonable for your system. (a) Does the sum of kinetic and potential energy remain constant as a function of time? What if you vary the step size (which varies the accuracy of the numerical integration)? (b) Increase the energy by increasing the initial speed. How does the amplitude A change? How does the angular frequency ω change? 7.P.67 Modify your numerical integration of the spring–mass system to include the effects of friction. Such a system is called a “damped oscillator.” Simulate the effect of all three kinds of friction: sliding friction (independent of v), viscous friction (proportional to v), and air resistance for large blunt objects (proportional to v 2). Choose friction to be small enough that you get at least two full cycles of oscillation, but large enough that you see an effect. If the friction is so large that the graph never crosses the axis, the system is said to be “over-damped.” For each kind of friction, display the following: 1. Display an animation of the motion of the system. 2. Plot a graph of position vs. time. 3. Plot graphs of K, U, and (K + U) vs. time. On the energy graphs, when is the energy loss rate large and small, and why? 7.P.68 Use a computer to carry out the following step-by-step quantitative calculations for a baseball. Remember that the direction

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of the air resistance force on an object is opposite to the direction of the object's velocity. A baseball has a mass of 155 grams and a diameter of 7 cm. The drag coefficient C for a baseball is about 0.35, and the density of air at sea level is about 1.3 × 10−3 grams/cm 3, or 1.3 kg/m3. (a) Including the effect of air resistance, compute and plot the trajectory of a baseball thrown or hit at an initial speed of 100 miles per hour, at an angle of 45 to the horizontal. How far does the ball go? Is this a reasonable distance? (A baseball field is about 400 feet from home plate to the fence in center field. An outfielder cannot throw a baseball in the air from the fence to home plate.) (b) Plot a graph of (K + U g) vs. time. (c) Temporarily neglect air resistance (set C = 0 in your computations) and determine the range of the baseball. Check your computations by using the analytical solution for motion of a projectile without air resistance. Compare your result with the range found in part (a), where the effect of air resistance was taken into account. Is the effect of air resistance significant? (The effect is surprisingly large—about a factor of 2!) (d) Neglecting air resistance, plot a graph of (K + U g) vs. time. (e) In Denver, a mile above sea level, the air is about 83% as dense as the air at sea level. Including the effect of air resistance, use your computer model to predict the trajectory and range of a baseball thrown in Denver. How does the range compare with the range at sea level? 7.P.69 Terminal speed for a falling sky diver has been measured to be about 60 m/s. Using your computer model from Problem 7.P.68, determine how far (in meters) and how long (in seconds) a sky diver falls before reaching terminal speed. (You will need to think about the meaning of “terminal speed” and how to test for it.) Plot graphs of speed vs. time and position vs. time. 7.P.70 In Problem 7.P.60 you experimented with falling coffee filters. It was important to let the coffee filters fall for a while to reach terminal speed before starting to time their fall, in order to measure the terminal speed. Model the fall of the coffee filters computationally, starting from rest, using your computer model from Problem 7.P.68. 7.P.71 You can study resonance in a driven oscillator in detail by modifying your computation for the spring–mass system with friction. Let one end of the spring be moved back and forth sinusoidally by a motor, with a motion given by Dsin(ω Dt) (see Figure 7.59). Here D is the amplitude of the motor motion and ω D is the angular frequency of the motor, which can be of the spring–mass system has a fixed value, determined by varied. (The free-oscillation angular frequency the spring stiffness k s and the mass m.)

Figure 7.59 A driven spring–mass system. We need an expression for the stretch s of the spring in order to be able to calculate the force −k ss of the spring on the mass. We have to do a bit of geometry to figure out the length of the spring when the mass is displaced a distance x from the equilibrium position, and the motor has moved the other end of the spring by an amount Dsin(ω Dt). In the figure, the spring gets longer when the mass moves to the right (+x), but the spring gets shorter when the motor moves

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to the right (−D sin(ω Dt)). The new length of the spring is L+ x − D sin(ω Dt), and the net stretch of the spring is this quantity minus the unstretched length L, yielding (s = x − Dsin(ω Dt)). A check that this is the correct expression for the net stretch of the spring is that if the motor moves to the right the same distance as the mass moves to the right, the spring will have zero stretch. Replace x in your computer computation with the quantity [x − Dsin(ω Dt)]. (a) Use viscous damping (friction proportional to v). The damping should be small. That is, without the motor driving the system (D = 0), the mass should oscillate for many cycles. Set ω D to 0.9ωF (that is, 0.9 times the freeoscillation ), and let x 0 be 0. Graph position x vs. time for enough cycles to show that there is angular frequency, a transient buildup to a “steady state.” In the steady state the energy dissipation per cycle has grown to exactly equal the energy input per cycle. Vary ω D in the range from 0 to 2.0ωF , with closely spaced values in the neighborhood of 1.0 ωF . Have the computer plot the position of the motor end as well as the position of the mass as a function of time. For each of these driving frequencies, record for later use the steadystate amplitude and the phase shift (the difference in angle between the sinusoids for the motor and for the mass). If a harmonic oscillator is lightly damped, it has a large response only for driving frequencies near its own free-oscillation frequency. (b) Is the steady-state angular frequency equal to ω F or ω D for these various values of ω D? (Note that during the transient buildup to the steady state the frequency is not well-defined, because the motion of the mass isn't a simple sinusoid.) (c) Sketch graphs of the steady-state amplitude and phase shift vs. ω D. Note that when ω D = ω F the amplitude of the mass can be much larger than D, just as you observed with your hand-driven spring–mass system. Also note the interesting variation of the phase shift as you go from low to high driving frequencies—does this agree with the phase shifts you observed with your hand-driven spring–mass system? (d) Repeat part (c) with viscous friction twice as large. What happens to the resonance curve (the graph of steadystate amplitude vs. angular frequency ω D)?

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Energy Quantization

KEY IDEAS A photon is a particle of light with a certain amount of kinetic energy and zero rest energy. The internal energy of microscopic systems is quantized; that is, the internal energy of the system can have only certain specific values (“energy levels”). The system is never observed to have energies between these levels. An atom in a higher energy level can drop to a lower energy level, and emit a photon whose energy is the difference in the energy of the two energy levels. An atom in the lowest-energy state (called the ground state) can absorb a photon and jump to an excited state, if the photon has just the right energy (the difference in the two energy levels). Collisions with energetic electrons can excite atoms to higher energy states, leaving the electron with less kinetic energy than it had initially. The energy of an atom plus its electrons, the vibrational and rotational energies of molecules, the energy of a nucleus, and the energy of the system of quarks making up a proton or neutron, are all quantized.

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PHOTONS A photon is a particle that has only kinetic energy—its rest energy is zero. A photon can never be at rest; it may be thought of as a packet of energy, traveling at the speed of light. A microscopic object such as an atom can gain energy by absorbing a photon; if this happens, all of the photon's energy is transferred to the object. Conversely, an object can lose energy by emitting a photon with a particular amount of energy. (Photons also have momentum, and the Momentum Principle also applies to absorption and emission of photons.) The general term applied to photons is “electromagnetic radiation” (or simply “light”), and in later chapters on electricity and magnetism we will go more deeply into the electric and magnetic aspects of electromagnetic radiation, and we will also discuss its wavelike aspects. Historically, photons of different energies have been given different names, usually because the instruments needed to detect and measure electromagnetic radiation depend on the photon energies involved. The entire range of possible photon energies is called the electromagnetic spectrum (Figure 8.1). Only photons whose energies fall between about 1.8 eV (red light) and 3.1 eV (violet light) can be detected by the human eye (Figure 8.2), so photons in this energy range are referred to as visible light. Colloquially we often refer to all photons as “light,” whether they are high-energy gamma rays or low-energy radio waves.

Figure 8.1 The electromagnetic spectrum.

Figure 8.2 The visible portion of the electromagnetic spectrum.

Although earlier scientists, including Isaac Newton, had speculated that light might be made up of discrete particles, it was not until early in the 20th century that clear evidence for the particle nature of light was obtained. As in the case of atoms, the noncontinuous, lumpy nature of light was not noticed for a long time because the amount of energy carried by one photon in an ordinary beam of light is extremely small compared to the total energy of the beam. It has now become routine to detect individual photons in a very weak beam of light and to measure the energy of an individual photon. Observations of the interaction of photons and ordinary matter have provided key insights into the internal structure of molecules, atoms, and nuclei. One can separate a beam of visible light into its component energies by passing a narrow beam of light through a prism. You have probably seen the continuous rainbow spectrum produced when ordinary white light goes through a prism, as in Figure 8.3. Your eye reacts differently to photons of different energies, which is why you see a colorful spectrum.

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Figure 8.3 A prism separates a narrow beam of white light into a continuous rainbow of colors.

8.X.1 The photon energy for green light lies between the values for red and violet light. What is the approximate energy of the photons in green light?

Answer

8.X.2 The intensity of sunlight above the Earth's atmosphere is about 1400 watts (J/s) per square meter. That is, Answer when sunlight hits perpendicular to a square meter of area, about 1400 watts of energy can be absorbed. Using the average photon energy found in the previous exercise, about how many photons per second strike an area of one square meter? (This is why the lumpiness of light was not noticed for so long.)

Absorption and Emission of Photons A microscopic object such as an atom or a molecule can gain energy by absorbing a photon. When a photon is absorbed, nothing is left over—the photon no longer exists, and the object that absorbed the photon simply has more energy. We can describe the absorption of a photon by a hydrogen atom in this compact way (the asterisk is used to denote an atom with higher internal energy): Applying the Energy Principle to this process, and choosing all particles as the system, we can write this equation:

Likewise, an atom, molecule, or solid can lose energy by emitting a photon, which comes into existence specifically to carry away a particular amount of energy. For a hydrogen atom, emission of a photon can be described this way:

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ELECTRONIC ENERGY LEVELS The term “quantized” is often used colloquially as a synonym for “discrete.” We will use it this way, although the technical meanings of the two words are not identical.

One of the most exciting discoveries of the early 20th century was that the internal energy of a small system such as an atom can have only certain discrete values, called energy “levels.” One piece of evidence leading to this realization was the fact that a particular atom or molecule emitted only photons with particular kinetic energies. For example, in contrast to the continuous spectrum of photon energies in white light (Figure 8.3), the visible light emitted by hydrogen atoms contains photons with only a few specific energies (Figure 8.4). In a spectrum, light of a particular color and energy is called a “spectral line” due to the linelike shape of the light coming through a slit, as seen in Figure 8.4.

Figure 8.4 The light emitted by excited atomic hydrogen contains only a few separate colors. In a spectrum, light of a particular color and energy is called a “spectral line” due to the linelike shape of the light coming through a slit as seen here.

An example of discrete energy levels is found in the “electronic” energy associated with the motion of electrons around the nucleus of an atom. You probably know from your study of chemistry that electronic energy levels in an atom are discrete—only certain values of K + U are possible for the system of nucleus plus electrons. The electronic energy levels in atomic hydrogen are shown in Figure 8.5 (atomic hydrogen is a single H atom, composed of one proton and one electron, not the usual diatomic molecule H 2).

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Figure 8.5 Electronic energy levels for a hydrogen atom. The potential energy curve describes the potential energy of the electron–proton system as a function of separation.

The diagram shows the familiar electric potential energy of the proton–electron system. The horizontal lines represent the possible bound energy states for the hydrogen atom; bound states with other values of (K + U) are not observed. Unbound states (K + U ≥ 0), however, are not quantized. The electric potential energy of an electron and a proton bound together in a hydrogen atom is negative for all separations:

A quantum mechanical calculation of the energy levels of a system involves solving a differential equation called the Schrödinger equation, a form of the Energy Principle consistent with the nature of microscopic systems. This equation includes both a potential energy term and a kinetic energy term. Solving this equation for the hydrogen atom predicts that the discrete energy levels for bound states of hydrogen (ignoring the rest energies of the proton and the electron) will be these:

ELECTRONIC ENERGY LEVELS OF A HYDROGEN ATOM

This arrangement of hydrogen energy levels has been verified by observations of the interaction of light and hydrogen, as we will see. These energy levels denote different bound states of the electron + proton system. Recall that a bound state has a negative value of K + U, since a potential energy of zero is associated with a very large separation between the electron and the proton. We have seen that bound gravitational orbits are also associated with negative values of K + U. As you can see from the energy diagram (Figure 8.5), in higher-energy states the average location of the electron is farther from the proton. Note that there is a lowest-energy state, called the “ground state”; a hydrogen atom cannot have K + U less than −13.6eV.

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Before the development of the fully quantum mechanical Schrödinger equation, Niels Bohr made a simple model for the hydrogen atom that also yielded the energy formula given above. We will study the Bohr model in a later chapter.

EXAMPLE Raising the Energy of a Hydrogen Atom A hydrogen atom is originally in its ground (lowest) electronic energy state (N = 1). It absorbs a photon, which raises the hydrogen atom's energy to the next energy level (N = 2). What was the energy of the photon, in eV?

Solution System: All particles Surroundings: Nothing significant Energy Principle:

It is not possible to add less than 10.2 eV to a hydrogen atom in its lowest energy level; any smaller amount fails to be absorbed.

8.X.3 How much energy in electron volts is required to ionize a hydrogen atom (that is, remove the electron from the proton), if initially the atom is in its lowest energy level?

Answer

Emission Spectra When a hydrogen atom drops from a higher energy level E4 to a lower energy level E1 it emits a photon with the associated energy E 4 − E 1 (Figure 8.6). A collection of hydrogen atoms can emit photons whose energies correspond to the differences in energy between any two electronic energy levels of the hydrogen atom. The light emitted by excited atomic hydrogen is called the “hydrogen emission spectrum,” and the bright lines of light of particular energies in the spectrum are called “spectral lines.” The observed energies of light emitted by excited atomic hydrogen are consistent with the energy level differences predicted by quantum mechanics. Other atoms have different quantized energy levels, and a line spectrum can be used to identify the emitting atoms by the pattern of emitted photon energies.

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Figure 8.6 An atom can drop to a lower energy level by emitting a photon whose energy corresponds to the difference between the energies of the levels. By convention a squiggly line with an arrowhead indicates a photon.

If an isolated hydrogen atom is initially in a high energy level, over time it will spontaneously emit photons and drop to lower and lower states, eventually ending up in the lowest energy level, which is called the “ground state.” Once an atom has dropped to the ground state, it cannot lose any more energy, or emit photons, because there is no lower energy level than the ground state.

EXAMPLE Emission from an Atom Different atoms have different quantized energy levels. Consider a particular atom (not hydrogen) whose first excited state is 6.0 eV above the ground state, and the second excited state is 8.5 eV above the ground state (Figure 8.7). What is the energy of a photon emitted when such an atom drops from the second excited state to the first excited state?

Figure 8.7 Emission from an atom.

Solution System: atom and photon Principle: Energy

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FURTHER DISCUSSION The Energy Principle is valid even for atomic systems with quantized energy. Writing down the Energy Principle explicitly helps to avoid the common mistake of confusing the energy levels of the atom with the energy of the photon. Note that in the example above the photon energy (2.5 eV) is the difference between the energies of the second and third energy levels.

EXAMPLE Emission from Hydrogen Atoms Consider a collection of many hydrogen atoms. Initially each individual atom is in one of the lowest three energy states (N = 1, 2, 3). What will be the energies of photons emitted by this collection of hydrogen atoms?

Solution There are three possibilities: transitions from N = 3 to N = 2, from N = 2 to N = 1, or from N = 3 to N = 1. The photon energies for the three transitions are these:

8.X.4 How many different photon energies would emerge from a collection of hydrogen atoms that occupy the lowest four energy states (N = 1, 2, 3, 4)? (You need not calculate the energies of these transitions.)

Answer

Absorption Spectra An atom can drop from a higher energy level to a lower energy level and emit a photon. An atom can also absorb a photon of an appropriate energy and jump from a lower energy level to a higher energy level (Figure 8.8). If an atom is in its ground state, only photon absorption is possible; photon emission cannot occur. Energy absorption can occur if the photon energy corresponds exactly to the difference between two energy levels of the atom. If we shine white light (containing a mixture of all photon energies in the visible spectrum) through a collection of atoms (such as a container of hydrogen gas), only photons whose energy is the difference of atomic energy levels will be strongly absorbed, while other photons can pass right through the material. In this case, the spectral

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lines are dark lines (missing energies) on a light background, instead of the bright lines on a dark background observed in an emission spectrum. This is called an “absorption spectrum” (Figure 8.9).

Figure 8.8 An atom can gain energy by absorbing a photon whose energy corresponds to the difference between two energy levels.

Figure 8.9 Measuring an absorption spectrum. A sample of a material is placed in a transparent container. Some energies of incoming light are absorbed. The outgoing light is depleted of the absorption energies (in this case E2 and E6 ).

Photon absorption may be followed almost immediately by emission from the excited state to a lower state, so at any instant there are few atoms in excited states that could absorb photons. The result is that in dark-line absorption spectra only absorption from the ground state is observed, not from higher-level states. The atoms that have jumped to higher energy levels through photon absorption will eventually drop to lower energy levels with the emission of photons. However, these photons are emitted in all directions, not just in the direction of the original beam of light, so the light that has passed through the material will have little intensity at photon energies corresponding to the strongly absorbed energies.

EXAMPLE Why Is Hydrogen Gas Transparent?

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At room temperature in a gas of atomic hydrogen almost all of the atoms are in the ground state. Why might you expect the gas to be completely transparent to visible light?

Solution As we saw in an earlier example, the energy required to go from the ground state of atomic hydrogen to the first excited state is 10.2 eV, but visible light consists of photons whose energies lie between 1.8 eV and 3.1 eV. So visible light cannot be absorbed by ground-state hydrogen atoms. (In a collision between a photon of visible light and an atom, the atom can acquire some kinetic energy, but no internal energy change is possible.)

The energy absorbed by the sample may be quickly emitted as photons with energies E 2 and E 6. However, these emitted photons go in all directions in 3D, so almost none of them reach the detector.

8.X.5 Suppose you had a collection of hypothetical quantum objects whose individual energy levels were −4.0eV, Answer −2.3eV, and −1.6eV. If nearly all of the individual objects were in the ground state, what would be the energies of dark spectral lines in an absorption spectrum if visible white light (1.8 to 3.1 eV) passes through the material?

Electron Excitation In order for an atom to emit a photon, it first has to absorb enough energy to raise the atom to an excited state, above the ground state. As we will see in a later section, collisions with other atoms do not provide enough energy unless the collection of atoms is at a very high temperature (for example, on the surface of the Sun). A common nonthermal way to excite atoms in a gas is to produce a beam of fast-moving electrons and send the beam through the gas. Because electrons are (negatively) charged electrically, they interact strongly with atoms, and an electron that passes near an atom may excite the atom from the ground state to an excited state whose internal energy is ΔE atom higher, with a corresponding loss of kinetic energy of the electron. Unlike photon absorption, which almost always occurs only if the photon energy exactly matches the difference between energy levels, an electron can give up some kinetic energy and have some kinetic energy left over. The process is illustrated in Figure 8.10 for an electron colliding with a mercury atom, whose first excited state is 4.9 eV above the ground state.

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Figure 8.10 An electron with initial kinetic energy of more than 4.9 eV can excite a mercury atom (Hg) from the ground state to the first excited state. The electron loses 4.9 eV of kinetic energy.

QUESTION For example, suppose an electron with kinetic energy of 6 eV collides with a mercury atom that is in the ground state. What can happen?

The first excited state of mercury is 4.9 eV above the ground state. This 6 eV electron has enough kinetic energy to excite the mercury atom to the first excited state, giving up 4.9 eV of energy. The electron moves away with 1.1 eV of kinetic energy remaining. The mercury atom will quickly drop back down to the ground state, emitting a 4.9 eV photon. If a beam containing a large number of energetic electrons passes through a gas of mercury vapor, there will be continual electron excitation and continual photon emission. However, as soon as you turn off the electron beam, the mercury will stop emitting photons.

QUESTION What happens if you send a beam of electrons with kinetic energy of 4 eV through the mercury vapor?

4 eV isn't enough energy to go from the ground state to the first excited state in mercury, so the electrons can pass right through the gas without changing the electronic energy states of the atoms. (An electron may collide with an atom, but because the mass of an electron is several thousand times smaller than the mass of a mercury atom, the electron will bounce off the atom like a Ping-Pong ball bouncing off a bowling ball—the kinetic energy of the atom will not change significantly.) The German physicists James Franck and Gustav Hertz performed an electron excitation experiment in 1914 that provided early dramatic evidence for quantized energies in atoms. Electrons with varying amounts of initial kinetic energy were sent through a gas of mercury vapor, and the number of electrons per second making it through the gas was measured. When the initial kinetic energy of the electrons was less than 4.9 eV, many electrons reached the detector. However, when the initial kinetic energy reached 4.9 eV, few electrons reached the detector. The Franck–Hertz experiment was correctly interpreted to mean that the first excited state of a mercury atom is 4.9 eV above the ground state, and an electron with kinetic energy of 4.9 eV can excite the atom, leaving the electron with almost no kinetic energy, so the number of electrons that made it through the gas became very small.

QUESTION What happens if the initial kinetic energy of the electron is 9.8 eV?

In this case the electron could excite one mercury atom and still have 4.9 eV of kinetic energy available to excite a second mercury atom. In the Franck–Hertz experiment it was observed that the number of electrons reaching the detector again dropped sharply when the initial kinetic energy of the electrons was increased past 9.8 eV.

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Every element has a different set of quantized energies. The first excited state in mercury is 4.9 eV above the ground state, whereas the first excited state in hydrogen is 10.2 eV above the ground state. These differences make it possible to identify elements from emission or absorption spectra, or measurements of the Franck–Hertz type.

QUESTION What happens if you send a beam of electrons with kinetic energy of 25 eV through a gas of atomic hydrogen?

Because the ground state of hydrogen has K + U = −13.6eV, only 13.6 eV is needed to raise a hydrogen atom to any of its excited states, or even to break the atom apart into a proton and an electron. Therefore a beam containing lots of 25 eV electrons passing through the atomic hydrogen gas will keep many excited states populated, and you'll observe many different photon energies being emitted as atoms drop from excited states to lower excited states or to the ground state. 8.X.6 The first excited state of a mercury atom is 4.9 eV above the ground state. A moving electron collides with a mercury atom, and excites the mercury atom to its first excited state. Immediately after the collision the kinetic energy of the electron is 0.3 eV. What was the kinetic energy of the electron just before the collision?

Answer

Summary: Three Processes Electron excitation: A fast moving electron can transfer all or part of its kinetic energy to an atom through a collision, if K electron is greater than or equal to the energy difference between two atomic states. Therefore, a beam containing many electrons with high kinetic energy can maintain a gas of atoms in many excited states. Photon emission: Atoms in excited states emit photons as the atoms drop from higher-energy states to lower-energy states. Photon absorption: Atoms in cold material, in the absence of an electron beam, can absorb photons and go from the ground state to an excited state, but only if the photon's energy exactly matches the energy difference between the excited state and the ground state. Absorption leads to a “dark-line” spectrum.

EXAMPLE Observed Photon Emission You observe photon emissions from a collection of quantum objects, each of which is known to have just four quantized energy levels. The collection is continually bombarded by an electron beam, and you detect emitted photons using a detector sensitive to photons in the energy range from 2.5 eV to 30 eV. With this detector, you observe photons emitted with energies of 3 eV, 6 eV, 8 eV, and 9 eV, but no other energies. (a) It is known that the ground-state energy of each of these objects is −10eV. Propose two possible arrangements of energy levels that are consistent with the experimental observations. Explain in detail, using diagrams. (b) You obtain a second detector that is sensitive to photon energies in the energy range from 0.1 eV to 2.5 eV. What additional photon energies do you observe to be emitted? Explain briefly. (c) You turn off the electron beam so that essentially all the objects are in the ground state. Then you send a beam of photons with a wide range of energies through the material. Using both detectors to determine the absorption spectrum, what are the photon energies of the dark lines? How can this information be used to choose between your two proposed energy level schemes? Explain briefly.

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Fundamental principle: The Energy Principle System: Everything—some unknown object and a photon Surroundings: Nothing significant Initial state: Object in excited state (no photon) Final state: Object in lower energy state, plus photon Energy Principle: E obj, f + E photon = E obj, i

The energy of any emitted photon must be equal to the difference in energy between an excited state and a lower-energy state. (a) Figure 8.11 shows a possible energy-level scheme that fits the emission data. The highest photon energy (9 eV) tells us the difference between the highest and lowest states. The rest is trial and error, checking that the correct photon energies would be produced. The arrows are labeled with the energies of photons emitted in transitions between the two states indicated. The 1 eV and 2 eV emissions would not be detected by a detector that is sensitive to the photon energy range 2.5 eV to 30 eV.

Figure 8.11 A possible energy-level scheme that fits the emission data. Lighter arrows indicate photons not detectable by this detector.

Figure 8.12 shows another possible scheme that fits the emission data. This is somewhat less likely, because it is more common to find the energy level spacing decreasing at higher energies, as shown in Figure 8.11. (b) With this detector we observe the 1 eV and 2 eV photon emissions. (c) With the electron beam turned off, almost all of the objects are in the ground state. If the energy levels are as shown in Figure 8.11, the dark lines will be at 6 eV, 8 eV, and 9 eV, corresponding to absorption in the ground state. If the energy levels are as shown in Figure 8.12, the dark lines will be at 1 eV, 3 eV, and 9 eV. Therefore the absorption spectrum allows us to distinguish between the two possible energy level schemes.

Note that since this was an energy analysis, we didn't need to know what kind of object this was, or whether the energy levels were electronic, vibrational, or rotational energy levels.

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FURTHER DISCUSSION Figure 8.12 is an example of an “energy level” diagram, with photon emissions indicated by arrows that start on the upper level and point downward, with arrowheads on the lower level. Photon absorptions are indicated by arrows pointing up from the ground state to an upper level.

Figure 8.12 Another possible energy level scheme that fits the emission data.

8.X.7 The energy levels of a particular quantum object are −8eV, −3eV, and −2eV. If a collection of these objects is bombarded by an electron beam so that there are some objects in each excited state, what are the energies of the photons that will be emitted?

Answer

Selection Rules For simplicity, we've been assuming that photon emissions can occur between any discrete energy level to a lower level. Real atomic and molecular systems may be more complex, and some photon emissions are “forbidden.” In a later chapter we will study angular momentum, a quantity that is related to rotational motion and is a conserved quantity, like momentum and energy. Photon emission and absorption, like all other physical processes, must conserve angular momentum, and as a result transitions between two particular energy levels may be forbidden. Such constraints are called “selection rules” and govern photon emission and absorption (electron excitation is not constrained by selection rules). Selection rules are beyond the scope of this introductory course, but you will learn about them if you take a later course in quantum mechanics. For the purposes of this introduction to quantized energy, we will ignore selection rules and assume that all energetically possible transitions are allowed. We also ignore questions of how probable an allowed transition is, which requires full quantum mechanics to calculate. Suppose that an object is in the second excited state, from which it could drop either to the first excited state or directly to the ground state. One of these two transitions could be much more probable than the other one, even if the low-probability transition isn't forbidden by a selection rule.

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THE EFFECT OF TEMPERATURE When we observe atoms we are usually observing a collection of atoms, not an isolated single atom. Each atom in the collection (the “sample”) interacts with other atoms and perhaps with a container. If an atom absorbs radiation coming from elsewhere, or absorbs energy from being jostled by a neighboring atom, it may jump to a higher energy level, and eventually emit photons and drop again to a lower level (Figure 8.13 shows emissions as downward transitions and absorptions as upward transitions for a collection of hydrogen atoms). In thermal equilibrium with its surroundings, an atom's energy continually goes up and down, but with an average energy determined by the temperature of the collection of atoms.

Figure 8.13 A dot represents the energy state of one hydrogen atom in a group of atoms. Energy is exchanged with the surroundings through photon emission and absorption.

In Figures 8.13 and 8.14 each dot represents one hydrogen atom in a container of atomic hydrogen. If three dots are on a line, this means three atoms are in this particular state. Each hydrogen atom has the same allowed energy levels, and at any particular instant some of the atoms will be in the ground state, some in the first excited state, and so on.

Figure 8.14 At a low temperature (system on the left) all atoms are in low-energy states. At a high temperature (system on the right) some atoms are in higher-energy states, though the largest fraction will still be in the ground state.

For a particular temperature, at any instant there is a certain probability that one particular atom will be in state 1, 2, 3, and so on. At low temperature almost all atoms are in the ground state (Figure 8.14). At high temperature the various atoms are in various states, from the ground state up to a high energy level, though the largest fraction will still be in the ground state. As we will see in a later chapter on statistical mechanics, in a collection of atoms the fraction of the atoms in a state whose energy is E above the ground state is proportional to the “Boltzmann factor” e −E/kB T, where k B is the Boltzmann constant (1.38 × 10−23 J/K) and T is the absolute temperature in kelvins (K), where 0° Celsius is 273.15 K.

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QUESTION Can you observe an emission spectrum if the material is very cold? Can you observe an absorption spectrum if the material is very cold?

Very cold matter will normally be in the ground state and will not emit photons. However, an absorption spectrum will show gaps (dark lines) corresponding to absorption of photons of energies appropriate to lift an atom from its ground state to an excited state. (In order to observe an absorption spectrum you would of course need to use a beam of photons whose energies are sufficiently high to raise the system energy at least to the first excited state above the ground state.)

Cold Samples In a sample at room temperature (about 300 K), almost all atoms are in their electronic ground state. For example, if the first excited state is 1 eV above the ground state, the Boltzmann factor is

This extremely tiny number represents the fact that at room temperature an insignificant fraction of the atoms will be excited thermally to 1 eV above the ground state. Therefore, it is reasonable to assume that essentially all the atoms in an ordinary roomtemperature absorption measurement start out in the electronic ground state. On the other hand, the surface temperature of the Sun is about 5000 K, and with E = 1 eV the Boltzmann factor is (−E/k B T) = exp(−2.3) = 0.1, so a sizable fraction of atoms can be in an excited state 1 eV above the ground state, due to thermal excitation.

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VIBRATIONAL ENERGY LEVELS For concreteness we have described energy quantization for the electronic states of atomic hydrogen. Many other kinds of microscopic systems have quantized energy. We will next discuss vibrational energy quantization in atomic oscillators.

Vibrational Energy in a Classical Oscillator We have modeled a solid as a network of classical harmonic oscillators, balls and springs (Figure 8.15). Although this classical model explains many aspects of the behavior of solids, the predictions it makes regarding the thermal properties of solids do not correspond to experimental measurements at low temperatures. The classical model also fails to explain some aspects of the interaction of light with solid matter. To explain these phenomena we need to make our model of a solid more sophisticated, by incorporating the quantum nature of these atomic oscillators into our model.

Figure 8.15 Model of a solid as a network of balls and springs.

The vibrational energy of a classical (nonquantum) spring–mass system can have any value within the colored region in Figure 8.16. Equivalently, the amplitude A (maximum stretch) can have any value within the allowed range. We can express the energy as E = k s A 2, because at the maximum stretch A the speed is zero, so the kinetic energy is zero, and the energy K + U is just the potential energy at that instant. Classically, the amplitude A can have continuous rather than quantized values.

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Figure 8.16 A classical harmonic oscillator (spring–mass system) can vibrate with any amplitude, and hence can have any energy. K + U of the system could have any value in the shaded region.

Vibrational Energy in a Quantum Mechanical Oscillator Historically, the first hint that energy might be quantized came from an attempt by the German physicist Max Planck in 1900 to explain some puzzling features of the spectrum of light emitted through a small hole in a hot furnace (so-called “blackbody radiation”). Planck found that he could correctly predict the properties of blackbody radiation if he assumed that transfers of energy in the form of electromagnetic radiation were quantized—that is, an energy transfer could take place only in fixed amounts, rather than having a continuous range of possible values. Later, with the full development of quantum mechanics, it was recognized that one could model an atom in the solid wall of the furnace as a tiny spring–mass system whose energy levels are discrete (Figure 8.17). Classically, the energy of a mass on a spring is related to the amplitude of its oscillation (E = k s A 2), so saying that a spring–mass oscillator can only have certain discrete amounts of energy is equivalent to saying that only certain amplitudes are allowed. To see this, note that we have

since when the stretch s is equal to the amplitude A (the maximum stretch), the momentum is zero.

Figure 8.17 A quantum harmonic oscillator can have only certain energies, indicated by the horizontal lines on the graph. This idealized potential energy well represents the situation near the bottom of a real well (which is at a negative energy).

By solving the quantum mechanical Schrödinger equation for a system involving an atom and a chemical bond, one can show that

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the spacing ΔE between the allowed, quantized energies has the following value:

In this case, the quantity k s = k s, i is the interatomic spring stiffness, which we estimated in Chapter 4, m = m a is the mass of one is the angular frequency of the oscillator (in radians per second), as defined in Chapter 4. The quantity ħ, atom, and pronounced “h-bar,” is equal to h, “Planck's constant,” divided by 2π. Planck's constant h has an astoundingly small value:

The quantized energies for an atomic harmonic oscillator are these:

QUANTIZED VIBRATIONAL ENERGY LEVELS

In other words, the energy can have only the following values: For an atom and a springlike interatomic bond: k s = k s, i (interatomic spring stiffness) and m = m a (mass of an atom)

It is an unusual and special feature of the quantized harmonic oscillator that the spacing between adjacent energy levels is the same for all levels (“uniform spacing”). This is quite different from the uneven spacing of energy levels in the hydrogen atom. These energy levels are predicted by quantum mechanics, and experiments confirm this prediction. It is important to understand that in both the quantum and the classical picture, the angular frequency

of a particular oscillator is fixed and is the same whether the energy of the oscillator is large or small. A useful way to think about the situation is to say that the frequency is determined by the parameters k s and m, while the energy is determined by the initial conditions (or equivalently by the amplitude A). In the quantum world, it is as though the amplitude can have only certain specific values. 8.X.8 Show that

has units of joules.

8.X.9 Suppose that a collection of quantum harmonic oscillators occupies the lowest four energy levels, and the Answer spacing between levels is 0.4 eV. What is the complete emission spectrum for this system? That is, what photon energies will appear in the emissions? Include all energies, whether or not they fall in the visible region of the electromagnetic spectrum.

Spring Stiffness ks The energy quantum depends on the stiffness k s of the “spring.” A stiffer spring (larger value of k s, Figure 8.18) means a steeper, narrower potential curve and larger spacing between the allowed energies. A smaller value of k s (Figure 8.19) means a shallower, wider potential curve and smaller spacing between the allowed energies.

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Figure 8.18 A quantum oscillator with a large ks has widely spaced energy levels.

Figure 8.19 A quantum oscillator with a small ks has closely spaced energy levels.

8.X.10 You may have measured the properties of a simple spring–mass system in the lab. Suppose that you found k s Answer = 0.7 N/m and m = 0.02 kg, and you observed an oscillation with an amplitude of 0.2 m. What is the approximate value of N, the “quantum number” for this oscillator? (That is, how many levels above the ground state is this oscillator?) 8.X.11 What is the energy of the photon emitted when a harmonic oscillator with stiffness k s and mass m when it drops from energy level 5 to energy level 2?

Answer

The Ground State of a Quantum Oscillator The lowest possible quantized energy E0 is not zero relative to the bottom of the potential curve, but is

above the

bottom of the well. This is closely related to the “Heisenberg uncertainty principle.” When applied to the harmonic oscillator, this principle implies that if you knew that the mass on the spring were exactly at the center of the potential, its speed would be completely undetermined, whereas if you knew that the speed of the mass were exactly zero, the position of the mass could be literally anywhere! The minimum energy

above the bottom of the well can be viewed as a kind of a compromise,

with neither the speed nor the position fixed at zero. The minimum energy is called the ground state of the system. In our work with the quantized oscillator we will measure energy from the ground state rather than from the bottom of the potential well, so the offset from the bottom of the well won't come into our calculations.

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Energy Levels for the Interatomic Potential Energy If we model the atoms in a solid as simple harmonic oscillators, whose potential energy curve is a parabola, then the energy levels of these systems are uniformly spaced. However, if we use a more realistic interatomic potential energy such as is shown in Figure 8.20, the energy levels of the system follow a more complex pattern. The lowest energy levels are nearly evenly spaced, corresponding to the fact that for small amplitude the bottom of the potential energy curve can be fit rather well by a curve of the form k sS2. However, for large energies the energy spacing is quite small. Of course there are no bound states for energies larger than the binding energy. The spacing of the vibrational energy levels of a diatomic molecule (Figure 8.21) typically corresponds to photons in the infrared region of the spectrum (see Problem 8.P.35).

Figure 8.20 Energy levels for the interatomic potential energy, which describes the interaction of two neighboring atoms. This diagram is not to scale; the number of vibrational energy levels in the uniform region may be very large.

Figure 8.21 The vibrational energy of a diatomic molecule is quantized.

Energy Spacing in Various Systems Most quantum objects have energy-level spacings that decrease as you go to higher levels. The harmonic oscillator potential energy is unusual in leading to uniform energy-level spacing. Even less common is the situation where the energy-level spacing actually increases with higher energy. One such exception is the so-called “square well” with very steep sides and a flat bottom. The energy levels in a square well actually get farther and farther apart for higher energies.

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ROTATIONAL ENERGY LEVELS A molecule in the gas phase has not only electronic energy and vibrational energy but rotational energy as well. Like electronic and vibrational energy levels, rotational energy levels are also discrete—only certain rotational energies are permitted. As we will see in a later chapter, the angular momentum of a molecule is quantized, and this quantization gives rise to discrete rotational energy levels. In this case all the energy of rotation is kinetic energy—there is no potential energy involved. Rotational energy levels for a diatomic molecule such as N 2 are shown in Figure 8.22. Note that the spacing between levels actually increases with increasing energy, which is the opposite of the trend seen in many atomic systems. Rotational energy levels are typically close together; transitions between these levels usually involve photons in the microwave region of the electromagnetic spectrum. Microwave ovens excite rotational energies of water molecules in food. These excited molecules then give up their energy to the food, making the food hotter.

Figure 8.22 Rotational kinetic energy levels of a diatomic molecule.

Case Study: A Diatomic Molecule Analyzing a diatomic molecule such as N 2 provides a good review of three important types of quantized energy levels: electronic, vibrational, and rotational. Only certain configurations of the electron clouds in a diatomic molecule are possible, corresponding to discrete electronic energy levels. The spacings of the low-lying molecular electronic energy levels are typically in the range of 1 eV or more, like the spacings of the low-lying energy levels in the hydrogen atom. For each major configuration of the electron clouds, many different vibrational energy states are possible, with energy spacings of the order of 1 × 10−2 eV (see Problem 8.P.35). For each major configuration of the electron clouds and each vibrational energy state, many different rotational kinetic energy states are possible, with energy spacings of the order of 1 × 10−4 eV, as we will see in a later chapter. These three different families of energy levels, with their very different energy level spacings, give rise to energy “bands” and a distinctive “band” spectrum of emitted photons. See Figure 8.23, which is not to scale due to the large differences in level spacing for the three kinds of quantized energy levels. An emitted photon can have the energy of an electronic transition plus a smallerenergy vibrational transition plus an even smaller-energy rotational transition. The various possible energy differences give rise to “bands” in the observed emission spectrum.

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Figure 8.23 The energy bands of a diatomic molecule (not to scale). This energy level structure gives rise to a band spectrum of emitted photons.

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OTHER ENERGY LEVELS It is not only electronic, vibrational, and rotational energies in atoms and molecules that are quantized. Small objects such as nuclei and hadrons (particles made of quarks) also have discrete energy levels.

Nuclear Energy Levels Just as atoms have discrete states corresponding to allowed configurations of the electron cloud, so nuclei have discrete states corresponding to allowed configurations of the nucleons (protons and neutrons). An excited nucleus can drop to a lower state with the emission of a photon. The energy spacing of nuclear levels is very large, of the order of 1 × 106 eV (1 MeV), so the emitted photons have energies in the range of millions of electron volts. Such energetic photons are called “gamma rays.” 8.X.12 A gamma ray with energy greater than or equal to 2.2 MeV can dissociate a deuteron (the nucleus of deuterium, heavy hydrogen) into its constituents, a proton and a neutron. What is the nuclear binding energy of the deuteron?

Answer

Hadronic Energy Levels Atoms have energy levels associated with the configurations of electrons in the atom, and nuclei have energy levels associated with the configurations of nucleons in the nucleus. Hadrons are particles made of quarks. In a simple model we represent baryons (particles like protons) as a combination of three quarks, and mesons (particles like pions) as a combination of a quark and an antiquark. Different configurations of quarks correspond to different energy levels of the multiquark system. These different configurations are the different hadrons we observe, and the corresponding energy levels are the observed rest energies of these particles. The energy spacing is of the order of hundreds of MeV, or about 1 × 108 eV. (Leptons, which include electrons, muons, and neutrinos, are not made of quarks.) An example of a hadron is the Δ + particle, which has a rest energy of 1232 MeV. It can be considered to be an excited state of the three-quark system whose ground state is the proton, whose rest energy is 938 MeV. The Δ + particle can decay into a proton with the emission of a very high energy photon. However, this “electromagnetic” decay occurs in only about 0.5% of the decays. In most cases the Δ + decays into a proton plus a pion, a strong-interaction decay that has a much higher rate, corresponding to the much stronger interaction. An excited electronic state of a hydrogen atom is still called hydrogen, but for historical reasons we give excited quark states distinctive names, such as Δ +. 8.X.13 What is the approximate energy of the photon emitted when the Δ + decays electromagnetically?

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Answer

COMPARISON OF ENERGY LEVEL SPACINGS Figure 8.24 is a summary of typical energy scales for various quantum objects. It is important to distinguish among hadronic, nuclear, electronic, vibrational, rotational, and other kinds of quantized energy situations. We can generalize the main issues to any kind of generic quantum object. The key features of all these different kinds of objects are these: discrete energy levels discrete emissions whose energies equal differences in energy levels Some of the homework problems at the end of the chapter ask you to analyze generic objects, independent of the details of what gave rise to the particular discrete energy levels.

Figure 8.24 Typical energy level spacings in various quantum objects.

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*CASE STUDY: HOW A LASER WORKS A laser offers an interesting case study of the interaction of light and atomic matter. We will give a basic introduction to lasers, but we can only scratch the surface of a vast topic that continues to be an active field of research and development. Laser action depends on a process called “stimulated emission” and on creating an “inverted population” of quantum states, which can lead to a chain reaction and produce a special kind of light, called “coherent” light.

Stimulated Emission If an atom can be placed into an excited state—that is, a state of higher energy than the ground state—normally it drops quickly to a lower-energy state and emits a photon in the process. For example, in a neon sign high-energy electrons collide with ground-state neon atoms and raise them to an excited state. An excited neon atom quickly drops back to the ground state, emitting a photon whose energy is the difference in energy between the two quantized energy levels, and you see the characteristic orange color corresponding to these photons. We call this emission process “spontaneous emission.” Spontaneous emission happens at an unpredictable time. All that quantum mechanics can predict is the probability that the excited atom will drop to the ground state in the next small time interval. It is truly not possible to predict the exact time of the emission, just as it is impossible to predict exactly when a free neutron will decay into a proton, electron, and antineutrino. A collection of excited atoms such as those in a neon sign emit spontaneously, at random times. In the full quantum-mechanical description of light, light has both the properties of a particle (the photon picture) sand the properties of a wave; see Section 8.9. Emissions at random times have random phases in a wave description: sine, cosine, and all phases between a sine and a cosine. We say that such light is “incoherent.” “Stimulated emission” of light is a different process that produces coherent light. Consider an atom in an energy level E above the ground state. A photon of this same energy E can interact with the atom in a remarkable way, causing the atom to drop immediately to the ground state with the emission of a second, “clone” photon. Where there had been one photon of energy E, there are now two photons of exactly the same energy E, and in a wave description they have exactly the same phase (Figure 8.25). That is, the two waves are both sines, or both cosines, or both with exactly the same phase somewhere between a sine and a cosine. Such light is called coherent light, and it has valuable properties for such applications as making holograms.

Figure 8.25 Stimulated emission produces two identical photons from one.

Chain Reaction The phenomenon of stimulated emission can be exploited to create a chain reaction. Start with one photon of the right energy E to match the energy difference between the ground state and an excited state of the atoms in the system. If that photon causes stimulated emission from an excited atom, there are now two identical photons, both capable of triggering stimulated emission of other excited atoms, which yields four identical photons, then 8, 16, 32, 64, 128, 256, 512, 1024, and so on.

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However, the chain reaction won't proceed if all of the photons escape from the system, so some kind of containment mechanism is required. In a common type of gas laser the gas is in a long tube with mirrors at both ends (Figure 8.26). Photons that happen to be going in the direction of the tube are continually reflected back and forth through the tube, which increases the probability that they will interact with excited atoms and cause stimulated emission. Since the stimulated-emission photons are clones, they too are headed in the direction of the tube and they too are likely to interact. One of the mirrors is deliberately made to be an imperfect reflector, so a fraction of the photons leaks out and provides a coherent beam of light.

Figure 8.26 A gas laser with mirrors at the ends. Photons leak through one of the mirrors.

Unfortunately, the stimulated-emission photons also have just the right energy E to be absorbed by an atom in the ground state, in which case they can no longer contribute to the chain reaction. For this reason we need an “inverted population,” with few atoms in the ground state to absorb the photons and stop the chain reaction. We discuss this issue next.

Inverted Population As we have discussed earlier in this chapter, in a normal collection of atoms more atoms are in the ground state than in any other state, and the number of atoms decreases in each successively higher state. In fact, for a collection of atoms in thermal equilibrium the fraction of atoms in a particular energy state decreases exponentially with the energy above the ground state, as we will see in a later chapter. By clever means it is possible to invert this population scheme, so that there are more atoms in an excited state than in the ground state (this is not a state of thermal equilibrium). An inverted population is crucial to laser action, so that stimulated emission with cloning of photons will dominate over simple absorption of photons by atoms in the ground state. There are many different schemes for creating an inverted population. For example, the scheme used in the common helium–neon laser is to “pump” atoms to a high energy level (Figure 8.27). There are diverse energy input schemes, including electric discharges and powerful flashes of ordinary light. Atoms in this high energy level rapidly drop to an intermediate energy level, higher than the ground state. This intermediate state happens to have a rather long lifetime for spontaneous emission, so that it is possible to sustain an inverted population, with more atoms in the intermediate state than in the ground state. Stimulated emission drives these excited atoms down to the ground state, from which they are again pumped up to the high-energy state. Spontaneous emission from the high-energy state to the intermediate state maintains the inverted population.

Figure 8.27 A three-state scheme for sustaining an inverted population.

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*WAVELENGTH OF LIGHT In this chapter we have emphasized the particle model of light, in terms of photons and their energy. However, as we have hinted earlier, some aspects of the behavior of light (of any energy) can best be understood by considering light to be a wave. For completeness we will mention a connection between the particle model and the wave model of light. Light consisting of photons whose energy is E can also be treated as a wave, and the energy E and wavelength λ are related as follows:

ENERGY AND WAVELENGTH OF LIGHT

where h is Planck's constant.

A model for light that incorporates both the features of a wave and a localized particle is a “wave packet,” a wave whose magnitude is nonzero only in a small region of space (Figure 8.28). We will return to this topic in a later chapter on waves and particles.

Figure 8.28 A wave packet is a traveling wave of finite spatial extent.

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SUMMARY Light consists of particles of light called photons with energy and momentum but no rest mass, so E = K. Their speed is always 3 × 108 m/s. At the microscopic level the energy of bound states is discrete, not continuous. (Unbound states have a continuous range of energies.) Quantized systems can have only certain energies. They can emit or absorb photons when they change energy levels. The temperature is a measure of the average energy of a collection of objects, so high temperature is associated with significant probability of finding particles in any of many states, from the ground state up to a high energy level. Very low temperature is associated with little probability of the particles being in any state other than the ground state. Discrete hydrogen atom energy levels E N = −(13.6eV)/N 2, where N is a nonzero positive integer (1, 2, 3 …) Discrete harmonic oscillator energy levels

Planck's constant h = 6.6 × 10−34 joule · second

Photon energy and wavelength:

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EXERCISES AND PROBLEMS Sections 8.1, 8.2, 8.3 8.X.14 How much energy in electron volts is required to ionize a hydrogen atom (that is, remove the electron from the proton), if initially the atom is in the state N = 2? (Remember that N = 1 if the atom is in the lowest energy level.) 8.X.15 A certain laser outputs pure red light (photon energy 1.8 eV) with power 700 milliwatts (0.7 watts). How many photons per second does this laser emit? 8.X.16 N = 1 is the lowest electronic energy state for a hydrogen atom. (a) If a hydrogen atom is in state N = 4, what is K + U for this atom (in eV)? (b) The hydrogen atom makes a transition to state N = 2. Now what is K + U in electron volts for this atom? (c) What is the energy (in eV) of the photon emitted in the transition from level N = 4 to N = 2? (d) Which of the arrows in Figure 8.29 represents this transition?

Figure 8.29

8.X.17 Match the description of a process with the corresponding arrow in Figure 8.30: (a) Absorption of a photon whose energy is E 1 − E 0 (b) Absorption from an excited state (a rare event at low temperatures) (c) Emission of a photon whose energy is E3 − E 1 (d) Emission of a photon whose energy is E2 − E 0 (e) In drawing arrows to represent energy transitions, which of the following statements are correct? (1) It doesn't matter in which direction you draw the arrow as long as it connects the initial and final states. (2) For emission, the arrow points down. (3) For absorption, the arrow points up. (4) The tail of the arrow is drawn on the initial state. (5) The head of the arrow is drawn on the final state.

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(6) It is not necessary to draw an arrowhead.

Figure 8.30

8.X.18 When starlight passes through a cold cloud of hydrogen gas, some hydrogen atoms absorb energy, then reradiate it in all directions. As a result, a spectrum of the star shows dark absorption lines at the energies for which less energy from the star reaches us. How does the spectrum of dark absorption lines for very cold hydrogen differ from the spectrum of bright emission lines from very hot hydrogen? 8.X.19 The Franck–Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts of kinetic energy loss by the electrons. Suppose that instead a similar experiment is done with a very cold gas of atomic hydrogen, so that all of the hydrogen atoms are initially in the ground state. If the kinetic energy of an electron is 11.6 eV just before it collides with a hydrogen atom, how much kinetic energy will the electron have just after it collides with and excites the hydrogen atom? 8.P.20 Hydrogen atoms: (a) What is the minimum kinetic energy in electron volts that an electron must have to be able to ionize a hydrogen atom that is in its ground state (that is, remove the electron from being bound to the proton)? (b) If electrons of energy 12.8 eV are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that can be emitted by the excited gas? (c) If instead of electrons, photons of all energies between 0 and 12.8 eV are incident on a gas of hydrogen atoms in the ground state, what are the energies at which the photons are absorbed? 8.P.21 Suppose we have reason to suspect that a certain quantum object has only three quantum states. When we excite such an object we observe that it emits electromagnetic radiation of three different energies: 2.48 eV (green), 1.91 eV (orange), and 0.57 eV (infrared). (a) Propose two possible energy-level schemes for this system. (b) Explain how to use an absorption measurement to distinguish between the two proposed schemes. 8.P.22 The eye is sensitive to photons with energies in the range from about 1.8 eV, corresponding to red light, to about 3.1 eV, corresponding to violet light. White light is a mixture of all the energies in the visible region. If you shine white light through a slit onto a glass prism, you can produce a rainbow spectrum on a screen, because the prism bends different colors of light by different amounts (Figure 8.31).

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Figure 8.31

If you replace the source of white light with an electric-discharge lamp containing excited atomic hydrogen, you will see only a few lines in the spectrum, rather than a continuous rainbow (Figure 8.32).

Figure 8.32

Predict how many lines will be seen in the visible spectrum of atomic hydrogen. Specify the atomic transitions that are responsible for these lines, given that the energies of the quantized states in atomic hydrogen are given by EN = −(13.6 eV)/N 2, where N is a nonzero positive integer (1, 2, 3, …). 8.P.23 Assume that a hypothetical object has just four quantum states, with the energies shown in Figure 8.33. (a) Suppose that the temperature is high enough that in a material containing many such objects, at any instant some objects are found in all of these states. What are all the energies of photons that could be strongly emitted by the material? (In actual quantum objects there are often “selection rules” that forbid certain emissions even though there is enough energy; assume that there are no such restrictions here.) (b) If the temperature is very low and electromagnetic radiation with a wide range of energies is passed through the material, what will be the energies of photons corresponding to missing (“dark”) lines in the spectrum? (Assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region.)

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Figure 8.33

8.P.24 A certain material is kept at very low temperature. It is observed that when photons with energies between 0.2 and 0.9 eV strike the material, only photons of 0.4 eV and 0.7 eV are absorbed. Next the material is warmed up so that it starts to emit photons. When it has been warmed up enough that 0.7 eV photons begin to be emitted, what other photon energies are also observed to be emitted by the material? Explain briefly. 8.P.25 Suppose we have reason to suspect that a certain quantum object has only three quantum states. When we excite a collection of such objects we observe that they emit electromagnetic radiation of three different energies: 0.3 eV (infrared), 2.0 eV (visible), and 2.3 eV (visible). (a) Draw a possible energy-level diagram for one of the quantum objects, which has three bound states. On the diagram, indicate the transitions corresponding to the emitted photons, and check that the possible transitions produce the observed photons and no others. Label the energies of each level (K + U, which is negative). The energy K + U of the ground state is −4 eV. (b) The material is now cooled down to a very low temperature, and the photon detector stops detecting photon emissions. Next a beam of light with a continuous range of energies from infrared through ultraviolet shines on the material, and the photon detector observes the beam of light after it passes through the material. What photon energies in this beam of light are observed to be significantly reduced in intensity (“dark absorption lines”)? (c) There exists another possible set of energy levels for these objects that produces the same photon emission spectrum. On an alternative energy-level diagram, different from the one you drew in part (a), indicate the transitions corresponding to the emitted photons, and check that the possible transitions produce the observed photons and no others. Label the energies of each level (K + U, which is negative). (d) For your second proposed energy-level scheme, what photon energies would be observed to be significantly reduced in intensity in an absorption experiment (“dark absorption lines”)? (Given the differences from part (b), you can see that an absorption measurement can be used to tell which of your two energy-level schemes is correct.) 8.P.26 Assume that a hypothetical object has just four quantum states, with the following energies: −1.0 eV (third excited state) −1.8 eV (second excited state) −2.9 eV (first excited state) −4.8 eV (ground state) (a) Suppose that material containing many such objects is hit with a beam of energetic electrons, which ensures that there are always some objects in all of these states. What are the six energies of photons that could be strongly emitted by the material? (In actual quantum objects there are often “selection rules” that forbid certain emissions even though there is enough energy; assume that there are no such restrictions here.) List the photon emission energies.

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(b) Next, suppose that the beam of electrons is shut off so that all of the objects are in the ground state almost all the time. If electromagnetic radiation with a wide range of energies is passed through the material, what will be the three energies of photons corresponding to missing (“dark”) lines in the spectrum? Remember that there is hardly any absorption from excited states, because emission from an excited state happens very quickly, so there is never a significant number of objects in an excited state. Assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region. List the dark-line energies. 8.P.27 Energy graphs: (a) Figure 8.34 shows a graph of potential energy vs. interatomic distance for a particular molecule. What is the direction of the associated force at location A? At location B? At location C? Rank the magnitude of the force at locations A, B, and C. (That is, which is greatest, which is smallest, and are any of these equal to each other?) For the energy level shown on the graph, draw a line whose height is the kinetic energy when the system is at location D.

Figure 8.34 (b) Figure 8.35 shows all of the quantized energies (bound states) for one of these molecules. The energy for each state is given on the graph, in electron volts (1eV = 1.6 × 10−19 J). How much energy is required to break a molecule apart, if it is initially in the ground state? (Note that the final state must be an unbound state; the unbound states are not quantized.)

Figure 8.35 (c) At high enough temperatures, in a collection of these molecules there will be at all times some molecules in each of these states, and light will be emitted. What are the energies in electron volts of the emitted light? (d) The “inertial” mass of the molecule is the mass that appears in Newton's second law, and it determines how much acceleration will result from applying a given force. Compare the inertial mass of a molecule in the top energy state and the inertial mass of a molecule in the ground state. If there is a difference, briefly explain why and calculate the difference. If there isn't a difference, briefly explain why not. 8.P.28 A bottle contains a gas with atoms whose lowest four energy levels are −12eV, −6eV, −3eV, and −2eV. Electrons run through the bottle and excite the atoms so that at all times there are large numbers of atoms in each of these four energy levels, but there are no atoms in higher energy levels. List the energies of the photons that will be emitted by the gas. Next the electron beam is turned off, and all the atoms are in the ground state. Light containing a continuous spectrum of photon energies from 0.5 eV to 15 eV shines through the bottle. A photon detector on the other side of the bottle shows that some photon energies are depleted in the spectrum (“dark lines”). What are the energies of the missing photons?

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8.P.29 Some material consisting of a collection of microscopic objects is kept at a high temperature. A photon detector capable of detecting photon energies from infrared through ultraviolet observes photons emitted with energies of 0.3 eV, 0.5 eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV. These are the only photon energies observed. (a) Draw and label a possible energy-level diagram for one of the microscopic objects, which has 4 bound states. On the diagram, indicate the transitions corresponding to the emitted photons. Explain briefly. (b) Would a spring–mass model be a good model for these microscopic objects? Why or why not? (c) The material is now cooled down to a very low temperature, and the photon detector stops detecting photon emissions. Next a beam of light with a continuous range of energies from infrared through ultraviolet shines on the material, and the photon detector observes the beam of light after it passes through the material. What photon energies in this beam of light are observed to be significantly reduced in intensity (“dark absorption lines”)? Explain briefly.

Sections 8.4, 8.5, 8.6, 8.7, 8.8 8.X.30 If you try to increase the amplitude of a quantum harmonic oscillator by adding an amount of energy

, the

amplitude doesn't increase. Why not? 8.X.31 If you double the amplitude, what happens to the frequency in a classical (nonquantum) harmonic oscillator? In a quantum harmonic oscillator? 8.X.32 Summarize the differences and similarities between different energy levels in a quantum oscillator. Specifically, for the first two levels in Figure 8.18, compare the angular frequency , the amplitude A, and the kinetic energy K at the same value of s. (In a full quantum-mechanical analysis the concepts of angular frequency and amplitude require reinterpretation. Nevertheless there remain elements of the classical picture. For example, larger amplitude corresponds to a higher probability of observing a large stretch.) 8.X.33 Which energy diagram in Figure 8.36 is appropriate for each of the following situations? (a) Hadronic (such as Δ +) (b) Vibrational states of a diatomic molecule such as O 2 (c) Idealized quantized spring–mass oscillator (d) Nuclear (such as the nucleus of a carbon atom) (e) Electronic, vibrational, and rotational states of a diatomic molecule such as O 2 (f) Rotational states of a diatomic molecule such as O 2 (g) Electronic states of a single atom such as hydrogen

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Figure 8.36 8.P.34 Consider a microscopic spring–mass system whose spring stiffness is 50 N/m, and the mass is 4 × 10−26kg. (a) What is the smallest amount of vibrational energy that can be added to this system? (b) What is the difference in mass (if any) of the microscopic oscillator between being in the ground state and being in the first excited state? (c) In a collection of these microscopic oscillators, the temperature is high enough that the ground state and the first three excited states are occupied. What are possible energies of photons emitted by these oscillators? 8.P.35 Molecular vibrational energy levels: (a) A HCl molecule can be considered to be a quantized harmonic oscillator, with quantized vibrational energy levels that are evenly spaced. Make a rough estimate of this uniform energy spacing in electron volts (where 1 eV = 1.6 × 10−19 J). You will need to make some rough estimates of atomic properties based on prior work. For comparison with the spacing of these vibrational energy states, note that the spacing between quantized energy levels for “electronic” states such as in atomic hydrogen is of the order of several electron volts. (b) List several photon energies that would be emitted if a number of these vibrational energy levels were occupied due to electron excitation. To what region of the spectrum (x-ray, visible, microwave, etc.) do these photons belong? (See Figure 8.1 at the beginning of the chapter.) 8.P.36 A hot bar of iron glows a dull red. Using our simple model of a solid (see Figure 8.15), answer the following questions, explaining in detail the processes involved. The mass of one mole of iron is 56 grams. (a) What is the energy of the lowest-energy spectral emission line? Give a numerical value. (b) What is the approximate energy of the highest-energy spectral emission line? Give a numerical value. (c) What is the quantum number of the highest-energy occupied state? (d) Predict the energies of two other lines in the emission spectrum of the glowing iron bar.

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(Note: Our simple model is too simple—the actual spectrum is more complicated. However, this simple analysis gets at some important aspects of the phenomenon.)

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Multiparticle Systems

KEY IDEAS The kinetic energy of a multiparticle system has two parts:

▪ Translational kinetic energy (due to motion of center of mass) ▪ Kinetic energy relative to the center of mass (rotation + vibration) Rotational kinetic energy can be calculated in terms of moment of inertia I and angular speed ω.

Treating a multiparticle system as if it were a point particle lets us calculate changes in translational kinetic energy. Analyzing the real system lets us calculate changes in the other energy terms (rotational, vibrational, potential).

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THE MOTION OF THE CENTER OF MASS Recall that in Chapter 3 we derived an equation that predicts the overall motion of a complex multiparticle system, in terms of the rate of change of the total momentum of the system :

THE MOMENTUM PRINCIPLE FOR MULTIPARTICLE SYSTEMS

(if v E 1, then you will sometimes find the system in one of its excited states, although it is still true that the system is most likely to be found in its ground state (Figure 12.38).

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Figure 12.38 At a low temperature it is unlikely to find a system in an excited state, but at a high temperature the likelihood increases.

The Boltzmann distribution has far-reaching consequences. For example, chemical and biochemical reaction rates typically depend strongly on temperature because with higher temperature the reactants are moving faster and may be found in excited states. Physical reaction rates are also affected. At the very high temperatures found in the interior of our Sun, kinetic energies are high enough to overcome the electric repulsion between nuclei and to allow the nuclei to come in contact, so that they can undergo thermonuclear fusion reactions. A gas becomes a plasma if the temperature is so high that k B T is comparable to the ionization energy. 12.X.13 At room temperature, show that

. It is useful to memorize this result, because it tells a lot about what

phenomena are likely to occur at room temperature. 12.X.14 A microscopic oscillator has its first and second excited states 0.05 eV and 0.10 eV above the ground state energy. Calculate the Boltzmann factor for the ground state, first excited state, and second excited state, at room temperature.

Answer

Many Observations vs. Many Systems If a microscopic system (such as a single oscillator) is in contact with a large system, the probability of finding a particular amount of energy in the microscopic system is governed by the Boltzmann distribution. One can think about this in two ways. The first is to imagine measuring the energy in this particular system repeatedly, over time, and recording the results. The fraction of the results that indicate a particular energy is predicted by the Boltzmann distribution. A second way to think about this is to imagine assembling a large number of identical systems, each in contact with a large system, and taking one “snapshot” in which the energy of each system is recorded simultaneously. The fraction of systems with a particular energy is predicted by the Boltzmann distribution. The second approach also gives us a way to predict the distribution of energy in a single large system. For example, consider a

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system consisting of all the air molecules in a room. Imagine that each molecule is identified by a letter: A, B, C … One could consider molecule A to be a microscopic system in contact with the large system consisting of all other molecules (B, C, …). However, one could also consider B to be the microscopic system, in contact with all other molecules (A, C, …). We can use the Boltzmann distribution to predict the fraction of all molecules in the room that have a particular energy. These two views of the Boltzmann distribution, one microscopic system observed repeatedly or a large number of microscopic systems observed once, complement each other. Sometimes one view is more helpful, sometimes the other.

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THE BOLTZMANN DISTRIBUTION IN A GAS The Boltzmann distribution applies to any kind of system—not just a solid. As a major application of the Boltzmann distribution, we will study a gas consisting of molecules that don't interact much with each other. Examples are the so-called “ideal gas” (with no interactions at all), and any real gas at sufficiently low density that the molecules seldom come near each other. In order to apply the Boltzmann distribution, we need a formula for the energy of a molecule in the gas. We will omit rest energy, and we will also omit nuclear energy and electronic energy from our total, because at ordinary temperatures there is not enough energy available in the surroundings to raise the molecule to an excited nuclear state or an excited electronic state. Also, for simplicity, instead of writing ΔE vib to represent an amount of vibrational energy above the ground vibrational state, we will simply write E vib . The energy of a single gas molecule in the gravitational field near the Earth's surface (excluding rest energy, nuclear energy, and electronic energy) is this: where E vib and E rot are the vibrational and rotational energies relative to the center of mass (if the molecule contains more than one atom, as in Figure 12.39) and y CM is the height of the molecule's center of mass above the Earth's surface. The mass of the molecule is M. For brevity we will speak of “the energy of the gas molecule,” but of course the gravitational potential energy really applies to the system of gas molecule plus Earth.

QUESTION For this expression for the energy to be accurate, why must the gas be an “ideal” gas (or a real gas at low density)?

Figure 12.39 Energy of an oxygen molecule.

In a real gas at somewhat high density we may not be able to neglect the potential energy associated with the intermolecular forces, in which case this expression for the energy would not be adequate. If the temperature T is the same everywhere in the (ideal) gas, the probability that a particular molecule will have a certain amount of energy is proportional to

We mentally divide the gas into two systems: one particular molecule of interest and all the rest of the molecules. These two systems are in thermal equilibrium with each other, because the gas molecules are continually colliding with each other and can share energy. The energy for the one particular molecule is then expected to follow the Boltzmann distribution. To avoid excess subscripts, in the following discussion we will simply write ν for the center-of-mass speed ν CM and y for the center-of-mass height y CM.

Separating the Various Factors

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It is useful to group the terms of the Boltzmann factor like this:

The first bracket is associated with the distribution of velocities, the second with the distribution of positions, the third with the distribution of vibrational energy, and the fourth with the distribution of rotational energy. We will discuss each of these individually.

Height Distribution in a Gas A striking property of Earth's atmosphere is that in high mountains the air density and pressure are significantly lower than at sea level. The air density is so low on top of Mount Everest that most climbers carry oxygen tanks. Can we explain this? In order to get at the main issue, the variation of density with height, we make the rough approximation that the temperature is constant— the same in the mountains as at sea level.

QUESTION How bad an approximation is it to consider the temperature to be the same at all altitudes?

Even in high mountains the temperature is typically above −29°C (−20°F), which is 244 K, and this is only 17% lower than room temperature of 293 K (20°C or 68°F). So maybe this approximation isn't too bad. We speak of the probability that the x component of the gas molecule's position lies between some x and x + dx, and similarly for y and z. Here dx is considered to be a very short distance, small compared to the size of the container but large compared to the size of a molecule. Focus just on that part of the Boltzmann distribution that deals with position, where the distribution is proportional to the probability of finding one particular molecule between x and x + dx, y and y + dy, and z and z + dz:

Evidently there's nothing very interesting about x and z (directions parallel to the ground). However, in the vertical direction there is an exponential fall-off with increasing height for the probability of finding a particular molecule at height y (Figure 12.40).

Figure 12.40 Number density vs. height in a constant-temperature atmosphere.

Looked at another way, our exponential formula tells us how the number density of the atmosphere depends on height, because in telling us about the behavior of one representative molecule, the formula also tells us something about all the molecules.

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We shifted gears from thinking about one molecule to thinking about many. Think again about one single air molecule. Suppose that you place it on a table, which is at room temperature. It is not a very unusual event for a thermally agitated atom in the table to hit our air molecule hard enough to send it kilometers high into the air! Actually, our particular molecule will very soon run into another air molecule and not make it very high in one great leap, but on average we do find lots of air molecules very high above sea level rather than finding them all lying on the ground. Notice again that k B T is the important factor in understanding the statistical behavior of matter. Here it sets the scale for the variation with height of the atmosphere's number density. At a height where the gravitational potential energy U g = mgy is equal to k B T, the number density has dropped by a factor of e −1 = 1/e = 0.37. Dry air at sea level is 78% nitrogen (one mole N 2 = 28 grams), 21% oxygen (one mole O 2 = 32 grams), about 1% argon, and 0.03% CO 2. An average mass of 29 grams per mole is good enough for most calculational purposes. 12.X.15 Suppose that you put one air molecule on your desk, so it is in thermal equilibrium with the desk at room Answer temperature. Suppose that there is no atmosphere to get in the way of this one molecule bouncing up and down on the desk. Calculate the typical height that the air molecule will be above your desk, so that Mgy ≈ k B T. 12.X.16 Approximately what fraction of the sea-level air density is found at the top of Mount Everest, a height of 8848 meters above sea level?

Answer

Distribution of Velocities in a Gas Next we look at the distribution of molecular speeds. When the gas is confined inside a finite container the momentum (or velocity) and height are quantized, but under almost all conditions the size of the energy quantum associated with momentum (or velocity) and position is so small compared to k B T that it is appropriate to take a nonquantum approach for these variables. We speak of the probability that the gas molecule has an x component of velocity within the range between some v x and v x + dv x , and similarly for v y and v z. Here dvx is considered to be a very small amount, small compared to the average speed of the molecules. Since we're explicitly interested in ν CM, it will be useful to express translational kinetic energy as this discussion. Since

rather than p2/(2M) in

, we can write the formula for the distribution of velocity in a gas as follows (remember

that we are simply writing v for the center-of-mass speed v CM):

The distribution for each velocity component is a bell-shaped curve, called a “Gaussian.” In Figure 12.41 we show the distribution of the x component of velocity for helium at room temperature.

QUESTION Judging from this graph, what is the average value for v x ? What are the average values of v y and v z? Why are these results plausible?

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Figure 12.41 Distribution of the x component of velocity for helium at room temperature.

Evidently the average value for the x component of the velocity is zero. This is reassuring, because a molecule is as likely to be headed to the right as it is to be headed to the left. Similarly, the average values for the y and z components of velocity are zero. It would not be particularly surprising to find a gas molecule with an x component of velocity such that

. On the other

hand, it would be very surprising to find a gas molecule with a value of v x many times larger. As usual, k B T sets the scale for thermal phenomena.

Distribution of Speeds in a Gas By converting from rectangular coordinates to “spherical” coordinates, the velocity-component distribution can be converted into a speed distribution. We won't go into the details, but the main idea is that the volume element in rectangular coordinates in “velocity space” dv x dv y dv z turns into the volume of a spherical shell with surface area 4πv 2, and thickness dv, as shown in Figure 12.42.

Figure 12.42 The volume of a shell with radius v and thickness dv is 4πν2 dv. Transferring to spherical coordinates, we have the following:

The only thing missing is a “normalization” factor in front to make the integral over all speeds from 0 to infinity be equal to 1.0 (since our one molecule must have a speed somewhere in that range). Here is the “Maxwell–Boltzmann” distribution for a lowdensity gas:

MAXWELL–BOLTZMANN SPEED DISTRIBUTION (LOW-DENSITY GAS)

The probability that a molecule of a gas has a center-of-mass speed within the range v to v + dv is given by P(v)dv.

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The distribution function for helium is shown for two different temperatures in Figure 12.43. You can see that the average speed of a helium atom is predicted to be about 1200 m/s at room temperature (20° Celsius, which is 293 Kelvin). The prediction was first made by Maxwell, in the mid-1800s, long before the development of quantum mechanics. In retrospect, this worked because the quantum levels of the translational kinetic energy are so close together as to be almost a continuum, so the classical model is a good approximation.

Figure 12.43 Distribution of speeds of helium atoms at two temperatures.

The average number of helium atoms in a container that have speeds in the range of 415.4 m/s to 415.7 m/s can be calculated by evaluating Maxwell's formula with v = 415.4 m/s and multiplying by dv = 0.3 m/s, which gives the probability of finding one molecule in this speed range. If there are N atoms in the container, multiplying by N gives the total number of molecules likely to be in that speed range at any given instant. The area of the vertical slice shown in Figure 12.44 represents the fraction of helium atoms that are likely to have speeds between 500 m/s and 600 m/s. At higher temperatures the distribution shifts to higher speeds (Figure 12.43).

Figure 12.44 Fraction of helium atoms at 293 K with speeds between 500 and 600 m/s.

Measuring the Distribution of Speeds in a Gas The actual distribution of speeds of molecules in any gas can be measured by an ingenious experiment. Make a tiny hole in a container of gas and let the molecules leak out into a vacuum (that is, a region from which the air is continually pumped out). The moving molecules run through collimating holes and then through a slot in a rapidly rotating drum and strike a row of devices at the other side of the drum that can detect gas molecules (Figure 12.45). Such measurements confirm the Maxwell prediction.

Figure 12.45 Apparatus for measuring the distribution of speeds of gas molecules.

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Fast molecules strike soonest, followed by medium-speed molecules, and finally slow molecules, so molecules with different speeds strike at different locations along the rotating drum. The number of molecules striking various locations along the drum is a direct indication of the distribution of speeds of molecules in the gas. (A correction has to be applied to these data to obtain the speed distribution inside the gas, because high-speed molecules emerge through the hole more frequently than low-speed molecules do, even if their numbers are the same in the interior of the gas. Also, the apparatus measures the distribution of v x rather than v.) How might we display the results of this experiment? We could present the data as a graph (Figure 12.46) of the fraction ΔN/N of the N helium atoms that have speeds between 0 and 500 m/s, between 500 and 1000 m/s, between 1000 and 1500 m/s, and so on.

Figure 12.46 Distribution showing fraction of atoms with speeds in a given range.

Average Translational Kinetic Energy in a Gas For a gas confined inside a container, quantum mechanics predicts that the kinetic energy

is quantized, but if the container is

of ordinary macroscopic size, the energy quanta are extremely small compared to the average kinetic energy, even at very low temperatures. Most gases liquefy before the temperature drops so low as to invalidate the pre-quantum analysis for the velocity distribution. Whenever the energy has a term containing a square of a position or momentum component, such as x 2 or predicts an associated average energy of

, pre-quantum theory

. This result is valid in quantum mechanics for high temperatures, where k B T is large

compared to the quantum energy spacing. This result follows from using an integral to take the average value of a quadratic term when the distribution is a Gaussian involving that quadratic term. That is, using integral tables (or integrating by parts), you can show that the average value of a quadratic term is (a bar over a quantity means “average value.”):

The average value is obtained by weighting the values of w 2 by the probability of finding that value; the denominator takes care of normalizing the distribution properly. As an example, when we did quantum-based calculations for the Einstein solid, we found that at high temperatures the specific heat capacity on a peratom basis was 3k B , which implies an average energy of 3k B T. We modeled an atomic oscillator as three onedimensional oscillators, each having an average energy of potential energy

corresponding to kinetic energy p2/(2M) and another

. This is a total of six quadratic terms, implying an average energy per atom of

, or 3k B T. We

therefore expect 3k B T the heat capacity per atom in a solid at high temperature to be 3k B , which is indeed what is observed.

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for the

HIGH-TEMPERATURE AVERAGE ENERGY If k B T >> the energy quantum, the average energy associated with a quadratic energy term is

.

The number of quadratic terms in the expression for the energy is often called the “degrees of freedom.”

QUESTION What is the average translational kinetic energy

in terms of k B T? What is the associated contribution to the specific

heat capacity of the gas?

Since there are three quadratic terms in the translational kinetic energy, three times

, the average value is

. We have the following important result:

AVERAGE

FOR AN IDEAL GAS

The contribution to the specific heat capacity for a gas is

.

Root-Mean-Square Speed The square root of

is called the “root-mean-square” or “rms” speed:

With this definition of v rms, we write

.

12.X.17 Calculate v rms for a helium atom in the room you're in (whose temperature is probably about 293 K). 12.X.18 Calculate v rms for a nitrogen molecule (N2; molecular mass 28) in the room you're in (whose temperature is probably about 293 K). Air is about 80% nitrogen.

Answer

Answer

Average Speed vs. RMS Speed The way you find the average speed of molecules in a gas is to weight the speed by the number of molecules that have that speed, and divide by the total number of molecules:

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For a continuous distribution of speeds this turns into an integral, where the weighting factors are given by the Maxwell speed distribution, which gives the probability that a molecule has a center-of-mass speed in the range v to v + dv. Using integral tables one finds this:

This shows that the average speed is smaller than the rms speed (0.92v rms). This same calculational scheme can be used to determine any average. For example, if you calculate the average value of v 2 by this method, you find

, as expected.

Earlier we commented that the average speed of helium atoms at room temperature is about 1200 m/s. In Exercise 12.X.17 you found that v rms at room temperature is 1350 m/s. This is an example of the fact that the rms speed is higher than the average speed. 12.X.19 The rms speed is somewhat higher than the average speed due to the averaging of squared speeds. Calculate Answer the average of the numbers 1, 2, 3, and 4, then calculate the rms average (the square root of the average of their squares), and show that the rms average is larger than the simple average. Squaring gives extra weight to larger contributions.

Application: Retaining a Gas in the Atmosphere Since

, helium atoms typically travel much faster than nitrogen molecules in our atmosphere, due to the small

mass of the helium atoms. Some few helium atoms will be going much faster than the average and may attain a high enough speed to escape from the Earth entirely (escape velocity from the Earth is about 1.1 × 104 m/s). This leads to a continuous leakage of highspeed helium atoms and other low-mass species such as hydrogen molecules (Figure 12.47). Other processes may also contribute to the flow of helium away from the Earth.

QUESTION Why doesn't the Moon have any atmosphere at all?

Figure 12.47 Low-mass atoms or molecules in our atmosphere may have speeds high enough to escape from the Earth entirely.

There's almost no helium in our atmosphere. Where do we obtain helium for party balloons and low-temperature

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refrigeration and scientific experiments? It is extracted from natural gas when the gas is pumped out of the ground. Heavy radioactive elements such as uranium in the Earth's crust emit alpha particles (helium nuclei), which capture electrons and become helium atoms. These atoms are trapped with natural gas in underground cavities. When the helium-bearing natural gas is pumped to the surface, the helium is extracted (at some cost).

The Moon's gravitational field is so weak that escape speed is quite small, and all common gases can escape, not just hydrogen and helium. 12.X.20 Calculate the escape speed from theMoon and compare with typical speeds of gas molecules. The mass of the Moon is 7 × 1022 kg, and its radius is 1.75 × 106 m.

Answer

Application: Speed of Sound Sound waves in a gas consist of propagation of variations in density (Figure 12.48), and the fundamental mechanism for this kind of wave propagation involves collisions between neighboring molecules, whose speeds are proportional to v rms (and roughly comparable to the speed of sound). For example, compare the v rms for nitrogen, 510 m/s, which you calculated in Exercise 12.X.18 with the speed of sound in air (which is mostly nitrogen) at 293 K, which is measured to be 344 m/s. (You may know the approximate rule that a 1-second delay between lightning and thunder indicates a distance of about 1000 feet, which is about 300 m.)

Figure 12.48 Periodic variations in gas density make a sound wave, which travels through a gas with speed v (the speed of sound).

12.X.21 Should the speed of sound in air increase or decrease with increasing temperature? What percentage change Answer would result from doubling the absolute temperature? (This effect is readily observed by measuring the speed of sound in a gas as a function of absolute temperature. The excellent agreement between theory and experiment provides additional evidence for our understanding of gases.)

Vibrational Energy in a Diatomic Gas Molecule We have treated the distribution of velocity and position. Next we discuss the distribution of vibrational energy, with the Boltzmann factor For a monatomic gas such as helium, there is no vibrational energy term. However, for a diatomic molecule such as N 2 or HCl, the vibrational energy is

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where k s is the effective spring stiffness corresponding to the interatomic electric force (not to be confused with k B , the Boltzmann constant), and s is the stretch of the interatomic bond. This formula for the vibrational energy is essentially the same as the formula for a one-dimensional spring–mass oscillator, which we have studied in detail (Figure 12.49).

Figure 12.49 The quantized vibrational energy levels of a diatomic oscillator.

Since p1 = p2 for momenta p1 and p2 relative to the center of mass, we can write

This equation has a form exactly like that for a spring–mass oscillator, with a different mass. Earlier we modeled a solid as a large number of isolated threedimensional atomic oscillators (each corresponding to three onedimensional oscillators, because there are springlike interatomic forces on an atom from neighboring atoms in all directions). This is an overly simplified model of a solid, because in a solid the atoms interact with each other. For example, if an atom moves to the left this affects the atoms to the right and to the left. In a low-density gas, however, the vibrational oscillators really are nearly independent of each other, because the gas molecules aren't even in contact with each other except when they happen to collide. So the analysis we carried out for the Einstein model of a solid applies even better to the vibrational portion of the energy in a real gas than it does to a real solid. Among the results that apply immediately are that the specific heat capacity associated with the vibrational energy of one oscillator is k B at high temperatures. The specific heat capacity decreases at very low temperatures, where k B T is small compared to the energy spacing of the quantized oscillator energies. Just as it was a surprise when the specific heat capacity of metals was found to decrease at low temperatures, there was a similar surprise in the measurements of the specific heat capacity of diatomic gases at low temperatures, because the contribution of the vibrational motion vanished.

QUESTION There are two quadratic terms in the vibrational energy (kinetic and spring). Therefore, at high temperatures what is the average vibrational energy in terms of k B T? What is the associated contribution to the specific heat capacity of the gas?

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At high temperatures the vibrational motion of a diatomic molecule (a one-dimensional oscillator) is expected to have an average energy that is approximately

, and contributes k B to the specific heat capacity.

Rotational Energy in a Diatomic Gas Molecule Finally we consider the rotational-energy portion of the distribution, for which the Boltzmann factor is For a monatomic gas there is no rotational term, but for a diatomic gas there is rotational kinetic energy associated with rotation of the “dumbbell” consisting of the two nuclei. The rotational kinetic energy is this (Figure 12.50):

This corresponds to rotational angular momenta Lrot,x and Lrot,y about the x axis and about the y axis. Note that only the nuclei contribute significantly to the rotational kinetic energy, because the electrons have much less mass.

Figure 12.50 Rotational kinetic energy levels of a diatomic molecule.

Rotation about the z axis connecting the two nuclei is irrelevant, for a somewhat subtle reason. The angular momentum is quantized, which leads to energy quantization. For rotations around the z axis (Figure 12.51) the energy is , where Lrot,z is the z component of the rotational angular momentum. Since I about the z axis for the tiny nuclei is extremely small compared with the moment of inertia about the x and y axes of the diatomic molecule, the rotational energy of the associated first excited state is enormous, and this state is not excited at ordinary temperatures.

Figure 12.51 A diatomic molecule can rotate around the x or the y axis. We say there are two rotational degrees of freedom.

Since there are two quadratic energy terms associated with rotation, we conclude that at high temperatures the rotational motion of a diatomic molecule has an average energy that is approximately

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, and contributes k B to the specific heat capacity.

The spacing between quantized rotational energies for a diatomic molecule is even smaller than the vibrational energy quantum. As a result, the gas must be cooled to a very low temperature before the pre-quantum results become invalid. Many gases liquefy before this low-temperature regime is reached.

Specific Heat Capacity of a Gas We are now in a position to discuss the specific heat capacity of a gas as a function of temperature. This property is important in calculating the thermal interactions of a gas. Historically, measurements of the specific heat capacity of gases were also important in testing theories of statistical mechanics. We'll concentrate on calculating C V , the specific heat capacity at constant volume (meaning no mechanical work is done on the gas). The associated experiment would be to add a known amount of energy to a gas in a rigid container and measure the temperature rise of the gas. Consider the average energy of a diatomic molecule such as N 2 or HCl, consisting of the translational kinetic energy associated with the motion of the center of mass, plus the vibrational energy, plus the rotational energy, plus the gravitational energy of the molecule and the Earth:

QUESTION At high temperature, what should the specific heat capacity at constant volume be?

There are seven quadratic terms in the expression for the energy, so if the temperature is very high, we expect an average energy of and a specific heat capacity at constant volume

.

The discrete energy levels for a diatomic molecule include electronic, vibrational, and rotational energy levels (Figure 12.52). The electronic energy levels correspond to particular configurations of the electron clouds, and the spacing between these levels is typically 1 eV or more. Since at room temperature k B T is about , the electronic levels do not contribute to the specific heat capacity at ordinary temperatures. The vibrational energy spacing is much smaller, and the rotational energy spacing is smaller still. A diatomic gas at high temperature has a bandlike spectrum, since transitions between electronic levels may be accompanied by one or more vibrational or rotational quanta.

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Figure 12.52 Electronic, vibrational, and rotational energy levels for a diatomic molecule.

What About Mgy? What about the Mgycm term in the energy? Remember that we are trying to predict the specific heat capacity of a gas at constant volume (so that no mechanical work is done on the system). We could measure C V by having a known amount of energy transfer Q due to a temperature difference into a closed container of gas and observing the temperature rise of the gas. During this process there is an extremely small increase in y CM of the gas in the container due to the temperature dependence of the height distribution, so the associated Mgy contribution to the specific heat capacity is negligible.

Specific Heat Capacity vs. Temperature Figure 12.53 is a graph of the specific heat capacity (at constant volume) of a diatomic gas as a function of temperature. In the following exercises, see whether you can explain these values of the specific heat capacity in terms of our analyses of the various contributions to the energy of a diatomic gas molecule.

Figure 12.53 Specific heat capacity CV of a diatomic gas vs. temperature.

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Remember that k B T is about

at room temperature.

12.X.22 What is the specific heat capacity on a per-molecule basis (at constant volume) of a monatomic gas such as helium or neon? Why doesn't the specific heat capacity depend on temperature? 12.X.23 Next consider a diatomic molecule.

Answer

Answer

(a) At high temperatures, what is the specific heat capacity (at constant volume) on a per-molecule basis of a diatomic gas such as oxygen or nitrogen? (b) Suppose that we lower the temperature of a diatomic gas to a point where k B T is small compared to the energy of the first excited vibrational state but still large compared to the energy of the first excited rotational state. Now what is the specific heat capacity (at constant volume)? (c) Lower the temperature even more, so that k B T is small compared to the energy of the first excited rotational state (some gases liquefy before this low temperature is reached). What is the specific heat capacity (at constant volume) at this low temperature? (d) As the temperature is decreased, the specific heat capacity (at constant volume) for H 2 eventually decreases to , before the gas liquefies. Does this transition to happen at a higher or lower temperature for D 2 (deuterium, whose nuclei each contain a proton plus a neutron)? Why? (Hint: Remember that it is the rotational angular momentum that is quantized.)

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SUMMARY The fundamental assumption of statistical mechanics is that, over time, an isolated system in a given macrostate is equally likely to be found in any of its possible microstates. Two blocks in thermal contact evolve to that division of energy that has associated with it the largest number of ways Ω = Ω 1Ω 2 of arranging the total energy among the atoms (largest number of microstates for a macrostate of given energy). Second law of thermodynamics: The entropy of a closed system never decreases. Only in a reversible process does the entropy of a closed system stay constant. Einstein solid: Each atom is modeled as three one-dimensional quantized oscillators. Number of ways to arrange q quanta of energy among N one-dimensional oscillators

Entropy: S

k B ln Ω, where k B = 1.38 × 10−23 J/K

Temperature: A small flow of energy Q into a system raises entropy by

.

, where

Specific heat capacity per atom:

The probability of finding energy E in a small system in contact with a large reservoir is proportional to Ω(E) is the number of microstates corresponding to energy E. For each energy term involving a quadratic term such as x 2 or k B T the average energy is

(“degree of freedom”), if the average energy is large compared to

and the contribution to the specific heat capacity is

. This contribution decreases at low

temperatures. At high temperatures, specific heat capacity on a per-atom basis in a solid ≈ 3k B . Speed distribution for a gas:

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EXERCISES AND PROBLEMS Sections 12.1, 12.2 12.X.24 List explicitly all the ways to arrange 2 quanta among 4 one-dimensional oscillators. 12.X.25 How many different ways are there to get 5 heads in 10 throws of a true coin? How many different ways are there to get no heads in 10 throws of a true coin? 12.X.26 How many different ways are there to arrange 4 quanta among 3 atoms in a solid? 12.X.27 A carbon nanoparticle (very small particle) contains 6000 carbon atoms. According to the Einstein model of a solid, how many oscillators are in this block? 12.X.28 In order to calculate the number of ways of arranging a given amount of energy in a tiny block of copper, the block is modeled as containing 8.7 × 105 independent oscillators. How many atoms are in the copper block? 12.X.29 In Chapter 4 you determined the stiffness of the interatomic “spring” (chemical bond) between atoms in a block of lead to be 5 N/m, based on the value of Young's modulus for lead. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective “interatomic spring stiffness” for an oscillator is 4 × 5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms). What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead? 12.X.30 Consider an object containing 6 one-dimensional oscillators (this object could represent a model of 2 atoms in an Einstein solid). There are 4 quanta of vibrational energy in the object. (a) How many microstates are there, all with the same energy? (b) If you examined a collection of 48,000 objects of this kind, each containing 4 quanta of energy, about how many of these objects would you expect to find in the microstate 000004? 12.P.31 The reasoning developed for counting microstates applies to many other situations involving probability. For example, if you flip a coin 5 times, how many different sequences of 3 heads and 2 tails are possible? Answer: 10 different sequences, such as HTHHT or TTHHH. In contrast, how many different sequences of 5 heads and 0 tails are possible? Obviously only one, HHHHH, and our formula gives 5!/[5!0!] = 1, using the standard definition that 0! is defined to equal 1. If the coin is equally likely on a single throw to come up heads or tails, any specific sequence like HTHHT or HHHHH is equally likely. However, there is only one way to get HHHHH, while there are 10 ways to get 3 heads and 2 tails, so this is 10 times more probable than getting all heads. Use the formula 5!/[N!(5 − N)!] to calculate the number of ways to get 0 heads, 1 head, 2 heads, 3 heads, 4 heads, or 5 heads in a sequence of 5 coin tosses. Make a graph of the number of ways vs. the number of heads.

Sections 12.3, 12.4, 12.5, 12.6 12.X.32 Energy conservation for two blocks in contact with each other is satisfied if all the energy is in one block and none in the other. Would you expect to observe this distribution in practice? Why or why not? 12.X.33 What is the advantage of plotting the (natural) logarithm of the number of ways of arranging the energy among the many atoms (natural logarithm of the number of microstates)? 12.X.34 Which has a higher temperature: a system whose entropy changes rapidly with increasing energy or one whose entropy changes little with increasing energy? 12.X.35 Explain why it is a disadvantage for some purposes that the specific heat capacity of all materials decreases at low temperatures. 12.X.36 Two blocks with different temperatures had entropies of 10 J/K and 35 J/K before they were brought in contact. What can you say about the entropy of the combined system after the two came in contact with each other? 12.X.37 Consider two blocks of copper. Block A contains 600 atoms and initially has a total of 20 quanta of energy. Block B contains 400 atoms and initially has 80 quanta of energy. The two blocks are placed in contact with each other, inside an

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insulated container (so no thermal energy can be exchanged with the surroundings). After waiting for a long time (for example, an hour), which of the following would you expect to be true? (A) Approximately 60 quanta of energy are in block A, and approximately 40 quanta of energy are in block B. (B) Approximately 50 quanta of energy are in block A, and approximately 50 quanta of energy are in block B. (C) The entropy of block A is equal to the entropy of block B. (D) The temperature of block A and the temperature of block B are equal. 12.X.38 Object A and object B are two identical microscopic objects. Figure 12.54 below shows the number of ways to arrange energy in one of these objects, as a function of the amount of energy in the object. (a) When there are 1.0 × 10−20 joules of energy in object A, what is the entropy of this object? (b) When there are 1.4 × 10−20 joules of energy in object B, what is the entropy of this object? (c) Now the two objects are placed in contact with each other. At this moment, before there is time for any energy flow between the objects, what is the entropy of the combined system of objects A and B?

Figure 12.54 12.P.39 The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic “oscillators” (masses on springs), note that since each oscillator is attached to two “springs,” and each “spring” is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. (Planck's constant divided by 2π), Avogadro's

Use these precise values for the constants:

number = 6.0221 × 1023 molecules/mole, k B = 1.3807 × 10−23 J/K (the Boltzmann constant). (a) What is one quantum of energy for one of these atomic oscillators? (b) Figure 12.55 contains the number of ways to arrange a given number of quanta of energy in a particular block of tungsten. Fill in the blanks to complete the table, including calculating the temperature of the block. The energy E is measured from the ground state. Nothing goes in the shaded boxes. Be sure to give the temperature to the nearest 0.1 Kelvin.

Figure 12.55 12.P.40 The interatomic spring stiffness for copper is determined from Young's modulus measurements to be 28 N/m. The mass of one mole of copper is 0.064 kg. If we model a block of copper as a collection of atomic “oscillators” (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two “springs,” and each “spring” is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above.

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Use these precise values for the constants: = 1.0546 × 10−34 J · s (Planck's constant divided by 2π), Avogadro's number = 6.0221 × 1023 molecules/mole, k B = 1.3807 × 10−23 J/K (the Boltzmann constant). (a) What is one quantum of energy for one of these atomic oscillators? (b) Figure 12.56 is a table containing the number of ways to arrange a given number of quanta of energy in a particular block of copper. Fill in the blanks to complete the table, including calculating the temperature of the block and the per-atom heat capacity C. The energy E is measured from the ground state. Nothing goes in the shaded boxes. Be sure to give the temperature to the nearest 0.1 Kelvin.

Figure 12.56 (c) There are 100 atoms in this object. What is the heat capacity on a per-atom basis? (Note that at high temperatures the heat capacity on a per-atom basis approaches the classical limit of 3k B = 4.2 × 10−23 J/K/atom.) 12.P.41 A nanoparticle containing 6 atoms can be modeled approximately as an Einstein solid of 18 independent oscillators. The evenly spaced energy levels of each oscillator are 4 × 10−21 J apart. (a) When the nanoparticle's energy is in the range 5 × 4 × 10−21 J to 6 × 4 × 10−21 J, what is the approximate temperature? (In order to keep precision for calculating the heat capacity, give the result to the nearest tenth of a Kelvin.) (b) When the nanoparticle's energy is in the range 8 × 4 × 10−21 J to 9 × 4 × 10−21 J, what is the approximate temperature? (In order to keep precision for calculating the heat capacity, give the result to the nearest tenth of a degree.) (c) When the nanoparticle's energy is in the range 5 × 4 × 10−21 J to 9 × 4 × 10−21 J, what is the approximate heat capacity per atom?

Note that between parts (a) and (b) the average energy increased from “5.5 quanta” to “8.5 quanta.” As a check, compare your result with the high temperature limit of 3k B . 12.P.42 For a certain metal the stiffness of the interatomic bond and the mass of one atom are such that the spacing of the quantum oscillator energy levels is 1.5 × 10−23 J. A nanoparticle of this metal consisting of 10 atoms has a total thermal energy of 18 × 10−23 J. (a) What is the entropy of this nanoparticle? (b) The temperature of the nanoparticle is 87 K. Next we add 18 × 10−23 J to the nanoparticle. By how much does the entropy increase? 12.P.43 A 50 gram block of copper (one mole has a mass of 63.5 grams) at a temperature of 35° C is put in contact with a 100 gram block of aluminum (molar mass 27 grams) at a temperature of 20° C. The blocks are inside an insulated enclosure, with little contact with the walls. At these temperatures, the hightemperature limit is valid for the specific heat capacity. Calculate the final temperature of the two blocks. Do NOT look up the specific heat capacities of aluminum and copper; you should be able to figure them out on your own. 12.P.44 Young's modulus for copper is measured by stretching a copper wire to be about 1.2 × 1011 N/m2. The density of copper is 3

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about 9 grams/cm , and the mass of a mole is 63.5 grams. Starting from a very low temperature, use these data to estimate roughly the temperature T at which we expect the specific heat capacity for copper to approach 3k B . Compare your estimate with the data shown on a graph in this chapter. 12.P.45 The goal of this experiment is to understand, in a concrete way, what specific heat capacity is and how it can be measured. You will need a microwave oven, a styrofoam coffee cup, and a clock or watch. In the range of temperature where water is a liquid (0 C to 100 C), it is approximately true that it takes 4.2 J of energy (1 calorie) to raise the temperature of 1 gram of water through 1 degree Kelvin. To measure this specific heat capacity of water, we need some way to raise a known mass of water from a known initial temperature to a final temperature that can also be measured, while we keep track of the energy supplied to the water. One way to do this, as discussed in the text, is to put water in a wellinsulated container within which a heater, whose power output is known, warms up the water. In this experiment, instead of a wellinsulated box with a heater, we will use microwave power, which preferentially warms up water by exciting rotational modes of the water molecules, as opposed to burners or heaters that warm up water in a pan by first warming up the pan. Use a styrofoam coffee cup of known volume in which water can be warmed up. The density of water is 1 gram/cm3. It is a good idea not to fill the cup completely full, because this makes it more likely to spill.

One method of recording the initial temperature of the water is to get water from the faucet and wait for it to equilibrate with room temperature (which can either be read off a thermostat or estimated based on past experience). After waiting about a half hour for this to happen, place the cup in the microwave oven and turn on the oven at maximum power. The cup needs to be watched as it warms up, so that when the water starts to boil, the elapsed time can be noted accurately. BE CAREFUL! A styrofoam cup full of hot liquid can buckle if you hold it near the rim. Hold the cup near the bottom. If the cup is full, do not attempt to move the cup while the water is hot. A spill can cause a painful burn.

On the back of the microwave oven (or inside the front door), there is usually a sticker with specifications that says “Output Power = …Watts” which can be used to calculate the energy supplied. If there is no indication, use a typical value of 600 watts for a standard microwave oven. Using all the quantities measured above and knowing the temperature interval over which you have warmed up the water, you can calculate the specific heat capacity of water. (a) Show and explain all your data and calculations, and compare with the accepted value for water (4.2 J/K per gram). (b) Discuss why your result might be expected to differ from the accepted value. For each effect that you consider, state whether this effect would lead to a result that is larger or smaller than the accepted value. 12.P.46 A nanoparticle consisting of four iron atoms (object 1) initially has 1 quantum of energy. It is brought into contact with a nanoparticle consisting of two iron atoms (object 2), which initially has 2 quanta of energy. The mass of one mole of iron is 56 grams. (a) Using the Einstein model of a solid, calculate and plot ln Ω 1 vs. q1 (the number of quanta in object 1), ln Ω 2 vs. q1, and ln Ω total vs. q1 (put all three plots on the same graph). Show your work and explain briefly. (b) Calculate the approximate temperature of the objects at equilibrium. State what assumptions or approximations you made. 12.P.47 Figure 12.57 shows a one-dimensional row of 5 microscopic objects each of mass 4 × 10−26 kg, connected by forces that can be modeled by springs of stiffness 15 N/m. These objects can move only along the x axis. (a) Using the Einstein model, calculate the approximate entropy of this system for total energy of 0, 1, 2, 3, 4, and 5 quanta. Think carefully about what the Einstein model is, and apply those concepts to this one-dimensional situation. (b) Calculate the approximate temperature of the system when the total energy is 4 quanta.

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(c) Calculate the approximate specific heat capacity on a per-object basis when the total energy is 4 quanta. (d) If the temperature is raised very high, what is the approximate specific heat capacity on a per-object basis? Give a numerical value and compare with your result in part (c).

Figure 12.57 12.P.48 A box contains a uniform disk of mass M and radius R that is pivoted on a low-friction axle through its center (Figure 12.58). A block of mass m is pressed against the disk by a spring, so that the block acts like a brake, making the disk hard to turn. The box and the spring have negligible mass. Astring is wrapped around the disk (out of the way of the brake) and passes through a hole in the box. A force of constant magnitude F acts on the end of the string. The motion takes place in outer space. At time ti the speed of the box is v i , and the rotational speed of the disk is ω i . At time tf the box has moved a distance x, and the end of the string has moved a longer distance d, as shown. (a) At time tf , what is the speed v f of the box? (b) During this process, the brake exerts a tangential friction force of magnitude f. At time tf , what is the angular speed ω f of the disk? (c) At time tf , assume that you know (from part b) the rotational speed ω f of the disk. From time ti to time tf , what is the increase in thermal energy of the apparatus? (d) Suppose that the increase in thermal energy in part (c) is 8 × 104 J. The disk and brake are made of iron, and their total mass is 1.2 kg. At time ti their temperature was 350 K. At time tf , what is their approximate temperature?

Figure 12.58

Sections 12.7, 12.8 12.X.49 Consider the exponential function e −x . Evaluate this function for x = 1, 10,000, and 0.01. 12.X.50 At room temperature (293 K), calculate k B T in joules and eV. 12.X.51 Sketch and label graphs of specific heat capacity vs. temperature for hydrogen gas (H2) and oxygen gas (O2), using the same temperature scale. Explain briefly. 12.X.52 Which has more internal energy at room temperature—a mole of helium or a mole of air? 12.X.53 Which of the following are true about the Boltzmann factor (A)

is small when ΔE is large.

(B)

is small at low temperature T.

(C)

tends to zero as T gets larger and larger.

, where ΔE is the energy above the ground state?

(D) Even at high temperature, when many energy levels are excited, the ground state (ΔE = 0) is the most populated state. 12.X.54 Explain qualitatively the basis for the Boltzmann distribution. Never mind the details of the math for the moment. Focus on

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the trade-offs involved with giving energy to a single oscillator vs. giving that energy to a large object. 12.X.55 Many chemical reactions proceed at rates that depend on the temperature. Discuss this from the point of view of the Boltzmann distribution. 12.X.56 Figure 12.59 shows is the distribution of speeds of atoms in a particular gas at a particular temperature. Approximately what is the average speed? Is the rms (root-mean-square) speed bigger or smaller than this? Approximately what fraction of the molecules have speeds greater than 1000 m/s?

Figure 12.59 12.X.57 The temperature of the surface of a certain star is 7000 K. Most hydrogen atoms at the surface of the star are in the electronic ground state. What is the approximate fraction of the hydrogen atoms that are in the first excited state (and therefore could emit a photon)? The energy of the first excited state above the ground state is (−13.6/22eV) − (−13.6eV) = 10.2eV = 1.632 × 10−18 J. (In this estimate we are ignoring the fact that there may be several excited states with the same energies—for example, the 2s and 2p states in hydrogen—because this makes only a small difference in the answer.) 12.X.58 A gas is made up of diatomic molecules. At temperature T1, the ratio of the number of molecules in vibrational energy state 2 to the number of molecules in the ground state is measured, and found to be 0.35. The difference in energy between state 2 and the ground state is ΔE. Which of the following conclusions is correct? (A) ΔE ≈ k B T1 (B) ΔE > k B T1 At a different temperature T2, the ratio is found to be 8 × 10−5. Which of the following is true? (D) ΔE ≈ k B T2 (E) ΔE > k B T2 12.X.59 Marbles of mass M = 10 grams are lying on the floor. They are of course in thermal equilibrium with their surroundings. What is a typical height above the floor for one of these marbles? That is, for what value of y is Mgy ≈ k B T? 12.X.60 Viruses of mass M = 2 × 10−20 kg are lying on the floor at room temperature (about 20°C = 293 K). They are of course in thermal equilibrium with their surroundings. What is a typical height above the floor for one of these viruses? That is, for what value of y is Mgy ≈ k B T? 12.X.61 How does the speed of sound in a gas change when you raise the temperature from 0°C to 20°C? Explain briefly. 12.P.62 Calculate the temperature rise of a gas. (a) You have a bottle containing a mole of a monatomic gas such as helium or neon. You warm up this monatomic gas with an electrical heater, which inputs Q = 580 joules of energy. How much does the temperature of the gas increase? (b) You have a bottle containing a mole of a diatomic gas such as nitrogen (N2) or oxygen (O2). The initial temperature

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is in the range where many rotational energy levels are excited but essentially no vibrational energy levels are excited. You warm up this diatomic gas with an electrical heater, which inputs Q = 580 joules of energy. How much does the temperature of the gas increase? (c) You have a bottle containing a mole of a diatomic gas such as nitrogen (N2) or oxygen (O2). The initial temperature is in the range where not only many rotational energy levels are excited but also many vibrational energy levels are excited. You warm up this diatomic gas with an electrical heater, which inputs Q = 580 joules of energy. How much does the temperature of the gas increase? 12.P.63 In studying a voyage to theMoon in Chapter 3 we somewhat arbitrarily started at a height of 50 kilometers above the surface of the Earth. (a) At this altitude, what is the density of the air as a fraction of the density at sea level? (b) Approximately how many air molecules are there in one cubic centimeter at this altitude? (c) At what altitude is air density one-millionth (1 × 106) that at sea level? 12.P.64 At sufficiently high temperatures, the thermal speeds of gas molecules may be high enough that collisions may ionize a molecule (that is, remove an outer electron). An ionized gas in which each molecule has lost an electron is called a “plasma.” Determine approximately the temperature at which air becomes a plasma. 12.P.65 100 joules of energy transfer due to a temperature difference are given to air in a 50 liter rigid container and to helium in a 50 liter rigid container, both initially at STP (standard temperature and pressure). (a) Which gas experiences a greater temperature rise? (b) What is the temperature rise of the helium gas? 12.P.66 It is possible to estimate some properties of a diatomic molecule from the temperature dependence of the specific heat capacity. (a) Below about 80 K the specific heat capacity at constant volume for hydrogen gas (H ) is 2 higher temperatures the specific heat capacity increases to

per molecule, but at

per molecule due to contributions from rotational

energy states. Use these observations to estimate the distance between the hydrogen nuclei in an H 2 molecule. (b) At about 2000 K the specific heat capacity at constant volume for hydrogen gas (H ) increases to 2

per molecule

due to contributions from vibrational energy states. Use these observations to estimate the stiffness of the “spring” that approximately represents the interatomic force. 12.P.67 In 1988, telescopes viewed Pluto as it crossed in front of a distant star. As the star emerged from behind the planet, light from the star was slightly dimmed as it went through Pluto's atmosphere. The observations indicated that the atmospheric density at a height of 50 km above the surface of Pluto is about onethird the density at the surface. The mass of Pluto is known to be about 1.5 × 1022 kg and its radius is about 1200 km. Spectroscopic data indicate that the atmosphere is mostly nitrogen (N2). Estimate the temperature of Pluto's atmosphere. State what approximations and/or simplifying assumptions you made. 12.P.68 Buckminsterfullerene, C 60, is a large molecule consisting of 60 carbon atoms connected to form a hollow sphere. The diameter of a C 60 molecule is about 7 × 10−10 m. It has been hypothesized that C 60 molecules might be found in clouds of interstellar dust, which often contain interesting chemical compounds. The temperature of an interstellar dust cloud may be very low, around 3 K. Suppose you are planning to try to detect the presence of C 60 in such a cold dust cloud by detecting photons emitted when molecules undergo transitions from one rotational energy state to another. Approximately, what is the highest-numbered rotational level from which you would expect to observe emissions? Rotational levels are l = 0, 1, 2, 3, ….

Computational Problems It is important to work through the first four computational problems in detail in order to make the ideas concrete.

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12.P.69 Calculate the number of ways to arrange energy in an Einstein solid. (a) Model a system consisting of two atoms (three oscillators each), among which 4 quanta of energy are to be distributed. Write a program to display a histogram showing the total number of possible microstates of the twoatom system vs. the number of quanta assigned to atom 1. Compare your histogram to the one shown in Figure 12.15 (you should get the same distribution). (b) Model a system consisting of two solid blocks, block 1 containing 300 oscillators and block 2 containing 200 oscillators. Find the possible distributions of 100 quanta among these blocks, and plot number of microstates vs. number of quanta assigned to block 1. Compare your histogram to the one shown in Figure 12.21. Determine the distribution of quanta for which the probability is half as large as the most probable 60-40 distribution. (c) Do a series of calculations distributing 100 quanta between two blocks whose total number of oscillators is 500, but whose relative number of atoms varies. For example, consider equal numbers of oscillators, and ratios of 2:1, 5:1, and so on. Describe your observations. 12.P.70 Start with your solution to Problem 12.P.69 (b). For the same system of two blocks, with N 1 = 300 oscillators and N 2 = 200 oscillators, plot ln(Ω1), ln(Ω2), and ln(Ω1Ω 2), for q1 running from 0 to 100 quanta. Your graph should look like the one in Figure 12.25. Determine the maximum value of ln(Ω1Ω 2) and the value of q1 where this maximum occurs. What is the significance of this value of q1? 12.P.71 Modify your calculations from Problem 12.P.70 to plot the temperature in kelvins of block 1 as a function of the number of quanta q1 present in the first block. On the same graph, plot the temperature in kelvins of block 2 as a function of q1 (of course, q2 = qtot − q1). In order to plot the temperature in kelvins, you must determine the values of ΔE and ΔS that correspond to a onequantum change in energy. Consider the model we are using. The energy of one quantum, in joules, is . The increment in entropy corresponding to this increment in energy is ΔS = kΔ(ln Ω). Assume that the blocks are made of aluminum. In Chapter 4 you made an estimate of the interatomic spring constant k s for aluminum. In Section 12.6 we show that the effective k s for oscillations in the Einstein solid is expected to be about 4 times the value obtained from measuring Young's modulus. What is the significance of the value of q1 (and of q2) where the temperature curves for the two blocks cross (the temperatures are equal)? 12.P.72 Modify your analysis of Problem 12.P.71 to determine the specific heat capacity as a function of temperature for a T(K)

C, Al (J/K/mole)

C, Pb (J/K/mole)

20

0.23

11.01

40

2.09

19.57

60

5.77

22.43

80

9.65

23.69

100

13.04

24.43

150

18.52

25.27

200

21.58

25.87

250

23.25

26.36

300

24.32

26.82

400

25.61

27.45

single block of metal. In order to see all of the important effects, consider a single block of 35 atoms (105 oscillators) with up to 300 quanta of energy. Note that in this analysis you are calculating quantities for a single block, not two blocks in contact.

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To make specific comparison with experimental data, consider the cases of aluminum (Al) and lead (Pb). For each metal, plot the theoretical specific heat capacity C per atom vs. T(K), with dots showing the actual experimental data given in the following table, which you should convert to the same per-atom basis as your theoretical calculations. Adjust the interatomic spring stiffness k s until your calculations approximately fit the experimental data. What value of k s gives a good fit? Compare with the estimated values of k s obtained in Chapter 4 based on Young's modulus, but remember that those estimated values are probably 4 times smaller than the effective spring constant for oscillations. Show from your graph that the high-temperature limit of the specific heat capacity is about 3k B per atom. The rise above this limit at high temperatures may be due to the fact that the assumed uniform spacing of the quantized oscillator energy levels isn't a good approximation for highly excited states. See Chapter 8. 12.P.73 For some examples of your choice, demonstrate by carrying out actual computer calculations that the “square-root” rule holds true for the fractional width of the peak representing the most probable arrangements of the energy. Warning: Check to see what is the largest number you can use in your computations; some programs or programming environments won't handle numbers bigger than about 1 × 10307, for example, and larger numbers are treated as “infinite.” 12.P.74 Create a computational model of one atom of the Einstein solid as shown in Figure 12.4. Let the atom move under the influence of the six springlike forces that act on the atom. Use the classical Momentum Principle, although the actual behavior is governed by quantum mechanics, with quantized energies. (a) Experiment with different initial displacements of the atom away from the equilibrium position. (b) In many materials the effective stiffness of the interatomic bonds is different in the x, y, and z directions; see how making the stiffnesses different affects the motion.

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Gases and Engines

KEY IDEAS We can model a gas as tiny balls in rapid random motion. Applying statistical ideas to this microscopic model allows us to predict macroscopic relationships among pressure, volume, and temperature, including the ideal gas law. The definition of temperature in terms of entropy is consistent with the everyday temperature scale. Both work and energy transfer due to a temperature difference can change the temperature and/or volume of a gas in a chamber with a movable piston. The attainable efficiency of engines that convert thermal energy into useful work is constrained by the second law of thermodynamics.

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GASES, SOLIDS, AND LIQUIDS In contrast to a solid, a gas has no fixed structure. The gas molecules are not bound to each other but move around very freely, which is why a gas does not have a well-defined shape of its own; it fills whatever container you put it in (Figure 13.1). Think, for example, of the constantly shifting shape of a cloud, or the deformability of a balloon, in contrast with the rigidity of a block of aluminum. On average, gas molecules are sufficiently far apart that most of the time they hardly interact with each other. This low level of interaction is what makes it feasible to model a gas in some detail, using relatively simple concepts.

Figure 13.1 Molecules in a gas.

In a gas the molecular motion must be sufficiently violent that molecules can't stay stuck together. At high enough temperatures, any molecules that do manage to bind to each other temporarily soon get knocked apart again by high-speed collisions with other molecules. However, at a low enough temperature, molecules move sufficiently slowly that collisions are no longer violent enough to break intermolecular bonds. Rather, more and more molecules stick to each other in a growing mass as the gas turns into a liquid or, at still lower temperatures, a solid.

Liquids Are More Complex A liquid is intermediate between a solid and a gas. The molecules in a liquid are sufficiently attracted to each other that the liquid doesn't fly apart like a gas (Figure 13.2), yet the attraction is not strong enough to keep each molecule near a fixed equilibrium position as in a solid. The molecules in a liquid can slide past each other, giving liquids their special property of fluid flow (unlike solids) with fixed volume (unlike gases).

Figure 13.2 Molecules in a liquid.

The analysis of liquids in terms of atomic, microscopic models is quite difficult compared with gases, where the molecules only rarely come in contact with other atoms, or compared with solids, where the atoms never move very far away from their equilibrium positions. For this reason, in this introductory textbook with its emphasis on atomic-level description and analysis we concentrate mostly on understanding gases and solids. An active field of research, called “molecular dynamics,” models liquids, gases, and solids by computational modeling using the Momentum Principle and appropriate forces between the molecules. The difference between this work and the computational models you have made is that sophisticated programming techniques are required to deal with very large numbers of molecules, sometimes as many as a million, in a reasonable amount of computer time.

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GAS LEAKS THROUGH A HOLE In the previous chapter we used statistical mechanics to determine the average speed of a gas molecule. Here we'll see some interesting phenomena in which the average speed plays a role. We will model the gas molecules as little balls that don't attract each other and interact only in brief elastic collisions. We will frequently use the symbol n = N/V to stand for the number of gas molecules per unit volume, which we will express in SI units as number per cubic meter:

NUMBER DENSITY: NUMBER PER CUBIC METER

N is the number of gas molecules in the volume V. The units of n are molecules per cubic meter.

Warning: You may be familiar from chemistry with the “ideal gas law” written in the form PV = nRT, where n is the number of moles. Here “n” means something else—the number of molecules per cubic meter, N/V.

EXAMPLE n at “Standard Temperature and Pressure” “Standard Temperature and Pressure” (or STP) are defined for a gas to be 0°C = 273 K and the average air pressure at sea level. Under STP conditions the ideal gas law can be used to show that one mole of a gas will occupy a volume of 22.4 liters = 22.4 × 103 cm 3 = 22.4 × 10−3 m3. What is the number density n of a gas at STP?

Solution One mole consists of 6.02 × 1023 molecules, so

One-Directional Gas We will calculate the leakage rate of a gas through a small hole in a container filled with the gas. First we'll consider a simplified one-directional example, in order to understand the basic issues before stating the results for a real threedimensional gas. The chain of reasoning that we follow is basically geometric. Consider a situation in which many gas molecules are all traveling to the right inside a tube. For the moment, temporarily assume that they all have the same speed v (Figure 13.3). The cross-sectional area of the tube is A (Figure 13.4).

Figure 13.3 Side view of molecules all traveling to the right with speed v inside a tube. There are n molecules per cubic meter inside the tube.

Figure 13.4 End view of molecules all traveling the same direction in a tube of cross-sectional area A.

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Because eventually we want to be able to calculate how fast a gas will leak through a hole in a container, we will calculate how many molecules leave this tube in a short time interval Δt. For a molecule that is traveling at speed v to be able to reach the right end of the tube in this time interval Δt, it must be within a distance vΔt of the end (Figure 13.5).

Figure 13.5 A molecule that will leave the tube in a time Δt must be within a distance v Δt from the end. Since the cross-sectional area of the tube is A, the volume V of the tube that contains just those molecules that will leave the tube during the time interval Δt is simply A(vΔt); see Figure 13.6.

Figure 13.6 Volume containing the molecules that will leave the tube in time Δt. There are n molecules per m3 inside the tube, and N = nV molecules inside the volume V. So the number N of molecules that will leave in the time interval Δt is

Dividing by the time interval Δt, we find the following result: This is for the one-directional case; all molecules have the same speed.

QUESTION Does this formula make sense? What would you expect if you increased the number of molecules per cubic meter, or the crosssectional area, or the speed?

The formula does make sense. The more molecules per unit volume (n = N/V), the more molecules will reach the right end of the tube per unit time. The bigger the cross-sectional area (A), the more molecules that will pass through that area per unit time. The faster the molecules are moving (v), the more molecules from farther away that can reach the end of the tube in a given time interval. The units are right: (molecules/m 3)(m 2) (m/s) = molecules/s.

Effect of Different Speeds We need to account for the fact that the gas molecules don't all have the same speed v. Suppose that n1 molecules per unit volume have speeds of approximately v 1, n2 molecules per unit volume have speeds of approximately v 2, and so on. The number of molecules crossing an area A per second is The average speed of all the molecules is by definition the following, where we weight each different speed by the number of molecules per cubic meter that have that approximate speed:

Therefore, A horizontal bar over a symbol is a standard notation for “average,” and v means average speed. Finally, we have a valid formula for the number of molecules in a onedirectional flow leaving the right end of the tube, even in the situation where they have a distribution of different speeds:

NUMBER OF MOLECULES CROSSING AREA A PER SECOND (ONE-DIRECTIONAL) LibraryPirate

This is for one-directional flow; molecules have various speeds.

A “one-directional” gas may sound a bit silly, but this formula does apply to real one-directional flows such as the flow of water or gas through a pipe or the flow of electrons through a copper wire in an electric circuit.

A Three-Dimensional Gas A more realistic model of a gas in a closed container would have approximately equal numbers of molecules heading to the left as well as to the right in the tube, in which case the number of molecules leaving the right end of the tube would be only . In a real three-dimensional gas, molecules are moving in all directions. Only those molecules that are headed in the +x direction can pass through a hole located to the right, not those moving in the −x direction or in the ±y or ±z directions. The molecules are headed randomly in all six directions, so we might expect our formula would have a factor of 1/6. However, the actual factor is 1/4, which comes from detailed averaging over all directions and is related to our use of the average speed (magnitude of velocity), rather than averages of velocity components v x or v y or v z. We don't want to get bogged down in the rather heavyweight mathematics required to prove this, so we just state that the factor is 1/4 rather than 1/6. (You can find a full derivation in books on statistical mechanics.)

NUMBER OF MOLECULES CROSSING AREA A PER SECOND (THREE-DIMENSIONAL)

This is for three-dimensional flow; molecules have various velocities.

EXAMPLE Leak Through a Hole in a Balloon At standard temperature and pressure, at what rate will helium escape from a balloon through a hole 1 millimeter in diameter? As we saw in the previous chapter, the average speed v of helium atoms at ordinary temperatures is about 1200 m/s.

Solution Earlier we calculated that at STP n = 2.68 × 1025 molecules per cubic meter. Therefore the leak rate is

Cooling of the Gas There is an interesting and important effect of gas escaping through a hole. If you look back over the derivations of the formulas, you can see that faster molecules escape disproportionately to their numbers. Faster molecules can be farther away from the hole than is true for slower molecules and still escape through the hole in the next short time interval. As a result, the distribution of speeds of molecules inside the container becomes somewhat depleted of high speeds. 13.X.1 What can you say about the temperature of the gas inside the container as the gas escapes? Why?

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Answer

EXAMPLE Leakage from a Balloon A party balloon filled with helium is about one foot in diameter (about 30 centimeters). The volume of a sphere is

, where r is the radius

of the sphere. In the previous example we calculated that the initial leak rate through a circular hole 1 millimeter in diameter is 6.3 × 1021 molecules/s. (a) If the helium were to escape at a constant rate equal to the initial rate, about how long would it take for all the helium to leak out? (b) As the helium escapes, the balloon shrinks, and the number of helium atoms per cubic meter n will stay roughly constant, in which case our analysis is pretty good. However, remember that the temperature of the gas drops due to a preferential loss of high-speed atoms. Would this effect make the amount of time to empty be more or less than the value you calculated in part (a)?

Solution When you blow up an ordinary rubber balloon the pressure is higher than one atmosphere, which means higher number density n but also a correspondingly higher leak rate, which is proportional to n. To a first approximation the density doesn't matter in this estimate of the time to empty. Also note that often when you puncture a balloon the balloon rips, creating a large opening; this is not the case we are considering. (a) The leak rate was 6.3 × 1021 atoms per second. Calculate the number of helium atoms in the balloon originally:

Assuming a constant rate:

(b) With the preferential loss of high-speed atoms, the speed distribution inside the balloon shifts to lower speeds (corresponding to a lower temperature). If the average speed is lower, the leak rate is lower, and it should take longer for the balloon to empty than we calculated in part (a), where we assumed a constant leak rate.

We've implicitly done the analysis in vacuum. If the balloon is in air, air molecules enter the balloon through the same hole.

13.X.2 A spacecraft containing air at STP is struck by a micrometeor that makes a hole 2 mm in diameter. Calculate the initial rate at which air escapes through the hole, in number of air molecules leaving the spacecraft per second. In the previous chapter we found that the average speed of air molecules at STP is about 500 m/s.

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Answer

MEAN FREE PATH Suppose that we place a special molecule somewhere in the helium gas, one whose movements we can trace. For example, it might be a molecule of perfume. On average, how far does this molecule go before it runs into a molecule of the gas? The average distance between collisions is called the “mean free path.” It plays an important role in many phenomena, including the creation of electric sparks in air (discussed in Volume II of this textbook). An approximate calculation of the mean free path depends on a simple geometrical argument. Draw a cylinder along the direction of motion of the special molecule, with length d and radius R + r, where R is the radius of the molecule, and r is the radius of a gas molecule, as shown in Figure 13.7 and Figure 13.8.

Figure 13.7 If a “special molecule” (blue) enters a cylinder of length d and radius R+ r, it will collide with a gas molecule (green).

Figure 13.8 End view of cylinder.

The geometrical significance of this cylinder is that if the path of the special molecule comes within one molecular radius of a gas molecule, there will be a collision, so any gas molecule whose center is inside the cylinder will be hit. How long should the cylinder be for there to be a collision? We define d to be the average distance the special molecule will travel before colliding with another molecule, so the cylinder drawn in Figure 13.7 should contain on average about one gas molecule. The cross-sectional area of the cylinder is A ≈ π(R + r)2 and the volume of the cylinder is Ad. If n stands for the number of gas molecules per cubic meter, we can write a formula involving the mean free path:

MEAN FREE PATH

A ≈ π(R + r)2; see Figure 13.7.

We are ignoring some subtle effects in this calculation, but this analysis gives us the main picture: the mean free path is shorter for higher density or larger molecular size.

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EXAMPLE Mean Free Path in Air To get an idea of the order of magnitude of a typical mean free path, assume that an N 2 molecule in the air has a radius of approximately 2 × 10−10 m (the radius of one of the N atoms being about 1 × 10−10 m), and calculate approximately the mean free path d of an N 2 molecule moving through air.

Solution

13.X.3 It is interesting to compare the mean free path of about 7 × 10−8 m to the average spacing L between air Answer molecules, which is the cube root of the volume occupied on average by one molecule (Figure 13.9). Calculate L. You may be surprised to find that the mean free path d is much larger than the average molecular spacing L. The molecules represent rather small targets.

Figure 13.9 The volume of air occupied on average by one molecule.

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PRESSURE AND TEMPERATURE We can use statistical ideas to relate what we know about the motion of molecules in a gas to the pressure that the gas exerts on its container. We'll consider a closed container, with the energy of the gas not changing. In a party balloon, helium atoms are continually hitting the rubber walls of the balloon. In an earlier example we found that at STP (standard temperature and pressure) about 6 × 1021 helium atoms per second escape through a 1-mm-diameter section of the balloon. That means that in a closed container 6 × 1021 helium atoms per second strike every 1-mm-diameter portion of the container's walls. The time- and space-averaged effect of this bombardment is an average force exerted on every square millimeter of the container. This average force per unit area is called the “pressure” P and is measured in N/m2, also called a “pascal.” We are going to calculate how big the pressure is in terms of the average speed of the helium atoms, thus building a link between the microscopic behavior of the helium atoms and the macroscopic time- and space-averaged pressure. On average, a helium atom bounces off the wall with no change of kinetic energy. We emphasize that this is the average behavior. Any individual atom may happen to gain or lose energy in the collision with a vibrating atom in the container wall, but if the container is not being warmed up or cooled down, the velocity distribution in the gas does not change, and on average the helium atoms rebound from the wall with the same kinetic energy they had just before hitting the wall (Figure 13.10).

Figure 13.10 On average, an atom bounces off the wall without changing speed.

QUESTION If the kinetic energy doesn't change when an atom bounces off the wall, is there any change in the atom's momentum?

Unlike speed or kinetic energy, momentum is a vector quantity, and there is a large change in the x component of momentum, from +px to −px . (There is no change in the y component.) Therefore the momentum change of the helium atom is Δpx,helium = −2px .

QUESTION What caused this change in the momentum of the helium atom?

A force is required to change the momentum of an object. In this case the force was applied by the wall of the container (or in more detail, by an atom in the wall of the container). By the principle of reciprocity (Newton's third law), the helium atom must have applied an equal and opposite force to the wall (or more precisely, to an atom in the wall of the container). Therefore the wall must have acquired an amount of momentum Δpx, wall = 2px . If we could calculate the average time Δt between collisions of helium atoms with a small area A of the wall, we could express the pressure (force per unit area) as follows, since :

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Assuming One Direction and One Speed We already know that if there are n+ helium atoms per unit volume that have an x component of velocity equal to v x > 0 (that is, moving in the +x direction toward the wall), the number of such atoms that hit an area A of the wall in a time Δt is n+Avx Δt. Therefore the average time between collisions is and we find

An Alternative Argument Alternatively, we can say that n+Avx helium atoms with this x component of velocity hit the area A every second, and each of them delivers 2px of momentum to the wall, so that the total momentum transfer to the wall per second is Divide by A to get the force per unit area, and you again get

Taking the Velocity Distribution into Account Actually, this is just the contribution to the pressure made by those atoms that happened to have this particular value of v x > 0. On average, only half the atoms are headed in the +x direction, so we replace n+ by n/2, where n is the number of atoms per cubic meter (going in either the +x or −x direction).

We also need to average over the slow and fast atoms:

We'll explain what we mean by this average. Suppose that n1 atoms per unit volume have x components of momentum of approximately px1, n2 atoms per unit volume have x components of momentum of approximately px2, and so on. The pressure is

The average value of for all the atoms is by definition the following, where we weight each different value of of atoms per cubic meter that have that approximate value:

so we have

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. Therefore we can write this:

by the number

Taking Direction into Account We can re-express the pressure in terms of the magnitude of momentum p rather than px by the following steps: First, note that

, or

.

Second, since the atoms are flying around in random directions, there should be no difference in averaging in the x, y, or z direction, so , which implies that .

Taking these factors into account, instead of

we can write the following important result:

GAS PRESSURE IN TERMS OF ATOMIC QUANTITIES

QUESTION Look back over the line of reasoning that led us to this result, and reflect on the nature of the argument. Stripped of the details, try to summarize the major steps leading to this result.

We reached this result by combining two effects: the number of molecules hitting an area per second is proportional to v, and the momentum transfer is also proportional to v. Hence the force per unit area is proportional to v 2.

The Ideal Gas Law It is useful to rewrite our result for the pressure, factoring out the term

, which is the average translational kinetic energy

K trans of a molecule of mass m:

The pressure of a gas is proportional to the number density (number of molecules per cubic meter, N/V), and proportional to the average translational kinetic energy of the gas molecules. This gives us a connection to temperature, because we found in the previous chapter that .

Substituting into the pressure formula, the “ideal gas law”:

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, which relates pressure to temperature and is called

MICROSCOPIC VERSION OF THE IDEAL GAS LAW

13.X.4 We found that at STP, n = 2.68 × 1025 molecules per m3. Calculate the value of one atmosphere of pressure in units of N/m2.

Answer

The Macroscopic Ideal Gas Law We can compare our microscopic version of the gas law with experiments that measure the macroscopic properties of gases. For gases with fairly low densities, measurements of the pressure for all gases (helium, oxygen, nitrogen, carbon dioxide, etc.) are well summarized by the macroscopic “ideal gas law,” which is probably familiar to you from chemistry, and which describes the observed behavior of any low-density gas:

MACROSCOPIC VERSION OF THE IDEAL GAS LAW

R is the “gas constant” (8.3 J/K/mole).

To compare with our microscopic prediction, we can convert this macroscopic version of the ideal gas law to a version involving microscopic quantities:

where N is the number of molecules in the gas. Therefore we have

This is simply P = nk B T, where n = N/V and R/6.02 × 1023 is the Boltzmann constant k B :

Warning about the Meaning of “n” Sometimes the ideal gas law is written in the form PV = nRT, where n is the number of moles rather than the number of molecules per cubic meter, n = N/V. On the rare occasions when we need to refer to the number of moles, we'll write it out as “number of moles.”

Temperature from Entropy or from the Ideal Gas Law LibraryPirate

In the previous chapter we found the relationship

, based on the Boltzmann distribution as applied to a low-density

gas. The Boltzmann distribution in turn was based on the statistical mechanics definition of temperature in terms of entropy as 1/T = dS/dE. When we inserted

into the kinetic theory result for pressure,

, we obtained the molecular version

of the ideal gas law, P = nk B T, which we showed was equivalent to the macroscopic version of the ideal gas law, P = (number of moles)RT/V. Low-density gases are described well by the ideal gas law, and for that reason gases are used to make accurate thermometers. You measure the pressure P and volume V of a known number of moles of a low-density gas and determine the temperature from the macroscopic gas law:

The fact that we could start from 1/T = dS/dE and derive the ideal gas law proves that the temperature measured by a gas thermometer is exactly the same as the “thermodynamic temperature” defined in terms of entropy.

Real Gases Our analysis works well for low-density gases. For high-density gases, there are two major complications. First, the molecules themselves take up a considerable fraction of the space, so the effective volume is less than the geometrical volume, and V in the gas law must be replaced by a smaller value. Second, at high densities the short-range electric forces between molecules have some effect. In this context these intermolecular forces are called “van der Waals” forces, which are the gradient of the interatomic potential energy discussed in Chapter 7. In a lowdensity gas almost all of the energy is kinetic energy, but in a high-density gas some of the energy goes into configurational energy associated with the interatomic potential energy. The effect is that in the gas law P must be replaced by a smaller value. When these two effects are taken into effect, the resulting “van der Waals” equation fits the experimental data quite well for all densities of a gas, although the corrections are different for different gases due to differences in molecular sizes and intermolecular forces.

Energy of a Diatomic Gas The formula for the average translational kinetic energy

is valid even for a gas with multiatom molecules such as

nitrogen (N2) or oxygen (O2). However, as you will recall from Chapter 9, this translational kinetic energy is only a part of the total energy of the molecule. In addition to translational energy, a diatomic molecule can have rotational and vibrational energy relative to the center of mass (Figure 13.11), so the energy contains additional terms. We write the energy of a diatomic molecule in the following way: The key point, as we discussed in the previous chapter, is that the average energy of a diatomic molecule is greater than the average energy of an atom in a monatomic gas at the same temperature. For a monatomic gas such as helium the average total energy per molecule is just .

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Figure 13.11 Energy of an oxygen molecule.

Application: Weight of a Gas in a Box The exponential dependence of number density

and therefore of pressure on height makes it possible to

understand something that otherwise might be rather odd about weighing a box that contains a gas (Figure 13.12). Let's do the weighing in a vacuum so we don't have to worry about buoyancy forces due to surrounding air. The box definitely weighs more on the scales if there is gas in it than if there is a vacuum in it. In fact, if the mass of the gas is M, the additional weight is Mg, because the Momentum Principle for multiparticle systems refers to the net force on a system, and the gravitational contribution to the net force is the sum of the gravitational forces on each individual gas molecule in the box, Mg.

Figure 13.12 Weighing a box containing a gas.

That seems reasonable until you think about the details of what is going on inside the box. At any given instant, the vast majority of the gas molecules are not touching the box! Also, some of the gas molecules are colliding with the top of the box, exerting an upward force on the box. How can these molecules possibly contribute to the weight measured by the scales? What is the mechanism for the Momentum Principle working out correctly for this multiparticle system? The number density n at the bottom of the box is slightly larger than the number density at the top of the box, if all the gas is at the same temperature, which is a good assumption in most situations. Therefore the pressure, P = nk B T, is larger at the bottom than at the top. Let's calculate the y components of the forces associated with these slightly different pressures, where A is the area of the top (and bottom) of the box, and h is the height of the box (Figure 13.13). where ΔP is the pressure difference from bottom to top inside the box. We can calculate ΔP directly, by starting from the fact that for a small height change Δy, we have this:

If we can evaluate dP/dy, we can determine the small pressure difference ΔP. Since P = nk B T, the height dependence of the pressure is the same as for the number density (for constant temperature), and we can write this:

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because the factor is very close to 1. The result is negative because the force the gas exerts on the top of the box is smaller than the force on the bottom. Choosing Δy = h, and writing Pbottom = nk B T, we have this result for ΔP:

Now that we know the pressure difference, we can calculate the y component of the net force:

Figure 13.13 The force due to gas pressure on the top of the box differs slightly from the force on the bottom of the box.

QUESTION Ah is the volume V of the box, and n = N/V is the number of molecules per m3 in the box, so what does this formula reduce to?

We have the striking result that the air inside the box pushes down on the box with a force equal to the combined weight Mg of all N molecules in the box: We have shown that the difference in the time- and space-averaged momentum transfers by molecular collisions to the top and bottom of the box is equal to the weight of the gas in the box, as predicted by the Momentum Principle for a multiparticle system. The pressure difference is very slight, but then the weight of the gas is very small, for that matter. What is surprising is that at any particular instant, relatively few of the molecules are actually in contact with the box, yet the effects of these relatively few molecules is the same as though they were all sitting on the bottom of the box. You could think of a box full of water in the same way. The water pressure is larger at the bottom than at the top, but with water this difference is quite large, corresponding to the much higher density of water. In fact, a column of water only 10 meters high makes a pressure equal to that produced by the many kilometers of atmosphere. Consider a column of water 10 meters high with cross-sectional area A. The density of water is 1 gram/cm3, which is 1 × 103 kg/m3. The total mass of the column is the density (kg/m 3) times the volume A(10 m), so the pressure on the bottom of the column is

This is the same result we found for atmospheric pressure at STP.

QUESTION In the preceding discussion the box was in vacuum, but under normal conditions the box we were weighing would have been surrounded by air. Suppose that the box has thin walls and is initially open to the air. Then we close the lid, trapping air inside

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it. Consider all forces including buoyancy forces, and determine what the scales will read. Will it be the weight of the box alone or the box plus air?

The scales will measure the weight of the box alone, because the weight of the air is Mairg and the buoyancy force is Mairg upward.

Application: Weight of a Bouncing Molecule There is a related calculation that is amusing. You may have done a related homework problem earlier on collisions, in which case you already went through a similar calculation. Consider a single molecule (otherwise in vacuum) that bounces up and down on a scale without losing significant energy. Figure 13.14 shows a snapshot at the instant that the molecule is just bouncing up after hitting the scale.

Figure 13.14 The ball has just hit the scale and has rebounded upward.

What do the scales read? If the scales can respond quickly, we will see brief spikes each time the molecule strikes it (Figure 13.15). Let's determine what the time-averaged force is.

Figure 13.15 Impacts of the ball on the scale.

QUESTION How long does it take for the molecule to reach the top of its trajectory, h, starting with a speed v? What then is the time Δt between impacts? How much momentum transfer is there to the scale on each impact?

It takes a time interval v/g to go up (for the speed to decrease from v to 0 with acceleration −g) and another time interval v/g to come down, so the time between impacts is Δt = 2v/g. Each impact transfers an amount of momentum ΔP = 2mv. Therefore the timeaveraged force is

If the scale is sluggish, and can't respond in a time as short as Δt = 2v/g, the scale will simply register the value mg, just as though the molecule were sitting quietly on the scale.

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ENERGY TRANSFERS In this section we offer an analysis of energy transfers between a gas and its surroundings. We will address such questions as these: How much energy transfer Q due to a temperature difference is required to raise the temperature of a gas by one degree (the heat capacity of a gas)? How does the temperature of a gas change when you compress it quickly? Do the answers to these questions depend on what kind of gas is involved? We need a device that lets us control the flow of energy into and out of a gas, in the form of work W or energy transfer Q. We will use a system consisting of a cylindrical container containing gas that is enclosed by a piston that can move in and out of the cylinder with little friction but which fits tightly enough to keep the gas from leaking out (Figure 13.16). This is similar to a cylinder in an automobile engine, into which is sprayed a mixture of gasoline vapor and air. The mixture is ignited by a spark, and the chemical reactions raise the temperature and pressure very high very quickly. The piston is pushed outward, which turns a shaft that ultimately drives the wheels.

Figure 13.16 A cylinder with a piston that can move vertically.

Force and Pressure Figure 13.17 shows a cylinder with a vertical-running piston on which we can load varying amounts of sand, in order to be able to control the pressure of the contained gas, and to be able to do controlled amounts of work on the gas. We will also put the cylinder in contact with hot or cold objects and allow energy transfer into or out of the cylinder.

Figure 13.17 The piston can be loaded with varying amounts of sand.

QUESTION Consider the piston plus sand as the system of interest for a moment and think about what forces act on this system.

A free-body diagram for the piston + sand system (Figure 13.18) includes the downward gravitational forces on the piston (Mg) and on the sand (mg), an upward time- and space-averaged force due to the pressure of the enclosed gas on the lower surface of the piston, and a downward time- and spaceaveraged force due to the pressure of the outside atmosphere on the upper surface of the system.

Figure 13.18 Forces on the piston.

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Since pressure is force per unit area, the upward force is PA, where A is the surface area of the bottom of the piston. Similarly, the downward force contributed by the outside atmosphere is PairA. These pressure-related forces are not actually continuous, since they are the result of collisions of individual gas molecules with the piston and sand. However, the rate of collisions is so extremely high over the area of the piston that the force seems essentially constant. For example, in Exercise 13.X.2 you found that on average, at STP (standard temperature and pressure) about 6 × 1021 helium atoms strike a tiny 1-mm-diameter section every second.

QUESTION In mechanical equilibrium (velocity of piston not changing), solve for the gas pressure inside the cylinder.

The Momentum Principle tells us that in equilibrium the net force on the piston + sand system must be zero, from which we are able to deduce that the pressure of the gas inside the cylinder is

A Sudden Change By varying the amount of sand (m) we can vary the pressure of the gas under study.

QUESTION If you suddenly add a lot of sand, what happens?

If you suddenly add a lot of sand to the piston, there is suddenly a sizable nonzero net downward force on the piston + sand system: Fnet = AP air + Mg + mg − AP ≠ 0. The piston starts to pick up speed downward. As it does so, it runs into gas molecules and tends to increase their speeds.

QUESTION What happens to the temperature of the gas in the cylinder?

Since higher average speed means higher temperature, the gas temperature starts to increase at the same time that the volume of the gas is decreasing. This is a double whammy: both increased temperature and decreased volume contribute to increased pressure, since P = (N/V)k B T. Therefore the pressure in the gas quickly rises, and eventually there will be a new equilibrium with a lower piston (supporting more sand) and a higher gas pressure in the cylinder. However, getting to that new equilibrium is pretty complicated. If there is no friction or other energy dissipation, the piston will oscillate down and up, with the gas pressure going up and down. It is even possible to determine an effective “spring stiffness” for the gas and calculate the frequency of the oscillation. However, in any real system there will be some friction, so we know that the system will eventually settle down to a new equilibrium configuration.

Quasistatic Processes To avoid these complicated (though interesting) transient effects, we will study what happens when we add sand very carefully and very slowly, one grain at a time, and we assume that the new equilibrium is established almost immediately. This is called a “quasistatic compression” because the system is at all times very nearly in equilibrium (Figure 13.19). Similarly, if we slowly remove one grain at a time, we can carry out a “quasistatic expansion.” Note in particular that at no time does the piston have any significant amount of kinetic energy, and macroscopic kinetic energy is essentially zero at all times. Of course there is plenty of microscopic kinetic energy in the gas molecules and the outside air molecules and in the thermal motion of atoms in the cylinder walls, piston, and sand.

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Figure 13.19 By adding or removing one grain of sand at a time we effect a “quasistatic” compression or expansion.

Suppose we carry out a lengthy, time-consuming quasistatic compression by adding lots of sand, one grain at a time, very slowly. The piston goes down quite a ways, the pressure in the gas is a lot higher, and the volume of the gas is a lot smaller. If we know the pressure and volume, we can calculate the temperature by using the ideal gas law P = (N/V)k B T (assuming the gas density isn't too high to make this invalid). Can we predict how low the piston will go for a given amount of added sand? Oddly enough, no—not without knowing something more about this device. There are two extreme cases that are both important in practice and calculable for an ideal gas. If the apparatus is made of metal (a very good thermal conductor) and is in good thermal contact with a large object at temperature T, the process will proceed at nearly constant temperature, with energy transfer from the gas into the surroundings. If the apparatus is made of glass (a very poor thermal conductor), the temperature of the gas inside the cylinder will rise in a predictable way, with negligible energy transfer to the surroundings in the form of energy transfer Q. We will analyze both kinds of processes: constant temperature processes and no-Q processes.

Constant-Temperature (Isothermal) Compression Suppose that the cylinder is made of metal (which is a very good thermal conductor) and is sitting in a very big tub of water whose temperature is T, as shown in Figure 13.20.

Figure 13.20 A cylinder of gas immersed in a large volume of water.

As we compress the gas, the temperature in the gas starts to increase. However, this will lead to energy flowing out of the gas into the water, because whenever the temperatures differ in two objects that are in thermal contact with each other, we have seen that there is a transfer of energy from the hotter object into the colder object. In fact, for many materials the rate of energy transfer is proportional to the temperature difference—double the temperature difference, double the rate at which energy transfers from the hotter object into the colder one. The mechanism for energy transfer due to a temperature difference is that atoms in the hotter object are on average moving faster than atoms in the colder object, so in collisions between atoms at the boundary between the two systems it is likely that energy will be gained by the colder object and lost by the hotter object. Energy transfer out of the gas will lower the temperature of the gas, since the total energy of the gas is proportional to the temperature. Quickly the

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temperature of the gas will fall back to the temperature of the water. The temperature of the big tub of water on the other hand will hardly change as a result of the energy added to it from the gas, because there is such a large mass of water to warm up. Therefore the entire quasistatic compression takes place essentially at the temperature of the water, and the final temperature of the gas is nearly the same as the initial temperature of the gas. This is a constant-temperature compression (also called an “isothermal” compression, which just means constant temperature). Similarly, we can slowly remove sand from the piston and carry out a quasistatic constant-temperature expansion.

QUESTION Suppose that the original gas pressure and volume were P1 and V 1, and as the result of a constant-temperature process the final volume is V2. What is the final pressure P 2?

Since the temperature hasn't changed, from the ideal gas law P = (N/V)k B T we can deduce that P 1V 1 = P 2V 2, and therefore P 2 = P 1(V 1/V 2).

Energy in a Constant-Temperature Compression A more difficult question we can ask (and answer!) is this: How much energy was added to the water in the constant-temperature compression?

QUESTION What energy inputs and outputs were made to the gas? What energy change occurred in the gas?

The piston did work on the gas, and there was energy transfer Q out of the gas (and into the water). The Earth's gravitational force did work on the gas (since the center of mass of the gas went down), but this is negligibly small compared to the work done by the piston (the lowering of the heavy piston involves much more gravitational energy than the lowering of the low-mass gas).

QUESTION Did the total energy of the gas change (ignoring the small gravitational energy change)?

Since the total energy of an ideal gas is proportional to temperature (including rotational and vibrational energy if the gas is not monatomic), and we made sure that we kept the temperature constant, the total energy of the gas did not change. Therefore we have this energy equation for the open system that is the gas: Symbolically, ΔE gas = W + Q = 0, where Q is negative (transfer out of system). In thermal processes of this kind, the energy equation is called “the first law of thermodynamics.” If we can calculate the work done by the piston, we can equate that result to the amount of energy transfer Q into the water. The piston exerts a (timeand space-averaged) force PA on the gas, where P is the pressure in the gas and A is the cross-sectional area of the piston (Figure 13.21).

Figure 13.21 Force on the gas by the piston.

QUESTION If the piston drops a distance h, is the work it does PAh?

As the piston drops, the pressure in the gas increases, so the force is not constant. We have to integrate the variable force through the distance h in order to determine the work. If we measure x downward from the initial piston position, at each step dx of the way the increment of work done is PA

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dx. However, A dx is an increment of volume (base area A time altitude dx), and the change in the volume is negative. Putting this all together, we have this:

WORK DONE BY A PISTON ON A GAS

Check the sign: If the volume decreases, the integral will be negative, and the minus sign in front of the integral makes the work done on the gas be positive, which is correct. Conversely, in an expansion of a gas, the integral is positive, and the work is negative, because the gas is doing work on the piston rather than the other way round.

QUESTION In a constant-temperature (isothermal) compression, work is done on the gas. Where does this energy flow go? How does the temperature stay constant?

As the piston moves down, it increases the average energy of the gas molecules that run into it. Therefore the input work starts to raise the temperature of the gas (higher speeds), but this higher temperature causes energy transfer Q out of the cylinder, into the (very slightly) lower-temperature water. The net effect is for energy to flow into the gas in the form of mechanical work, and out of the gas into the water in the form of energy transfer Q. There is no change in the total energy of the gas, because the temperature of the gas didn't change. The outward energy transfer brings the gas temperature back to what it was before the falling piston tried to increase the temperature. Now we can calculate quantitatively the work done by the piston in the constant-temperature compression, which is equal to the energy transfer from the gas to the surrounding water. We add lots of sand, one grain at a time (to maintain quasistatic equilibrium), and we compress the gas from an initial volume V 1 to a final volume V 2 (Figure 13.22).

Figure 13.22 The gas is compressed at a constant temperature T (the cylinder is immersed in a large tub of water at temperature T).

Replace P by Nk B T/V, where T is a constant:

Since the process was a compression, V 2 is smaller than V 1. Since −ln(V 2/V 1) = ln(V 1/V 2), we have

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Just as the symbol W is normally used for “work,” so the symbol Q is normally used for “energy transfer due to a temperature difference.” Here Q is negative (energy output).

Graphical Representation There is a useful graphical representation of this constant-temperature process. We plot the pressure P vs. the volume V, as shown in Figure 13.23, and the process is represented by a curve of constant T (which is also a curve of constant PV, since PV = Nk B T). The area under the process curve represents the work W, which is also the energy transfer Q that flows to the surroundings. In a compression the volume decreases, so we move to the left on the graph. In an expansion the volume increases, and we move to the right on the graph.

Figure 13.23 If T is kept constant, then the product PV is also constant.

The First Law of Thermodynamics For historical reasons, in thermal processes the Energy Principle is called “the first law of thermodynamics.”

THE FIRST LAW OF THERMODYNAMICS

In the particular case of constant-temperature (isothermal) compression of an ideal gas, we have where E int is the “internal” energy of the gas (the sum of the translational, rotational, vibrational, and other energy terms of all the molecules). 13.X.5 A mole of nitrogen is compressed (by piling lots of sand on the piston) to a volume of 12 liters at room temperature (293 K). Answer The cylinder is placed on an electric heating element whose temperature is maintained at 293.001 K. A quasistatic expansion is carried out at constant temperature by very slowly removing grains of sand from the top of the piston, with the temperature of the gas staying constant at 293 K. When the volume is 18 liters, how much energy transfer Q has gone from the heating element into the gas? How much work W has been done on the piston by the gas? How much has the energy of the gas changed? (You must assume that there is no energy transfer from the gas to the surrounding air, and no friction in the motion of the piston, all of which is pretty unrealistic in the real world! Nevertheless there are processes that can be approximated by a constant-temperature expansion. This problem is an idealization of a real process.)

Heat Capacity We were able to calculate the important properties of a constant-temperature compression, where the apparatus was in good thermal contact with its surroundings. Before analyzing the opposite extreme, where the apparatus is thermally insulated from its surroundings and no energy transfer due to a temperature difference is involved, we will review the concept of heat capacity in this context.

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Lock the piston in position so that the volume of the gas won't change (Figure 13.24), and so no work is done (W = 0). Put the apparatus in a big tub of water whose temperature is higher than the gas temperature, so that energy transfer will go from the water to the gas. Allow an amount of energy transfer Q due to the temperature difference to enter the gas and observe how much temperature rise ΔT has occurred in the gas. We define the “specific heat capacity C V at constant volume” on a per-molecule basis in the following way, where N as usual is the total number of molecules in the gas: Since W = 0, this is also equal to Q. The larger the specific heat capacity, the smaller the temperature rise ΔT for a given energy input Q.

Figure 13.24 If the piston is locked so the volume of the gas cannot change, we can measure CV of the gas.

QUESTION What is C V for a monatomic gas such as helium?

Since the total energy in a monatomic gas is , so

, the energy transfer Q will increase the total energy of the gas by an amount

. Experimental measurements of the specific heat capacity of monatomic gases agree with this prediction.

For gases that are not monatomic, such as nitrogen (N2), C V can be larger than

because the total energy can be larger than

due to

rotational and vibrational energy relative to the center of mass of the molecule. Before quantum mechanics, theoretical predictions for the contribution of the rotational and vibrational energies to the specific heat capacity of gases did not agree with experimental measurements. This was one of the puzzles that was eventually resolved by quantum mechanics, as we found in the previous chapter.

SPECIFIC HEAT CAPACITY AT CONSTANT VOLUME ON A PER-MOLECULE BASIS

Heat Capacity at Constant Pressure If we don't lock the piston but let the gas expand at constant pressure, the incoming energy transfer Q due to the temperature difference not only raises the energy of the gas by an amount ΔE thermal = NCV ΔT but also raises the piston, which involves an amount of work W (Figure 13.25). We define the “specific heat capacity C P at constant pressure” on a per-molecule basis as follows:

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Evidently C P is bigger than C V , because NCP ΔT = NCV ΔT + W. We can calculate the work done on the piston by the gas as the gas expands, which is the negative of the work done by the piston on the gas:

Therefore for an ideal gas there is a simple relationship between C V and C P :

Figure 13.25 If the piston is free to move, the pressure inside the cylinder remains the same, and we can measure Cp of the gas.

13.X.6 What is the specific heat capacity at constant pressure on a permolecule basis C P for helium?

Answer

Molar Specific Heat Capacity Specific heat capacity is often given on a per-mole basis rather than a permolecule basis. The molar specific heat capacity at constant volume for an , where R is the molar gas constant (8.3 J/K), which is 6 × 1023k B . The molar specific heat capacity at constant pressure is C P = ideal gas is C V +R for an ideal gas.

No-Q (Adiabatic) Compression With the apparatus made of metal and sitting in a big tub of water, the temperature during the compression didn't change. Now we analyze the opposite extreme. We make the cylinder and piston out of glass or ceramic (which are poor conductors for energy transfer due to a temperature difference) to minimize energy transfer out of the gas as the gas gets hotter during the compression. In fact, we make the approximation that there is no energy transfer at all (Figure 13.26). Such a no-Q process is also called “adiabatic” (which means “no energy transfer due to a temperature difference”).

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Figure 13.26 If the cylinder and piston are made of insulating material, no energy transfer Q can occur between the gas and its surroundings. This is called an adiabatic compression.

How realistic is such a no-Q approximation? For many real situations this can be a rather good approximation if the compression or expansion is fast, so that there isn't enough time for significant energy transfer due to a temperature difference. Such a flow of energy from one object to another is a rather slow process. For example, a cup of coffee sitting on a table may stay quite hot for ten minutes or more. However, didn't we say that we were going to carry out compressions and expansions very slowly, “quasistatically”? Yes, so there is a contradiction. However, it is often the case that motion may be slow enough to be a good approximation to a quasistatic process and nevertheless may also be fast compared to the time required for significant energy transfer due to a temperature difference. Again, think of how long it takes a cup of coffee to cool off. We again consider the gas in the cylinder as the system of interest, and the work done by the piston is equal to the increase in energy of the gas:

QUESTION Why did we use C V in this equation? The volume isn't constant in this compression!

An ideal gas is unique among materials in that its total energy is entirely determined by the temperature, independent of volume or pressure. So the change of energy of the gas itself is just NCV ΔT, even when the volume is not constant. As a practical matter, real gases behave this way as long as the density is not too high. At high densities the energy of the gas includes a significant term associated with the interatomic forces. We can work through the integration, starting from a “differential” form of the equation shown above: Use the gas law P = (N/V)k B T to substitute for the pressure:

Divide through by k B T and rearrange:

Now integrate:

Rewrite, using the properties of logarithms:

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Therefore we have

13.X.7 Use the ideal gas law to eliminate T from this expression, and show that PV γ = constant in a no-Q process, where the Answer parameter γ is defined as the ratio of the constant-pressure specific heat capacity to the constantvolume specific heat capacity, γ ≡ C P /C V . (Since C P is greater than C V , γ is greater than 1.) In particular, if the initial pressure and volume are P1 and V 1, and the final pressure and volume are P 2 and V 2, then

. Alternatively, we also have

.

Graphical Representation Again, there is a useful graphical representation of this no-Q process. In Figure 13.27 we plot the pressure P vs. the volume V, and the process is represented by a curve along which PV γ = constant. The area under the process curve represents the work W. For comparison we also show the curve for a constanttemperature process (T and PV constant). The no-Q curve is much steeper than the constant-temperature curve.

Figure 13.27 PV curves for a no-Q (adiabatic) process and a constant temperature (isothermal) process.

Work vs. Q We have studied the response of a gas to energy inputs and outputs, both mechanical W (work) and thermal Q (energy transfer due to a temperature difference). Work represents organized, orderly, macroscopic energy input. Energy transfer due to a temperature difference represents disorganized, disorderly, microscopic energy input. There is randomness at the atomic level in the collisions of atoms, which is the basic mechanism for energy transfer due to a temperature difference. In the previous chapter we saw deep consequences of the distinction between work W and energy transfer Q.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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FUNDAMENTAL LIMITATIONS ON EFFICIENCY An important technology involves the conversion of energy transfer Q due to a temperature difference into useful work W (Figure 13.28). For example, in a steam-powered electricity generating plant, coal is burned to warm up water that drives a steam engine to turn a generator, which converts the work done by the steam engine into electric energy. Energy conservation of course puts limits on how much useful work you can get from the burning of the coal—but there is a further limitation due to the second law of thermodynamics. It turns out that in a practical generating plant only about one-third of the energy of the coal can be turned into useful work! The fundamental problem is that energy transfer due to a temperature difference is a disorderly energy transfer, and the second law of thermodynamics takes that into account.

Figure 13.28 An engine converts net energy transfer Q into net work W.

We will find that the most efficient processes are “reversible” processes—that is, processes that produce no change in the entropy of the Universe (a reversed movie of such a process looks possible). In the next sections we discuss two processes, mechanical friction and finite-temperature-difference energy transfer, which are major contributors to entropy production and whose effects must be minimized in order to obtain the most efficient performance in a mechanical system. Where we are headed is to establish limits on how efficiently thermal energy can be turned into mechanical energy in an engine, as a consequence of the second law of thermodynamics. This is a revealing example of the power of the second law of thermodynamics to set limits on possible phenomena.

Friction and Entropy Production The second law of thermodynamics says that the entropy of the Universe never decreases. Portions of the Universe may experience a decrease of entropy, but only if the entropy of other portions increases at least as much (and typically more). An example of a process that increases the total entropy of the Universe is sliding friction. A block sliding across a table slows down as kinetic energy associated with the overall motion of the block is dissipated into random (thermal) energy inside the block and table, with an increase in entropy. This friction process is irreversible. We would be astonished if after coming to rest the block suddenly started picking up speed back toward its starting position, although this would not violate conservation of energy. However, the probability is vanishingly small that the dissipated energy in the block and table could concentrate back into an orderly motion of the block.

Energy Transfer and Entropy Production Mechanical friction contributes to increasing the entropy of the Universe. We will show that energy transfer between two objects that have different temperatures also makes the total entropy of the Universe increase. This is then a process to avoid, if possible, in an efficient engine. Connect a metal bar between a large block at a high temperature TH and another large block at a low temperature TL (Figure 13.29). We write the rate of energy transfer from the high-temperature block (the “source”) to the lower-temperature block (the “sink”) as over the letter Q means “rate” or “amount per second.” Since Q has units of joules,

has units of joules/s, or watts. This energy transfer

rate is proportional to the “thermal conductivity” σ of the bar (metals have higher thermal conductivity than glass or plastic),

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. The dot

proportional to the temperature difference (twice the temperature difference, twice the rate of energy transfer), proportional to the cross-sectional area A of the bar (twice the crosssectional area is like having two bars), and inversely proportional to the length L of the bar (twice the length of the bar, half the energy transfer rate):

RATE OF ENERGY TRANSFER DUE TO A TEMPERATURE DIFFERENCE

Units are J/S, or watts.

is a positive quantity and is the absolute value of the rate of energy flow into or out of a system

due to a temperature difference.

Figure 13.29 Energy flows at a rate

from high temperature to low.

The quantity (TH − TL)/L is called the “temperature gradient.” The larger the gradient (the more rapidly the temperature changes with distance along the bar), the larger the number of joules per second of energy transfer. We want to show you that energy transfer between different temperatures increases the total entropy of the Universe. Suppose that the source and sink are so large that for a while the energy transfer doesn't change their temperatures much. Since 1/T = (dS)/(dE), it follows that a small energy input (dE) into a system, of amount Q, leads to an entropy change of that system:

Suppose that the energy transfer rate is energy

. In a short time interval Δt, the high-temperature source at temperature TH loses an amount of

.

QUESTION In this short time interval Δt, does the entropy of the source increase or decrease? By how much?

Since ΔSH = (ΔE)/TH, and the high-temperature source loses energy, we have

.

In this same time interval, the low-temperature sink at temperature TL gains the same amount of energy,

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.

QUESTION In this short time interval Δt, does the entropy of the sink increase or decrease? By how much?

Since the low-temperature sink gains energy, we have

.

There is a crucial and perhaps not entirely obvious point: Does the entropy of the bar change during this time interval? No. The blocks are so large that their temperatures don't change much in a short time interval, so the temperatures at the ends of the bar, and all along the bar, are not changing. Every second, energy enters the bar, but the same amount of energy leaves the bar, so the energy of the bar isn't changing. Nothing about the bar is changing. It is merely a conduit for the energy transfer, but its state is not changing. Therefore the change in the total entropy of the Universe in a time interval Δt is

QUESTION Is this quantity positive or negative? Is this consistent with the second law of thermodynamics?

Since TL < TH, 1/TL > 1/TH, and the entropy change of the Universe in this process is positive. This is consistent with the second law of thermodynamics, which states that the entropy of the Universe should never decrease.

Reversible and Irreversible Processes An increase in the total entropy of the Universe is associated with an irreversible process, because to return to the earlier state would require that the total entropy of the Universe actually decrease, which won't happen with macroscopic systems. We would be astonished if the energy transfer suddenly reversed and ran from the colder block thermally back “uphill” to the hotter block! There is only one way to carry out energy transfer due to a temperature difference reversibly or nearly so (that is, with little or no change in the total entropy of the Universe)—do it between systems whose temperatures are very nearly equal to each other (TH ≈ TL). There's a practical problem, however.

QUESTION If the two temperatures are nearly equal (to avoid increasing the entropy of the Universe), at what rate is energy transferred from the (slightly) hotter source to the (slightly) cooler sink?

Energy transfer will flow extremely slowly, because

is proportional to the temperature difference (or more specifically to the temperature

gradient). If there is hardly any temperature difference, the energy transfer rate will be very small. So we can carry out such energy transfers nearly reversibly, hardly changing the entropy of the Universe, but only if we do it so slowly as to be of little practical use in driving some kind of mechanical engine that converts thermal energy into work. 13.X.8 The thermal conductivity of copper (a good thermal conductor) is 400 watts/K/m (for comparison, the thermal conductivity of iron is about 70 watts/K/m, and that of glass is only about 1 watt/K/m). One end of a copper bar 1 cm on a side and 30 cm long is immersed in a large pot of boiling water (100°C), and the other end is embedded in a large block of ice (0°C). It takes about 335 joules to melt one gram of ice. How long does it take to melt one gram of ice? Does the entropy of the Universe increase, decrease, or stay the same? Answer

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A MAXIMALLY EFFICIENT PROCESS Despite the serious practical problem that reversible energy transfer due to a nearly zero difference in temperature proceeds infinitesimally slowly, we will analyze reversible “engines” in which (nearly) reversible energy input with tiny temperature differences is used (very slowly!) to do something mechanically useful, such as lift a weight or turn an electric generator. We will also assume that we can nearly eliminate sliding friction. The idea that we will pursue is to see how efficiently we can convert disorderly thermal energy into orderly mechanical energy (a lifted weight). We expect that reversible processes, which don't increase the total entropy of the Universe, should be the most efficient processes, though we recognize that these most efficient processes must proceed excruciatingly slowly. Since we will use nearly reversible processes to do work, we could run the engine backward. Such a backward-running engine turns out to be a refrigerator—an engine in which work input to the engine can lead to extracting thermal energy from something to make it colder or to keep it cold. The conception of a theoretically most efficient possible engine, and the recognition that this ideal engine would have to be reversible and not increase what we now call the total entropy of the Universe, was due to a young French engineer, Sadi Carnot, in 1824. His achievement is all the more remarkable because it came before the principle of energy conservation was established! After using the second law of thermodynamics to determine the maximum possible efficiency of infinitely slow engines, in the last part of the chapter we will analyze engines that run at useful speeds. We will find as expected that they are even less efficient than the infinitely slow engines.

A Cyclic Process of a Reversible Engine Some of the first practical engines drove pumps to pump water out of deep mines in England, repeatedly lifting large amounts of water large distances. The energy came from burning coal, which boiled water to make steam in a cylinder that pushed up on a piston and operated the pump. People started wondering how efficient such an engine could be. What limits the amount of useful work you can get from burning a ton of coal? We describe a scheme for running an engine in a reversible way, which ought to be as efficient as possible. For concreteness, our engine is a device consisting of a cylinder containing an ideal gas, with a piston and some sand on the piston to adjust the pressure on the gas (Figure 13.30). This is a simple and familiar device that permits energy exchanges in the form of work and energy transfer due to a temperature difference. As we'll see, the actual construction details of the engine don't matter for the theoretical question regarding the maximum possible efficiency, though they matter very much in the actual construction of a useful engine.

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Figure 13.30 Constant-temperature expansion; gas temperature just slightly lower than T H.

A Constant-Temperature (Isothermal) Expansion In addition to the engine itself, we need a large high-temperature source, large enough that extracting some energy from it will cause only a negligible decrease in its temperature. Start by arranging that the engine (the gas cylinder with its piston) has a temperature very slightly lower than the temperature TH of the high-temperature source. We connect the engine to the hightemperature source and perform a reversible constant-temperature (isothermal) expansion of the engine, doing some work on the surroundings in the process (Figure 13.30). We lift the piston by slowly shifting some sand sideways from the piston onto nearby shelves. This requires almost no work on our part but results in the lifting of some weight. We make sure that the temperature of the gas doesn't change during this process. There is energy transfer into the engine because it is very slightly cooler, but the temperature difference is so slight that there is negligible change in the entropy of the Universe.

Entropy and Energy in Lifting the Piston Remember that the entropy change of a system when there is energy transfer Q is ΔS = Q/T. In lifting the piston the entropy of the high-temperature source has changed by an amount ΔSH = −Q H/TH, where Q H is a positive quantity. The entropy of the gas has changed by an amount ΔSgas = +Q H/TH. The entropy of the Universe hasn't changed at all. (The increased entropy of the gas is associated with the fact that there are more ways to arrange the molecules in a large volume than in a small volume.) Because the temperature hasn't changed, there is no change in the energy of the ideal gas. Recall that the energy of an ideal gas, or a low-density real gas, is a function solely of the temperature, not the volume. (A dense real gas is more complicated, because the electric potential energy for pairs of molecules depends on distance, and therefore the energy of the gas depends on volume as well as temperature.) We have succeeded in converting all of the energy transfer from the hightemperature source into useful work (lifting the piston), because none of the input energy went into changing the energy of the gas. This is 100% efficiency in converting thermal energy into useful work on the surroundings. What's the problem?

The Need for a Cycle The problem is this: We need to be able to do this again, and again, and again. For example, we want to keep pumping water out of the deep mine, over and over. However, to repeat the lifting process, we have to return to the original state, with the gas compressed. We could simply reverse the process, doing work on the gas (constant-temperature compression), with energy transfer from the gas into the high-temperature source. The net effect, though, would be that there was zero net energy transfer to the gas, and zero net work done on the surroundings. We need to run our engine in a nontrivial repetitive “cycle,” where we can lift the piston repeatedly. Each time we lift the piston and do useful work, we need to get the piston back down again without having to do the same amount of work to push it down. If we can get the gas back to its original state with net work done on the surroundings, we will have a useful device. One possibility for bringing the piston down would be to let the gas cool down. However, to lower the temperature we would have to place the gas cylinder in contact with a large object at some low temperature TL, called the “sink” because, as we'll see, we will dump some energy into it. We can't place the gas cylinder in contact with the sink immediately, because the gas is at a high temperature TH, and placing the hot gas in contact with the cold sink would mean that there would be a large temperature difference, and there would be a large increase in the entropy of the Universe associated with the energy transfer from the hot gas to the cold sink.

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A No-Q (Adiabatic) Expansion To avoid this large production of entropy, we need to bring the temperature of the gas down almost to TL before making contact with the sink. To do this, we disconnect the gas cylinder from the high-temperature source and perform a reversible no-Q (adiabatic) expansion, which does some more work on the surroundings and lowers the temperature of the engine (Figure 13.31). This is accomplished by slowly removing some weight from the piston, allowing the gas to expand and the temperature to fall. We stop the expansion when the temperature of the gas is just slightly higher than the low temperature TL of the sink.

Figure 13.31 No-Q expansion; the gas temperature drops to T L .

A Constant-Temperature (Isothermal) Compression We can now safely place the gas cylinder in contact with the low-temperature sink. This is okay as far as entropy production is concerned, because we made sure that the temperature of the gas is only very slightly higher than the sink temperature TL. The piston is now quite high, and we need to bring it down. We do this by slowly adding some weight to the piston, performing a constanttemperature (isothermal) compression (Figure 13.32). There is energy transfer into the sink. Because the temperature of the ideal gas is nearly TL at all times, there is no energy change in the gas. Therefore the work that we do is equal to the energy transfer Q L into the sink.

Figure 13.32 Constant-temperature compression; the gas temperature is just slightly higher than T L . There is an increase in the entropy of the sink ΔSL = +Q L/TL, and there is an entropy decrease in the entropy of the gas ΔSgas = −Q L/TL. The entropy of the Universe doesn't change. The decreased entropy of the gas is associated with the fact that there are fewer

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ways to arrange the molecules in a small volume than in a large volume.

A No-Q (Adiabatic) Compression We stop the compression at a particular state of the gas chosen with care so that we can do the following: We disconnect the engine from the lowtemperature sink and perform a no-Q compression that raises the temperature of the gas to a temperature just barely less than TH, the temperature of the high-temperature source. We do some work to perform this compression (Figure 13.33).

Figure 13.33 No-Q compression; gas temperature rises to T H. The engine is now back to its original state, ready to begin another cycle.

This compression brings the system back to its original state (density, pressure, temperature), so we can repeat the cycle of four processes all over again. The possible usefulness of this engine is that we can run it repeatedly, over and over. What remains to be analyzed is how much work is done on the surroundings in one cycle, and how much energy input we need to supply. Of course if we find that the net work is zero, the engine won't be useful. However, explicit calculations for an ideal gas show that at least in that case there is net work done on the surroundings. We will find that this is true for any reversible engine run in such a cycle. This reversible cycle of two constant-temperature (isothermal) processes and two no-Q (adiabatic) processes running between a hightemperature source and a low-temperature sink is called a “Carnot cycle.”

Summary of the Cycle Figures 13.34 and 13.35 summarize the four processes of the engine cycle. Step 1: Isothermal expansion in contact with TH Step 2: Adiabatic expansion (no contact); temperature falls to TL Step 3: Isothermal compression in contact with TL Step 4: Adiabatic compression (no contact); temperature rises to TH The four processes return the gas to its original temperature and volume. The question is, did we get any net work out of this cycle? Figure 13.34 shows the pressure vs. volume during the cycle. Earlier in this chapter we saw that the work done on the gas is , so the work done by the gas on the surroundings is

, which is the area under one of these curves.

Because V increases during steps 1 and 2, the work done on the surroundings is positive, but this work is negative in steps 3 and 4 because V decreases. The net work done on the surroundings is therefore represented by the shaded area shown in Figure 13.34. There is net work, related to the fact that the PV curves for the adiabatic processes (shown in blue and given by PV γ = constant) are steeper than the PV curves for the isothermal processes (shown in red and given by PV = constant).

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Figure 13.34 P vs. V for the engine cycle. The shaded area represents the net work W done by the engine on the surroundings.

Figure 13.35 The engine cycle.

Entropy in a Cycle of a Reversible Engine Let Q H be the (absolute value of the) energy extracted from the source at temperature TH during the constant-temperature expansion, and let Q L be the (absolute value of the) energy dumped into the sink at temperature TL during the constant-temperature compression. Remember that the entropy change of a system when there is energy transfer Q is ΔS = Q/T.

QUESTION Is there any entropy change in the engine in one cycle?

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The entropy change of the engine in one cycle is zero, because the gas is brought back to its original state.

QUESTION Therefore, in one cycle, how much entropy change is there in the surroundings as a result of the work done on and by the surroundings?

Only the energy transfer due to a temperature difference affects the entropy during one cycle (work doesn't contribute), and we calculated these through the formula ΔS = Q/T. We were careful to run the engine in a reversible way, avoiding making any increase in the total entropy of the Universe, so in one cycle we have

Therefore we can write this important relationship for one cycle of the engine:

ONE CYCLE OF A REVERSIBLE ENGINE

This is a surprisingly simple result for such a complicated process. What does this simple result depend on? It depends solely on the second law of thermodynamics, that statistically the total entropy of the Universe will essentially never decrease. We illustrated the processes in the cycle with a cylinder filled with an ideal gas, but the result above doesn't actually depend at all on the details of how the engine is constructed. In particular, it doesn't matter whether the engine contains helium gas, or a solid block of rubber, or a liter of liquid alcohol. There would be an easily visible difference between using a solid or liquid rather than a gas in the engine, because the distance through which the piston would move would be much smaller than with a gas. However, the simple result shown above would remain the same.

Energy in a Cycle of a Reversible Engine Let's review the energy changes in the engine in one cycle. The source inputs an amount of energy Q H. The sink removes an amount of energy Q L. The working substance returns to its original state, so it undergoes no change of energy. Is there anything else? Yes, it may be that there has been net work W done on the surroundings. How can we tell? We can use the fact that the internal energy of the engine does not change in one complete cycle:

QUESTION Use the result that Q H/TH = Q L/TL (because the total entropy of the Universe doesn't change), to determine whether W is positive or negative.

Because Q H = (TH/TL)Q L > Q L, we have W = Q L − Q H < 0. The fact that W is negative means that our engine does net work on the surroundings in one cycle. Figure 13.36 outlines the basic scheme. In one cycle the net effect is that the engine takes in energy Q H from a hot source, does some work on something (for example, turns an electric generator), and exhausts the remaining energy Q L into a cold sink. The magnitude of the work done is |W| = Q H−Q L.

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Figure 13.36 Conservation of energy lets us determine the net work output.

The crucial issue is that the engine will not run in repeatable cycles without exhausting some energy to the low-temperature sink. You cannot convert all of the energy Q H into useful work; a portion of Q H goes into the low-temperature sink (Q L). We need the low-temperature block to allow us to begin the constant-temperature compression phase of the cycle in a way that avoids any energy transfer with significant temperature difference, which would increase the total entropy of the Universe. We have to pay for the high-temperature energy Q H that we use, and the loss of some of the input energy in the form of Q L is an unfortunate fact of life. We pay for coal or fuel oil or electricity to warm something up to the high temperature TH, from which we can draw energy to do work for us. For example, in an old railroad steam engine a fire maintained water at a high temperature to constitute the high-temperature source. In an automobile engine we burn gasoline to create a high temperature and push the pistons. In an electricity generating station we burn coal or fuel oil, or use nuclear fission reactions, to create a high-temperature source from which to drive the generators. We'd like to get our full money's worth (Q H), but we don't. We only get |W| = Q H − Q L, having “lost” some of the energy to a lowtemperature sink, which typically is our ordinary surroundings at around 20°C (293 K).

Efficiency This leads to the question, what fraction of the energy Q H that we pay for turns into useful work W? This fraction is called the “efficiency” of the engine:

QUESTION Given this definition of efficiency, prove the following result:

THE EFFICIENCY OF A REVERSIBLE ENGINE

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We have |W|/Q H = (Q H − Q L)/Q H = 1 − Q L/QH = 1 − TL/TH. This may seem surprising. The efficiency of a reversible engine depends solely on the ratio of the absolute temperatures of the source and sink. We emphasize that it doesn't matter how the engine is designed or what kind of materials it is made of. This is the highest efficiency we can ever achieve for converting thermal energy into useful work. 13.X.9 What is the efficiency of a reversible engine if the source is a large container of boiling water and the sink is a large block of melting ice? 13.X.10 An actual electricity generating plant is powered by a nuclear reactor, with a high temperature of 300°C and low temperature of 25°C (near room temperature). What is the efficiency, assuming that we can treat the processes as being reversible?

Answer

Answer

No Other Engine Can Be More Efficient We can show that no engine running between temperatures TH and TL can be more efficient than a reversible engine. The proof is a “proof by contradiction.” Suppose that an inventor claims to have invented some cleverly designed new kind of (cyclic) engine, with a higher efficiency than (1−TL/TH) when run between these same two temperatures. In that case, for a given Q H the new engine would exhaust less Q L than is exhausted from a reversible engine, and the entropy change of the Universe with this new engine would be negative instead of zero:

This would be a violation of the second law of thermodynamics, so it is impossible. We can be quite sure that the inventor's claims are not valid. Note carefully that the inventor's claims don't violate energy conservation. Nothing about energy conservation forbids converting 100% of the input energy Q H into work W in a cycle. The impossibility lies, rather, in the massive improbability of seeing the total entropy of the Universe decrease. We took great pains to make a reversible engine, for which the total entropy change of the Universe was zero. No other engine can be more efficient. In fact, any real engine will have some friction and some energy transfer across differing temperatures, in which case the entropy change of the Universe will be positive. So the second law of thermodynamics leads to the following, where “= 0” applies only to ideal, reversible engines, and “> 0” applies to real engines:

ENTROPY CHANGE OF THE UNIVERSE FOR REAL ENGINES

We should not underestimate the need for the ingenuity of inventors, however. The reversible engine made with an ideal gas cylinder is pretty useless for practical purposes. The challenge to ingenuity is to design engines that are practical, and this involves many kinds of engineering design decisions and trade-offs. However, no matter how ingenious the design, the second law of thermodynamics puts a rigid limitation on how efficient any engine can be.

QUESTION If ΔSUniverse > 0 in one cycle of an engine, show that the efficiency is less than the ideal efficiency (1 − TL/TH) obtained with

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a reversible engine.

Running the Engine Backward: A Refrigerator or Heat Pump We could run our engine backward, since we took care to make all aspects of the cycle reversible. Starting from our original starting point, we would disconnect from the hot block, perform a no-Q expansion to lower the temperature to that of the cold block, connect to the cold block, do a constanttemperature expansion, then disconnect from the cold block. Next we do a no-Q compression to raise the temperature to that of the hot block, connect to the hot block, and do a constant-temperature compression back to the original state. The net effect in one cycle is that the low-temperature block gives some energy Q L to the engine, the high-temperature block absorbs some energy Q H from the engine, and we do some (positive) work on the system (instead of the engine doing work on the surroundings). Here is the Energy Principle when we run this reversed engine:

QUESTION Use the result that Q H/TH = Q L/TL (which is still valid for our reversed engine, as you can convince yourself) to determine whether W is positive or negative.

Since Q H = (TH/TL)Q L > Q L, we have W = Q H − Q L > 0. This seems an odd kind of “engine”: it doesn't do any work for us—we have to do work on it. Is that of any use? Yes! At the cost of doing some work, we extract energy from a low-temperature block and exhaust it into a high-temperature block. This is a refrigerator (Figure 13.37).

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Figure 13.37 Run the engine backward and you have a refrigerator.

Consider how we keep food cold in an ordinary home refrigerator. A reversed engine maintains the inside of the refrigerator at a low temperature TL, while the room is at a higher temperature TH. Although the door and walls are heavily insulated, some energy does leak through from the room, and this energy, Q L, must be removed to keep the food at a constant low temperature TL. We exhaust an amount of energy Q H into the room at temperature TH. It takes an amount of work W to achieve this effect of moving energy out of the cold region and into the warm region. In the most favorable case (no increase in total entropy) we have the following two conditions stemming from entropy and energy considerations (the high-temperature exhaust energy must equal the sum of the low-temperature energy plus the work we put in to drive the cycle):

Here our view of “efficiency” shifts a bit. What we care about is how much work W we have to do to remove an amount of leakage energy Q L from the inside of the refrigerator.

QUESTION Show that the energy removal we get per amount of work done is as follows:

Isn't this result somewhat surprising? Are we getting something for nothing when Q L is bigger than the work W that we do? No, there's no violation of energy conservation. It's just that Q L plus W equals the output energy Q H, and we only have to add a modest amount of work to drive the energy “uphill.”

Heat Pump A related device is the “heat pump” used in some areas to warm homes by driving low-temperature energy “uphill” into a highertemperature house. Figure 13.38 shows a house whose interior is maintained at a temperature TH despite continual leakage Q H into the outside air. Energy Q L at low temperature TL is pumped out of the ground into the house, by the addition of some work W that

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we do in an engine called a heat pump. This is somewhat similar to the refrigerator situation, except that now what we care about is how much high-temperature energy Q H we get per amount of work W done by the heat pump.

Figure 13.38 Run the engine backward and you have a heat pump.

QUESTION Show that this ratio is given by the following formula for the heat pump:

Again, this feels like we're getting something for nothing, since Q H is greater than the work W that we do. After all, if you warm the house directly with gas or oil or electric energy, you have to pay for every joule of leakage, not some small fraction of the leakage. So why aren't heat pumps much more commonly used than they are? Basically because we've been calculating the best deal we can possibly obtain (the case of the reversible engine—that is, one whose operation leads to no increase in the total entropy of the Universe). Real engines running either forward or reversed don't attain the theoretical maximum performance, as we will see in the next sections. 13.X.11 Suppose that there is a leakage rate of 50 watts through the insulation into a refrigerator, which we maintain Answer at 3°C. What is the minimum electric power required to continually remove this leakage energy, to maintain the low temperature? In that case, what is the rate at which energy is exhausted into the room? Room temperature is about 20°C. 13.X.12 Suppose that the temperature underground from where we draw low-temperature energy for a heat pump is about 5°C, and we want to keep the house at a temperature of 20°C. How many joules of work must our heat pump supply for every joule of leakage of energy there is out of the house?

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Answer

*WHY DON'T WE ATTAIN THE THEORETICAL EFFICIENCY? In Exercise 13.X.10 you calculated the reversible-engine efficiency of an actual nuclear-powered electricity generating plant to be 48%, but the observed efficiency of this plant is only 30%. Real engines do not attain the efficiency predicted by the second law of thermodynamics for ideal reversible engines. In fact, the real efficiency is often only about half of the theoretical efficiency. The reason for less than optimum performance in real engines is partly due to mechanical friction, but this effect can be minimized by good design and proper lubrication. The main limitation on performance comes from the practical necessity of incorporating energy transfers between parts of the system that are at significantly different temperatures, which leads to sizable increases in the total entropy of the Universe, and to much reduced efficiency. The problem is speed. As we saw earlier in the chapter, the rate of energy transfer

in joules per second between two objects (such

as the hot or cold block and the engine) at temperatures TH and TL connected by some conducting material of length L and crosssectional area A, with thermal conductivity σ is this:

QUESTION What is the rate of energy transfer in a reversible engine when in contact with the hot block or the cold block?

Alas, a reversible engine is totally impractical when it comes to getting anything done in a finite amount of time, because the rate of energy transfer is zero. If an engine is perfectly reversible, a cycle takes an infinite amount of time! Consider the portion of an engine's cycle where the working substance is in thermal contact with the high-temperature source. In order to carry out the expansion quickly, there must be a high rate of energy transfer from the source into the working substance. That means we need a large contact area A, a short distance L, and a high thermal conductivity σ. Most significant of all, the temperature of the source (TH) must be considerably higher than the temperature of the engine, leading to irreversibility and increase of the entropy of the Universe.

The Efficiency of a Nonreversible Engine It is actually possible to calculate the effect of such nonreversible energy transfer, as was pointed out in an article titled “Efficiency of a Carnot engine at maximum power output,” by F. L. Curzon and B. Ahlborn, American Journal of Physics, volume 43, pages 22–24 (January 1975). Suppose that when the engine is in contact with the high-temperature block at temperature TH, the engine is at a lower temperature joules per second, where b is a constant that T1, so that there is a finite energy transfer rate into the engine of lumps together the factors of thermal conductivity, cross-sectional area, and length (σA/L). Similarly, assume that when the engine is in contact with the low-temperature block at temperature TL, the engine is at a higher temperature T2, so that there is a finite energy transfer rate into the working substance of

joules per second. We're making the factor b be the same for

both the high-temperature and the lowtemperature contacts. This simplifies the algebra, and it turns out that using different b factors doesn't affect the final result anyway. Now the energy flow diagram looks like Figure 13.39, with an ideal reversible engine running between the temperatures T1 and T2, but with irreversible finite-rate energy transfer between this reversible engine and the source and sink at temperatures TH and TL.

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Figure 13.39 An engine that runs at a nonzero rate.

Energy enters the engine from the high-temperature source at a rate low-temperature sink at a rate

, work is done at a rate W˙, and energy is dumped into the

. All of these quantities are measured in watts (joules per second). From energy conservation we

have

Curzon and Ahlborn pointed out that although this engine will necessarily be less efficient than a reversible engine (due to the irreversible energy transfers), it might be useful to ask the question, “What is the maximum output power

per input power

?”

This question comes down to the question of what value of b will maximize the output power. This is essentially an engineering design question, yet there is an unusually simple answer to this question about the maximum output power.

There Must Be a Maximum Power Output By considering two extreme designs (Figure 13.40) it is easy to see that there must be a maximum possible power output (per power input). One extreme set of operating conditions is the reversible engine we've already considered, in which all the energy flow rates are zero, including the output power .

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Figure 13.40 Two extreme designs where there is zero power output. Another extreme is to arrange conditions in such a way that T1 ≈ T2, so that we have essentially connected the source and sink by a conducting bar. In that case the input power

is equal to

, and the useful output power

is again zero.

Somewhere between these two extremes, for both of which there is zero output power, we expect to find a maximum output power. We now look for the conditions that maximize the output power. The search will be carried out by expressing the output power in terms of T1, the higher of the two temperatures to which the engine is directly exposed in Figure 13.39. Then we will vary T1 (by adjusting b in Figure 13.39) until we maximize the power output. The details are given in a derivation at the end of the chapter. We obtain the following surprisingly simple result:

THE EFFICIENCY OF A MAXIMUM-POWER ENGINE

This formula gives us the upper bound on the fraction of the thermal input power that can end up as output mechanical power if we maximize the output power. The formula looks quite similar to the reversible-engine zero-power efficiency, but the square root makes a big difference.

Comparison with the Real World Curzon and Ahlborn give an example of a plant powered by a nuclear reactor, with high temperature 300°C and low temperature 25°C. Using the formula given above, the calculated efficiency is 28%, and the actual observed efficiency is 30% (see Exercise 13.X.13 at the end of this section). This confirms the estimate that Curzon and Ahlborn made that the main source of irreversibility is in finite-temperature-difference energy transfer, and

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power plants will normally be run in such a way as to maximize output power. There is an economic issue here. Suppose that you are responsible for building and operating an electricity generating plant, and you must supply one megawatt of power (one million joules per second). If the operating expense of buying fuel is extremely high but capital investment in generators is inexpensive, it makes sense to build lots of generators and run them very slowly, nearly reversibly, to make the required megawatt of electric power. If the operating expense of buying fuel is relatively low but capital investment in generators is expensive, it makes sense to build few generators and go for maximum power, even though the fuel is not used very efficiently in producing the megawatt of electric power. Similar issues of maximum efficiency apply to refrigerators and heat pumps. Energy leaks into a refrigerator at some rate (joules per second) and must be removed by “pushing it thermally uphill” to a higher temperature, the temperature of the room. However, in order to have a nonzero rate of energy flow from the food to the engine, the engine must reach an even lower temperature than the food, with entropy-producing energy transfer from the food into the engine. Also, in order for there to be a nonzero rate of energy flow from the engine to the room air, the engine must reach an even higher temperature than the air, with entropy-producing energy transfer from the engine into the air. Schematically the situation looks like Figure 13.41.

Figure 13.41 A refrigerator with nonzero cooling rate.

Often a refrigerator or freezer has exposed coils (a “heat exchanger”) where the energy transfer occurs between the working substance circulating in the coils and the air. If you touch the coils, you find that they are indeed much hotter than room temperature, in order to drive a sufficiently high rate of energy transfer into the air. Notice not only that the heat exchanger must be hotter than the air, but also that it is a rather large and costly device because it has to have a large surface area to get a large rate of energy transfer. In the case of heat pumps, where the heat pump picks up low-temperature energy underground, the heat pump must reach an even lower temperature in order to get a nonzero rate of energy flow from the ground into the engine. Also, in order for there to be a nonzero rate of energy flow into the house, the engine must reach an even higher temperature than the air, with entropyproducing energy transfer from the engine into the air. The radiators (heat exchangers) in the house must be considerably hotter than the air. There is considerable expense in all the metal in the ground and in the house that enables adequate energy transfer rate. Schematically the situation looks like Figure 13.42.

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Figure 13.42 A heat pump with nonzero energy transfer rate due to a temperature difference.

So although a heat pump does have a theoretical advantage in warming a house in part from energy in the cold ground, the expense of the heat exchangers and the problems of going to rather low temperatures in the ground are practical limitations, especially in very cold climates. Heat pumps are more useful in climates where the winters are not too severe. 13.X.13 Curzon and Ahlborn give an example of a plant powered by a nuclear reactor, with high temperature 300°C Answer and low temperature 25°C. In Exercise 13.X.10 you calculated the reversible-engine zero-power efficiency to be 48%. Now calculate the non-reversible-engine maximumpower efficiency for this power plant and compare it with the observed efficiency, which is 30%.

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*APPLICATION: A RANDOM WALK If you would find it useful to study one more application of the basic statistical concepts underlying our analysis of a gas, here is an interesting one. If we could watch one special molecule wandering around in the air, colliding frequently with air molecules, its path would look something like Figure 13.43. This kind of motion is called a “random walk.” It is somewhat surprising to find that despite the random nature of this motion we can calculate something significant about the motion, using simple statistical reasoning.

Figure 13.43 A 2D random walk, generated by a computer program, using random numbers and many steps.

One Dimension For simplicity, first we'll consider just the x component of the motion. Pick the origin of the x axis to be at the original position of the special molecule. The first thing that happens is that the special molecule moves until it collides with an air molecule. We call this first x component of the displacement Δx 1. This component of the displacement may be to the right (+x direction) or to the left (−x direction). As a result of the collision, the special molecule may change direction, and change speed. The next displacement before another collision we call Δx 2, and so on. After N collisions, the net displacement Δx away from the origin is After N collisions, what is the average (most probable) position of the special molecule? On average, it is just as likely that it has moved to the right as moved to the left, so the average x component of displacement ought to be zero. It is easy to see that this will be the case, by taking the average value: Each of the individual x displacements are equally likely to be to the left or the right, so the average value of the nth x displacement (for n = 1, 2, 3, … N) is zero. Therefore the average value of the net x component of displacement is also zero. On average, after N collisions the special molecule ends up to the left of the origin as often as it ends up to the right. However, we can ask the question, “On average, how far away from the origin (along the x axis) does the special molecule get, no matter whether it ends up to the left or the right?” A good indicator of this distance is the average value of the square of the net x displacement, (Δx)2, because that's always a positive number. We can calculate this average. Since

, we have

In this square of the net x displacement, there are two kinds of terms: squares of individual x displacements such as (Δx 2)2 and “cross terms” like 2Δx 1Δx 3. Given the random nature of the process, we expect that one-fourth of the time both Δx 1 and Δx 3 are positive (product is positive), one-fourth of the time they're both negative (product is positive), one-fourth of the time Δx 1 is positive with Δx 3 negative (product is negative), and one-fourth of the time Δx 1 is negative with Δx 3 positive (product is negative).

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Therefore, the average value of each cross term

is zero.

As for the other terms, the squares of individual x displacements such as (Δx 2)2, the average value of each of these terms is some number dx 2, related to d2, the square of the mean free path that we discussed in Section 13.3.

QUESTION Count up how many of these square terms there are for N collisions and calculate the average value of the square of the net displacement:

There are N of these terms, and the square root of this quantity is the “root-mean-square” or “rms” value of the net displacement:

We can also write this formula in terms of time. If we let v be the average speed of the special molecule between collisions and we let T be the average time between collisions, we have v = d/T. Also, the total time t for the N collisions is t = NT.

QUESTION Rewrite the formula for the rms displacement in terms of just dx , v, and t (that is, eliminate N and T): Δx rms =?

We find that

. This is a somewhat curious result. For ordinary motion at constant

speed, displacement increases proportional to time: double the time, double the displacement. In contrast, the rms displacement in a random walk grows with the square root of the time: on average, the rms displacement doubles when the time quadruples. This calculation was done for the x component of the motion, but we can generalize our result to real three-dimensional motion in a gas. Figure 13.44 shows a three-dimensional displacement involving Δx, Δy, and Δz.

Figure 13.44 Pythagorean theorem in three dimensions.

You can see in Figure 13.44 a three-dimensional version of the Pythagorean theorem for triangles. We can assume that the motion in each dimension is independent of the motions in the other two dimensions, and we have the following result:

(Here, dx is the component in the x direction of the three-dimensional mean free path d.)

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Writing the rms displacement in terms of the average speed v, the time t, and the mean free path d, we have

This result is somewhat unusual, because it predicts (correctly, it turns out) that for this random process the displacement from the starting location is proportional to the square root of the time rather than to the time. 13.X.14 As you have calculated, the average speed of an air molecule at room temperature is about 500 m/s, and we saw in Section 13.3 that the mean free path is about 7 × 10−8 m. What is Δr rms for an air molecule after one second?

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Answer

*DERIVATION: MAXIMUM-POWER EFFICIENCY In this appendix we find the conditions under which the power output is maximized for an engine that connects to high- and lowtemperature reservoirs through finite-temperature-difference connections.

Expressing the Power in Terms of T 1 We will express the power output in terms of T1, the higher temperature to which the reversible engine is subjected. We assume that the engine, the part that runs between the temperatures T1 and T2, can be considered to be a reversible engine because there are no different-temperature energy transfers within it, and we're assuming that we can make mechanical friction negligibly small.

QUESTION In terms of the temperatures T1 and T2, what does the second law of thermodynamics say about the relationship between and

?

Since we have a reversible engine, we know that

. Using this relationship, we can write a formula for the

energy transfer into the lowtemperature sink in two different ways:

Therefore we have

With a bit of manipulation we can solve this equation for T2 in terms of T1:

For the inner, reversible engine, the input power is of course

, and using the result for the efficiency of a

reversible engine, we have

Using our solution for T2 in terms of T1, this becomes

Varying the Thermal Conduction to Maximize the Power We consider the source and sink temperatures TH and TL to be fixed, so the only temperature that is unspecified in this equation is T1, which we will vary until we maximize the output power

. Physically, this corresponds to varying the parameter b that

by differentiating it with respect to T1, then set this derivative equal to zero (corresponding to a maximum, where the slope is zero): determines the energy transfer rate. As usual, we maximize

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Solving for T1 and simplifying, we get a quadratic equation:

There are two solutions to the quadratic equation:

Only the “+” sign makes physical sense in this situation, because if TL were nearly as large as TH, the solution with the “−” sign would make T1 nearly 0 instead of lying between TL and TH. We conclude that if

, the output power

will be maximized. Since

T1 determines the value of b (= σA/L) that will make T1 come out right to maximize the power output for a given Finally, we can plug our value for T1 back into the expression for

, our value for .

to find the efficiency at maximum power:

Here is the final result, where we write “≤” because we have calculated an upper bound on the efficiency (having taken into account issues of energy transfer rate due to a temperature difference but not mechanical friction):

THE EFFICIENCY OF A MAXIMUM-POWER ENGINE

This gives us the fraction of the thermal input power that ends up as output mechanical power if we maximize the output power. The formula looks quite similar to the reversible-engine zero-power efficiency, but the square root makes a big difference. For example, in the case where the absolute temperatures in kelvins of TL and TH are not very different, it's easy to see that the maximum-power engine has an efficiency only one-half that of the reversible engine:

In the case of the nuclear power station running between 25°C and 300°C, we saw that a reversible engine would have had an efficiency of 48%, but the efficiency of a maximum-power engine would be only 28% (the actual power plant efficiency was 30%).

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SUMMARY Mean free path d: n[π(R + r)2(d)] ≈ 1 Root-mean-square speed:

Work done on a gas: First law of thermodynamics: ΔE sys = W + Q Molecular specific heat capacity at constant volume C V : Q = NCV ΔT Molecular specific heat capacity at constant pressure C P : Q = NCP ΔT Number of gas molecules hitting an area A per second

(3-D; various speeds)

where n = N/V (number of molecules per cubic meter) For a multiatom gas molecule,

for a monatomic gas (He, etc.) for other gases (N2, etc.) C P = C V + k B (molecular specific heat capacity) C P = C V + R (molar specific heat capacity) In a constant-temperature (isothermal) compression,

In a no-Q (adiabatic) compression, PV γ = constant, where γ = C P /C V , and also

k B = 1.38 × 10−23 J/K R = (6.02 × 1023)k B = 8.3 J/K At “standard temperature and pressure” (STP, which means a temperature of 0°Celsius or 273 Kelvin, and a pressure of 1

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atmosphere), one mole of a gas occupies 22.4 liters (a liter is 1000 cubic centimeters). Sea-level pressure (1 atm) is 1 × 105 N/m2. Rate of energy transfer due to a temperature difference:

For a cyclic engine running between high temperature TH and low temperature TL we have: Entropy change of the universe for real engines

The efficiency of a real engine

The “=” sign applies only if the engine is reversible (extremely slow processes, no friction). A heat engine run in reverse is a refrigerator (or a heat pump). If we maximize power output, we have the efficiency of a maximum-power engine

Random walk root-mean-square displacement:

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EXERCISES AND PROBLEMS Sections 13.1 13.X.15 Gas leaks at a rate L (in molecules per second) through a small circular hole. If the density of the gas is doubled, the Kelvin temperature is doubled, and the radius of the hole is doubled, what is the new leak rate? 13.X.16 Suppose that we make a circular hole 4 millimeters in diameter in a balloon. Calculate the initial rate at which chlorine escapes through the hole (at 0 C), in number of chlorine atoms leaving the balloon per second. 13.X.17 How does the mean free path of an atom in a gas change if the temperature is increased, with the volume kept constant? 13.X.18 What is the approximate time between collisions for one particular air molecule? 13.P.19 In an example problem in Section 13.2 we considered leakage from a flexible party balloon. If the leakage is from a rigid container (a metal storage tank, for example), the number of atoms per cubic meter, n, will decrease with time t. If the total volume of the tank is V, there are N = nV atoms in the tank at any instant, so we can write the following “differential” equation (that is, an equation that involves derivatives):

This says, “the rate of change of the number of atoms in the tank is equal to the (negative) of the rate at which atoms are leaving the tank.” (a) Show that the differential equation is satisfied if the number density

where t is the time elapsed since the hole was made. Just plug this function of n, and its derivative, into the equation and show that the two sides of the equation are equal for all values of the time t. Also show that the initial particle density n is equal to ninitial. (b) Despite the fancy math, this solution is really only approximate, because the average speed isn't a constant but is decreasing. Suppose, however, that we use a heater to keep the container and the gas at a nearly constant temperature, so that the average speed does remain nearly constant. Suppose that the (rigid) container is again a sphere 30 cm in diameter, with a circular hole 1 millimeter in diameter. About how long would it take for most of the helium to leak out? Explain your choice of what you mean by “most.” 13.P.20 Natural uranium ore consists mostly of the isotope U-238 (92 protons and 146 neutrons), but 0.7% of the ore consists of the isotope U-235 (92 protons and 143 neutrons). Because only U-235 fissions in a reactor, industrial processes are used to enrich the uranium by enhancing the U-235 content. One of the enrichment methods is “gaseous diffusion.” The gas UF6, uranium hexafluoride, is manufactured from supplies of natural uranium and fluorine (each of the six fluorine atoms has 9 protons and 10 neutrons). A container is filled with UF6 gas. There are tiny holes in the container, and gas molecules leak through these holes into an adjoining container, where pumps sweep out the leaked gas. (a) Explain why the gas that initially leaks into the second container has a slightly higher fraction of U-235 than is found in natural uranium. (b) Estimate roughly the practical change in the concentration of U-235 that can be achieved in this single-stage separation process. Explain what approximations or simplifying assumptions you have made to obtain your estimate. (c) A typical nuclear reactor requires uranium that has been enriched to the point where about 3% of the uranium is U235. Estimate roughly the number of stages of gaseous diffusion required (that is, the number of times the gas must be allowed to leak from one container into another). Note that the effects are multiplicative.

This is why a practical gaseous diffusion plant has a large number of stages, each operating at high pressure, which makes this an expensive process. The first large gaseous diffusion plant was constructed during World War II at Oak Ridge,

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Tennessee, and used inexpensive Tennessee Valley Authority electricity. 13.P.21 You are on a spacecraft measuring 8 m by 3 m by 3 m when it is struck by a piece of space junk, leaving a circular hole of radius 4 mm, unfortunately in a place that can be reached only by making a time-consuming spacewalk. About how much time do you have to patch the leak? Explain what approximations you make in assessing the seriousness of the situation.

Section 13.4 13.P.22 A rigid, thermally insulated container with a volume of 22.4 liters is filled with one mole of helium gas (4 grams per mole) at a temperature of 0 Celsius (273 K). The container is sitting in a room, surrounded by air at STP. (a) Calculate the pressure inside the container in N/m2. (b) Calculate the root-mean-square average speed of the helium atoms. (c) Now open a tiny square hole in the container, with area 1 × 10−8 m2 (the hole is 0.1 mm on a side). After 5 seconds, how many helium atoms have left the container? (d) Air molecules from the room enter the container through the hole during these 5 seconds. Which is greater, the number of air molecules that enter the container or the number of helium atoms that leave the container? Explain briefly. (e) Does the pressure inside the container increase slightly, stay the same, or decrease slightly during these 5 seconds? Explain briefly. If you have to make any simplifying assumptions, state them clearly. 13.P.23 A roughly spherical meteor made mainly of iron (density about 8 grams per cubic centimeter) is hurtling downward through the air at low altitude. At an instant when its speed is 1 × 104 m/s, calculate the approximate rate of change of the meteor's speed. Do the analysis for two different meteors—one with a radius of 10 meters and one with a radius of 100 meters. Start from fundamental principles. Do not try to use some existing formula that applies to a very different situation. Follow the kind of reasoning used in this chapter, applied to the new situation, rather than trying to use the results of this chapter. A major difference from our earlier analyses is that the meteor is traveling much faster than the average thermal speed of the air molecules, so it is a good approximation to consider the air molecules below the meteor to be essentially at rest, and to assume that no air molecules manage to catch up with the meteor and hit it from behind. The meteor drills a temporary hole in the atmosphere, a vacuum, that gets filled explosively by air rushing in after the meteor has passed. There is good evidence that a very large meteor, perhaps 10 kilometers in diameter, hit the Earth near the Yucatan Peninsula in southern Mexico 65 million years ago and caused so much damage that the dinosaurs became extinct. See the excellent account in T. rex and the Crater of Doom, by Walter Alvarez (Princeton University Press, 1997). Alvarez is the geologist who made the first discoveries leading to our current understanding of this cataclysmic event.

Section 13.5 13.X.24 If you expand the volume of a gas containing N molecules to twice its original volume, while maintaining a constant temperature, how much energy transfer due to a temperature difference is there from the surroundings? 13.X.25 If you compress a volume of helium containing N atoms to half the original volume in a well-insulated cylinder, what is the ratio of the final pressure to the original pressure? 13.P.26 Two moles of nitrogen is compressed (by piling lots of sand on the piston) to a volume of 11 liters at room temperature (293 K). The cylinder is placed on an electric heating element whose temperature is maintained at 293.001 K. A quasistatic expansion is carried out at constant temperature by very slowly removing grains of sand from the top of the piston, with the temperature of the gas staying constant at 293 K. (You must assume that there is no energy transfer due to a temperature difference from the gas to the surrounding air, and no friction in the motion of the piston, all of which is pretty unrealistic in the real world! Nevertheless there are processes that can be approximated by a constant-temperature expansion. This problem is an idealization of a real process.) (a) When the volume is 20 liters, how much energy transfer Q has gone from the heating element into the gas? (b) How much work W has been done on the piston by the gas? (c) How much has the energy of the gas changed?

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13.P.27 Ahorizontal cylinder 10 cm in diameter contains helium gas at room temperature and atmospheric pressure. A piston keeps the gas inside a region of the cylinder 20 cm long. (a) If you quickly pull the piston outward a distance of 12 cm, what is the approximate temperature of the helium immediately afterward? What approximations did you make? (b) How much work did you do, including sign? (Note that you need to consider the outside air as well as the inside helium in calculating the amount of work you do.) (c) Immediately after the pull, what force must you exert on the piston to hold it in position (with the helium enclosed in a volume that is 20 + 12 = 32 cm long)? (d) You wait while the helium slowly returns to room temperature, maintaining the piston at its current location. After this wait, what force must you exert to hold the piston in position? (e) Next, you very slowly allow the piston to move back into the cylinder, stopping when the region of helium gas is 25 cm long. What force must you exert to hold the piston at this position? What approximations did you make? 13.P.28 Atmospheric pressure at sea level is about 1 × 105 N/m2, which is about 15 pounds per square inch (psi). A bicycle tire typically is pumped up to 50 psi above atmospheric pressure (psi “gauge”), for an actual pressure of about 65 psi. In rapidly pumping up a bicycle tire, starting from atmospheric pressure, about how high does the temperature of the air rise? Explain what approximations and simplifying assumptions you make.

Sections 13.6 13.X.29 An aluminum bar 30 cm long and 3 cm by 4 cm on its sides is connected between two large metal blocks at temperatures of 135°C and 20°C, and energy is transferred from the hotter block to the cooler block at a rate

1. If instead the two blocks

were connected by an aluminum bar that is 20 cm long and 3 cm by 2 cm on its sides, what would be the energy transfer rate? 13.X.30 A (nearly) reversible engine is used to melt ice as well as do some useful work. If the engine does 1000 joules of work and dumps 400 joules into the ice, what is the temperature of the high-temperature source? 13.X.31 For engines that have high and low temperatures not very different from each other, show that the efficiency of a maximum-power engine is about half of the efficiency of a reversible engine. 13.P.32 In one cycle of a reversible engine running between a high-temperature source and a low-temperature sink, a consequence of the second law of thermodynamics is that Q H/TH = Q L/TL. This result is independent of what kind of material the engine contains. It is instructive to check this general result for a specific model where we can calculate everything explicitly. Consider a reversible cycle of an ideal gas of N molecules, starting at high temperature TH and volume V 1. (a) Perform a constant-temperature (isothermal) expansion to volume V 2, and calculate the associated energy transfer Q H into the gas. (b) Next perform a no-Q (adiabatic) expansion to volume V 3 and temperature TL. (c) Next perform a constant-temperature (isothermal) compression to volume V4, and calculate the associated energy transfer Q L out of the gas. (d) Finally perform a no-Q (adiabatic) compression back to the starting state, temperature TH and volume V 1.

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Electric Field

KEY IDEAS A charged particle makes an electric field at every location in space (except its own location). The electric field due to one particle affects other charged particles. The electric force on a charged particle is proportional to the net electric field at the location of that particle. The Superposition Principle:

▪ The net electric field at any location is the vector sum of the individual electric fields of all charged particles at other locations (called the “source charges”).

▪ The field due to one charged particle is not changed by the presence of other charged particles. An electric dipole consists of two particles with charges equal in magnitude and opposite in sign, separated by a short distance. Changes in electric fields travel at the speed of light (“retardation”).

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NEW CONCEPTS Two important new ideas will form the core of our study of electric and magnetic interactions. The first is the concept of electric and magnetic fields. This concept is more abstract than the concept of force, which we used extensively in our study of modern mechanics. The reason we want to incorporate the idea of “field” into our models of the world is that this concept turns out to be a very powerful one, which allows us to explain and predict important phenomena that would otherwise be inaccessible to us. The second important idea is a more sophisticated and complex model of matter. In our previous study of mechanics and thermal physics it was usually adequate to model a solid as an array of electrically neutral microscopic masses (atoms) connected by springs (chemical bonds). As we consider electric and magnetic interactions in more depth, we will find that we need to consider the individual charged particles—electrons and nuclei—that make up ordinary matter. The material in this chapter lays the foundation for all succeeding chapters, so it is worth taking time to understand it thoroughly. Do the in-line exercises! It is necessary to practice using basic concepts before getting to complex and challenging problems.

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ELECTRIC CHARGE AND FORCE In this section we briefly review concepts familiar to you from your previous studies.

Point Charges There are two kinds of electric charge, which are called positive and negative. Particles with like charges repel each other (two positive or two negative particles); particles with unlike charges (positive and negative) attract each other. By “point particle,” we mean an object whose radius is very small compared to the distance between it and all other objects of interest, so we can treat the object as if all its charge and mass were concentrated at a single mathematical point. Small particles such as protons and electrons can almost always be considered to be point particles.

The Coulomb Force Law for Point Particles The electromagnetic interaction is one of the four fundamental physical interactions (see Chapter 3, The Fundamental Interactions). The electric force law, called Coulomb's law, describes the magnitude of the electric force between two pointlike electrically charged particles:

where Q 1 and Q 2 are the magnitudes of the electric charge of objects 1 and 2, and r is the distance between the objects. As indicated in Figure 14.1: The electric force acts along a line between two pointlike objects. Like charges repel; unlike charges attract. Two charged objects interact even if they are some distance apart.

Figure 14.1 Two protons repel each other; two electrons repel each other; a proton and an electron attract each other.

Units and Constants The SI unit of electric charge is the coulomb, abbreviated “C.” The charge of one proton is e is often used to represent this amount of positive charge:

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in SI units. The constant

. An electron has a charge of

.

The constant has the value . We write this constant in this way, instead of using the letter k, for two reasons. Since the letter k is often used for other quantities, we avoid confusion by not using it here. Also, the constant will appear by itself in important situations.

Charged Particles There are many microscopic particles, some of which are electrically charged, and hence interact with each other through the electric interaction. The characteristics of some charged particles are shown in the table below. This is not an exhaustive list; you may also have learned about other charged particles, such as the W and Δ particles, and other particles that are short-lived and not commonly encountered in everyday circumstances. Positrons and antiprotons are antimatter. A positron is an antielectron; if a positron and an electron encounter each other they will annihilate, releasing all their energy as highenergy photons. Antimatter is therefore not found in ordinary matter. Particle

Mass

Charge

Radius

electron

? (too small to measure)

positron

?

proton antiproton muon pion

or

?

or

Ordinary matter is composed of protons, electrons, and neutrons (which have about the same size and mass as protons but are uncharged). However, some other charged particles do play a role in everyday processes, as we will see, for example, in our study of sparks in a later chapter.

Size and Structure of Atoms As you have learned in chemistry or from other sources, matter is made of tiny atoms. In a solid metal, atoms are arranged in a regular three-dimensional array, called a lattice (Figure 14.2). A cubic centimeter of solid metal in which atoms are packed right next to each other contains around atoms, which is an astronomically large number.

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Figure 14.2 A metal lattice. The radius of a single atom is on the order of there are on the order of

m, and in a cube that is 1 cm on a side,

atoms!

A neutral atom has equal numbers of protons and electrons. The protons and neutrons are all found in the nucleus at the center of the atom. The electrons are spread out in a cloud surrounding the nucleus. Each atom consists of a cloud of electrons continually in motion around a central “nucleus” made of protons and neutrons. If we imagine taking a “snapshot” of an iron atom, with a nucleus of 26 protons and 30 neutrons surrounded by 26 moving electrons, it might look something like Figure 14.3. In this figure the nucleus is hardly visible, because it is much smaller than the electron cloud, whose radius is on the order of .

Figure 14.3 The electron cloud of an iron atom. The radius of the electron cloud is approximately

meter. On this

scale the tiny nucleus, located at the center of the cloud, would not actually be visible.

The nucleus of the iron atom, depicted in Figure 14.4, has a radius of roughly meter, about 25,000 times smaller than the tiny electron cloud. If an iron atom were the size of a football field, the nucleus would have a radius of only 4 millimeters! Yet almost all of the mass of an atom is in the nucleus, because the mass of a proton or neutron is about 2000 times larger than the mass of an electron.

Figure 14.4 The nucleus of a iron atom contains 26 protons and 30 neutrons. Its radius is approximately

14.X.1 What is the approximate radius of the electron cloud of a typical atom? 14.X.2 Which of the following charged particles are constituents of ordinary matter? Protons, positrons, electrons, antiprotons, muons

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meter.

Answer

Answer

THE CONCEPT OF “ELECTRIC FIELD” Consider the following thought experiment. Having evacuated the air from the room (to avoid collisions with air molecules), you hold a proton in front of you and release it. There are no other objects nearby. You observe that the proton begins to move downward, picking up speed (accelerating) at the rate of each second (Figure 14.5). Recall that at speeds much less than the speed of light:

QUESTION What do you think is responsible for this change in the velocity of the proton? You probably inferred that the gravitational interaction of the Earth and the proton caused the downward acceleration of the proton. This is a reasonable explanation for this observation.

Figure 14.5 Acceleration of a proton.

Now suppose that at a later time, you release another proton in the same location. This time you observe that the proton begins to move to the right, picking up speed at a rate of (Figure 14.6).

QUESTION What might be responsible for this change in the velocity of the proton? This acceleration cannot be due to a gravitational interaction of the proton with the Earth, since the magnitude of the effect is too large, and the direction is not appropriate. Could it be due to a gravitational interaction with a nearby black hole? No, because if there were a black hole very nearby we would not be here to observe anything! Could it be an interaction via the strong (nuclear) force? No, because the strong force is a very short range force, and there are no other objects near enough.

Figure 14.6 Acceleration of another proton at a later time.

It is, however, plausible that the interaction causing the acceleration of the proton could be an electric interaction, since electric interactions can have large effects and can occur over rather large distances.

QUESTION What charged objects might be responsible for this interaction, and where might they be?

There are many possible configurations of charged particles that might produce the observed effect. As indicated in Figure 14.7, there might be positively charged objects to your left. Alternatively, there could be negative charges to your right. Perhaps there are both— you can't draw a definite conclusion from a single observation. There are many possible arrangements of charges in space that could produce the observed effect.

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Figure 14.7 Three possible arrangements of charged particles that might be responsible for the observed high acceleration of a proton.

(If, however, you made several observations of the proton over some time period, you might note a change in the proton's acceleration. For example, if you noticed that the acceleration of the proton increased as it moved to the right, you might suspect that there was a negative charge to your right, which would have an increasingly large effect on the proton as it moved closer to the charge.)

QUESTION On the basis of your observations so far, can you predict what you would observe if you released an electron instead of a proton at the same location (Figure 14.8)?

Figure 14.8 A (negatively charged) electron.

An electron would accelerate to the left rather than to the right, since it has a negative charge. Its acceleration would be greater than , because the mass of the electron is much less than the mass of the proton.

QUESTION What would happen if you placed an alpha particle (a helium nucleus, with charge 14.9)?

) at the observation location (Figure

Figure 14.9 An alpha particle with two (positively charged) protons and two (uncharged) neutrons.

Since the alpha particle has a positive charge, it would accelerate to the right as the proton did. However, because the charge of the alpha particle is twice the charge of the proton, and its mass is about four times that of a proton, the magnitude of its acceleration would be one-half of .

QUESTION If you released a neutron at the observation location, what would you expect to observe (Figure 14.10)?

Figure 14.10 An (uncharged) neutron.

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Since the neutron has no electric charge, it would experience no electric force, and should simply accelerate toward the Earth at .

QUESTION Finally, suppose we do not put any particle at the observation location. Is there anything there? Since we know that if we were to put a charged particle at that location, it would experience a force, it seems that in a certain sense there is something there, waiting for a charged particle to interact with. This “virtual force” is called “electric field.” The electric field created by a charge is present throughout space at all times, whether or not there is another charge around to feel its effects. The electric field created by a charge penetrates through matter. The field permeates the neighboring space, lying in wait to affect anything brought into its web of interaction. We define the electric field at a location by the following equation:

DEFINITION OF ELECTRIC FIELD

This equation says that the force on particle 2 is determined by the charge of particle 2 and by the electric field other charged particles in the vicinity, as shown in Figure 14.11.

Figure 14.11 A particle with charge q 2 experiences a force

because of its interaction with the electric field

made by all

produced

by all other charged particles in the vicinity.

How might we measure the electric field at a given location in space? We can't see it, but we can measure it indirectly. If we place a charge at that location in space, we can measure the force on the charge due to its interaction with the electric field at that location. We can determine the magnitude and direction of by measuring a force on a known charge q:

Electric field therefore has units of newtons per coulomb. Note that it doesn't matter whether the affected charge q is positive or negative. If points north, a positive charge would experience a force to the north, while a negative charge would experience a pointing north. Though we will force to the south (Figure 14.12); dividing this force by a negative q would still yield a vector usually write

for the value of the electric field at a particular location and time, this is really shorthand for

the electric field has a value at all locations and times.

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, since

Figure 14.12 A positive charge experiences a force in the direction of the electric field at its location, while a negative charge experiences a force in the opposite direction.

Drawing Fields: Tail of Arrow at Observation Location In diagrams we will use arrows to represent the electric field (a vector) at a specific observation location. By convention, such an arrow is always drawn with its tail at the location where the field is measured (the “observation location”). The length of the arrow is proportional to the magnitude of the field.

EXAMPLE Electric Force and Electric Field The charge of an alpha particle (a helium nucleus, consisting of two protons and two neutrons) is . An alpha particle at a particular location experiences a force of

.

What is the electric field at that location? If the alpha particle were removed and an electron were placed at that location, what force would the electron experience?

Solution To find the field we divide the force on the alpha particle by the charge of the alpha particle:

To find the force on the electron, we multiply the electric field by the charge on the electron:

FURTHER DISCUSSION Since the electron's charge is negative, the force on the electron is in a direction opposite to the direction of the electric field. Note

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that we did not need to know the source of the electric field in order to calculate this force. 14.X.3 What force would a proton experience if placed at the same location (see preceding example)?

Answer

14.X.4 What force would a neutron at the same location experience?

Answer

No “Self-Force” It is important to note that a point charge is not affected by its own electric field. A point charge does not exert a force on itself! In the next section we will see that mathematically this is reassuring, because at the location of a point charge its own electric field would be infinite

Physically, this makes sense too, since after all, the charge can't start itself moving, nor is there any way

to decide in what direction it should go. When we use the electric field concept with point charges we always talk about a charge q1 (the source charge) making a field . and a different charge q2 in a different place being affected by that field with a force

,

The Physical Concept of “Field” The word “field” has a special meaning in mathematical physics. A field is a physical quantity that has a value at every location in space. Its value at every location can be a scalar or a vector. For example, the temperature in a room is a scalar field. At every location in the room, the temperature has a value, which we could write as , or as if it were changing with time. The air flow in the room is a vector field. At every location in the room, air flows in a particular direction with a particular speed. Electric field is a vector field; at every location in space surrounding a charge the electric field has a magnitude and a direction.

QUESTION Think of another example of a quantity that is a field. The field concept is also used with gravitation. Instead of saying that the Earth exerts a force on a falling object, we can say that the mass of the Earth creates a “gravitational field” surrounding the Earth, and any object near the Earth is acted upon by the gravitational field at that location (Figure 14.13). Gravitational field has units “newtons per kilogram.” At a location near the Earth's surface we can say that there is a gravitational field pointing downward (that is, toward the center of the Earth) of magnitude newtons per kilogram.

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Figure 14.13 Gravitational field of the Earth, in a plane cutting through the center of the Earth.

A falling object of mass experiences a gravitational force of magnitude the object is moving at a speed slow compared to the speed of light, we can calculate its acceleration, .

newtons. If

Of course, farther away from the surface of the Earth, the gravitational field of the Earth is smaller in magnitude, just as the electric field of a charge decreases in magnitude as the distance from the charge increases. The field concept also applies to magnetism. As we will see in a later chapter, electric currents, including electric currents inside the Earth, create “magnetic fields” that affect compass needles.

QUESTION Find an example of a quantity that is not a field. In contrast, many quantities are not fields. The position and velocity at one instant of a car going around a race track are not fields. The distance between the Earth and the Sun is not a field. The rate of change of the momentum of the Moon is not a field.

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THE ELECTRIC FIELD OF A POINT CHARGE By asking what expression for

satisfies the equation

for the force on one point charge by another, we can find an

algebraic expression for the electric field at a location in space called the “observation location”—the location where we detect or measure the field—due to a charged particle q1 (the “source charge”) at the source location. The electric field at the observation location (marked by an “×” in Figure 14.14) is:

ELECTRIC FIELD OF A POINT CHARGE

where q1 is the source charge, is the relative position vector giving the position of the observation location relative to the source charge, and is the unit vector in the direction of . , the magnitude of , is the distance from the source location to the observation location.

Figure 14.14 The relative position vector

extends from the source charge to the observation location, marked by “×”. is a

unit vector with the same direction as the vector away from the source charge.

. Here the source charge is positive, so the electric field points

This equation for electric field of a point charge may look complex, but we can make sense of it by taking it apart and looking at one piece at a time.

Direction of Electric Field The direction of the electric field at the observation location depends on both the direction of

and the sign of the source charge q1,

which may be positive or negative (Figure 14.15). If the source charge q1 is positive, the quantity If the source charge q1 is negative, the direction of 14.16).

is in the direction of , pointing away from the source charge (Figure 14.14). is in the direction of

, pointing toward the source charge (Figure

Figure 14.15 The direction of the electric field depends on both the unit vector and the sign of the source charge.

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Figure 14.16 Here the source charge is negative, so the electric field points toward the source charge, in the direction of

.

Distance Dependence of Electric Field As in the gravitational force law and the electric force law, the square of the distance appears in the denominator (Figure 14.17). This means that the electric field of a point charge depends very strongly on distance. For example, if you double the distance between the source location and the observation location, the only thing that changes is the denominator, which gets four times bigger, so the electric field is only as large as before.

QUESTION If you move the source charge 5 times farther away from the observation location, how does the electric field change? The magnitude of the electric field decreases by a factor of 25.

Figure 14.17 The magnitude of the electric field is inversely proportional to the square of the distance from source charge to observation location.

Magnitude of Charge Because q1, the charge of the source particle, appears in the numerator (Figure 14.18), the larger the charge of the source particle, the larger the magnitude of the electric field at any location in space.

QUESTION At a particular observation location, the magnitude of the electric field due to a point charge is found to be 10 N/C. If the source charge were replaced by a different particle whose charge was 7 times larger, how would this change the electric field at the observation location?

The electric field would now have a magnitude of 70 N/C.

Figure 14.18 The magnitude of the electric field is proportional to the magnitude of the source charge.

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This may seem an odd way to write a single constant; this is done because it is based on the constant , which is in a sense a more fundamental constant, as we will see in a later chapter.

Magnitude of Electric Field The scalar quantity

is not necessarily equal to the magnitude of the electric field. Because the charge q1 can be negative or positive, the expression above may be negative or positive. The magnitude of a vector is a positive quantity. To get the magnitude of a field it is necessary to take the absolute value of the signed scalar quantity above. (Remember that the sign of q1 helps determine the direction of the field.)

Patterns of Electric Field Near Point Charges The electric field of a point charge is spherically symmetric. That is, the magnitude of the field at any location a distance r from the source charge is the same, and at each location the field points either toward the source charge or away from it, depending on the sign of the source charge. Figure 14.19 shows the electric field due to a positive source charge, at various locations in a plane containing the charge. The tail of each arrow is placed at the location where the electric field was measured. Notice that at every observation location the electric field points radially away from the positive source charge, and that the magnitude of the field decreases rapidly with distance from the charge.

Figure 14.19 Electric field of a positive point charge, at various observation locations in a plane containing the charge.

Figure 14.20 shows the electric field due to a negative source charge. At each observation location, the electric field points radially ), so a positively charged particle would experience a force toward inward toward the negative source charge (in the direction of the negative charge. The magnitude of the field decreases rapidly with distance from the charge. We will frequently use twodimensional diagrams because they are easy to draw, but it is important to remember the 3-D character of electric fields in space.

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Figure 14.20 The electric field of a negative point charge, at various observation locations in a plane containing the charge.

EXAMPLE Field of a Particle A particle with charge this particle at a location

(a nanocoulomb is

) is located at the origin. What is the electric field due to ?

Solution

The electric field points away from the positively charged particle, as shown in Figure 14.21.

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Figure 14.21 The electric field calculated in the example.

EXAMPLE Source Location The electric field at location

is found to be

N/C. The only charged

particle in the surroundings has charge −3 nC. What is the location of this particle?

Solution Draw a diagram (Figure 14.22) to find the relative position of the source and the observation location. Remember that the electric field of a negatively charged particle points toward the particle. Find the magnitude of the electric field, and use this to determine the distance from the source to the observation location:

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Figure 14.22 Draw this diagram, using the fact that the electric field due to a negative charge points toward the source charge.

Find the unit vector . From the diagram, note that in this case (because the source charge is negative)

:

Find , and use this to find the source location, where location 2 is the observation location and location 1 is the source location:

14.X.5 A particle with charge

(a nanocoulomb is field due to this particle at a location ?

) is located at the origin. What is the electric

14.X.6 A charged particle located at the origin creates an electric field of

at location

Answer

Answer

. What is the particle's charge?

The Electric Field of a Uniformly Charged Sphere In Chapter 16 we will apply the superposition principle to calculate the electric field of macroscopic objects with charge spread out

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over their surfaces. One of the results of such a calculation is sufficiently useful that we present it now, without going through the calculations involved. As shown in Figure 14.23, a spherical object of radius R with charge Q uniformly spread out over its surface produces an electric is a vector from the center of the sphere to the observation location: field with the following distance dependence, if

In other words, at locations outside the charged sphere, the electric field due to the sphere is the same as if all its charge were located at its center. Inside the sphere, the electric field due to all the charge on the surface of the sphere adds up to zero!

Figure 14.23 The electric field of a uniformly charged sphere, at locations outside and inside the sphere.

Informally, we may say that a uniformly charged sphere “acts like” a point charge, at locations outside the sphere. This is not only because the electric field due to the sphere is the same as the electric field of a point charge located at the center of the sphere, but also because the charged sphere responds to applied electric fields (due to other charges) in the same way as would a point charge located at its center. 14.X.7 A sphere with radius 1 cm has a charge of

spread out uniformly over its surface. What is the magnitude of the electric field due to the sphere at a location 4 cm from the center of the sphere?

Answer

14.X.8 A sphere with radius 2 cm is placed at a location near a point charge. The sphere has a charge of Answer spread uniformly over its surface. The electric field due to the point charge has a magnitude of 500 N/C at the center of the sphere. What is the magnitude of the force on the sphere due to the point charge?

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SUPERPOSITION OF ELECTRIC FIELDS An important property of electric field is a consequence of the superposition principle for electric forces: the net electric field due to two or more charges is the vector sum of each field due to each individual charge. Superposition as a vector sum holds true for forces in general, so it also holds true for fields, because field is force per unit charge. We state the superposition principle here in terms of electric field, in a form intended to emphasize an important aspect of the principle:

THE SUPERPOSITION PRINCIPLE The net electric field at a location in space is the vector sum of the individual electric fields contributed by all charged particles located elsewhere. The electric field contributed by a charged particle is unaffected by the presence of other charged particles.

This may seem obvious, but it actually is a bit subtle. The presence of other particles does not affect the fundamental electric interaction between each pair of particles! The interaction doesn't get “used up”; the interaction between one particle and another is unaffected by the presence of a third particle. A significant consequence of the superposition principle is that matter cannot “block” electric fields; electric fields “penetrate” through matter.

Applying the Superposition Principle Consider three charges q1, q2, and q3 that interact with each other, as shown in Figure 14.24. We could calculate the net force that q1 and q2 exert on q3 by using Coulomb's law and vector addition.

Figure 14.24 Three interacting charged particles.

However, let's use the field concept to find the force on q3 due to the other two charges. We pretend that q3 isn't there, and we find due to q1 and q2, at the location where q3 would be. We can then calculate the force on q3 as q3 . In the electric field Figure 14.25 the scheme is shown graphically. is the vector sum of (due to q1) and (due to q2), at the location of q3.

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Figure 14.25 Electric field at the location of q 3 .

Advantage of Electric Field in Calculations We could just as easily have calculated the force on q3 by using Coulomb's law directly, without using the more abstract concept of electric field. What advantage is there to using electric field? One of the advantages is that once you have calculated the field due to q1 and q2, you can quickly calculate the force on any amount of charge Q placed at that location, not just the force on q3. The “force per unit charge” idea simplifies these kinds of calculations. If we are interested in the effect that all three charges would have on a fourth charge q4 placed at some location, we would include all three charges as the source of the electric field at that location, but exclude q4, as shown in Figure 14.26.

Figure 14.26 Electric field at the location of q 3 . Knowing the electric field would simply be

at a particular location, we know what force would act on a charge q4 placed at that location: it .

EXAMPLE Electric Field and Force Due to Two Charges A small object with charge

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is located at the origin. A second small object with charge

is

located at . What is the net electric field at location due to Q 1 and Q 2? If a small object with a charge of were placed at location A, what would the force on this object be?

Solution Diagram: Figure 14.27

In Figure 14.28 the charge

has been placed at the observation location.

The individual contributions and the net field are shown in Figure 14.27. When Q 3 is placed at location A, it experiences a force in the direction shown in Figure 14.28. The force on Q 3 is in a direction opposite to the direction of the net electric field at A, because Q 3 is negative.

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Figure 14.27 Two point charges make an electric field at location A.

Figure 14.28 When a negative charge is placed at location A, it experiences a force in a direction opposite to the net field at that location.

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THE ELECTRIC FIELD OF A DIPOLE Neutral matter contains both positive and negative charges (protons and electrons). The simplest piece of neutral matter that we can analyze in detail is an “electric dipole” consisting of two equally but oppositely charged pointlike objects, separated by a distance s, as shown in Figure 14.29. Dipoles occur frequently in nature, so this analysis has practical applications. For example, a single molecule of HCl is an electric dipole: the H end is somewhat positive and the Cl end is somewhat negative. Moreover, the analysis of a single dipole provides a basis for analyzing more complicated forms of matter.

Figure 14.29 An electric dipole.

By applying the superposition principle we can calculate the electric field due to a dipole at any location in space. We will be particularly interested in the electric field in two important locations: at a location along the axis of the dipole and at a location along the perpendicular bisector of the axis.

Along the Dipole Axis Figure 14.30 shows a horizontal dipole with charge and , and separation s (here q is a positive number, so is a positive charge, and is a negative charge). Its center is at the origin. We will apply the superposition principle to find the electric field at an observation location on the axis of the dipole (in this case, the x axis), due to the dipole.

Figure 14.30 Electric field of a dipole at a location on the dipole axis (in the case shown, this is the x axis).

For the orientation shown, we can see that the net field will be in the direction, because at this location the field of the closer charge is larger than the field of the more distant charge. Note that if the dipole were oriented with its negative end on the right,

would point in the opposite direction. For the orientation shown, the electric field the positive end of the dipole axis is:

Similarly, the electric field

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due only to the negative end of the dipole is:

due only to

We add the two contributions to get the net field at the observation location:

Because the dipole could be oriented along any axis (see Figure 14.31), we can make this result more general by changing the x to an r, representing the distance from the center of the dipole to an observation location on the axis of the dipole, however the dipole may be oriented. Because the direction of the electric field at a location on the dipole axis depends on the orientation of the dipole, it is more useful to write an algebraic expression for the magnitude of the dipole, and to figure out the direction by looking at the particular situation.

QUESTION If the dipole in Figure 14.31 were rotated through an angle of 180°, so the negative end was closer to the observation location, what would be the direction of the electric field at the observation location? If the dipole were rotated, the electric field would now point toward the dipole.

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Figure 14.31 The electric field due to a dipole, at a location on the axis of the dipole, for a dipole with an arbitrary orientation.

Approximation: Far from the Dipole This is still a fairly messy expression, and it does not give us much information about how fast the magnitude of the field decreases as we get farther away from the dipole. Let's try to simplify it. If we are far from the dipole, as is normally the case when we interact with molecular dipoles, then the dipole separation s is very small compared to the distance r from the dipole, so:

Using this approximation, we can simplify the expression:

ELECTRIC FIELD OF A DIPOLE, ON DIPOLE AXIS at a location on the dipole axis, if q is the magnitude of one of the charges. s is the separation between the charges. r is the distance from the center of the dipole to the observation location. The direction is determined by looking at the particular situation.

Since a dipole can have any orientation, we give only the magnitude of the field. For a particular dipole you will need to determine the direction of the field at a particular observation location by looking at the orientation of the dipole.

Is This Result Reasonable? QUESTION LibraryPirate

Should we have expected the magnitude of the electric field of the dipole to be proportional to

?

The electric field of a single point charge is proportional to , but when we add the electric fields contributed by more than one charge, the result may have quite a different distance dependence.

QUESTION Does our result have the correct units? The units of electric field come out to

, as they should.

EXAMPLE HCl Molecule An HCl molecule is a dipole, which can be considered to be a particle of charge (a H + ion) and a particle of charge (a Cl − ion) separated by a distance of about . What are the magnitude and direction of the electric field due to the HCl molecule at the location indicated in Figure 14.32, which is 2 ( ) from the center of the molecule, on the molecular axis?

Figure 14.32 An HCl molecule.

Solution

Because the positive end of the dipole is closer to the observation location than the negative end, the electric field points away from the dipole, along the dipole axis.

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Next consider the electric field at an observation location on the axis perpendicular to the axis of the dipole. For a dipole aligned as shown in Figure 14.33, this is the y axis. First we need to find and , the vectors from each source charge to the observation location.

The magnitudes of

The net field

and

are equal, because each charge is the same distance from the observation location.

is the vector sum of

and

:

The y components of the two fields cancel, and the net field come as a bit of a surprise.

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is horizontal to the left, as shown in Figure 14.33, which may

Figure 14.33 Electric field of a dipole at a location on the perpendicular axis.

Again, since in general a dipole could have any orientation (for example, see Figure 14.34), it is appropriate to replace the “y” with an “r” representing the distance from the center of the dipole to a location on the axis perpendicular to the dipole, and to give only the magnitude of the electric field at this location.

Figure 14.34 The electric field of a dipole at a location on the axis perpendicular to the dipole.

If we are far from the dipole, as is normally the case when we interact with molecular dipoles, the magnitude of the field can be further simplified. You should be able to show that in this important case , by making an appropriate approximation, we obtain:

ELECTRIC FIELD OF A DIPOLE, ON

AXIS

on a location on the perpendicular axis, if q is the magnitude of one of the charges. s is the separation between the charges. r is the distance from the center of the dipole to the observation location. The direction is determined by looking at the particular situation.

Note that the magnitude of the electric field at a location along the y axis is half of the magnitude of the electric field at a location the same distance away on the x axis.

Along the Other Perpendicular Axis

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QUESTION What would the electric field of a dipole be at a location on the other perpendicular axis?

By looking at the symmetry of the system, we can conclude by inspection that the electric field along both perpendicular axes should have exactly the same distance dependence.

Other Locations By applying the superposition principle we can calculate the electric field of a dipole at any location, though we may not get a simple algebraic expression. Figure 14.35 shows the electric field of a dipole at locations in two perpendicular planes containing the dipole, calculated numerically. Figure 14.36 shows only locations in the vertical plane, displaying the symmetry of the field pattern. Problem 14.P.77 is a computer exercise that allows you to visualize the electric field of a dipole at any location.

Figure 14.35 Electric field of a dipole at locations in two planes. Included are locations on the dipole axis and the perpendicular axis, as well as other locations.

Figure 14.36 Electric field of a dipole, at locations in a plane containing the dipole.

The Importance of Approximations

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By considering the common situation where , we were able to come up with a simple algebraic expression for the electric field of a dipole at locations along the x and y axes, which gives us a clear insight into the distance dependence of the dipole field. If we had not made this simplification, we would have been left with a messy expression, and it would not have been easy to see that the electric field of a dipole falls off like . (This turns out to be true for locations not on the axes as well, but the algebra required to demonstrate this is more complicated.)

QUESTION How would we calculate the electric field of a dipole at a location very near the dipole, where

?

Since the approximation we made in the calculations above is no longer valid, we would have to go back to adding the fields of the two charges exactly, as we did in the first step of our work above. 14.X.9 A dipole is located at the origin, and is composed of charged particles with charge a distance along the x axis.

and

, separated by

(a) Calculate the magnitude of the electric field due to this dipole at location

.

(b) Calculate the magnitude of the electric field due to this dipole at location

.

Answer

Interaction of a Point Charge and a Dipole Since we have analytical expressions for the electric field of a dipole at locations along the x, y, or x axes, we can calculate the force exerted on a point charge that interacts with the field of the dipole. Figure 14.37 shows a dipole acting on a point charge that is a distance from the center of the dipole.

EXAMPLE Force of a Dipole on a Point Charge What are the direction and the magnitude of the force exerted by the dipole on the point charge shown in Figure 14.37?

Figure 14.37 A dipole and a point charge.

Solution The electric field due to the dipole at the location of the point charge points in the dipole exerts a force on Q toward the dipole:

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direction (Figure 14.39). The

QUESTION Does the direction of this electric force make sense?

Yes. The negative end of the dipole is closer to the location of interest, so its contribution to the net electric field is larger than the contribution from the positive end of the dipole.

QUESTION What are the magnitude and direction of the force exerted on the dipole by the point charge?

By the principle of reciprocity of electric forces (Newton's third law), the force exerted by Q on the dipole must be equal in magnitude and opposite in direction to the force on the point charge by the dipole.

QUESTION How is this possible? Since the particles making up the dipole have equal and opposite charges, why isn't the net force on the dipole zero? The key lies in the distance dependence of the electric field of the point charge. Although the two ends of the dipole are very close together, the positive end is very slightly farther away from the point charge. Hence, the magnitude of the point charge's electric field is slightly less at this location, and the resulting repulsive force on the positive end of the dipole is slightly less than the attractive force on the negative end of the dipole. The net force is small, but attractive (Figure 14.38).

Figure 14.38 The electric field of the point charge is slightly larger in magnitude at the location of the negative end of the dipole. Hence, the net force on the dipole is to the right. The arrows representing field and force vectors are offset slightly from the location of the charges for clarity.

14.X.10 Consider the situation in Figure 14.39.

Answer

(a) If we double the distance d, by what factor is the force on the point charge due to the dipole reduced? (b) How would the magnitude of the force change if the point charge had a charge of (c) If the charge of the point charge were

?

, how would the force change?

Figure 14.39 The electric field of the dipole at the location of the point charge, and the force on the point charge due to this field.

14.X.11 Refer to your answer to Exercise 14.X.9 part (a). Calculate the magnitude of the force on a proton placed at

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Answer

the location where you calculated the electric field.

Electric Dipole Moment The electric field of a dipole is proportional to the product

, called the “electric dipole moment” and denoted by p.

Far away from a dipole, the same electric field could be due to a small q and a large s, or a large q and a small s. The only thing that matters is the product . Moreover, this is the quantity that is measurable for molecules such as HCl and that are permanent dipoles, not the individual values of q (the amount of charge on an end) and s (the separation of the charges). We can rewrite our expressions for the electric field of a dipole in terms of the dipole moment p:

Dipole Moment as a Vector The dipole moment can be defined as a vector that points from the negative charge to the positive charge, with a magnitude (Figure 14.40). Note that the electric field along the axis of the dipole (outside the dipole) points in the same direction as the dipole moment , which is one of the useful properties of the vector .

Figure 14.40 Electric field of a dipole on the axis and perpendicular axis, and the electric dipole moment vector.

In Chapter 18 we will see a similar pattern of magnetic field around a magnetic dipole characterized by a magnetic dipole moment that is a vector.

A Dipole in a Uniform Electric Field We will see in Chapter 16 that it is possible to arrange a collection of point charges in a configuration that produces an electric field nearly uniform in magnitude and direction within a particular region. What would be the force on a dipole in such a uniform electric field?

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The net force on the dipole would be zero, since the forces on the two ends would be equal in magnitude and opposite in direction, as shown in Figure 14.41. However, there would be a torque on the dipole about its center of mass, and the dipole would begin to rotate.

Figure 14.41 A dipole in a uniform electric field would experience no net force. It would experience a torque about its center of mass (in this example

is out of the page, and the dipole would begin to rotate counterclockwise).

14.X.12 What would be the equilibrium position of the dipole shown in Figure 14.41? In this equilibrium position, compare the orientation of the vector dipole moment with the direction of the applied electric field.

EXAMPLE Dipole and Charged Ball A dipole consisting of point charges and separated by a distance of 2 mm is centered at location (the size of the dipole is exaggerated in the diagram). A hollow plastic ball with radius 3 cm and charge distributed uniformly over its surface is centered at location , as shown in Figure 14.42. What is the net electric field at location C, ?( .)

Figure 14.42 A dipole and a charged ball. Dipole size is exaggerated (not to scale).

Solution Apply the superposition principle:

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Answer

The distance from the center of the dipole to the observation location is

Because location C is reasonably far from the dipole, along the dipole axis, we can approximate the magnitude of the dipole field as:

By looking at the diagram we see that the direction of the dipole field is (along the positive y axis, because the negative end is closer to location C). The electric field at location C due to the dipole is:

The electric field due to the ball is the same as if the ball were a point charge located at its center. The vector the center of ball to location C is:

The magnitude of

is

from

, so the unit vector is:

The electric field due to the charged ball is then

The net electric field is:

FURTHER DISCUSSION As shown in Figure 14.43, both the x and y components of are positive, which is correct, since the ball is negatively charged. Arrows representing each contribution to the electric field, and the net electric field, are shown in Figure 14.43.

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Figure 14.43 Contributions of dipole and ball, and net electric field at location C.

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CHOICE OF SYSTEM In our previous study of mechanics, when we analyzed multiparticle systems in terms of momentum or energy we had to decide what objects to include in our “system” and what objects to consider as external to the system. Similarly, when we use the field concept to calculate a force rather than using Coulomb's law directly, we split the Universe into two parts: the charges that are the sources of the field, and the charge that is affected by that field. This is particularly useful when the sources of the field are numerous and fixed in position. We calculate the electric field due to such sources throughout a given region of space, and then we can predict what would happen to a charged particle that enters that region. (It is of course true that the moving charged particle exerts forces on all the other charges, but often we're not interested in those forces, or the other charges are not free to move.) For example, in an oscilloscope, charges on metal plates are the sources of an electric field that affects the trajectory of single electrons (Figure 14.44). The electric field has nearly the same magnitude and direction everywhere within a box (this can be achieved with a suitable configuration of positive charges below and negative charges above). The trajectory of an electron outside this box, where the field is nearly zero, is a straight line. Upon entering the box (through a small hole) the electron experiences a downward force (it has negative charge, hence the force on it is opposite to the field) and has a trajectory as shown.

Figure 14.44 Path of an electron through a region of uniform electric field.

14.X.13 If the uniform upward-pointing electric field depicted in Figure 14.44 has a magnitude of 5000 N/C, what is the magnitude of the force on the electron while it is in the box? 14.X.14 If a different particle experiences a force of the previous exercise, what is the charge of the particle?

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when passing through the region identified in

Answer

Answer

IS ELECTRIC FIELD REAL? Reflecting on what we have done so far, we can give two reasons for why we introduced the concept of electric field: 1. Once we know the electric field at some location, we know the electric force acting on any charge q that we place there. 2. In terms of electric field, we can describe the electric properties of matter, independent of how the applied electric field is produced.

Later we'll see that if the electric field in air exceeds about

N/C, air becomes a conductor, no matter how this electric field is produced. This is an example of the value of the electric field concept for parameterizing the behavior of matter. It would be very awkward to describe the properties of matter if we didn't split the problem into two parts using the field concept, in which we talk about the field made by other charges and the effect that a field has on a particular kind of matter. Even with these advantages of the field concept, it may seem that it doesn't matter whether we think of electric field as merely a calculational convenience or as something real. However, Einstein's special theory of relativity, which has been experimentally verified in a wide variety of situations, rules out instantaneous action at a distance and implies that the field concept is necessary, not merely convenient, as we will show next.

Retardation: Electric Field and the Speed of Light Special relativity predicts that nothing can move faster than the speed of light, not even information, and no one has ever observed a violation of this prediction. The speed of light is , or about one foot (30 cm) per nanosecond, where a nanosecond is seconds. Consider a charge that is 30 feet away (about 9 meters), as shown in Figure 14.45. It takes light 30 nanoseconds (30 ns) to travel from the charge to you.

Figure 14.45 Electric field of a point charge.

Suppose that the charge is suddenly moved to a new location, as shown in Figure 14.46. You cannot observe change in the electric field until 30 nanoseconds have elapsed! If the field could change instantaneously, that would provide a mechanism for sending signals faster than the speed of light.

Figure 14.46 At

charge is moved, but the electric field at the observation location does not change for 30 ns.

For a while (30 nanoseconds), the electric field has some kind of reality and existence independent of the original source of the field. If you place a positive charge at your observation location during this 30 ns interval, you will see it accelerate to the right, in the direction of the old electric field that is still valid.

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At the end of the 30-nanosecond delay, the field finally changes to correspond to the place to which the charge was moved 30 nanoseconds ago, as shown in Figure 14.47. You see the new position of the charge at the same instant that you notice the change in the electric field, because of course light itself travels at the speed of light.

Figure 14.47 At

ns, the field at the observation location finally changes.

What If the Source Charges Vanish? An even more dramatic example concerns the electric field made by a remote electric dipole. Suppose that an electron and a positron are separated by a small distance s, making a dipole, and creating an electric field throughout space. The electron and positron may suddenly come together and react, annihilating each other and releasing a large amount of energy in the form of high-energy photons (gamma rays), which have no electric charge: At a time shortly after the annihilation has occurred, there are no longer any charged particles in the vicinity. However, if you monitor the electric field in this region, you will keep detecting the (former) dipole's field for a while even though the dipole no longer exists! If you are a distance r away from the original location of the dipole, you will still detect the dipole's electric field for a time , which is the same as the time it would take light to travel from that location to your position.

Coulomb's Law Is Correct Only at Low Speeds This interesting behavior, called “retardation,” means that Coulomb's law is not completely correct, since the formula

doesn't contain time t or speed of light c. Coulomb's law is an approximation that is quite accurate as long as charges are moving slowly. However, when charges move with speeds that are a significant fraction of the speed of light, Coulomb's law is not adequate. Similarly, the formula for the electric field of a point charge,

is only an approximation, valid if the charge is moving at a speed that is small compared with the speed of light. Later we will consider the fields of moving charges, which include magnetic fields as well as electric fields. We will also consider the fields of accelerated charges, which produce a very special kind of electric field—a component of the electromagnetic field that makes up radio and television waves and light. Evidently the electric field isn't just a calculational device. Space can be altered by the presence of an electric field, even when the source charges that made the field are gone.

Relativistic Electric Field It turns out that for a charged particle moving at high speed in the unchanged. However, a factor of

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direction, the x component of electric field at any location is

appears in the y and z components of electric field, and they are much larger, as shown in Figure 14.48. The fields of relativistic particles are discussed further in Chapter 21.

Figure 14.48 The pattern of electric field of a positively charged particle moving at high speed in the spherically symmetric.

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direction is no longer

SUMMARY At a location where there is an electric field

, a point particle with charge q2 experiences a force

(Figure 14.49).

Figure 14.49

A point particle with charge q1 makes an electric field at all locations in space (Figure 14.50) except its own location:

where q1 is the source charge, is the relative position vector giving the position of the observation location relative to the source charge, is the unit vector in the direction of , and , the magnitude of , is the distance from the source location to the observation location.

Figure 14.50

A point charge is not affected by its own electric field. The source location is the location of the charge(s) that create an electric field. The observation location is the location at which the electric field is detected.

Figure 14.51 Electric field at selected locations in a plane containing a positive or negative point charge.

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The Superposition Principle The net electric field at a particular location due to two or more charges is the vector sum of each field due to each individual charge. Each individual contribution is unaffected by the presence of the other charges. Thus, matter cannot “block” electric fields.

Retardation Changes in electric fields (due to changes in source charge distribution) propagate through space at the speed of light. Therefore the electric field at a distant point does not change instantaneously when the source charge distribution changes.

Electric dipole An electric dipole consists of two point particles having equal and opposite charges and electric dipole moment of a dipole is defined as . The electric dipole moment vector toward the positive charge.

, separated by a distance s. The points from the negative charge

Electric field at a location on the dipole axis:

Figure 14.52 The pattern of electric field around an electric dipole. The direction is determined by looking at the particular situation.

Electric field at a location on the perpendicular axis:

The direction is determined by looking at the particular situation.

For a spherical object with a total charge Q distributed uniformly over its surface (a “uniform spherical shell of charge”): at a location inside the shell,

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at a location outside the shell,

(

extends from the center of sphere to the observation location; Q is the total charge on the sphere).

Constants:

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EXERCISES AND PROBLEMS Section 14.3 14.X.15 What is the relationship between the terms “field” and “force”? What are their units? 14.X.16 An electron in a region in which there is an electric field experiences a force of magnitude

N. What is the

magnitude of the electric field at the location of the electron? 14.X.17 At location A there is an electric field in the direction shown by the orange arrow in Figure 14.53. This electric field is due to charged particles that are not shown in the diagram. (a) If a proton is placed at location A, which of the arrows (a–h) best indicates the direction of the electric force on the proton? (b) If the proton is removed and an electron is placed at location A, which of the arrows in Figure 14.53 best indicates the direction of the electric force on the electron?

Figure 14.53 14.X.18 The electric field at a particular location is measured to be placed at this particular location?

N/C. What force would a positron experience if

14.X.19 An electron in a region in which there is an electric field experiences a force of magnitude

. What is the

magnitude of the electric field at the location of the electron? 14.X.20 If the particle in Figure 14.54 is a proton, and the electric field the force

has the value

, what is

on the proton?

Figure 14.54 14.X.21 An electron in a region in which there is an electric field experiences a force of . What is the electric field at the location of the electron? 14.P.22 In the region shown in Figure 14.55 there is an electric field due to a point charge located at the center of the dashed circle. The arrows indicate the magnitude and direction of the electric field at the locations shown.

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Figure 14.55 (a) What is the sign of the source charge? (b) Now a particle whose charge is the

C is placed at location B. What is the direction of the electric force on

C charge?

(c) The electric field at location B has the value at this location? (d) What is the electric force on the C charge?

N/C. What is the unit vector in the direction of

(e) What is the unit vector in the direction of this electric force? 14.P.23 In the region shown in Figure 14.56 there is an electric field due to charged objects not shown in the diagram. A tiny glass ball with a charge of

coulomb placed at location A experiences a force of

, as

shown in the figure.

Figure 14.56 (a) Which arrow in Figure 14.57 best indicates the direction of the electric field at location A?

Figure 14.57 (b) What is the electric field at location A? (c) What is the magnitude of this electric field? (d) Now the glass ball is moved very far away. A tiny plastic ball, with charge

coulombs is placed at location A. Which arrow in Figure 14.57 best indicates the direction of the electric force on the negatively charged

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plastic ball? (e) What is the force on the negative plastic ball? (f) You discover that the source of the electric field at location A is a negatively charged particle. Which of the numbered locations in Figure 14.58 shows the location of this negatively charged particle, relative to location A?

Figure 14.58

14.P.24 An electron is observed to accelerate in the

direction with an acceleration of . Explain how to use the definition of electric field to determine the electric field at this location, and give the direction and magnitude of the field.

14.P.25 A proton is observed to have an instantaneous acceleration of

. What is the magnitude of the electric field

at the proton's location? 14.P.26 You are the captain of a spaceship. You need to measure the electric field at a specified location P in space outside your ship. You send a crew member outside with a meter stick, a stopwatch, and a small ball of known mass M and net charge (held by insulating strings while being carried). (a) Write down the instructions you will give to the crew member, explaining what observations to make. (b) Explain how you will analyze the data that the crew member brings you to determine the magnitude and direction of the electric field at location P.

Section 14.4 14.X.27 Criticize the following statement: “A proton can never be at rest, because it makes a very large electric field near itself that accelerates it.” 14.X.28 A particle with charge

(a nanocoulomb is m?

particle at a location

C) is located at the origin. What is the electric field due to this

14.X.29 What is the electric field at a location =

, due to a particle with charge

14.X.30 In a hydrogen atom in its ground state, the electron is on average a distance of about

located at the origin? m from the proton. What

is the magnitude of the electric field due to the proton at the location of the electron? 14.X.31 In a hydrogen atom in its ground state, the electron is on average a distance of about is the magnitude of the electric field due to the proton at a location 14.X.32 In Figure 14.59 a proton at location A makes an electric field experiences a force

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.

m from the proton. What

m away?

at location B. A different proton, placed at location B,

Figure 14.59

Now the proton at B is removed and replaced by a lithium nucleus, containing three protons and four neutrons. (a) Now what is the value of the electric field at location B due to the proton? (b) What is the force on the lithium nucleus? (c) The lithium nucleus is removed, and an electron is placed at location B. (d) Now what is the value of the electric field at location B due to the proton? (e) What is the magnitude of the force on the electron? (f) Which arrow in Figure 14.60 best indicates the direction of the force on the electron due to the electric field?

Figure 14.60

14.X.33 A sphere with radius 1 cm has a charge of

C spread out uniformly over its surface. What is the magnitude of the electric field due to the sphere at a location 4 cm from the center of the sphere?

14.X.34 A sphere with radius 2 cm is placed at a location near a point charge. The sphere has a charge of

C spread uniformly over its surface. The electric field due to the point charge has a magnitude of 470 N/C at the center of the sphere. What is the magnitude of the force on the sphere due to the point charge?

14.X.35 What is the magnitude and direction of the electric field at location

at location

if there is a negative point charge of

? Include units.

14.P.36 You want to calculate the electric field at location .

, due to a particle with charge

(a) What is the source location? (b) What is the observation location? (c) What is the vector (d) What is

?

(e) What is the vector ? (f)

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that points from the source location to the observation location?

located at

What is the value of

?

(g) Finally, what is the electric field, expressed as a vector? 14.P.37 An electron is located at the electron.

. You need to find the electric field at location

, due to

(a) What is the source location? (b) What is the observation location? (c) What is the vector (d) What is

?

?

(e) What is the vector ? (f) What is the value of

?

(g) Finally, what is the electric field at the observation location, expressed as a vector? 14.P.38 A charged particle located at the origin creates an electric field of

N/C at a location

.

What is the particle's charge? 14.P.39 At a particular location in the room there is an electric field = point particle of charge 1 C in order to produce this electric field?

N/C. Where would you place a single negative

14.P.40 The electric field at a location C points north, and the magnitude is

N/C. Give numerical answers to the following

questions: (a) Where relative to C should you place a single proton to produce this field? (b) Where relative to C should you place a single electron to produce this field? (c) Where should you place a proton and an electron, at equal distances from C, to produce this field? 14.P.41 You want to create an electric field =

N/C at location

.

(a) Where would you place a proton to produce this field at the origin? (b) Instead of a proton, where would you place an electron to produce this field at the origin? (Hint: This problem will be much easier if you draw a diagram.) 14.P.42 A

(“pi-minus”) particle, which has charge

, is at location

.

(a) What is the electric field at location

, due to the

(b) At a particular moment an antiproton (same mass as the proton, charge moment what is the force on the antiproton, due to the ? 14.P.43 What is the electric field at a location

particle?

) is at the observation location. At this

, due to a particle with charge

located at the

origin? 14.P.44 At a particular location in the room there is an electric field

. Figure out where to place a single

positive point particle, and how much charge it should have, in order to produce this electric field (there are many possible answers!). Do the same for a single negatively charged point particle. Be sure to draw diagrams to explain the geometry of the situation. 14.P.45 Where must an electron be to create an electric field of

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at a location in space? Calculate its

displacement from the observation location and show its location on a diagram. 14.P.46 The electric field at a location C points north, and the magnitude is

. Give numerical answers to the

following questions: (a) Where relative to C should you place a single proton to produce this field? (b) Where relative to C should you place a single electron to produce this field? (c) Where should you place a proton and an electron, at equal distances from C, to produce this field? 14.P.47 A lithium nucleus consisting of 3 protons and 4 neutrons accelerates to the right due to electric forces, and the initial magnitude of the acceleration is meters per second per second. (a) What is the direction of the electric field that acts on the lithium nucleus? (b) What is the magnitude of the electric field that acts on the lithium nucleus? Be quantitative (that is, give a number). (c) If this acceleration is due solely to a single helium nucleus (2 protons and 2 neutrons), where is the helium nucleus initially located? Be quantitative (that is, give a number). 14.P.48

(a) On a clear and carefully drawn diagram, place a helium nucleus (consisting of 2 protons and 2 neutrons) and a proton in such a way that the electric field due to these charges is zero at a location marked ×, a distance from the helium nucleus. Explain briefly but carefully, and use diagrams to help in the explanation. Be quantitative about the relative distances. (b) On a clear and carefully drawn diagram, place a helium nucleus and an electron in such a way that the electric field due to these charges is zero at a location marked ×. Explain briefly but carefully, and use diagrams to help in the explanation. Be quantitative about the relative distances.

Section 14.5 14.X.49 Draw a diagram showing two separated point charges placed in such a way that the electric field is zero somewhere, and indicate that position. Explain your reasons. 14.P.50 At a particular moment, three charged particles are located as shown in Figure 14.61. Your answers to the following questions should be vectors. ( .)

Figure 14.61 (a) Find the electric field at the location of

, due to

.

(b) Find the electric field at the location of

, due to

.

(c) Find the net electric field at the location of (d) Find the net force on

.

.

(e) Find the electric field at location A due to

.

(f) Find the electric field at location A due to

.

(g) Find the electric field at location A due to

.

(h) What is the net electric field at location A?

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(i) If a particle with charge

were placed at location A, what would be the force on this particle?

14.P.51 An

ion is located 400 nm ( (ions not shown to scale).

, about 4000 atomic diameters) from a

ion, as shown in Figure 14.62

Figure 14.62 (a) Determine the magnitude and direction of the electric field

at location A, 100 nm to the left of the

(b) Determine the magnitude and direction of the electric field

at location A, 100 nm to the right of the

ion. ion.

(c) If an electron is placed at location A, what are the magnitude and direction of the force on the electron? 14.P.52 At a particular moment, one negative and two positive charges are located as shown in Figure 14.63. Your answers to each part of this problem should be vectors. ( .)

Figure 14.63 (a) Find the electric field at the location of Q 1, due to Q 2 and Q 3. (b) Use the electric field you calculated in part (a) to find the force on Q 1. (c) Find the electric field at location A, due to all three charges. (d) An alpha particle (

, containing two protons and two neutrons) is released from rest at location A. Use your answer from the previous part to determine the initial acceleration of the alpha particle.

14.P.53 A hollow ball with radius 2 cm has a charge of

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spread uniformly over its surface, as shown in Figure 14.64.

Figure 14.64

The center of the ball is at

. A point charge of

(a) What is the net electric field at location

is located at

.

?

(b) Draw an arrow representing the net electric field at that location. Make sure that the arrow you drew makes sense.

Section 14.6 14.X.54 We found that the force exerted on a distant charged object by a dipole is given by

In this formula, what is the meaning of the symbols q, Q, s, and r? 14.X.55 At a given instant in time, three charged objects are located near each other, as shown in Figure 14.65. Explain why the formula

cannot be used to calculate the electric force on the ball of charge

.

Figure 14.65 14.X.56 Where could you place one positive charge and one negative charge to produce the pattern of electric field shown in Figure 14.66? (As usual, each electric field vector is drawn with its tail at the location where the electric field was measured.) Briefly explain your choices.

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Figure 14.66 14.X.57 Consider Figure 14.67.

Figure 14.67 (a) Which arrow (a–j) in Figure 14.68 best indicates the direction of the electric field at location A due to the dipole? (b) Which arrow (a–j) best indicates the direction of the electric field at location B due to the dipole?

Figure 14.68 14.X.58 Consider Figure 14.69.

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Figure 14.69 (a) If an electron were at location A, which arrow (a–j) in Figure 14.70 would best indicate the direction of the electric force on the electron by the dipole? (b) If an electron were at location A, which arrow (a–j) would best indicate the direction of the electric force on the electron by the dipole?

Figure 14.70 14.X.59 A dipole is located at the corner of a square whose side has length d, as shown in Figure 14.69. (a) Which arrow in Figure 14.70 best represents the direction of the electric field at location A due to the dipole? (b) Which arrow best represents the direction of the electric field at location B, due to the dipole? 14.X.60 A dipole is located at the origin and is composed of charged particles with charge and , separated by a distance along the x axis. The charge is on the axis. Calculate the electric field due to this dipole at a .

location

14.X.61 A dipole is located at the origin and is composed of charged particles with charge and , separated by a distance along the x axis. Calculate the magnitude of the electric field due to this dipole at a location m. 14.X.62 Which of these statements about a dipole are correct? Select all that are true. (A) At a distance d from a dipole, where field due to the dipole is proportional to

(the separation between the charges), the magnitude of the electric .

(B) A dipole consists of two particles whose charges are equal in magnitude but opposite in sign. (C)

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The net electric field due to a dipole is zero, since the contribution of the negative charge cancels out the contribution of the positive charge. (D) At a distance d from a dipole, where field due to the dipole is proportional to

(the separation between the charges), the magnitude of the electric .

(E) The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge. 14.X.63 A dipole consists of two charges

and

, held apart by a rod of length 3 mm, as shown in Figure 14.71.

Figure 14.71 (a) What is the magnitude of the electric field due to the dipole at location A, 5 cm from the center of the dipole? (b) What is the magnitude of the electric field due to the dipole at location B, 5 cm from the center of the dipole? 14.X.64 If we triple the distance d, by what factor is the force on the point charge due to the dipole in Figure 14.72 reduced? (Note that the factor is smaller than one if the force is reduced and larger than one if the force is increased.)

Figure 14.72 14.X.65 The distance between the dipole and the point charge in Figure 14.72 is d. If the distance between them were changed to , by what factor would the force on the point charge due to the dipole change? Express your answer as the ratio (magnitude of new force / magnitude of ). 14.X.66 If the charge of the point charge in Figure 14.72 were

(instead of Q):

(a) By what factor would the magnitude of the force on the point charge due to the dipole change? Express your answer as the ratio (magnitude of new force / magnitude of ). (b) Would the direction of the force change? 14.X.67 Draw a diagram like the one in Figure 14.73.

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Figure 14.73

On your diagram, draw vectors showing: (a) The electric field of the dipole at the location of the negatively charged ball (b) The net force on the ball due to the dipole (c) The electric field of the ball at the center of the dipole (d) The net force on the dipole due to the ball 14.X.68 If the distance between the ball and the dipole in Figure 14.73 were doubled, what change would there be in the force on the ball due to the dipole? 14.X.69 The dipole moment of the HF (hydrogen fluoride) molecule has been measured to be dipole as having charges of

and

. If we model the

separated by a distance s, what is s? Is this plausible?

14.P.70 A dipole is located at the origin and is composed of charged particles with charge and , separated by a distance m along the y axis. The charge is on the axis. Calculate the force on a proton due to this dipole at a location

.

14.P.71 A dipole is located at the origin and is composed of charged particles with charge and , separated by a distance m along the x axis. The charge is on the axis. Calculate the force on a proton at a location due to this dipole. 14.P.72 A dipole is centered at the origin and is composed of charged particles with charge and , separated by a distance m along the y axis. The charge is on the axis, and the charge is on the axis. (a) A proton is located at (b) An electron is located at

. What is the force on the proton due to the dipole? . What is the force on the electron due to the dipole?

(Hint: Make a diagram. One approach is to calculate magnitudes, and get directions from your diagram.) 14.P.73 Two dipoles are oriented as shown in Figure 14.74. Each dipole consists of two charges and , held apart by a rod of length s, and the center of each dipole is a distance d from location A. If nC, mm, and cm, what is the electric field at location A? (Hint: draw a diagram and show the direction of each dipole's contribution to the electric field on the diagram.)

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Figure 14.74 14.P.74 Two dipoles are oriented as shown in Figure 14.75. Each dipole consists of charges held apart by a short rod (not shown to scale). What is the electric field at location A? Start by drawing a diagram that shows the direction of each dipole's contribution to the electric field at location A.

Figure 14.75 14.P.75 A charge of

( ) and a dipole with charges location A that is zero, as shown in Figure 14.76.

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and

separated by 0.3 mm contribute a net field at

Figure 14.76 (a) Which end of the dipole is positively charged? (b) How large is the charge q?

Section 14.8 14.P.76 You make repeated measurements of the electric field due to a distant charge, and you find it is constant in magnitude your partner moves the charge. The electric field doesn't change for a while, but at time and direction. At time nanoseconds you observe a sudden change. How far away was the charge originally?

Computational problems 14.P.77 Consider a dipole comprising two singly charged ions (with charge as shown in Figure 14.77).

and

, respectively) that are 0.2 nm apart (

Figure 14.77 (a) Write a computer program to find the magnitude and direction of the electric field due to this dipole at eight locations arranged as shown, and eight other locations not in the xy plane (if you are working in a 3-D environment). Display a vector of appropriate length and direction indicating the field at each location. Make sure the display is comprehensible by scaling the vectors so they are easily visible but do not overlap. (b) Now place a proton at location nm and release it. Before writing the program to do this, predict qualitatively what you expect its motion to be. Using your expression for electric field from part (a), compute and animate the trajectory of the proton, leaving a trail to show the proton's path (that is, use the field concept in your calculations.) You may wish to start with . (c) Simultaneously compute and plot a graph showing the potential energy U, kinetic energy K, and vs. time . Your graph will be more useful as a computational diagnostic tool if you for the entire system do not include the potential energy associated with the interaction of the pair of charges making up the dipole, which does not change. (d) Explain the shape of the K and U graphs. 14.P.78 Write a computer program to calculate and plot a graph of the magnitude of the electric field of a dipole at locations on the x axis as a function of distance from the center of the dipole. Vary x from 0.2 nm ( ) to 0.5 nm from the center of the dipole. Do the calculation two different ways, and put both plots (in different colors) on the same axes: (a) Calculate the electric field exactly as the superposition of the fields due to the individual charges. (b) Calculate the electric field using the approximate formula for the dipole field derived in Section 14.6.

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Rescale your graph if necessary so you can see differences in the two values. (c) Comment on the validity of the approximate formula. How close to the dipole (compared to s, the dipole separation) do you have to get before the approximate formula no longer gives good results? What is your criterion for “good results”?

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Electric Fields and Matter

KEY IDEAS The net charge of an object is the sum of the charges of all its constituent charged particles.

▪ A neutral object has a net charge of zero. Charge is conserved: The change in the net charge of a system plus the change in the net charge of the surroundings is zero. An object containing charged particles that are freely mobile is called a conductor. An object containing charged particles that can move only very small distances is called an insulator. Contact between two objects can sometimes result in a transfer of charge from one to the other. Charged objects can interact electrically with neutral objects. The distribution of positive and negative charges within an object can be altered by an electric field due to an external object. This is called polarization.

▪ Polarization occurs extremely rapidly. ▪ A polarized object, even if neutral, can create a nonzero electric field in the surrounding space.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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CHARGED PARTICLES IN MATTER Since ordinary matter is composed of charged particles, electric fields can affect matter. In order to understand the effect of electric fields on matter, in this chapter we will extend our microscopic model of matter to include the fact that matter contains charged particles: protons and electrons.

Net Charge

DEFINITION OF NET CHARGE The net charge of an object is the sum of the charges of all of its constituent particles. An object with a net charge of zero is called “neutral.” An object with a nonzero net charge (either positive or negative) is called “charged.”

Elementary particles such as protons and electrons are electrically charged. If a proton and an electron combine to form a hydrogen atom, however, the hydrogen atom is electrically “neutral”—its net charge is the sum of the charges of its constituent particles, which in this case is zero: A sodium atom has 11 protons in its nucleus and 11 electrons surrounding the nucleus, so it has a net charge of zero and is electrically neutral. However, a sodium atom can lose an electron, becoming a sodium ion, Na +.

QUESTION What is the net charge of a sodium ion, Na + (Figure 15.1)?

Figure 15.1 A sodium ion Na + consists of a sodium atom that has lost an electron.

A sodium ion has 11 protons and 10 electrons, so its net charge is

Ordinary matter is electrically neutral. However, it is possible to remove or add charged particles, giving an object a nonzero net charge.

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Is a dipole electrically neutral?

A dipole is neutral, because the sum of its constituent charges is zero: Evidently even a neutral object can make a nonzero electric field in the surrounding space.

Conservation of Charge Conservation of charge is a fundamental principle, because it applies to every system in every situation.

In an extremely wide variety of experiments, no one has ever observed a change in the net charge of the universe. These results are summarized by the fundamental principle called “conservation of charge”: if the net charge of a system changes, the net charge of the surroundings must change by the opposite amount. For example, if your comb acquires negative charge, your hair acquires an equal amount of positive charge:

CONSERVATION OF CHARGE The net charge of a system plus its surroundings cannot change.

Consider the annihilation reaction between an electron and a positron:

QUESTION In this reaction an electron and a positron are destroyed, creating two high-energy photons (called gamma rays). Does this reaction violate the principle of conservation of charge?

No. The net charge of the system (electron plus positron) was initially zero; the charge of the two photons is also zero. Even though charged particles were destroyed, the net charge of the system did not change. 15.X.1 A carbon atom is composed of 6 protons, 6 neutrons, and 6 electrons. What is the net charge of this atom? 15.X.2 A neutral chlorine atom contains 17 protons and 17 electrons. When a chlorine atom gains one extra electron, it becomes a chloride ion. What is the net charge of a chloride ion?

Answer

Answer

Conductors and Insulators All materials are made of atoms that contain electrons and protons. However, at the microscopic level there can be differences in structure that lead to very different behavior when macroscopic objects are exposed to electric fields. In this chapter we will examine two different kinds of materials: conductors and insulators. (There are other classes of technologically important materials, such as semiconductors and superconductors, which we will discuss briefly in later chapters.)

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INSULATORS Many materials are made up of molecules that do not easily break apart, and whose electrons are tightly bound to the molecules and can move only very short distances (typically less than the diameter of an atom). These materials are called insulators because such materials can electrically “insulate” one charged object from another, since charged particles cannot flow through the material. You are familiar with many insulating materials, such as rubber, most plastics, wood, paper, and glass.

Related experiment: 15.EXP.23

CONDUCTORS Other materials contain charged particles, such as ions or electrons, that are free to move through the material. These materials are called conductors. Most metals, such as copper, silver, iron, aluminum, and gold, are excellent conductors; aqueous salt solutions are also conductors. In the next section we will focus on interactions involving materials that are insulators; we will discuss conductors, particularly metals, in a subsequent section.

DEFINITION OF “CONDUCTOR” AND “INSULATOR” A conductor contains mobile charged particles that can move throughout the material.An insulator has no mobile charged particles.

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HOW INSULATORS BECOME CHARGED Many of the interactions we observe in our everyday lives are electric in nature. By being systematic in observing the behavior of simple systems, and by thinking carefully in our analysis of this behavior, we can uncover some deep questions about the interaction of ordinary matter with electric fields. In the section “Basic Experiments” at the end of this chapter are some simple experiments that you can do to make the issues vivid. All you need is a roll of “invisible” tape such as Scotch® brand Magic™ Tape or a generic brand of frosted tape (Figure 15.2). When you pull a long piece of invisible tape off a roll, it often curls up or sticks to your hand, and this is due to electric interactions between the electrically charged tape and your hand. We encourage you to do these simple experiments as you study this chapter.

Figure 15.2 Pulling a piece of invisible tape off of another piece charges the tapes electrically. See Experiment 15.EXP.13.

In experimenting with charged objects such as invisible tape we find the following: There are two kinds of charge, (which are called “+” and “−”). Like charges repel (Figure 15.3), unlike charges attract.

Figure 15.3 Two pieces of invisible tape with like charges repel each other. The electric force

▪ acts along a line between the charges, ▪ decreases rapidly as the distance between the charges increases, and ▪ is proportional to the amounts of both charges.

Charging by Rubbing Related experiments: Experiments

A charged object that has a net negative charge has more electrons than protons. A positively charged object has fewer electrons than protons.

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QUESTION Normally most ordinary objects are neutral. How can they become charged?

It has been known for a long time that if you rub a glass rod with silk, the glass rod becomes positively charged and the silk negatively charged (Figure 15.4). If you rub a clear plastic object such as a pen through your hair (or with fur, wool, or even cotton), the plastic ends up having a negative charge and so repels electrons (Figure 15.5). A similar process occurs when you separate one piece of invisible tape from another. Many objects made of insulating material acquire a nonzero net charge through contact with other objects.

Figure 15.4 Rub silk on glass and the glass becomes positively charged.

Figure 15.5 Rub a plastic pen on your hair or shirt and the pen becomes negatively charged.

There are a variety of possible explanations for this phenomenon. Electrons could be removed from one object and transferred to the other object. Large organic molecules in the plastic or your hair may break at their weakest bond in such a way that negative ions (negatively charged fragments) are deposited on the plastic and/or positive ions (positively charged fragments) are deposited on your hair. It may be significant that almost the only materials that can be charged easily by rubbing are those that contain large organic molecules, which can be broken fairly easily. It is typically more difficult to pull single electrons out of atoms or molecules, although we cannot rule out the possibility of stripping a single electron out of a molecule. Glass (silicon dioxide) is one of the few common inorganic materials that can be charged easily by rubbing, with silk. It may be that positive ions break off the large organic molecules in the silk and are deposited on the glass, or that silk strips single electrons off of glass. Molecular breakage or electron transfer provides an explanation of our puzzle as to why tapes and combs get charged, but such details as to why the plastic rather than your hair becomes negative are the subject of continuing research by physicists, chemists, and materials scientists. Part of the complexity of these phenomena is due to the fact that they are surface phenomena. The special nature of intermolecular interactions at the surface of a solid are generally less well understood than those in the interior, and there is a great deal of current research on the properties of surfaces. Moreover, unless one takes extraordinary precautions, real surfaces are

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always “dirty” with various kinds of (possibly charged) contaminants, which further complicates any prediction about the effect of rubbing, which may remove or deposit charged contaminants. It is known that rubbing is not essential to transferring charge from one object to another. Mere contact is sufficient. However, rubbing produces many points of contact, which facilitates transfer.

Protons Are Not Removed from Nuclei One thing is certain: you cannot remove bare nuclei from inside the surface atoms or remove protons from inside the nuclei of the surface atoms by rubbing. The amount of energy required to do this would be enormous. Removing protons would amount to transmuting one element into another! The nucleus is buried deep inside the atom, and the protons are bound tightly in the nucleus. A much smaller force is required to remove one electron from an atom, or to break a chemical bond and transfer an entire ion to another object. Therefore the only charged objects that can be transferred by rubbing are positive or negative ions, or electrons. We can make an approximate comparison of the energy required to charge an object by different mechanisms: breaking a bond, removing an electron from an atom, or removing a proton from a nucleus. We saw in Chapter 8 (Energy Quantization) that the energy required to ionize a hydrogen atom (that is, to move the electron very far away from the nucleus) was about 14 eV (recall that 1 eV = 1.6 × 10−19 J), so we can estimate the energy required to remove an electron from any atom as about 10 eV. The energy required to break one of the oxygen–hydrogen bonds in water is about 450 kJ/mol, or about 4.6 eV per bond, so we can estimate the energy required to break an average chemical bond as about 5 eV. As we saw in Chapter 6 (The Energy Principle), an input of about 2.2 MeV (2.2 × 106 eV) is required to break apart the nucleus of a deuterium atom, which consists of one proton and one neutron, so we can estimate the energy required to remove a proton from a nucleus as about 1 × 106 eV.

Mechanism

Energy Required

Break chemical bond

≈ 5 eV

Remove electron

≈ 10 eV

Remove proton from nucleus

≈ 1 × 106 eV

It is clear that either breaking bonds or removing single electrons is a possible mechanism for charging a macroscopic object by rubbing, but removing protons from the atomic nuclei is not!

The Location of Charge Transfer Suppose that two neutral strips of invisible tape are stuck together, then pulled quickly apart, as shown in Figure 15.6. Both tapes will become charged as a result of this process. (The sign of the charge on each tape depends on the chemical composition of the tape and the glue, which varies from brand to brand.)

Figure 15.6 Two initially neutral tapes being pulled apart and becoming charged. Where are the regions with excess + and − charges?

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QUESTION If the upper tape ends up being negatively charged after the tapes are pulled apart, what is the charge of the lower tape?

The principle of conservation of charge requires that the net charge of the system remain unchanged. Since the tapes started out neutral, the lower tape must now be positively charged, and the absolute value of the charge of each tape must be the same. For example, if the charge of the upper tape is −1nC, the charge of the lower tape must be +1nC (1 nC = 1 × 10−9 C).

QUESTION After the tapes are separated, where are the excess charges on the negative upper tape located? Where are the excess charges on the positive lower tape located?

The tapes interact only along the bottom surface of the upper tape and the top surface of the lower tape. Any excess charges must end up on one of these surfaces, as shown in Figure 15.7. The parts of the tapes that have been separated now have an excess or a deficiency of negative charge. The particles that moved from one tape to the other may have been electrons, or they may have been negative or positive ions.

Figure 15.7 In this case the upper tape becomes negatively charged, so there must be excess electrons or negative ions on the part of its lower surface that is no longer touching the lower tape. The corresponding part of the lower tape has a deficiency of electrons, so there are now positive ions on it.

How Much Charge Is on a Charged Object? Objects like invisible tapes, plastic pens, Ping-Pong balls, balloons, and glass rods can be electrically charged by rubbing with an appropriate material, ripping apart, or other similar contact. It would be useful to know approximately how much excess charge is on an ordinary small object like these when it is charged. Even knowing an approximate order of magnitude would be useful—is it closer to 10 coulombs, 0.1 coulomb, or 1 × 10−10 coulombs?

EXAMPLE Approximate Amount of Charge on a Piece of Tape We can make a rough estimate of the amount of excess charge on a piece of tape by doing a simple experiment, illustrated in Figure 15.8. Prepare a pair of oppositely charged invisible tapes prepared by pulling a pair of initially neutral tapes apart, so that the magnitude of charge on the two tapes is the same. If we suspend the positively charged tape between two books and lower the negatively charged tape toward it, when the negative tape gets close enough we will see that the positive tape is lifted upward. If we know the mass of the tape, we can estimate the electric force on the tape, and from this and the distance between the tapes, estimate the charge on the tape.

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Figure 15.8 Tapes A and B are oppositely charged. When A gets sufficiently close, the electrical force on B is sufficient to lift it up.

Solution Momentum Principle:

In Figure 15.8, estimate the electric field due to A by approximating A as a point charge (a very rough approximation):

so

If the 20-cm-long piece of tape has a mass of about 0.15g, and the bottom tape starts to be lifted when the top tape is about 2.5 cm away, then since the magnitude of the charge on the tapes is the same,

FURTHER DISCUSSION This is a rough estimate, but it turns out to be a reasonable one—a small object charged by rubbing usually has a charge on the order of 10 nCs. In Chapter 16 we'll see how to calculate the electric field of an object that is not pointlike but has charge spread over its entire surface. Using these techniques, we find that the electric field at a location very close to the surface of the charged tape is around 2 × 105 N/C. Since the electric field required to ionize the air itself is about 3 × 106 NC, the field near the surface of the charged tape is pretty large.

Experiment 15.EXP.18 guides you through a more careful version of this experiment. Your results may be

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somewhat different, depending on the design of the experiment (there are other possible geometries), the length of your tapes, and other factors.

Fraction of Surface Atoms with Excess Charge Assuming that the net charge on the negative tape is distributed uniformly over its surface, we can estimate the fraction of those surface atoms that have gained an excess electron or negative ion.

EXAMPLE Fraction of Surface Atoms with Excess Charge What fraction of the atoms on the charged surface of the tape have gained or lost charge? (Assume that an atom gains or loses at most 1 electron charge.)

Solution Surface area of a 1-cm-wide, 20-cm-long tape: Approximate cross-sectional area occupied by an atom whose radius is approximately 1 × 10−10 m:

Number of atoms on the surface:

Number of excess electrons (or ions):

Fraction of surface atoms with excess charge:

Thus only about one in a million atoms on the surface of the tape has acquired an excess electron or lost an electron—a small fraction.

FURTHER DISCUSSION At the macroscopic level, the charge on the surface of the tape may appear to be distributed quite uniformly. At the atomic level, though, we see that the atoms with excess charge are sprinkled quite sparsely over the surface.

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POLARIZATION A positively charged object, like a positive piece of invisible tape, is attracted not only to a negative tape, but also to your hand, your desk, your book, and every other nearby neutral object (Figure 15.9). The same is true for a negatively charged object, such as a negative piece of invisible tape. Since

Figure 15.9 A charged tape, whether positive or negative, is attracted to your neutral hand.

this implies that your neutral hand must make a nonzero electric field. It is not obvious how or why this can happen.

QUESTION Why are charged objects attracted to neutral objects?

The attraction of both positively and negatively charged invisible tape to your hand, and to many other neutral objects, is deeply mysterious. The net charge of a neutral object is zero, so your neutral hand should not make an electric field that could act on a charged tape; nor should your neutral hand experience a force due to the electric field made by a charged tape. Nothing in our statement of the properties of electric interactions allows us to explain this attraction!

The Structure of an Atom An external charge can cause a shift in the position of the charges that make up a neutral atom or molecule. To see this clearly we need to look in more detail at the structure of atoms. We'll consider a hydrogen atom because it is the simplest atom, but the effects we discuss occur with other atoms as well. In Figure 15.10 we show a special kind of picture of a hydrogen atom, based on quantum mechanics, the theory that describes the detailed structure of atoms. A hydrogen atom consists of an electron and a nucleus normally consisting of one proton (and no neutrons). The lightweight electron doesn't follow a well-defined orbit around the heavy nucleus the way the Earth does around the Sun. Rather, there is only a probability for finding the electron in any particular place.

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Figure 15.10 A quantum-mechanical view of a hydrogen atom. The picture is a two-dimensional slice through a threedimensional spherical distribution. Each dot represents the location of the electron at the time of a multipleexposure photo. The tiny nucleus is shown as a red dot at the center of the electron cloud (actual nuclear radius is only about 1 × 10−15 m). A circle is drawn through regions of constant density.

Figure 15.10 shows this probability graphically. You can think of the picture as a multiple exposure. For each exposure, the position of the electron at that time is shown as a dot. Because the electron is most likely to be found near the nucleus, that part of the multiple exposure is so dark you can't see the individual dots. The electron is seldom found a long way from the nucleus, so as you get farther and farther from the nucleus the density of dots gets less and less. We call this probability distribution the “electron cloud.” In hydrogen the cloud consists of just one electron, but in other atoms the electron cloud is made up of many electrons. The average location of the electron is in the center, at the same location as the nucleus. You're just as likely to find the electron to the right of the nucleus as to the left of the nucleus. It is impossible to show the nucleus accurately on this scale. Although the mass of a proton is 2000 times the mass of an electron, the radius of the proton, about 1 × 10−15 cm, is only about 1/100,000 as big as the radius of the electron cloud, which is itself only about 1 × 10−10 meters! We used an oversize red dot to mark the position of the tiny nucleus in Figure 15.10. In the following exercise, remember that in the previous chapter we pointed out that the electric field produced by a uniformly distributed sphere of charge, outside the sphere, is the same as though all the charge were located at the center of the sphere (this will be discussed in more detail in Chapter 16). 15.X.3 A student asked, “Since the positive nucleus of the atom is hidden inside a negative electron cloud, why doesn't all matter appear to be negatively charged?” Explain to the student the flaw in this reasoning.

Answer

Polarization of Atoms If the electron cloud in an atom could be considered to be spherically uniform and always centered on the nucleus, a neutral atom would have no interaction with an external charge. If the electron cloud is centered on the nucleus, the electric field produced by the N electrons would exactly cancel the field produced by the N protons. However, the electron cloud doesn't always stay centered, as we'll see next. In an atom the electron cloud is not rigidly connected to the nucleus. The electron cloud and the nucleus can move relative to each other. If an external charge is nearby, it creates an electric field, which exerts forces on the electron cloud and on the nucleus. Under the influence of this “applied” electric field the electron cloud and the nucleus shift position relative to each other. Figure 15.11 shows the probability distribution or electron cloud for hydrogen when there is an external positive charge located somewhere to the left of the hydrogen atom. You can see that the cloud has been distorted, because the positive charge attracts the electron to the left and repels the nucleus to the right.

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Figure 15.11 A positive charge makes an electric field that shifts the electron cloud of the hydrogen atom to the left (and shifts the hydrogen nucleus to the right). An ellipse is drawn through regions of constant density. It is now more probable that the electron will be found to the left of the nucleus than to the right.

Average Location of the Electron The average location of the electron is now not at the center where the nucleus is located, but is displaced somewhat to the left of the nucleus. That is, each time you take a snapshot, you're more likely to find the electron to the left of the nucleus than to the right of the nucleus. The hydrogen atom isn't immediately torn apart, because the attraction between the nucleus and the electron is stronger than the forces exerted by the distant external charge. However, if the external charge gets very close the hydrogen atom may break up or react with the external charge. If the external charge were a proton, it could combine with the hydrogen atom to form ionized molecular hydrogen (H2+). You can see in Figure 15.11 that the outer regions of the cloud are affected the most by the external charge. This is because in the outer regions the electron is farther from the nucleus and can be influenced more by the external positive charge. In an atom containing several electrons, the outer electrons are affected the most. The picture is deliberately exaggerated to show the effect: unless the polarization is caused by charges only a few atomic diameters away, the shift in the electron cloud is normally too small to represent accurately in a drawing. An atom is said to be “polarized” when its electron cloud has been shifted by the influence of an external charge so that the electron cloud is not centered on the nucleus.

Diagrams of Polarized Atoms or Molecules For most purposes we can approximate the charge distribution of the polarized atom as consisting of an approximately spherical negative cloud whose center is displaced from the positive nucleus (Figure 15.12). A uniform spherical charge distribution acts as if it were a point charge located at the center of the sphere, both in the sense that it makes an electric field outside the sphere identical to the electric field of a point charge and that it responds to applied fields as though it were a point charge. It is therefore reasonable to model a polarized atom as a dipole, consisting of two opposite point charges separated by a small distance.

Figure 15.12 We can approximate a polarized atom as a roughly spherical electron cloud whose center is displaced from the positive nucleus.

To simplify drawing a polarized atom or molecule and to emphasize its most important aspects, we will usually represent it as an

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exaggeratedly elongated blob, with + and − at the ends (Figure 15.13).

Figure 15.13 A simplified representation of a polarized atom or molecule.

Induced Dipoles Are Created by Applied Electric Fields Figure 15.12 shows quite clearly that a polarized atom or molecule is a dipole, since there are two opposite charges separated by a distance. However, the polarized atom or molecule is not a permanent dipole. If the applied electric field is removed (for example, by removing the charges making that field), the electron cloud will shift back to its original position, and there will no longer be any charge separation. We call the polarized atom or molecule an “induced” dipole, because the dipole was induced (caused) to form by the presence of an applied electric field. An “induced dipole” is created when a neutral object is polarized by an applied electric field. The induced dipole will vanish if the applied field is removed. A “permanent dipole” consists of two opposite charges separated by a fixed distance, such as HCl or H 2O molecules, or the dipole that can be constructed out of + and − tapes (Experiment 15.EXP.20).

Polarizability It has been found experimentally that for almost all materials, the amount of polarization induced (that is, the dipole moment of the polarized atoms or molecules) is directly proportional to the magnitude of the applied electric field. This result can be written like this:

The magnitude of the electric dipole moment of a dipole is:

and the direction is from the negative charge toward the positive charge (see Chapter 14). The constant a is called the “polarizability” of a particular material. The polarizability of many materials has been measured experimentally, and these experimental values may be found in reference volumes. 15.X.4 In an induced dipole, is the distance between the charges fixed, or can it vary? Explain. 15.X.5 A typical atomic polarizability is 1 × 10−40 C · m/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying a large field of 3 × 106 N/C, which is large enough to cause a spark in air?

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Answer

Answer

In the previous chapter we found that since the electric field of a dipole was proportional to 1/r 3, the force exerted by a dipole on a point charge was also proportional to 1/r 3. Because of the reciprocity of the electric force, the force on the dipole by the point charge was therefore also proportional to 1/r 3. Let us extend this analysis by considering the case of a point charge q1 and a neutral atom. Even though the entire process happens very quickly, it is instructive to analyze it as if it occurred in several steps. (Of course, the process is not instantaneous, since information about changes in electric field takes a finite time to propagate to distant locations.) due to the point charge (Figure 15.14). This electric field affects Step 1: At the location of the atom there is an electric field both the nucleus and the electron cloud, both of which, due to their spherical symmetry, can be modeled as point charges. The force on the electron cloud and the force on the nucleus are in opposite directions. Since the electron cloud and the nucleus can move relative to each other, they shift in opposite directions, until a new equilibrium position is reached.

Figure 15.14 At the location of the atom there is an electric field

The atom is now polarized, with dipole moment

due to the point charge.

proportional to the applied electric field

.

. The atom, which is now an induced dipole, makes an electric field Step 2: The polarized atom now has a dipole moment at the location of the point charge (Figure 15.15). We can write an expression for the magnitude of :

Figure 15.15 The polarized atom makes an electric field Since we know

at the location of the point charge.

, the electric field of the point charge at the location of the dipole, we can put that into our equation:

QUESTION Step 3: What is the force exerted on the point charge by the induced dipole (Figure 15.16)?

Figure 15.16 The force on the point charge due to the electric field of the polarized atom is equal in magnitude to the force

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on the polarized atom due to the electric field of the point charge.

We find that the force on the point charge by the polarized atom is proportional to 1/r 5.

QUESTION What is the force on the neutral atom by the point charge?

Because of the reciprocity of the electric interaction (Newton's third law), the force on the neutral atom by the point charge is equal in magnitude and opposite in direction to the force on the point charge by the neutral atom:

so the force on a (polarized) neutral atom by a point charge also is proportional to 1/r 5. 15.X.6 Atom A is easier to polarize than atom B. Which atom, A or B, would experience a greater attraction to a point charge a distance r away? 15.X.7 If the distance between a neutral atom and a point charge is doubled, by what factor does the force on the atom by the point charge change?

Answer

Answer

Interaction of Charged Objects and Neutral Matter We are now in a position to explain why both positively and negatively charged objects (such as + and − tapes) are strongly attracted to neutral matter.

QUESTION Try to explain in detail what happens when a positively charged tape is brought near your hand. This is a complex process; consider all the interactions involved.

In considering the interactions of fields and matter, the following scheme is useful. (1) Identify any sources of electric fields. (2) Identify any charges at other locations that can be affected by these fields. (3) Redistribution of the affected charges may create an electric field at the location of the original source charges: are they affected? The positively charged tape makes an electric field, which points away from the tape. This electric field is present inside your hand, and affects atoms, molecules, and ions inside your hand. Figure 15.17 shows the polarization caused inside your finger by the electric field of the tape. The induced dipoles in your finger create an electric field at the location of the tape, which attracts the tape. You should be able to construct a diagram like the one in Figure 15.17 illustrating what happens when a negatively charged tape interacts with your finger.

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Figure 15.17 The electric field of a positively charged tape polarizes your finger. The induced dipoles in your finger create an electric field at the location of the tape, which attracts the tape.

Related experiment: 15.EXP.20

You may have noticed that the attraction between your neutral hand and a hanging charged invisible tape changes much more rapidly with distance (1/r 5) than does the interaction between two charged tapes (1/r 2). 15.X.8 Explain in detail, including diagrams, what happens when a negatively charged tape is brought near your finger.

Answer

Determining the Charge of an Object Suppose that you have a negatively charged tape hanging from the desk, and you rub a wooden pencil on a wool sweater and bring it near the tape.

QUESTION If the tape swings toward the pencil, does this show that the pencil had been charged positively by rubbing it on the wool?

Not necessarily. Even if the pencil is uncharged, the charged tape will polarize the pencil and be attracted by the induced dipoles.

QUESTION Can a charged object repel a neutral object? Why or why not? Draw diagrams to help you make your point.

Polarization always brings the unlike-sign charge closer, yielding a net attraction. Repulsion of an induced dipole can't happen. Therefore repulsion is the better test of whether an object is charged.

Electric Field Penetrates Intervening Matter The superposition principle states that the presence of matter does not affect the electric field produced by a charged object. Intervening matter does not “screen” or “shield” the electric field, just as your desk does not “screen” or “shield” your book from the gravitational field of the Earth.

Related experiment: 15.EXP.22

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You may have already observed one case of electric field passing through intervening matter. You see the same interaction between a charged invisible tape and your hand, or another tape, when approaching either side of a hanging tape, despite the charges being on just one side of the tape. The charges are initially on either the slick side or the sticky side (depending on whether it is the upper or lower tape in a pair of tapes), and one can show that the charges can't move through the tape to the other side (Experiment 15.EXP.23).

Intervening Matter and Superposition The fact that an electric field acts through intervening matter is another example of the superposition principle. It is true that the repulsion between two like-charged pieces of tape is weaker when a piece of paper is in the way (Experiment 15.EXP.22), but when viewed in terms of the superposition principle this reduction is not due to the paper partially “blocking” the field of the other tape. Rather, we say that the net field is due to the superposition of two fields: the same field that you would have had without the paper intervening, plus another field due to the induced dipoles in the paper. At this time we can't prove that this view is correct and that there is no “blocking” of electric field. However, we will find repeatedly that the superposition principle makes the right predictions for a broad range of phenomena and offers a simpler explanation than any kind of hypothetical “blocking” effect.

Figure 15.18 Two like charged tapes repel each other even if another object such as a piece of paper is placed between them. Since the tapes are also attracted to the paper, the net effect may be small.

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POLARIZATION OF INSULATORS In insulators, all of the electrons are firmly bound to the atoms or molecules making up the material. We have seen that an individual atom or molecule can be polarized by an applied electric field, producing an induced dipole of atomic or molecular dimensions. The electrons in an atom or molecule of an insulator shift position slightly, but remain bound to the molecule—no charged particles can move more than about one atomic diameter, or 1 × 10−10 m (most move much less than this distance; see Exercise 15.X.5). In Figure 15.19 we show a solid block of insulating material, each of whose molecules has been polarized by an applied electric field (that is, an electric field made by external charges—in this case a single positive charge). The molecules are of course not shown to scale! This is an example of “induced polarization”—the electric field has induced the normally unpolarized insulator to become polarized. In each molecule the electrons have moved a very short distance, and the molecules themselves are not free to move. However, the net effect can be very large because there are many molecules in the insulator to be affected. Note that the polarized molecules align with the electric field that is polarizing them, and that the stronger the electric field the larger the “stretch” of the induced dipole.

Figure 15.19 A block of insulating material (plastic, glass, etc.) polarized by an applied electric field. The molecules are not shown to scale!

Polarization Happens Very Rapidly Because the electron cloud is displaced only a tiny distance when an atom or molecule polarizes, this process happens extremely rapidly. The process can take much less than a nanosecond to complete.

Diagrams Showing Polarization of Insulators In diagrams of insulators we show polarized molecules exaggerated in size, to indicate that individual molecules in an insulator polarize, but the electrons remain bound to the molecule. We show the extent of polarization by the degree to which the molecule is “stretched.” Keep this diagrammatic convention in mind, and compare it to diagrams of polarized conductors in the following sections.

Charge on or in an Insulator

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Since there are no mobile charged particles in an insulator, excess charges stay where they are. Excess charge can be located in the interior of an insulator, or can be bound to a particular spot on the surface without spreading out along that surface (Figure 15.20).

Figure 15.20 In an insulator, charge can occur in patches on the surface, and there can be excess charge inside.

Low-Density Approximation When atoms or molecules in an insulator polarize in response to an electric field created by external charged objects, the polarized molecules themselves make electric fields that affect neighboring molecules. Because the effect of polarized molecules on each other is typically small compared to the effect of the original applied field, we will neglect this when discussing polarized insulators. In formal terms, when an electric field E applied is applied to a dense material (a solid or a liquid), the induced dipole moment of one of the atoms or molecules in the material isn't simply p = αE applied , but is really

, where E dipoles is

the additional electric field at the location of one of the molecules, due to all the other induced dipoles in the material. In this text we make the simplifying assumption of low density and assume that Edipoles is small compared to E applied . This is good enough for our purposes, but accurate measurements of polarizability must take this effect into account.

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POLARIZATION OF CONDUCTORS As we stated earlier, a conductor has some kind of charged particles that can move freely throughout the material. In contrast to an insulator, where electrons and nuclei can move only very small distances (around 1 × 10−10 m, or much less), the charged particles in a conductor are free to move large distances.

Ionic Solutions Ionic solutions are conductors, such as a solution of sodium chloride (table salt) in water. In salt water, the mobile charged particles are Na + ions and Cl − ions (Figure 15.21; there are also very small concentrations of H + and OH − ions, which are not shown).

Figure 15.21 A beaker containing an ionic solution (salt water).

QUESTION What happens when an electric field is created in the region of the beaker?

When an electric field is applied to a conductor, the mobile charged particles begin to move in the direction of the force exerted on them by the field. However, as the charges move, they begin to pile up in one location, creating a concentration of charge that itself creates an electric field in the region occupied by the remaining mobile charges. The net electric field in the region is the superposition of the applied (external) field and the electric field created by the relocated charges in the material. Figure 15.22 is a diagram of the polarization that occurs in the salt water. The ions (charged atoms or molecules) are in constant motion, so the actual situation isn't simple. Moreover, the interior of the liquid is full of positive and negative ions; there's just a slight excess concentration of ions near the sides of the beaker.

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Figure 15.22 Under the influence of an applied electric field (labeled

), the liquid polarizes. There is a slight excess ion

concentration at the two sides of the beaker. The electric field due to the redistributed ions is not shown.

Drift Speed and Applied Electric Field If a beaker of salt water is placed in a region where there is an electric field (due to charges outside the beaker), a sodium ion or a chloride ion will experience an electric force, and will begin to move in the direction of the force. However, even if the force remains constant, the ion will not keep accelerating, because it will collide with water molecules or with other ions. In effect, there is a kind of friction at the microscopic level. To keep the ions in a salt solution moving at a constant speed, a constant electric field must be applied to the solution. The speed at which mobile charges (in this case, sodium or chloride ions) move through a conductor is called the drift speed. Drift speed is directly proportional to the net electric field at the location of the charge. The proportionality constant is called the mobility of the mobile charges.

DRIFT SPEED

is the average drift speed of a mobile charge. u is the mobility of the charge. The units of mobility are

.

is the magnitude of the net electric field at the location of the mobile charge.

As implied by this equation, if the net electric field at the location of a mobile charge in a conductor is zero, the charge will stop moving. 15.X.9 An electric field of magnitude 190 N/C is applied to a solution containing chloride ions. The mobility of chloride ions in solution is 7.91 × 10−8 (m/s)/(N/C). What is the average drift speed of the chloride ions in the solution?

Answer

The Polarization Process in an Ionic Solution Polarization occurs very rapidly, but it is not instantaneous. Let's “slow down time” so we can talk about the process of polarization; we'll operate on a time scale of attoseconds (1 × 10−18 seconds!). To simplify our analysis, we'll imagine that we are able temporarily to “freeze” the ions in the salt water, and to release them after we have brought charges nearby to apply an electric field. Consider the net electric field at a location in the interior of the liquid, at a time attoseconds after polarization has begun, but long before the process has finished. The electric field E app due to external charges is shown in Figure 15.23, and also a smaller electric field E pol due to the polarization charges present at this time. The net electric field at a location in the middle of the liquid is now smaller than it was before polarization began.

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Figure 15.23 An intermediate stage in the polarization process, before polarization is complete.

QUESTION Will the polarization of the salt water increase beyond what it is now?

At this instant the net electric field in the solution still has magnitude greater than zero, so ions in the solution will still experience forces in the direction to increase the polarization. Because , the drift speed of the ions is not zero. More ions will pile up at the sides of the beaker, and the net electric field in the interior will be further weakened. Eventually the conductor will reach equilibrium on the microscopic level. Equilibrium at the microscopic level means that there is no net motion of mobile charges in any direction:

EQUILIBRIUM INSIDE A CONDUCTOR When a conductor is in equilibrium at the microscopic level: The average drift speed of the mobile charges inside the conductor is zero. There is no net flow of charges in any direction.

QUESTION How weak does the net electric field inside the conductor get? In the final state of equilibrium (when there is no further increase in polarization), how big is the net electric field in the interior of the liquid?

An Example of a “Proof by Contradiction” You may have correctly deduced that in the final state the net electric field in the conductor goes to zero at equilibrium. A rigorous way to reason about this using formal logic is to construct a “proof by contradiction.” In a proof by contradiction, we assume the opposite of what we want to prove, then, making valid logical deductions from this assumption, show that we reach a conclusion that is impossible or contradictory. We therefore conclude that the original assumption was wrong, and its opposite must be true. 1. Assume that in equilibrium the net electric field in the interior of an ionic solution is greater than zero. 2. Since E net >, all mobile ions in the solution will experience a nonzero force. Since u > 0, the average drift speed all ions move in the direction of the force. 3. Since , there is a net flow of charges. Therefore the system cannot be in equilibrium, because by definition in

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, and

equilibrium and there is no net flow of charges. This result, that above) that the ionic solution is in equilibrium.

, contradicts our original assumption (point 1

4. Because we have reached a contradiction, we must conclude that the original assumption (that the net electric field in the solution may be nonzero in equilibrium) is wrong. Thus, we conclude that the net electric field in an ionic solution in equilibrium must be zero.

This reasoning holds true for any conductor, including not only ionic solutions, but solid metal objects as well.

Superposition Note that the electric field inside the liquid is zero, not because of any “blocking” of fields due to external charges, but by the superposition of two effects: the effect of the external charges and the effect of the polarization charges. This is another example of the superposition principle in action. It is not true that the net electric field in a solution is zero at all times. While the ionic solution is in the process of polarizing, it is not in equilibrium; there is a nonzero electric field, and hence a nonzero force on an ion in the liquid, as you saw above. If electrodes are placed in the ionic solution and connected to a battery, the battery prevents the system from reaching equilibrium. In such a case (no equilibrium), there can be a field continuously acting on ions inside the liquid, resulting in continuous shifting of the ions through the liquid, constituting an electric current. Since there are a very large number of ions in the solution, none of them has to move very far during the polarization process. Even a tiny shift leads to the buildup of an electric field large enough to cancel out the applied electric field.

Polarization of Salt Water in the Body Your own body consists mainly of salt water, including the blood and the insides of cells. Look again at the diagrams in which you focused on the way an external charge polarizes individual molecules inside your finger. An additional effect is the polarization of the salt water inside your finger. As shown in Figure 15.22, there will be a shift of Na + and Cl − ions in the blood and tissues. This shift may be a larger effect than the molecular polarization. It is a bit unsettling to realize that a charged tape or comb messes with the inside of your body!

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A MODEL OF A METAL You probably know that metals are very good electrical conductors. In almost all metals, the mobile charged particles are electrons.

The Mobile Electron Sea The atoms in a solid piece of metal are arranged in a regular 3-D geometric array, called a “lattice” (Figure 15.24). The inner electrons of each metal atom are bound to the nucleus. Some of the outer electrons participate in chemical bonds between atoms (the “springs” in the ball-and-spring model of a solid). However, some of the outer electrons (usually one electron per atom) join a “sea” of mobile electrons that are free to move throughout the entire macroscopic piece of solid metal (Figure 15.25). In a sense, the entire hunk of metal is like one giant molecule, in which some of the electrons are spread out over the entire crystal. The electrons are not completely free; they are bound to the metal as a whole and are difficult to remove from the metal. (For example, electrons do not drip out when you shake a piece of metal!) Metals are excellent conductors because of the presence of these mobile electrons.

Figure 15.24 The ball-and-spring model of a solid.

Figure 15.25 A 2-D slice of an unpolarized metal: uniform mobile-electron sea (blue), positive atomic cores (red “+” symbols).

NO NET INTERACTION BETWEEN MOBILE ELECTRONS Although the roaming electrons repel each other strongly, this repulsion between electrons is neutralized on the average by the attractions exerted by the positive atomic cores (a “core” is a neutral atom minus its roaming electron, so it has a charge of +e). The effect is that on average, the net electric field inside a piece of metal in equilibrium is zero.

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Because of this, in some ways the mobile electrons look like an ideal gas: they move in a region free from the electric field, so they appear not to interact with each other or with the atomic cores. In fact, in some simple models of electron motion the mobile electron sea is treated as an ideal gas.

The Drude Model In a simple classical model of electron motion (called the “Drude model” after the physicist who first proposed it), a mobile electron in the metal, under the influence of the electric field inside the metal, does accelerate and gain energy, but then it loses that energy by colliding with the lattice of atomic cores, which is vibrating because of its own thermal energy and acquires more thermal energy due to the collisions of the electrons with the lattice. After a collision, an electron again gets accelerated, and again collides. This process is what makes the metal filament in a lightbulb get hot. Figure 15.26 shows a graph of this start–stop motion for a single electron.

Figure 15.26 A mobile electron speeds up under the influence of the electric field inside the wire, then collides with an atomic core and loses energy.

Drift Speed and Electron Mobility The average speed of an electron in this start–stop motion is called the “drift” speed , and we say that the electron “drifts” through the metal. Actually, the slow drift motion is superimposed on high-speed motion of the electrons in all directions inside the metal, much as the wind is a slow drift motion superimposed on the high-speed motion of air molecules in all directions. A full treatment of electrons in a metal, including the reason for the high-speed motion in all directions, requires quantum mechanics, but the simple classical Drude model allows us to understand most of the important aspects of circuits on a microscopic level. We express the momentum principle

in a form involving finite time steps for momentum in the direction of

,

where E net is the magnitude of the net electric field inside the wire, and Δt is the time between collisions. If we make the simplifying assumption that the electron loses all its momentum during each collision, we have

The speed of the electron (of mass m e ) at the time of collision turns out to be small compared to the speed of light, so we have

However, the time between collisions is not the same for all electrons. Some experience longer times between collisions, some shorter times. To get an average, “drift” speed for all electrons at a particular instant, we need the average time between collisions:

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, the average time between collisions of the electrons with the atomic cores, is determined by the high-speed random motion of the electrons and by the temperature of the metal. (At a higher temperature the thermal motion of the atomic cores is greater, and the average time between collisions is reduced, leading to a smaller drift speed for the same field E.)

QUESTION Is drift speed directly proportional to the magnitude of the electric field?

Assuming that increasing the electric field does not result in a significant change in temperature, then doubling the electric field E doubles the drift speed attained in that time; hence the drift speed is directly proportional to the electric field. The proportionality factor is called the electron “mobility” and is denoted by u (or by μ in some books). Evidently,

Different metals have different electron mobilities. The higher the mobility, the higher the drift speed for a given electric field. The direction of the drift velocity of a mobile (negatively charged) electron is opposite to the direction of the electric field. 15.X.10 The mobility of the mobile electrons in copper is 4.5 × 10−3 (m/s)/(N/C). How large an electric field would be required to give the mobile electrons in a block of copper a drift speed of 1 × 10−3 m/s?

Answer

Polarization Happens Very Quickly When an electric field (due to some external charges) is applied to a metal, the metal polarizes. We can describe the polarization of a metal as shifting the entire mobile electron sea relative to the fixed positive cores. In Figure 15.27, in response to an applied field, electrons have piled up on the left, creating a very thin negatively charged layer near the surface. There is a corresponding deficiency of electrons on the right, creating a very thin positively charged layer near the surface.

Figure 15.27 Polarized metal: mobile electron sea shifted left relative to the positive atomic cores, under the influence of an applied electric field. There is an excess of electrons on the left side and a deficiency of electrons on the right. These charges contribute to the net electric field inside the metal.

The shift in the electron sea is extremely small, much less than an atomic diameter! (See Exercise 15.X.5.) It is not necessary for

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electrons at one end of the block to move to the other end. Just displace the entire electron sea slightly and you have lots of excess electrons on one surface. Because the electron sea has to move only a tiny distance, this displacement can happen very rapidly—it can take much less than a nanosecond.

Diagrams Showing Polarization of Metals In Figure 15.28 we show a polarized metal in a simplified way that is both easier to draw and easier to interpret at a glance.

Figure 15.28 A simpler way to depict the polarization of a polarized metal. Excess charges are drawn outside the boundary lines, to indicate that they are on the surface.

We show − and + signs outside the surfaces to indicate which surfaces have thin layers of negative charge (electron excess) or positive charge (electron deficiency), as a result of shifts in the mobile electron sea. Note that by convention we draw + and − signs just outside the surface of a metal object to indicate that the excess charge (excess or deficiency of electrons) is on the surface of the object. If the charge is drawn inside the boundary, the diagram is ambiguous—it is not clear whether the charge is inside the object or on the surface. We do not show the positive atomic cores and the mobile electron sea inside the metal, because the interior is all neutral. The diagram is much easier to interpret if we do not clutter up neutral regions with charges that must be counted to see whether they balance.

Note that most charge buildup is typically on the ends of the metal, but that there is also a small amount of charge on the sides as well. Compare these conventions to the convention we used earlier to show the individual atoms or molecules polarize in an insulator.

Polarized and/or Charged Take care to use technical terms precisely. The metal block shown in Figure 15.28 is polarized. It is not charged; its net charge is still zero. On the other hand, a charged object can also be polarized. The positively charged metal block depicted schematically in Figure 15.29 is also polarized. “Polarized” and “charged” are not synonyms.

Figure 15.29 This metal block is both charged and polarized.

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Net Electric Field in a Metal Goes to Zero in Equilibrium The reasoning process that we went through when considering the polarization of ionic solutions applies equally well to metals, or to any conductor. We used proof by contradiction to demonstrate that in equilibrium the net electric field inside a conductor must be zero (because if it were not zero, mobile charged particles would move under the influence of the field, and the system would not be in equilibrium). It is intriguing that it is possible for mobile charges to rearrange themselves in such a way that the net electric field is zero not just at one single location, but also at every location inside the metal. It would be a very difficult problem for us to calculate exactly where to place charged particles to make a net field of zero inside a metal object, but in fact the many mobile charges do rearrange in just such a way as to accomplish this. It can be shown that it is because of the 1/r 2 distance dependence of the electric field that this is possible—if the exponent were not exactly 2.0, the world would be quite different. When equilibrium is reached in a metal, things are essentially unchanged in the interior of the metal. There is no excess charge—we still have a uniform sea of electrons filling the space around the positive atomic core. The net electric field inside the metal, which is the sum of the applied field and the field due to the charge buildup on the edges of the metal, is still zero.

At the surfaces there is some excess charge, so we can represent a polarized metal as having thin layers of charge on its surfaces but being unpolarized in the interior, unlike an insulator. The shifting of the mobile electron sea in metals is a much larger effect than occurs in insulators, where the polarization is limited by the fact that all the electrons, including the outermost ones, are bound to the atoms, unlike the situation in metals. A polarized insulator is a collection of tiny (molecule-sized) dipoles, whereas a polarized metal forms one giant dipole.

E is Not Always Zero inside a Metal Do not overgeneralize our previous conclusions. It is not true that the net electric field in a metal is zero at all times. While the metal is in the process of polarizing, the metal is not in equilibrium, and there is a nonzero electric field inside the metal, creating a nonzero force on electrons in the electron sea, as you saw above. In an electric circuit, the battery prevents the system from reaching equilibrium. In such a nonequilibrium situation, there can be an electric field inside the metal, and hence a force continuously acting on electrons in the mobile electron sea, resulting in continuous shifting of the electron sea around the closed circuit, constituting an electric current.

Excess Charges on Conductors Another important property of metals (and of the 1/r 2 property of the electric interaction, as we will see when we study Gauss's law in a later chapter) is that any excess charges on a piece of metal, or any conductor, are always found on an outer or inner surface. This makes intuitive sense, since any excess charges will repel each other and will end up as far apart as possible—on the surface of the conductor. Any multiatom region in the interior of the conductor has a net charge of zero. Moreover, the mutual repulsion among any excess charges makes the mobile electron sea redistribute itself in such a way that charge appears almost immediately all over the surface (Figure 15.30).

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Figure 15.30 In a metal, charge is spread all over the surface (not necessarily uniformly), and there is no excess charge inside.

Here is a summary of the behavior of conductors vs. insulators: Conductor

Insulator

Mobile charges

Yes

No

Polarization

Entire sea of mobile charges moves

Individual atoms or molecules polarize

Equilibrium

inside

nonzero inside

Location of excess charge

Only on surface

Anywhere on or inside material

Distribution of excess charge

Spread out over entire surface

Located in patches

EXAMPLE Plastic and Metal Rods A lightweight (conducting) metal ball hangs from a thread, to the right of an (insulating) plastic rod. Both are initially uncharged (Figure 15.31).

Figure 15.31 The plastic rod and ball are initially uncharged. (a) You rub the left end of the plastic rod with wool, depositing charged molecular fragments whose total (negative) charge is that of 1 × 109 electrons. You observe that the ball moves toward the rod, as shown in Figure 15.32.

Figure 15.32 The left end of the plastic rod was rubbed with wool and became negatively charged. Explain. Show all excess charged particles, polarization, and so on, clearly in a diagram. Make it clear whether charged particles that you show are on the surface of an object or inside it. (b) You perform a similar experiment with a (conducting) metal rod. You touch the left end of the rod with a charged metal object, depositing 1 × 109 excess electrons on the left end. You then remove the object. You see the ball deflect more than it did with the plastic rod in part (a), as shown in Figure 15.33.

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Figure 15.33 Replace the plastic rod with a neutral metal rod, then touch the left end with a negatively charged metal object. The ball deflects more than it did with the plastic rod.

Explain. Show all excess charged particles, polarization, and so on, clearly in a diagram. Make it clear whether charged particles that you show are on the surface of an object or inside it.

Solution (a) The plastic rod is an insulator, so the excess charge remains on the left-end surface of the rod. This charge polarizes the molecules inside the rod.The original charge plus the polarized molecules make a field that polarizes the neutral metal ball, as shown in Figure 15.34.

Figure 15.34 The field due to the plastic rod (charge on end plus polarized molecules) exerts a net force to the left on the polarized metal ball (Figure 15.35).

Figure 15.35

The excess charge on the polarized metal ball is on the surface of the ball, because the ball is a conductor. The interior of the ball is neutral.

Both objects are conductors, so all excess charge is on the surfaces. (b) The excess negative charge spreads all over the surface of the metal rod, which is a conductor. This excess negative charge polarizes the metal ball. The polarized metal ball in turn polarizes the negatively charged metal rod somewhat, as shown in Figure 15.36.

Figure 15.36 Polarization of the ball is greater in this case, because more of the original charge is closer to the ball (Figure 15.37). The net force on the ball is greater in this case than it was with the plastic rod.

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Figure 15.37

EXAMPLE A Ball and a Wire The center of a small spherical metal ball of radius R, carrying a negative charge −Q, is located a distance r from the center of a short, thin, neutral copper wire of length L (Figure 15.38). The ball and the wire are held in position by threads that are not shown. If R = 5 mm, Q = 1 × 10−9 C, r = 10 cm, and L = 4 mm, calculate the force that the ball exerts on the wire.

Figure 15.38

Solution The ball makes a field that polarizes the wire as shown in Figure 15.39. The polarized wire in turn makes a field that polarizes the ball, but let's assume that we can neglect this tiny effect, so we can model the ball as a point charge. The polarized wire will be attracted by the ball, and the ball will be attracted by the polarized wire.

Figure 15.39 The ball polarizes the wire. Neglect the effect of the wire on the ball.

We need to find the charge q on one end of the wire. We know that at any location inside the metal wire, in equilibrium (Figure 15.40). Consider a location in the center of the wire, and model the wire as though there were +q and −q on the ends, a distance L apart, ignoring the small amount of charge on the rest of the wire (here we can't use dipole formulas; we're between the charges).

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Figure 15.40 The net electric field inside the wire must be zero in equilibrium.

At a location outside the wire we model the wire as a dipole, so the force on the ball (which is equal in magnitude to the force on the wire) is this:

Not too surprisingly, we find a force proportional to 1/r 5. Now we can calculate numerical values:

which is a very small charge. This justifies our assumption that the polarized wire won't polarize the ball to any significant extent. The force is tiny:

If we double r, there is 1/4 as much q and 1/32 as much force. If we double Q, there is 2 times as much q and 4 times as much force.

15.X.11 An object can be both charged and polarized. On a negatively charged metal ball, the charge is spread uniformly all over the surface (Figure 15.41). If a positive charge is brought near, the charged ball will polarize.

Figure 15.41 This is a cross section of the metal ball. (a) Draw the approximate final charge distribution on the ball. (b) At the center, draw the electric field due to the external positive charge. (c) At the center, draw the electric field due to the charge on the surface of the ball.

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Answer

(d) At the center, draw the net electric field.

If any of these quantities is zero, state this explicitly. 15.X.12 A negatively charged plastic pen is brought near a neutral solid metal cylinder (Figure 15.42).

Answer

Figure 15.42 This is a cross section of the metal or plastic cylinder. (a) Show the approximate charge distribution for the metal cylinder. (b) Draw a vector representing the net force exerted by the pen on the metal cylinder, and explain your force vector briefly but completely, including all relevant interactions. (c) At the center, draw the electric field due to the external negative charge. (d) At the center, draw the electric field due to the charge on the surface of the ball. (e) At the center, draw the net electric field. (f) Replace the solid metal cylinder with a solid plastic cylinder. Show the approximate charge distribution for the plastic cylinder and draw a vector representing the net force exerted by the pen on the plastic cylinder; explain your force vector briefly but completely, including all relevant interactions.

If any of these quantities is zero, state this explicitly.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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CHARGING AND DISCHARGING We have seen that objects made of insulators can often acquire a nonzero net charge if they are rubbed by another insulator. You can charge initially neutral pieces of invisible tape by stripping them off other pieces of tape. You can charge an initially neutral pen by rubbing it on your hair. In both these cases, some kind of charged particle is added to or removed from a surface that was originally neutral. In Section 15.2 we discussed possible mechanisms for the transfer of charge between one insulating object and another, including the transfer of positive or negative ions and the transfer of electrons. Since a conducting object contains mobile charged particles, the process of charging or discharging a conductor involves a flow of charged particles from one conductor to another. For example, electrons from the mobile electron sea in a metal object can move into the mobile electron sea of a different metal object if the objects come into contact with each other. Although you may not have previously thought of yourself as a conductor, your own body plays an interesting role in some kinds of charging or discharging phenomena. In the following discussion we will see why.

Discharging by Contact If you exercise on a hot day, you sweat, and your body becomes covered with a layer of salt water. Even in a cool place, when you are not moving, there is usually a thin layer of salt water covering your skin. As we saw earlier, salt water is a conductor, so you have a conducting film all over the surface of your skin. When you approach a negatively charged surface, your body polarizes as shown in Figure 15.43. The polarization includes not only atomic or molecular polarization but also polarization of the blood and sweat, which are salt solutions.

Figure 15.43 The metal is charged, and the person is uncharged but slightly polarized. When you touch the charged object, the negatively charged object attracts positive Na + ions in the film of salt water on the skin. The Na + ions pick up an electron, partially neutralizing the excess negative charge of the object (Figure 15.44). The body acquires a net negative charge. (The Na atom can react with the water to form NaOH and hydrogen!) In the case of a small piece of metal, on which charge is free to redistribute itself, this process nearly neutralizes the metal, because the original net amount of charge is now spread out over the much larger area of metal plus human body.

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Figure 15.44 The net negative charge is distributed over a much larger area, nearly neutralizing the metal. Similarly, a positive metal surface would attract negative Cl − ions from your skin, which give up an electron to the metal. The body acquires a net positive charge. (Chlorine can be emitted in tiny quantities!)

Grounding Touching a small charged object is a pretty effective way to discharge the object, even though you're wearing shoes with insulating soles. An even better way to discharge a conducting object is to “ground” it by making a good connection to the earth or ground (typically through a water pipe that goes into the ground). Earth is a rather good conductor due to the presence of water containing ions. Grounding spreads charge throughout a huge region, neutralizing an object essentially completely.

Discharging an Insulator You can easily discharge a charged metal foil by briefly touching it anywhere, because it is a conductor. It is more difficult to discharge a charged strip of invisible tape, which is an insulator and does not allow charge to move through the tape. To discharge a charged piece of invisible tape, it turns out that one may simply rub one's fingers across the slick side of the tape, as shown in Figure 15.45. After doing this, one finds that the tape no longer interacts with neutral objects like your hand and appears to be neutral itself.

Figure 15.45 You run your finger along the slick side of the tape, and the tape seems to become neutralized.

QUESTION How is it possible to discharge a tape by rubbing the slick side even when it was the sticky side that got charged? Tape is an insulator, so charges can't move through the tape.

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Related experiment: 15.EXP.24 There are mobile charges on your skin. Positive ions from the salt solution on the skin are attracted to the negatively charged tape and are deposited on its slick surface, so the tape becomes neutral (net charge becomes zero) as shown in Figure 15.46. The + charges on the top and the − charges on the bottom actually make the tape into a dipole (and there are induced dipoles inside the tape), but these dipoles exert much weaker forces on other objects than the negatively charged tape did. Thus the tape acts like ordinary neutral matter.

Figure 15.46 Positive ions from the salt solution on your skin are attracted to the negatively charged tape.

Charging by Induction It is possible to make use of the polarizability of a conductor to make it acquire a net charge, without actually touching a charged object. The process, called “charging by induction,” is illustrated in the sequence in Figure 15.47. In this example, a piece of neutral aluminum foil hangs from a neutral insulating tape. You charge a plastic pen by rubbing it on wool, and bring it near the left side of the neutral foil.

Figure 15.47

Related experiment: 15.EXP.25

Effect of Humidity on Tapes Isolated atoms are always symmetrical and unpolarized unless an external charge shifts the electron cloud and makes an induced

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dipole. However, some molecules are permanently polarized even in the absence of an external charge, and this leads to important physical and chemical effects. For example, water molecules are permanently polarized. The water molecule (H2O) is not spherically symmetrical but has both hydrogen atoms off to one side of the oxygen atom. In Figure 15.48 the δ+ and δ− symbols are used to indicate that slight shifts of the electron clouds to the right leave the right side of the molecule a bit negative and the left side a bit positive, so the water molecule is a permanent dipole.

Figure 15.48 A water molecule is a permanent dipole.

Many of water's unusual chemical and physical properties are due to this structure. In particular, the charged ends can bind to ions, which is why many chemicals dissolve well in water.

Related experiment: 15.EXP.26

When water molecules in the air strike a surface they sometimes become attached to the surface, probably because the charged ends bind to the surface. A film of water builds up on all surfaces. Pure water is a very poor conductor but does contain small amounts of mobile H + and H − ions. More important, the water dissolves surface contaminants such as salt, and the impure water provides an effective path for charges to spread onto neighboring objects. After a while a charged surface loses its original charge, so experiments with charged objects work better when the humidity is low.

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WHEN THE FIELD CONCEPT IS LESS USEFUL The field concept and the idea of splitting of the universe into two parts are not very useful if the charge affected by the field significantly alters the original distribution of source charges. Consider a negatively charged metal sphere (Figure 15.49). The electric field due to the sphere points radially inward.

Figure 15.49 Electric field of a uniformly charged metal sphere.

If we place a particle with very little charge q near this charged sphere (a single proton, for example), it hardly alters the distribution , where is the electric field of charge on the metal sphere. We can reliably calculate the small force on the small charge as we calculated in the absence of the small additional charge q. However, if we place a particle with a big charge Q near the sphere, the sphere polarizes to a significant extent (Figure 15.50). We show the electric field due solely to the new charge distribution on the sphere (we don't show the large additional contribution to the net electric field due to Q). Clearly, the force on Q is not simply Q times the original , but Q times a significantly larger field.

Figure 15.50 Electric field due to the polarized sphere.

With these effects in mind, we need to qualify our previous method for measuring electric field, in which we measure the force exerted on a charge q and determine the force per unit charge:

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This procedure is valid only if q is small enough not to disturb the arrangement of other charges that create

.

Since no object can have a charge smaller than e (the charge of a proton), sometimes it is not possible to find a charge small enough that it doesn't disturb the arrangement of source charges. In this case, we can't measure the electric field without changing the field! On the other hand, if we know the locations of the source charges, we can calculate the electric field at a location, by applying the superposition principle and adding up the contributions of all the point charges that are the sources of the field:

If even the smallest possible charge e would disturb this arrangement of source charges, we can't use the calculated field to predict the force that would act on a charge placed at this location. However, we could use the calculated field to predict the polarization of a neutral atom placed at that location, because a neutral atom, even if (slightly) polarized, would disturb the existing arrangement of source charges much less than a charged object would. Another way to improve the measurement would be to measure

for a positive q and also measure for a negative q, and average the results. A negative charge would polarize the sphere in Figure 15.49 in such a way as to reduce rather than increase the , by pushing the negative charges on the sphere farther away. value of

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SUMMARY Net charge

The net charge of an object is the sum of the charges of all of its constituent particles. An object with a net charge of zero is called “neutral.” Conservation of charge The net charge of a system plus the net charge of its surroundings cannot change.

Polarization of an atom or molecule produces an induced dipole

(where p is the dipole moment and a is the atomic polarizability). Insulator An insulator has no mobile charges. Conductor A conductor contains mobile charges that can move through the material. The average drift speed of a mobile charged particle in a conductor is directly proportional to the magnitude of the net electric field inside the material. The proportionality constant u is called the “mobility” and has units of (m/s)/(N/C). Different materials have different mobilities. Metal A metal is a conductor. It has a mobile electron sea, spread throughout the object, like an ideal gas. Conductor

Insulator

Mobile charges

Yes

No

Polarization

Entire sea of mobile charges moves

Individual atoms or molecules polarize

Equilibrium

inside

nonzero inside

Location of excess charge

Only on surface

Anywhere on or inside material

Distribution of excess charge

Spread out over entire surface

Located in patches

inside a conductor in equilibrium. Excess charges move to the surface of a conductor. Force between a point charge and a neutral atom is proportional to 1/r 5:

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BASIC EXPERIMENTS Is Invisible Tape Electrically Charged? When you pull a long piece of invisible tape off a roll, it often curls up or sticks to your hand. We will do some simple experiments with invisible tape. Obtain a roll of invisible tape, such as Scotch® brand Magic™ Tape or a generic brand. It must be the kind of tape that almost disappears when you smooth it down on a surface, not ordinary cellophane tape. Our first task is to determine whether or not a piece of invisible tape might be electrically charged.

QUESTION How can we decide whether a piece of invisible tape is electrically charged?

If an object has a net electric charge, it should create an electric field in the surrounding space. Another charged object placed nearby should therefore experience an electric force. If we observe a change in an object's momentum, we can conclude that a force acts on the object. We know that the electric field of a point charge has these characteristics: The magnitude of

is proportional to the amount of charge.

The magnitude of

decreases with distance from the charge.

The direction of

is directly away from or toward the source charge.

Therefore, since

, the electric force on object 2 should have the same properties. In addition, we should be able to observe both attraction and repulsion, since charges of different sign will be affected differently by a particular field. We will observe the interactions of two pieces of invisible tape, and see whether they meet the criteria listed above. Preparing a U Tape Use a strip of tape about 20 cm long (about 8 inches, about as long as this paper is wide). Shorter pieces are not flexible enough, and longer pieces are difficult to handle. Fold under one end of the strip to make a nonsticky handle, as shown in Figure 15.51.

Figure 15.51 Fold under one end of a strip of tape to make a nonsticky handle.

HOW TO PREPARE A U TAPE Stick a strip of tape with a handle down onto a smooth flat surface such as a desk. This is a “base” tape. Smooth this base tape down with your thumb or fingertips. This base tape provides a standard surface to work from. (Without this base tape, you get different effects on different kinds of surfaces.) Stick another tape with a handle down on top of the base tape, as shown in Figure 15.52.

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Figure 15.52 The U tape lies on top of the base tape. Smooth the upper tape down well with your thumb or fingertips. Write U (for Upper) on the handle of the upper tape. With a quick motion, pull the U tape up and off the base tape, leaving the base tape stuck to the desk. Hang the U tape vertically from the edge of the desk, and bring your hand near the hanging tape, as shown in Figure 15.53.

Figure 15.53 Bring your hand near a hanging U tape, and observe what happens. If the tape is in good condition and the room is not too humid, you should find that there is an attraction between the hanging strip of tape and your hand when you get close to the tape. If there is no attraction, remake the U tape.

Experimental technique: try to handle the tapes only by their ends while you are doing an experiment. 15.EXP.13 Interaction of Two U Tapes (a) If U tapes are electrically charged, how would you expect two U tapes to interact with each other? Would you expect them to repel each other, attract each other, or not to interact at all? Make a prediction, and briefly state a reason. (b) Make two U (“upper”) tapes by following the procedure detailed above. Make sure that both tapes interact with your hand. Hang one on the edge of a desk. Bring the second U tape near the hanging U tape. Since the hanging tape is attracted to your hands, try to keep your hands out of the way. For example, you might approach the vertically hanging tape with the other tape oriented horizontally, held by two hands at its ends. What happens? You should have seen the two U tapes repel each other. If you did not observe repulsion, try remaking the U tapes (or making new ones, both from the same roll of invisible tape). It is important to see this effect before continuing further. Making a Tape Not Interact You may have already discovered that if you handle a U tape too much, it no longer repels another U tape. Next we will learn a systematic way for making this happen. Make sure that you have an active U tape, which is attracted to your hand.

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Holding onto the bottom of the U tape, slowly rub your fingers or thumb back and forth along the slick side of the tape (Figure 15.54).

Figure 15.54 Neutralizing a tape by rubbing the slick side. You should find that the U tape no longer interacts with your hand. If it still does, repeat the process.

This is a little odd; if the U tape was originally electrically charged, the charges would presumably have been on its sticky side. However, by running a finger along the other side (the slick side) we have apparently “neutralized” it—it now appears uncharged. It will be a while before we can explain this peculiar effect, but now we have a useful way to neutralize a U tape. 15.EXP.14 Is This an Electric Interaction? To decide whether the interaction between two U tapes is or is not an electric interaction, we will see whether it obeys the criteria for an electric interaction. (As is done throughout the scientific community, it is important to compare your results to the results of other experimenters.)

Figure 15.55 Two U tapes repel each other. (a) Does the force act along a line connecting the two tapes? Think of a way to determine whether or not the force between two tapes acts along a line drawn from one object to the other, and do the experiment. What did you find? (What would you see if this were not the case?) (b) Does the force decrease rapidly as the distance between the tapes increases? How can you determine this? (c) Is the force proportional to the amounts of both charges? Design and carry out an experiment to test this. One way to vary the amount of charge on a tape is to neutralize part of one of the tapes, by running your finger along the length of the slick side of the tape, being careful that your finger touches only a portion of the width of the tape. What do you observe?

The real world is messy! You may have noted several difficulties in making your measurements. For example, the tapes

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are both attracted to your hand, and also repel each other. If you tried to use a ruler to measure the distance between the tapes, you might have found that the tapes are attracted to the ruler, too. Unlike Charges So far we have observed that two U tapes repel each other, that the force acts along a line between the tapes, that the strength of the repulsion decreases as the tapes get farther away from each other, and that the strength of the interaction depends on the amount of charge on the tape. These observations are consistent with the hypothesis that the U tapes are electrically charged and that all U tapes have like electric charge.

QUESTION How could you prepare a tape that might have an electric charge unlike the charge of a U tape? Think of a plan before reading further.

Perhaps you reasoned along these lines: We don't know how the U tape became charged, but if the tapes started out neutral, maybe the U tape pulled some charged particles off of the lower tape (or vice versa). So now the lower tape should have an equal amount of charge, of the opposite sign. Making an L Tape Here is a reproducible procedure for making an L tape, whose charge is unlike the charge of a U tape: HOW TO PREPARE AN L TAPE Stick a strip of tape with a handle down onto a base tape, smooth this tape down thoroughly with your thumb or fingertips, and write L (for Lower) on the handle of this tape. Stick another tape with a handle down on top of the L tape, and write U (for Upper) on the handle of this tape. Smooth the upper tape down well with your thumb or fingertips. You now have three layers of tape on the desk: a base tape, an L tape, and a U tape (Figure 15.56).

Figure 15.56 Preparing an L tape: First, smooth down three layers of tape—a base tape, an L tape, and a U tape. Slowly lift the L tape off the base tape, bringing the U tape along with it (and leaving the bottom base tape stuck to the desk). Hang the double layer of tape vertically from the edge of the desk and see whether there is attraction between it and your hand (Figure 15.57). If so, get rid of these interactions (hold the bottom of the tape and slowly rub the slick side with your fingers or thumb).

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Figure 15.57 Preparing an L tape: Second, lift the upper two layers (the L and U tapes) and hang them from the desk. Make sure that they are not attracted to your hand! (Neutralize if necessary.)

Check that the tape pair is no longer attracted to your hand. This is important! Hold onto the bottom tab of the L tape and quickly pull the U tape up and off (Figure 15.58). Hang the U tape vertically from the edge of the desk, not too close to the L tape!

Figure 15.58 Preparing an L tape: Third, quickly pull the U tape off the L tape.

Repeating exactly the same procedure, make another pair of tapes so that you have at least two U tapes and two L tapes. Before separating the tapes from each other, always remember to make sure that the tapes are not attracted to your hand.

QUESTION An important step in preparing an L tape is to neutralize the L/U tape pair before separating the two tapes. Considering the principle of charge conservation, why is this step important? What can go wrong if this step is omitted?

The principle of conservation of charge states that if the pair has a total charge of zero before separation, the two tapes will have a total charge of zero after separation: one tape will have a charge of +q and the other a charge of −q. However, if the total charge before separation is nonzero and positive (say), the separated tapes could both have positive charge, as long as their individual charges add up to the original amount. 15.EXP.15 Observations of L and U Tapes You should now have two L tapes and two U tapes. Make sure that both the U tapes and the L tapes are active (attracted to your hand). (a) If an L tape is indeed electrically charged, and its charge is unlike the charge on a U tape, what interaction

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would you predict between an L tape and a U tape? (b) What interaction do you observe between an L tape and a U tape? (c) What interaction would you predict between two L tapes? (d) What interaction do you observe between two L tapes? (e) Is the pattern of interactions consistent with the statement: “Like charges repel; unlike charges attract”? A U Tape and an L Tape: Distance Dependence of Attraction If U and L tapes are electrically charged, then we would expect the strength of the attractive interaction to decrease as the distance between the tapes increases. Make the same sort of observations you made with two U tapes. 15.EXP.16 Distance Dependence of Force between U and L Tapes Move a U tape very slowly toward a hanging L tape. Observe the deflections of the tapes from the vertical, at several distances (for example, the distance at which you first see attraction, half that distance, etc.) The deflections of the tapes away from the vertical is a measure of the strength of the interaction. (a) Does the force decrease rapidly as the distance between the tapes increases? (b) Why is this measurement more difficult with a U and an L tape than with two U tapes? Summary and Conclusions: U and L Tapes Let's summarize the observations and try to conclude, at least tentatively, whether U and L tapes are electrically charged. Presumably you have observed the following: There are two kinds of charge, called “ + ” and “ − ”. Like charges repel, unlike charges attract. The electric force

▪ acts along a line between the charges, ▪ decreases rapidly as the distance between the charges increases, and ▪ is proportional to the amounts of both charges.

Our observations of U and L tapes seem to be consistent with a description of the electric interactions between charged objects. We tentatively conclude that U and L tapes are electrically charged, and have unlike charges. How a Plastic Comb or Pen Becomes Charged Charged objects, such as invisible tape, are negatively charged if they have more electrons than protons, and positively charged if they have fewer electrons than protons. Are U tapes positively or negatively charged? How can we tell? Charging an object in a standard manner gives us a “litmus test.” It is known that if you rub a glass rod with silk, the glass rod becomes positively charged and the silk negatively charged. Likewise, if you rub a clear plastic object such as a pen through your hair (or with fur, wool, or even cotton), the plastic ends up having a negative charge and so repels electrons. A similar process occurs when you separate one tape from another. See Section 15.2 for a discussion of how objects become charged through rubbing or contact. 15.EXP.17 Determining the Charge on U and L Tapes Prepare a U tape and an L tape, and hang them from your desk. Test them with your hand to make sure they are both charged. Rub a plastic pen or comb on your hair (clear plastic seems to charge best), or on a piece of cotton or wool, and

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bring it close to each tape. You should observe that one of the tapes is repelled by the pen, and one is attracted to it. Knowing that the plastic is negatively charged, what can you conclude about the sign of the electric charge on U tapes? On L tapes? These results may be reversed if you try a different brand of “invisible tape.” Be sure to compare your results with those of other students. Make sure you all agree on the assignment of “+” and “−” labels to your tapes. (If other groups are using different brands of tape, you may disagree on whether U tapes or L tapes are positive, but the electric interactions between your + tapes and their − tapes should be repulsive!) In any of your experiments, did you find any objects, other than tapes or a charged comb or pen, that repelled a U or L tape? If so, those objects must have been charged. List these objects and whether the charge was + or −. Amount of Charge on a Tape We have concluded that U tapes and L tapes are electrically charged, but we have no idea how much charge is on one of the charged tapes—we don't even know an approximate order of magnitude for this quantity. Even a rough measurement of the amount of charge on a tape would be useful, because it would give us a feel for the amount of charge there might be on an ordinary object that is observed to interact electrically with other objects. Therefore the following experiment is an important one. 15.EXP.18 Amount of Excess Charge on a Tape In this problem you will design and carry out an experiment to determine the approximate number of excess electron charges on the surface of a negatively charged tape. Initial estimates Since we do not know what order of magnitude to expect for our answer, it is important to put upper and lower bounds on reasonable answers. (a) What is the smallest amount of excess charge that a tape could possibly have? (b) What is the largest amount of excess charge a tape could possibly have? Design and perform an experiment A centimeter ruler is printed on the inside back cover of this textbook. A piece of half-inch-wide (1.2 cm) invisible tape, 20 cm long (8 inches), has a mass of about 0.16 grams. ((c)) Make a clear and understandable diagram of your experimental setup, indicating each quantity you measured. Report all measurements you made. Analyze the results ((d)) Clearly present your physical analysis of your data. Make an appropriate diagram, labeling all vector quantities. Reason from fundamental physics principles. Explicitly report any simplifying assumptions or approximations you have made in your analysis. Report two quantities:

▪ The amount of charge on a tape, in coulombs ▪ The number of excess electrons to which this charge corresponds

Present your analysis clearly. Your reasoning must be clear to a reader. ((e)) Estimate whether the true amount of excess charge is larger or smaller than the value you calculated from your

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experimental data. Explain your reasoning briefly. Is this a lot of charge? ((f)) What fraction of the molecules on the surface of the tape have gained an extra electronic charge? To estimate this, you may assume that molecules in the tape are arranged in a cubic lattice, as indicated in the accompanying figure, and that the diameter of a molecule in the tape is about 3 × 10−10 m.

Figure 15.59 Array of molecules on one surface of a piece of invisible tape.

Does your answer suggest that it is a common event or a rare event for a molecule to gain an extra electron? ((g)) If the electric field at a location in air exceeds 3 × 106 N/C, the air will become ionized and a spark will be triggered. In Chapter 16 we will see that the electric field in a region very close to a uniformly charged disk or plate depends approximately only on the charge Q per unit area A:

Use this model (or make a different but justifiable simplifying assumption) to calculate the magnitude of the electric field at a location in the air very close to your tape (less than 1 mm from the surface of the tape). How does it compare to the electric field needed to trigger a spark in the air? Interaction of Charges and Neutral Matter We have focused on the interactions of U and L tapes with other U and L tapes. Let's look more broadly at the interactions of charged tapes with other objects. 15.EXP.19 Interactions of U and L Tapes with Other Objects Which other objects (paper, metal, plastic, etc.) have an attractive interaction with a hanging U or L tape, and which objects have a repulsive interaction? Which objects have no interaction? Record the objects you try and the interactions observed. The attraction of both U and L tapes to your hand, and to many other objects, is deeply mysterious. The net charge of a neutral object is 0, so your neutral hand should not make an electric field that could act on a charged tape; nor should your neutral hand experience a force due to the electric field made by a charged tape. Nothing in our statement of the properties of electric interactions allows us to explain this attraction!

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Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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ADDITIONAL EXPERIMENTS Observing Interactions with Dipoles You can also use charged tapes to observe the behavior of a dipole.

QUESTION Consider the forces that a positive charge Q exerts on the charges making up a dipole and describe the main features of the resulting motion of the dipole.

Figure 15.60 A positive charge Q interacts with a dipole.

There is a twist (torque) that tends to align the dipole along the line connecting the charge Q and the center of the dipole, with the negative end of the dipole closer to the single positive charge. The dipole has a nonzero net force acting on it that makes it move toward the positive charge. 15.EXP.20 An Electric “Compass” Make a tall dipole and observe the motion. Take a + tape and a − tape and stick them together, overlapping them only enough to hold them together. Avoid discharging the tapes with too much handling. Hang the combination from a thread or a hair. Now approach the tapes with a charged object and admire how sensitive a charge detector you have made! Slowly move the charged object all around the dipole and observe how the dipole tracks the object. If you draw an appropriately labeled arrow on the tape, you have an electric “compass” that points in the direction of electric field. 15.EXP.21 Observing Attraction of Like-Charged Objects (!) Because of the very rapid 1/r 5 increase in the attraction to neutral matter at short distances, it is sometimes the case that at short distances the attractive effect can actually overcome the repulsion between like-charged objects. Specifically, it could be that

where q1 and q2 are the excess charges on the surfaces of the two tapes, and N is the total number of neutral atoms in tape 2 (since each neutral atom participates in the attraction.) As can be seen by dividing the inequality by q1, you can enhance the effect by making q1 be significantly larger than q2, so you might wish to partially discharge one of the tapes.

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Figure 15.61 A dipole made from U and L tapes.

Try it! Hold one of the U tapes horizontally, with its slick side facing away from you and toward the slick side of the hanging tape. Move toward the hanging tape and check that the hanging tape is repelled as you approach. Then move close enough so that the tapes touch each other (a partner may have to hold the bottom of the hanging tape in order to be able to get very close). You may be able to detect some slight attraction when the tapes are very close together or touch each other, despite the fact that the tapes repel at longer distances. Do you see such an effect? The effect is quite easy to see in the interaction between a highly charged Van de Graaff generator and a charged tape. 15.EXP.22 Interaction through a Piece of Paper Have a partner hold a piece of paper close to, but not touching, a hanging U tape. Bring another U tape toward the hanging tape from the other side of the paper, holding both ends of this tape so that it can't swing.

Figure 15.62 Attempting to observe an electric interaction (repulsion) through a piece of paper.

Can you observe repulsion occurring right through the intervening paper? This is difficult, because the paper attracts the hanging tape, which masks the repulsion due to the other tape. You can heighten the sensitivity of the experiment by moving the tape rhythmically toward and away from the hanging tape, as though you were pushing a swing. This lets you build up a sizable swing in the hanging tape even though the repulsive force is quite small, because you are adding up lots of small interactions. Using rhythmic movements, are you able to observe repulsion through the intervening paper? The effect is especially hard to observe if you have weak repulsion due to high humidity. The farther away you can

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detect repulsion, the better, because the competing attraction falls off rapidly with distance. Under good conditions of low humidity, when tapes remain strongly charged and repulsion is observable with the tapes quite far apart from each other, it is possible to see repulsion with the paper in place, showing that electric field does go right through intervening matter. You have seen evidence of this when you observed attraction between a tape and your hand even when you approached the slick side of the tape. 15.EXP.23 Is Tape a Conductor or an Insulator? Prepare a hanging tape that has the top half charged and the bottom half uncharged. After a second or two, check to see whether the bottom half of the tape has become charged. Based on this observation, is tape a conductor or an insulator? That is, are the charges free or bound? Explain fully and rigorously how your observations justify your conclusion. (Hint: Draw a diagram showing what effect the charges on the upper half of the tape have on each other and on charges inside the tape, and reason through what will happen if any of these charges are free to move.) 15.EXP.24 Discharging a Tape The previous exercise suggests that a key element in neutralizing a tape is the salt solution on the surface of your finger. Design one or more experiments you can do to confirm or reject this explanation of discharging a tape. 15.EXP.25 Charging by Induction Hang a short piece of aluminum foil (about the width of the tape and half as long as your thumb) from a tape, with another piece of tape added to the bottom of the foil as a handle.

Figure 15.63 Hang a short piece of aluminum foil from a neutral tape.

Now carry out the following operations exactly as specified: 1. Make sure that the tape and foil are uncharged (touch the foil, and rub the slick side of the tape). 2. Have a partner hold onto the bottom tape to keep the foil from moving. 3. Bring a charged plastic pen or comb very close to the foil, but don't touch the foil with the plastic. 4. While holding the plastic near the foil, tap the back of the foil with your finger. 5. Move your finger away from the foil, then move the plastic away from the foil.

You should find that the metal foil is now strongly charged. This process is called “charging by induction.” Now touch the charged aluminum foil with your finger and observe that this discharges the foil, as predicted by our earlier discussion. Complete a “comic strip” of diagrams illustrating the charging by induction process you carried out.

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Figure 15.64 Give a step-by-step explanation of charging by induction.

Make sure you have the sign of the charges right. In each diagram show charge distributions, polarization, movement of charges, and so on. For each frame, explain briefly what happens. Remember that excess charges on the metal foil can only be on the surface. This process is called charging by induction because the entire piece of foil becomes an induced dipole when it is polarized by the external charge. Charging by induction makes it possible to charge a metal without touching the external charge to the metal. Explain the process of discharging the metal foil by touching it, using the same kind of time-sequence “comic strip” diagrams you used in the previous exercise. Illustrate the important aspects of each step in the process. Include any changes to your body as well as to the foil. Be precise in your use of words. 15.EXP.26 A Water Film as a Conductor Prepare a hanging tape that has the top half charged and the bottom half uncharged. Let it hang while you do other work, but check every few minutes to see what has happened in the two halves. What do you predict will happen to the state of charge in the two halves? What do you observe over a period of many minutes? (If the room is very dry or very wet you may not be able to see this effect.) Suppose you were to breathe heavily through your mouth onto the slick and sticky sides of a short section in the middle of a long charged tape. Your breath is very moist. What do you predict you would find immediately afterward? Try the experiment—what do you observe? (Repeat if you see no effect.) 15.EXP.27 Transferring Charge by Contact Make two identical hanging foil arrangements, and charge one of the aluminum foils by induction. Discharge the other foil by touching it with your finger. Charge a tape or pen or comb, and note the approximate strength of the interaction between it and the charged foil.

Figure 15.65 Prepare two hanging pieces of aluminum foil.

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Next make the two foils touch each other, being careful not to touch either foil with your fingers. Note the approximate strength of the interaction that there is now between the plastic and each foil. Compared with the situation before the two foils touched, what sign of charge, and roughly how much, is there now on each foil? Discuss this fully with your partners, and convince yourselves that you understand the process. Make a written explanation, including appropriate diagrams. What would you expect to happen if one piece of foil were much larger than the other?

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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EXERCISES AND PROBLEMS Section 15.1 15.X.28 Which statements about a neutral atom are correct? Select all that apply. (A) A neutral atom is composed of both positively and negatively charged particles. (B) The positively charged particles in the nucleus are positrons. (C) The electrons are attracted to the positively charged nucleus. (D) Positively charged protons are located in the tiny, massive nucleus. (E) The radius of the electron cloud is twice as large as the radius of the nucleus. (F) The negatively charged electrons are spread out in a “cloud” around the nucleus.

Section 15.2 15.X.29 Which of the following could be reasonable explanations for how a piece of invisible tape gets charged? Select all that apply. (A) Protons are pulled out of nuclei in one tape and transferred to another tape. (B) Charged molecular fragments are broken off one tape and transferred to another. (C) Electrons are pulled out of molecules in one tape and transferred to another tape. (D) Neutrons are pulled out of nuclei in one tape and transferred to another tape. 15.X.30 Criticize the following statement: “Since an atom's electron cloud is spherical, the effect of the electrons cancels the effect of the nucleus, so a neutral atom can't interact with a charged object.” (“Criticize” means to explain why the given statement is inadequate or incorrect, as well as to correct it.) 15.X.31 Criticize the following statement: “A positive charge attracts neutral plastic by polarizing the molecules and then attracting the negative side of the molecules.” (“Criticize” means to explain why the given statement is inadequate or incorrect, as well as to correct it.) 15.X.32 Jill stuck a piece of invisible tape down onto another piece of tape. Then she yanked the upper tape off the lower tape, and she found that this upper tape strongly repelled other upper tapes and was charged positive. Jack ran his thumb along the slick (upper) side of the upper tape, and the tape no longer repelled other upper tapes. Jill and Jack explained this by saying that Jack rubbed some protons out of the carbon nuclei in the tape. Give a critique of their explanation. If Jill's and Jack's explanation is deficient, give a physically possible explanation for why the upper tape no longer repelled other upper tapes. Include explanatory diagrams.

Section 15.3 15.X.33 You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attracted to the pen. Assuming that the pen has a net negative charge, which of the following could be true? Select all that apply. (A) The tape might be negatively charged. (B) The tape might be positively charged. (C) The tape might be uncharged. (D) There is not enough information to conclude anything. 15.X.34 There is a region where an electric field points to the right, due to charged particles somewhere. A neutral carbon atom is placed inside this region. Draw a diagram of the situation, and use it to answer the following question: Which of the following statements are correct? Select all that apply.

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(A) Because the net charge of the carbon atom is zero, it cannot be affected by an electric field. (B) The electron cloud in the carbon atom shifts to the left. (C) The neutral carbon atom polarizes and becomes a dipole. (D) The nucleus of the carbon atom shifts to the left. 15.X.35 A charged particle with charge q1 is a distance r from a neutral atom, as shown in Figure 15.66.

Figure 15.66 (a) If q1 is negative which diagram (1–10) in Figure 15.67 best shows the charge distribution in the neutral atom in this situation?

Figure 15.67 (b) Which of the arrows (a–j) in Figure 15.68 best indicates the direction of the electric field at the location of the charged particle, made by the polarized neutral atom?

Figure 15.68 (c) Which of the arrows (a–j) in Figure 15.68 best indicates the direction of the force on the charged particle, due to the polarized neutral atom? (d) Which of the arrows (a–j) in Figure 15.68 best indicates the direction of the force on the polarized neutral atom, due to the charged particle? 15.X.36 Atom A is easier to polarize than atom B. Which atom, A or B, would experience a greater attraction to a point charge a distance r away? (A) A (B) B (C) Same (D) The polarization does not matter. 15.X.37 If the distance between a neutral atom and a point charge is tripled, by what factor does the force on the atom by the point charge change? Express your answer as a ratio: (new force/original force).

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15.X.38 Which observation provides evidence that two objects have the same sign charge? (A) The two objects repel each other. (B) The two objects attract each other. (C) The two objects do not interact at all. (D) The strength of the interaction between the two objects depends on distance. 15.X.39 Is the following statement true or false? If true, what principle makes it true? If false, give a counterexample or say why. See Figure 15.69.

Figure 15.69

“The electric field E point at the center of an induced dipole, due to the point charge, is equal in magnitude and opposite in direction to the electric field E dipole at the location of the point charge, due to the induced dipole.” 15.X.40 Explain briefly why the attraction between a point charge and a dipole has a different distance dependence for induced dipoles (1/r 5) than for permanent dipoles (1/r 3). (You need not explain either situation in full detail: just explain why there is this difference in their behavior.) 15.P.41 A large positive charge pulls on a distant electron. How does the net force on the electron change if a slab of glass is inserted between the large positive charge and the electron? Does the net force get bigger, smaller, or stay the same? Explain, using only labeled diagrams. (Be sure to show all the forces on the electron before determining the net force on the electron, not just the force exerted by the large positive charge. Remember that the part of the net force on the electron contributed by the large positive charge does not change when the glass is inserted: the electric interaction extends through matter.) 15.P.42 Try rubbing a plastic pen through your hair, and you'll find that you can pick up a tiny scrap of paper when the pen is about one centimeter above the paper. From this simple experiment you can estimate how much an atom in the paper is polarized by the pen! You will need to make several assumptions and approximations. Hints may be found at the end of the chapter. (a) Suppose that the center of the outer electron cloud (q = −4e) of a carbon atom shifts a distance s when the atom is polarized by the pen. Calculate s algebraically in terms of the charge Q on the pen. (b) Assume that the pen carries about as much charge Q as we typically find on a piece of charged invisible tape. Evaluate s numerically. How does this compare with the size of an atom or a nucleus? (c) Calculate the polarizability a of a carbon atom. Compare your answer to the measured value of 1.96 × 10−40C · m/(N/C) (T. M. Miller and B. Bederson, “Atomic and molecular polarizabilities: a review of recent advances,” Advances in Atomic and Molecular Physics, 13, 1–55, 1977). (d) Carefully list all assumptions and approximations you made. Hints (a) (a1) What must the force on a single carbon atom in the paper be at the moment the paper is lifted by the pen? (a2) You know how to calculate the force on a point charge due to a dipole. How does this relate to the force on the dipole by the point charge? In this problem, is there something you can model as a dipole and something else you can model as a point charge? (b) Note that the dipole moment (p = qs) of a polarized atom or molecule is directly proportional to the applied electric field. In this case the charged pen is generating the applied electric field.

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15.P.43 An electron and a neutral carbon atom are initially 1 × 10−6 apart (about 10000 atomic diameters), and there are no other particles in the vicinity. The polarizability of a carbon atom has been measured to be α = 1.96 × 10−40 C · m/(N/C). (a) Calculate the initial magnitude and direction of the acceleration of the electron. Explain your steps clearly. Pay particular attention to clearly defining your algebraic symbols. Don't put numbers into your calculation until the very end. (b) If the electron and carbon atom were initially twice as far apart, how much smaller would the initial acceleration of the electron be? 15.P.44 Two identical permanent dipoles, each consisting of charges +q and −q separated by a distance s, are aligned along the x axis, a distance r from each other, where r >> s. Show all of the steps in your work, and briefly explain each step. (a) Draw a diagram like the one shown in Figure 15.70. Draw vectors showing all individual forces acting on each particle, and draw heavier vectors showing the net force on each dipole.

Figure 15.70 (b) Show that the magnitude of the net force exerted on one dipole by the other dipole is

Section 15.4 15.X.45 Explain briefly why repulsion is a better test for the sign of a charged object than attraction is. 15.X.46 A solid plastic ball has negative charge uniformly spread over its surface. Which of the diagrams in Figure 15.71 best shows the polarization of molecules inside the ball?

Figure 15.71 15.X.47 A charged piece of invisible tape is brought near your hand, as shown in Figure 15.72. Your hand is initially neutral.

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Figure 15.72 (a) If the tape is negatively charged, which of the diagrams 1–10 in Figure 15.73 best shows the polarization of a neutral molecule in your hand?

Figure 15.73 (b) Which arrow in Figure 15.74 best indicates the direction (a–j) of the electric field at the location of the tape due to the large number of polarized molecules in your hand?

Figure 15.74 (c) Which arrow in Figure 15.74 best indicates the direction (a–j) of the force on the tape due to the polarized molecules in your hand? (d) Which arrow in Figure 15.74 best indicates the direction (a–j) of the force on your hand due to the charged tape? 15.P.48 A dipole consisting of two oppositely charged balls connected by a wooden stick is located as shown in Figure 15.75. A block of plastic is located nearby, as shown. Locations B, C, and D all lie on a line perpendicular to the axis of the dipole, passing through the midpoint of the dipole.

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Figure 15.75

Before selecting answers to the following questions, draw your own diagram of this situation, showing all the fields and charge distributions requested. Answer the following questions by selecting either a direction (a–j) or an orientation of a polarized molecule (1–10) from the diagrams in Figure 15.76.

Figure 15.76 (a) Which of the arrows (a–j) best indicates the direction of the electric field at location C due only to the dipole? (b) Which of the arrows (a–j) best indicates the direction of the electric field at location D due only to the dipole? (c) Which of the diagrams (1–10) best indicates the polarization of a molecule of plastic at location C? (d) Which of the diagrams (1–10) best indicates the polarization of a molecule of plastic at location D? (e) Which of the following statements is correct? (1) A molecule located at C would not be polarized at all. (2) The polarization of a molecule located at D would be the same as the polarization of a molecule located at C. (3) A molecule located at D would be polarized more than a molecule located at C. (4) A molecule located at D would be polarized less than a molecule located at C. (f) Which of the arrows (a–j) best indicates the direction of the electric field at location B due only to the dipole? (g) Which of the arrows (a–j) best indicates the direction of the electric field at location B due only to the plastic

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block? The magnitude of the electric field at B due to the plastic is less than the magnitude of the electric field at B due to the dipole. (h) Which of the arrows (a–j) best indicates the direction of the net electric field at location B? (i) Which of the following statements is correct? (1) The electric field at B due only to the dipole would be larger if the plastic block were not there. (2) The electric field at B due only to the dipole would be the same if the plastic block were not there. (3) The electric field at B due only to the dipole would be smaller if the plastic block were not there. (4) The electric field at B due only to the dipole would be zero if the plastic block were not there. (j) Using the diagrammatic conventions discussed in the text, a student drew the diagram in Figure 15.77 to help answer the questions asked above. Which of the following statements about the student's diagram are true? Check all that apply.

Figure 15.77 (1) The direction of polarization of the plastic block is wrong. (2) The diagram is correct; this is just a different way of drawing the polarization. (3) The diagram shows mobile charges; this is wrong because an insulator does not have mobile charged particles.

Sections 15.5 15.X.49 A negatively charged iron block is placed in a region where there is an electric field downward (in the −y direction) due to charges not shown. Which of the diagrams (a–f) in Figure 15.78 best describes the charge distribution in and/or on the iron block?

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Figure 15.78 15.X.50 A neutral copper block is polarized as shown in Figure 15.79, due to an electric field made by external charges (not shown). Which arrow (a–j) in Figure 15.79 best indicates the direction of the net electric field at location B, which is inside the copper block?

Figure 15.79 15.X.51 Two small, negatively charged plastic spheres are placed near a neutral iron block, as shown in Figure 15.80. Which arrow (a–j) in Figure 15.80 best indicates the direction of the net electric field at location A?

Figure 15.80 15.X.52

(a) Which of the diagrams (A–F) in Figure 15.81 correctly displays the polarization of a metal sphere by an electric field that points to the left, using the conventions discussed in this chapter?

Figure 15.81 (b) Which of the diagrams (A–F) in Figure 15.81 correctly displays the polarization of a plastic sphere by an electric

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field that points to the left, using the conventions discussed in this chapter? 15.X.53 Which of the following are true? Select all that apply. (A) In static equilibrium, there is no net flow of mobile charged particles inside a conductor. (B) The electric field from an external charge cannot penetrate to the center of a block of iron. (C) The net electric field inside a block of aluminum is zero under all circumstances. (D) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not at equilibrium. (E) The net electric field at any location inside a block of copper is zero if the copper block is at equilibrium. 15.X.54 Which of the following are true? Check all that apply. “Static equilibrium” means that no charges are flowing. (A) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not at static equilibrium. (B) The net electric field inside a block of aluminum is zero under all circumstances. (C) The net electric field at any location inside a block of copper is zero if the copper block is at static equilibrium. (D) The electric field from an external charge cannot penetrate to the center of a block of iron. (E) In static equilibrium, there is no net flow of mobile charged particles inside a conductor. 15.X.55 In a particular metal, the mobility of the mobile electrons is 0.0077 (m/s)/(N/C). At a particular moment the net electric field everywhere inside a cube of this metal is 0.053 N/C in the +x direction. What is the average drift speed of the mobile electrons in the metal at this instant? 15.X.56 An electric field is applied to a solution containing bromide ions. As a result, the ions move through the solution with an average drift speed of 3.7 × 10−7 m/s. The mobility of bromide ions in solution is 8.09 × 10−8 (m/s)/(N/C). What is the magnitude of the net electric field inside the solution? 15.X.57 Carbon tetrachloride (CCl 4) is a liquid whose molecules are symmetrical and so are not permanent dipoles, unlike water molecules. Explain briefly how the effect of an external charge on a beaker of water (H2O) differs from its effect on a beaker of CCl 4. (Hint: Consider the behavior of the permanent dipole you made out of U and L tapes.) 15.X.58 A positive charge is located between a neutral block of plastic and a neutral block of copper. Draw the approximate charge distribution for this situation.

Figure 15.82 15.X.59 Make a table showing the major differences in the electric properties of plastic, salt water, and copper. Include diagrams showing polarization by an external charge. 15.X.60 Figure 15.83 shows a neutral, solid piece of metal placed near two point charges. Copy this diagram.

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Figure 15.83 (a) On your diagram, show the polarization of the piece of metal. (b) Then, at location A inside the solid piece of metal, carefully draw and label three vectors: (1)

, the electric field due to −q1

(2)

, the electric field due to +q2

(3)

, the electric field due to all of the charges on the metal

(c) Explain briefly why you drew the vectors the way you did. 15.X.61 You place a neutral block of nickel near a small glass sphere that has a charge of 2 × 10−8 coulombs uniformly distributed over its surface, as shown in Figure 15.84.

Figure 15.84 (a) About how long do you have to wait to make sure that the mobile electron sea inside the nickel block has reached equilibrium? (1) Less than a nanosecond (1 × 10−9s) (2) Several hours (3) About 1 second (4) About 10 minutes (b) In equilibrium, what is the average drift speed of the mobile electrons inside the nickel block? (1) About 1 × 105 m/s (2) About 1 × 10-5 m/s (3) 0 m/s (c) In the equation

, what is the meaning of the symbol u?

(1) The density of mobile electrons inside the metal, in electrons/m 3. (2) The mobility of an electron inside the metal, in (m/s)/(N/C). (3) The time it takes a block of metal to reach equilibrium, in seconds.

15.X.62 This question focuses on reasoning about equilibrium inside the nickel block shown in Figure 15.84. Start with these premises: The definition of equilibrium inside a conductor The relationship between average drift speed and electric field in a conductor to reason about which situations are possible inside the nickel block at equilibrium. Some of the situations listed below are

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possible, some are ruled out by one premise, and some are ruled out by two premises. If a situation is ruled out by two premises, choose both. and E net = 0

Case 1:

(A) Possible (B) Not possible by definition of equilibrium (C) Not possible because

Case 2:

and

(A) Possible (B) Not possible by definition of equilibrium (C) Not possible because

Case 3:

and

(A) Possible (B) Not possible by definition of equilibrium (C) Not possible because

Case 4:

and

(A) Possible (B) Not possible by definition of equilibrium (C) Not possible because

Now that you have considered each case, in equilibrium, which one is the only situation that is physically possible? (A) Case 1 (B) Case 2 (C) Case 3 (D) Case 4 15.X.63 A positively charged sphere is placed near a neutral block of nickel, as shown in Figure 15.84. (a) Which of the diagrams in Figure 15.85 best represents the equilibrium distribution of charge on the neutral nickel block?

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Figure 15.85 (b) At location P inside the nickel block the electric field due to the charged sphere is equilibrium, which of the following statements must be true?

N/C. At

(1) It is not possible to determine the electric field at location P due only to charges on the surface of the nickel block. (2) The electric field at location P due only to charges on the surface of the nickel block is (3) Because the net electric field at location P is surface of the polarized nickel block must be

15.P.64

N/C.

N/C, the field at P due only to charges on the N/C.

(a) The positively charged particle shown in diagram 1 in Figure 15.86 creates an electric field Which of the arrows (a–j) in Figure 15.86 best indicates the direction of

at location A.

at location A?

Figure 15.86 (b) Now a block of metal is placed in the location shown in diagram 2 in Figure 15.86. Which of the arrows (a–j) in Figure 15.86 best indicates the direction of the electric field at location A due only to the charges in and/or on the metal block? (c)

is greater than

. With the metal block still in place, which of the arrows (a–j) in Figure 15.86 best

indicates the direction of the net electric field at location A? (d) With the metal block still in place, which of the following statements about the magnitude of

, the field due

only to the charged particle, is correct? (1)

is less than it was originally, because the block is in the way.

(2)

is the same as it was originally, without the block.

(3)

is zero, because the electric field due to the particle can't go through the block.

(e) With the metal block still in place, how does the magnitude of ?

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at location A compare to the magnitude of

(f) Which of the arrows (a–j) in Figure 15.86 best indicates the direction of the net electric field at the center of the metal block (inside the metal)? 15.P.65 Two identical metal spheres are suspended from insulating threads. One is charged with excess electrons, and the other is neutral. When the two spheres are brought near each other, they swing toward each other and touch, then swing away from each other. (a) Explain in detail why both these swings happen. In your explanation, include clear diagrams showing charge distributions, including the final charge distribution. (b) Next the spheres are moved away from each other. Then a block of plastic is placed between them as shown in Figure 15.87. The original positions of the spheres are indicated, before the plastic is placed between them. Sketch the new positions of the spheres. Explain, including charge distributions on the spheres.

Figure 15.87 (c) Show the polarization of a molecule inside the plastic at points A, B, C, D, and E. Explain briefly. 15.P.66 A small glass ball is rubbed all over with a small silk cloth and acquires a charge of +5 nC. The silk cloth and the glass ball are placed 30 cm apart. (a) On a diagram like that shown in Figure 15.88, draw the electric field vectors qualitatively at the locations marked ×. Pay careful attention to directions and to relative magnitudes. Use dashed lines to explain your reasoning graphically, and draw the final electric field vectors with solid lines.

Figure 15.88 (b) Next, a neutral block of copper is placed between the silk and the glass. On a diagram like that shown in Figure 15.89, carefully show the approximate charge distribution for the copper block and the electric field vectors inside the copper at the ×.

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Figure 15.89 (c) The copper block is replaced by a neutral block of plastic. Carefully show the approximate molecular polarization of the plastic block at the locations marked × in Figure 15.90.

Figure 15.90 (d) Even if you have to state your result as an inequality, make as quantitative a statement as you can about the electric field at the location of the glass ball and the net force on the ball when the plastic block is in place compared to when there is no block. Explain briefly. 15.P.67 A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1 × 1010 excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20 cm. (a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. If you have to make any simplifying assumptions, state them explicitly and justify them. (b) Now the balls are moved so that as they hang, the distance between their centers is only 5 cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42 = 16, but in real life the increase is a bit less than a factor of 16. Explain why, including a diagram. (Nothing but the distance between centers is changed —the charge on each ball is unchanged, and no other objects are around.) 15.P.68 As shown in Figure 15.91, an electroscope consists of a steel ball connected to a steel rod, with very thin gold foil leaves connected to the bottom of the rod (in good electric contact with the rod). The bottom of the electroscope is enclosed in a glass jar and held in place by a rubber stopper.

Figure 15.91 (a) The electroscope is brought near to but not touching a positively charged glass rod as shown in Figure 15.92. The foil leaves are observed to spread apart. Explain why in detail, using as many diagrams as necessary.

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Figure 15.92 (b) The electroscope is moved far away from the glass rod and the steel ball is touched momentarily to a metal block. The foil leaves spread apart and stay spread apart when the electroscope is moved away from the block. As the electroscope is moved close to but not touching the positively charged glass rod, the foil leaves move closer together. Is the metal block positive, negative, or neutral? How do you know? Explain. 15.P.69 A very thin spherical plastic shell of radius 15 cm carries a uniformly distributed negative charge of −8nC(−8 × 10−9 C) on its outer surface (so it makes an electric field as though all the charge were concentrated at the center of the sphere). An uncharged solid metal block is placed nearby. The block is 10 cm thick, and it is 10 cm away from the surface of the sphere. See Figure 15.93.

Figure 15.93 (a) Sketch the approximate charge distribution of the neutral solid metal block. (b) Draw the electric field vector at the center of the metal block that is due solely to the charge distribution you sketched (that is, excluding the contributions of the sphere). (c) Calculate the magnitude of the electric field vector you drew. Explain briefly. If you must make any approximations, state what they are. 15.P.70 Two plastic balls are charged equally and positively and held in place by insulating threads, as shown in Figure 15.94.

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Figure 15.94

They repel each other with an electric force of magnitude F. Then an uncharged metal ball is held in place by insulating threads between the balls, closer to the left ball. State what change (if any) there is in the net electric force on the left ball and on the net electric force on the right ball. Show relevant force vectors. Also show the charge distribution on the metal ball. Explain briefly but completely. 15.P.71 A thin, hollow spherical plastic shell of radius R carries a uniformly distributed negative charge −Q. A slice through the plastic shell is shown in Figure 15.95.

Figure 15.95

To the left of the spherical shell are four charges packed closely together as shown (the distance “s” is shown greatly enlarged for clarity). The distance from the center of the four charges to the center of the plastic shell is L, which is much larger than (L >> s). Remember that a uniformly charged sphere makes an electric field as though all the charge were concentrated at the center of the sphere. (a) Calculate the x and y components of the electric field at location B, a distance b to the right of the outer surface of the plastic shell. Explain briefly, including showing the electric field on a diagram. Your results should not contain any symbols other than the given quantities R, Q, q, s, L, and b (and fundamental constants). You need not simplify the final algebraic results except for taking into account the fact that L >> s. (b) What simplifying assumption did you have to make in part (a)? (c) The plastic shell is removed and replaced by an uncharged metal ball, as in Figure 15.96.

Figure 15.96 At location A inside the metal ball, a distance b to the left of the outer surface of the ball, accurately draw and label the electric field due to the ball charges and the electric field of the four charges. Explain briefly. (d) Show the distribution of ball charges. (e) Calculate the x and y components of the net electric field at location A.

Section 15.7 15.X.72 A student said, “When you touch a charged piece of metal, the metal is no longer charged: all the charge on the metal is neutralized.” As a practical matter, this is nearly correct, but it isn't exactly right. What's wrong with saying that all the charge on the metal is neutralized?

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15.X.73 You are wearing shoes with thick rubber soles. You briefly touch a negatively charged metal sphere. Afterward, the sphere seems to have little or no charge. Why? Explain in detail. 15.X.74 Criticize the following statement: “When you rub your finger along the slick side of a U tape, the excess charges flow onto your finger, and this discharges the tape.” Draw diagrams illustrating a more plausible explanation. 15.X.75 Can you charge a piece of plastic by induction? Explain, using diagrams. Compare with the amount of charging obtained when you charge a piece of metal by induction. 15.X.76 Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 5 nC. Using insulating handles, the blocks are moved so they touch each other. After touching for a few seconds, the blocks are separated (again using insulating handles). (a) What is the final charge of block A? (b) What happened while the blocks were in contact with each other? (1) Protons moved from block B to block A (2) Positrons moved from block B to block A (3) Electrons moved from block B to block A (4) Both protons and electrons moved. (5) No charged particles moved.

15.X.77 You run your finger along the slick side of a positively charged tape, and then observe that the tape is no longer attracted to your hand. Which of the following are not plausible explanations for this observation? Check all that apply. (A) Sodium ions (Na+) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (B) Electrons from the mobile electron sea in your hand move onto the tape, leaving the tape with a zero (or very small) net charge. (C) Chloride ions (Cl−) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (D) Protons are pulled out of the nuclei of atoms in the tape and move onto your finger. 15.X.78 You observe that a negatively charged plastic pen repels a charged piece of magic tape. You then observe that the same piece of tape is repelled when brought near a metal sphere. You are wearing rubber soled shoes, and you touch the metal sphere with your hand. After you touch the metal sphere, you observe that the tape is attracted to the metal sphere. Which of the following statements could be true? Check all that apply. (A) Electrons from the sphere traveled through your body into the Earth. (B) Electrons from the sphere moved into the salt water on your skin, where they reacted with sodium ions. (C) After you touched it, the metal sphere was very nearly neutral. (D) Chloride ions from the salt water on your hand moved onto the sphere. (E) The excess negative charge from the sphere spread out all over your body. (F) Electrons from your hand moved onto the sphere. (G) Sodium ions from the salt water on your hand moved onto the sphere. 15.P.79 The diagrams in Figure 15.97 show a sequence of events involving a small lightweight aluminum ball that is suspended from a cotton thread. In order to get enough information, you will need to read through the entire sequence of events described below before beginning to answer the questions. Before trying to select answers, you will need to draw your own diagrams showing the charge state of each object in each situation.

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Figure 15.97 (a) A small, lightweight aluminum ball hangs from a cotton thread. You touch the ball briefly with your fingers, then release it (Diagram 1 in Figure 15.97). Which of the diagrams in Figure 15.98 best shows the distribution of charge in and/or on the ball at this moment, using the diagrammatic conventions discussed in this chapter?

Figure 15.98 (b) A block of metal that is known to be charged is now moved near the ball (Diagram 2 in Figure 15.97). The ball starts to swing toward the block of metal, as shown in Diagram 3 in Figure 15.97. Remember to read through the

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whole sequence before answering this question: Which of the diagrams in Figure 15.98 best shows the distribution of charge in and/or on the ball at this moment? (c) The ball briefly touches the charged metal block (Diagram 4 in Figure 15.97). Then the ball swings away from the block and hangs motionless at an angle, as shown in Diagram 5 in Figure 15.97. Which of the diagrams in Figure 15.98 best shows the distribution of charge in and/or on the ball at this moment? (d) Finally, the block is moved far away. A negatively charged rod is brought near the ball. The ball is repelled by the charged rod, as shown in Diagram 6 in Figure 15.97. Which of the diagrams in Figure 15.98 best shows the distribution of charge in and/or on the ball at this moment? 15.P.80 You have three metal blocks marked A, B, and C, sitting on insulating stands. Block A is charged +, but blocks B and C are neutral.

Figure 15.99

Without using any additional equipment, and without altering the amount of charge on block A, explain how you could make block B be charged + and block C be charged −. Explain your procedure in detail, including diagrams of the charge distributions at each step in the process. 15.P.81 You have two identical metal spheres labeled A and B, mounted on insulating posts, and you have a plastic pen that charges negatively when you rub it on your hair.

Figure 15.100 (a) (+ and −) Explain in detail, including diagrams, what operations you would carry out to make sphere A have some positive charge and to make sphere B have an equal amount of negative charge (the spheres are initially uncharged). (b) (+ and +) Explain in detail, including diagrams, what operations you would carry out to make sphere A have some positive charge and to make sphere B have an equal amount of positive charge (the spheres are initially uncharged). 15.P.82 Here is a variant of “charging by induction.” Place two uncharged metal objects so as to touch each other, one behind the other. Call them front object and back object. While you hold a charged comb in front of the front object, your partner moves away the back object (handling it through an insulator so as not to discharge it). Now you move the comb away. Explain this process. Use only labeled diagrams in your explanation (no prose!). 15.P.83 Metal sphere A is charged negatively and then brought near an uncharged metal sphere B. Both spheres rest on insulating supports, and the humidity is very low.

Figure 15.101

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(a) Use +'s and −'s to show the approximate distribution of charges on the two spheres. (Hint: Think hard about both spheres, not just B.)

Figure 15.102 (b) A small, lightweight hollow metal ball, initially uncharged, is suspended from a string and hung between the two spheres.

It is observed that the ball swings rapidly back and forth hitting one sphere and then the other. This goes on for 5 seconds, but then the ball stops swinging and hangs between the two spheres. Explain in detail, step by step, why the ball swings back and forth and why it finally stops swinging. Your explanation must include good physics diagrams.

Section 15.8 15.X.84 Suppose that you try to measure the electric field at a location by placing a charge there and observing the force , so that you measure . Then you remove and place a much larger charge at the same location, and observe the force . This time you measure , though you expected to measure again. What's going on here? Why didn't you get in your second measurement? Sketch a possible situation that would lead to these measurements.

Section on Experiments 15.X.85 What evidence do you have that two U tapes have the same sign charge? (A) Two U-tapes repel each other. (B) A U-tape and an L-tape attract each other. (C) Two U-tapes are attracted to each other. (D) Two U-tapes do not interact at all. (E) The strength of the interaction between two U-tapes depends on distance. (F) Two U-tapes sometimes attract, sometimes repel, and sometimes do not interact. 15.X.86 How does the interaction between two U-tapes vary with distance? (A) Strength of interaction first increases, then decreases, as distance decreases. (B) Strength of interaction increases as distance decreases. (C) Strength of interaction does not depend on distance. (D) Strength of interaction decreases as distance decreases. 15.X.87 What physics principle or principles lead to a theoretical prediction that a U-tape and an L-tape should have opposite charges? Select all that apply.

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(A) Conservation of charge (B) Newton's second law (C) Coulomb's law (D) Retardation (E) The principle of relativity (F) The superposition principle 15.P.88 You take two invisible tapes of some unknown brand, stick them together, and discharge the pair before pulling them apart and hanging them from the edge of your desk. When you bring an uncharged plastic pen within 10 cm of either the U tape or the L tape you see a slight attraction. Next you rub the pen through your hair, which is known to charge the pen negatively. Now you find that if you bring the charged pen within 8 cm of the L tape you see a slight repulsion, and if you bring the pen within 12 cm of the U tape you see a slight attraction. Briefly explain all of your observations.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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Electric Field of Distributed Charges

KEY IDEAS To calculate the electric field at some observation location due to a macroscopic object that is covered with an approximately uniform layer of charge: (1) Divide the object's surface into many small pieces. (2) Approximate each piece by a point charge located at its center. (3) Add up the contributions of each piece. • The addition can be done numerically. • In some cases the addition can be done analytically, using calculus. (4) Check the answer. Many objects can be modeled as a uniformly charged object such as:

▪ ▪ ▪ ▪ ▪

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A thin rod A ring A disk (or a square flat plate) Two parallel disks (a capacitor) A hollow sphere (spherical shell)

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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OVERVIEW In this chapter we will study mathematical techniques for adding up the contributions to the electric field of large numbers of point charges distributed over large areas. The most general technique is to divide the charge distribution into a large but finite number of pieces, approximate each piece by a point charge, and use a computer to add up the contributions (“numerical integration”). This chapter introduces an approach to setting up and carrying out such a computation. In a few special but important cases we can get an analytical solution by using an integral to add up the contributions. We are able to do this for some locations near a charged rod, ring, disk, capacitor, and sphere. A major advantage of the analytical approach is that the result allows us to see how the field varies with distance from the charge distribution, which is important in many applications. These results turn out to be useful because it is often possible to model ordinary objects as combinations of spheres, rods, rings, and disks, and thereby to estimate their electric fields.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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A UNIFORMLY CHARGED THIN ROD As an example of how to calculate the electric field due to large numbers of charges, we'll consider the electric field of a uniformly charged thin rod (Figure 16.1). The rod might, for example, be a glass rod that was rubbed all over with silk, giving it a nearly uniform distribution of positive charge on its surface. We'll consider a thin rod of length L and total positive charge Q. To illustrate the process, we will pick an observation location on the midline of the rod; this simplifies the algebra we will need to do.

Figure 16.1 The electric field at locations near a positively charged rod. Near the middle of the rod the electric field vectors lie in a plane perpendicular to the rod.

QUESTION Before launching into a calculation, think about the pattern of electric field you would expect to observe around a rod. Viewed end-on, what would it look like?

The field of a point charge was spherically symmetric, since a point charge itself has spherical symmetry. Since a rod is shaped like a cylinder, we should expect cylindrical symmetry in the electric field of a uniformly charged rod. Near a positively charged rod, the electric field might look like Figure 16.1. If we took a slice perpendicular to the rod, near the middle of the rod, the pattern of field might look like Figure 16.2.

Figure 16.2 The electric field near a charged rod should be cylindrically symmetric. This is a representation of the field in a plane perpendicular to the rod, near the middle of the rod.

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The process of finding the electric field due to charge distributed over a macroscopic object has four steps: 1. Divide the charged object into small pieces. Make a diagram and draw the electric field 2. Choose an origin and axes. Write an algebraic expression for the electric field

contributed by one of the pieces.

due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically. 4. Check that the result is physically correct.

We will carry out this process to find the electric field of a uniformly charged thin rod, at a location on the midline.

Step 1: Divide the Distribution into Pieces; Draw To apply the superposition principle, we imagine cutting up the thin rod into very short sections each with positive charge ΔQ, as shown in Figure 16.3. The Greek capital letter delta (Δ) denotes a small portion of something or a change in something. Here ΔQ is a small portion of the total charge Q of the rod, which contributes to the net field at the observation location. We can treat the piece ΔQ as though it were a point charge, which should be a fairly good approximation as long as its size is small compared to the distance to the observation location.

Figure 16.3

is the contribution to the total field at location x, 0, 0 made by a small piece of the rod of length Δy.

In Figure 16.3 we have picked a representative piece of the rod that is not at a “special” location (in this case, neither at an end nor at the middle), and drawn its contribution to the net field at our chosen observation location.

Assumptions We assume that the rod is so thin that we can ignore the thickness of the rod. We choose the piece of charge ΔQ to be small enough that it can be modeled as a point charge.

Step 2: Write an Expression for the Electric Field Due to One Piece We will approximate each piece of the rod as a point charge. Therefore the expression for will have the familiar form of the equation for the electric field of a point charge. The challenge is to write this expression in terms of our chosen origin and axes, and to figure out a general expression for the amount of charge on each piece.

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In order to write an algebraic expression for the contribution to the electric field of one representative piece of the rod, we need to pick an origin and axes. It is possible to put the origin anywhere, but some choices make the algebra easier than others. Here we put the origin at the center of the rod; the x axis extends to the right, and the y axis extends up. The location of one piece of the rod (shown in Figure 16.4) depends on y.

Figure 16.4 The components of

may be calculated by multiplying the scalar quantity by the components of .

QUESTION What variables should remain in our answer?

The coordinates of the observation location should remain, but we will sum (integrate) over all pieces of the rod, so the coordinates of the piece of the rod should not remain. However, in the expression we want to integrate it is okay to have variables representing the location of the rod segment; they are called “integration variables” and will disappear after we do the sum. A piece of the rod is a distance y from the origin. The length of the piece is Δy, a small increment in the integration variable y, shown in Figure 16.4. Our invented variable y will not appear in the final result. If there is no Δ something in the expression (such as Δy), we can't do a sum or evaluate an integral. The vector points from the source (the representative little piece of charge ΔQ) to the observation location. We can read the from the diagram in Figure 16.4: components of

Note that −y, the y component of

, is negative (as it should be).

The scalar part of the field, due to this piece, is:

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We can now write the vector

The components of

(Figure 16.3):

(Figure 16.4) are then:

ΔQ and the Integration Variable QUESTION Which things are constants and which things are variables in the expressions for ΔE x and ΔE y ?

You should have identified y as a variable, since it differs for each small piece. We need to rewrite the expression in a form in which it is easy to add up all the ΔE x 's and ΔE y 's due to all the pieces of the rod. This means expressing everything in terms of one integration variable, related to the coordinates of the piece; in this case y is the integration variable. In particular, we need to express the charge ΔQ in terms of the integration variable y. The rod is uniformly charged with a total charge Q (positive or negative), so the amount of charge on a section of length Δy is equal to

since Δy/L is the fraction of the whole rod represented by Δy. Alternatively, there is a linear charge density of Q/L in coulombs/meter, so on a length of Δy meters there is an amount of charge ΔQ = (Q/L)Δy. The important thing is to realize that ΔQ is a small fraction of the total charge Q, and that you must express ΔQ in terms of Δy.

Algebraic Expression for Putting it all together, we have an expression for the x component:

A similar approach yields an expression for the y component:

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A Note on the Symbol Δ We have used the Greek letter Δ (capital delta) in two ways in setting up this problem. ΔE or ΔQ refers to a small contribution to a total quantity. Δy refers to a change in the integration variable y, which determines the location of the piece of the rod currently under consideration. You will need to use the symbol Δ in both these ways when setting up problems like this one.

Step 3: Add Up the Contributions of All the Pieces Before performing the actual summation, we should think about what we expect the answer to be. Since we have chosen an observation location on the midline of the rod, we should expect the y component of the final result to be zero, as indicated in Figure 16.5. In an optional section at the end of the chapter we discuss the more complex situation of an observation location not on the midline.

Figure 16.5 Along the midplane only the x component of the electric field is nonzero, because the y contributions cancel.

x Components Each ΔQ contributes ΔE x to the net field. If we number each piece 1, 2, 3, and so on, we have It is standard practice to write such sums in a more compact form using a Σ (the Greek capital sigma stands for “Summation”):

y Components

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At this point we have to decide how to add up these terms. Each term in this summation is different. Although we can choose to make all the pieces the same length Δy and take Δy out of the sum, y itself is different for every piece along the rod. How can we add up all the contributions of all the pieces?

Numerical Summation One way to add up all these different contributions would be to divide the rod into 10 slices as shown in Figure 16.6, calculate the contribution of each of those 10 slices (using our formula with Δy = L/10), and add up these 10 numbers. Of course that would be only an approximate result, because each slice isn't really a point particle, but it might be good enough for many purposes. Let's do this calculation for the case of L = 1 meter, Δy = 0.1 meter, Q = 1 nC, and x = 0.05 meter. For each slice we'll take y to be at the center of the slice, and we'll use a calculator to evaluate each term in the summation. The results are shown in Figure 16.7, including the sum for all pieces. As expected, we see that the sum of the y components is zero.

Figure 16.6 Division of the rod into 10 equal slices.

Figure 16.7 Contributions to

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.

It would be possible to get a more accurate value by cutting the rod into more slices, but it would be tedious to do the calculations by hand. We wrote a little computer program to do these calculations for various numbers of slices, and the results of these calculations are shown in Figure 16.8.

Figure 16.8 Effect of number of slices.

Judging from the computations summarized in Figure 16.8, our calculation with just 10 slices was not very accurate (though it might be good enough for some purposes), while taking more than 50 slices makes almost no difference. The accuracy depends on how good an approximation it is to consider one slice as a point charge. With 50 slices, each slice Δy is 1/50th of a meter long (0.02 meters), and the nearest slice is 0.05 meters away from the observation location. Apparently the 0.02-meter slices are adequately approximated by point charges at their centers, since using smaller 0.01-meter slices (100 slices) gives practically the same result. How do we know that our computer summations are correct? For 10 slices they agree with the calculator results. For larger and larger numbers of slices the computer results approach a constant value, which is expected. Later we will discuss additional ways to check such work.

Summation as an Integral A major disadvantage of adding up the contributions numerically is that we don't get an analytical (algebraic) form for the electric field. This means that we can't easily answer such questions as, “How does the electric field vary with r near a rod?” For a point charge we know that the field goes like 1/r 2, and far from a permanent dipole it goes like 1/r 3. Is there a way to do the summation to get an algebraic rather than a numerical answer for the rod? That is what integral calculus was invented to do. The key idea of integral calculus applied to problems like ours is to imagine taking not 50 or 100 slices but an infinite number of infinitesimal slices. We let the number of slices N = L/(Δy) increase without bound, and the corresponding slice length Δy = L/N decreases without bound. We take the limit as Δy gets arbitrarily small:

This limit is called a “definite integral” and is written like this:

The integral sign ∫ is a distorted S standing for Summation, just as the Greek sigma Σ stands for summation. It is important that in this context you think of an integral as a sum of many contributions. The integration variable y ranges from y = −L/2 (the bottom of the rod) to y = +L/2 (the top of the rod), so these are the limits on the definite integral. If you have drawn an appropriately labeled diagram, you should be able to read the correct limits of the integration variable off the diagram. This is another of the benefits of drawing a well-labeled physics diagram. Our small length Δy has now changed into dy, an “infinitesimal” increment in y that must appear in the integrand. This is why Δy must appear in the algebraic expression for the contribution of one piece of the charge distribution.

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Evaluating the Integral Most of the physics in this problem went into setting up the integral. Evaluating the integral is simply mathematics. In some cases it is easy to evaluate the integral. In this particular case, the integral is not a very simple one, but if you would like to exercise your skill at integration, give it a try! Otherwise, it can be found in tables of integrals in mathematical handbooks and in some calculus textbooks. Some calculators can do such integrals, or one can use a tool such as Maple or Mathematica. Looking up the result in a table of integrals, we get the following:

Note that as we expected this result does not contain the integration variable y, which was simply a variable referring to the coordinates of one piece of the rod and was necessary in setting up the summation. The y component of the electric field can also be found by integration. In this situation the result comes out to zero, as expected.

Replace x with r Because the rod, and its associated electric field, are cylindrically symmetric, the axis we called the x axis could have been rotated around the rod by any angle, and we would have obtained the same answer. To indicate this, we replace x with r in our result (you may recognize this as converting to cylindrical coordinates):

ELECTRIC FIELD OF A UNIFORMLY CHARGED THIN ROD

at a location a distance r from the midpoint of the rod along a line perpendicular to the rod. Q is the total charge on the rod, and L is the length of the rod. The direction is either radially away from or toward the rod (depending on the sign of Q). See Figure 16.9.

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Figure 16.9 Electric field of a uniformly charged thin rod, near the midpoint of the rod.

Step 4: Check the Result Because there are many opportunities to make mistakes in this procedure, it is extremely important to check the result in as many ways as you can. Different checks provide information about different kinds of possible errors. Direction: First, is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero. Units: Second, do we have the right units? The units should be the same as the units of the expression for the electric field for a single point particle:

We easily verify that our answer does have the right units, since

Special case: r >> L (very short rod, or very far from rod)

QUESTION Next let's try a special case for which we already know the answer. If r is verymuch larger than L, the distant rod looks almost like a point charge, so the net field ought to look like the field of a point charge. Does it?

Special case: L r.

QUESTION Before reading further, try to show that if L >> r, the electric field is approximately:

This distance dependence is illustrated in Figure 16.10. This also holds for a short rod if we are very close, so that L >> r.

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Figure 16.10 The electric field of a very long uniformly charged positive rod, shown at locations along a line very near the center of the rod. Only a small portion of the rod is visible.

APPROXIMATE ELECTRIC FIELD: VERY LONG ROD

at a location a perpendicular distance r 0)

Path perpendicular to

: Potential does not change (ΔV = 0)

Indicating the Path Direction To show the direction of the path, we use the following notation. As usual, note that the symbol Δ indicates “final – initial.”

(We will see later that the symbol VA actually has a meaning, and denotes the “potential at location A,” which is defined as the potential difference along a path starting infinitely far away and traveling to location A.) Because you can determine the sign of a potential difference on purely physical grounds, you should never get the wrong sign! In the following exercises you are asked to give the correct sign for the potential difference between two locations. Use the rule given above. 17.X.8 In Figure 17.14, what is the direction of the electric field? Is ΔV = V f − V i positive or negative?

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Answer

Figure 17.14 Exercise 17.X.8. 17.X.9 In Figure 17.15, what is the direction of the electric field? Is ΔV = V f − V i positive or negative?

Figure 17.15 Exercise 17.X.9.

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Answer

POTENTIAL DIFFERENCE IN A NONUNIFORM FIELD Two Adjacent Regions with Different Fields In the previous sections we considered only situations in which the electric field is uniform within a region. However, often the real world is more complex than this. We need to be able to calculate changes in electric potential and electric potential energy in regions of space in which the electric field is not uniform. Consider a situation in which the path from the initial location to the final location of interest passes through two regions of uniform but different electric field. In region 1 the electric field is , and in region 2 the electric field is , as shown in Figure 17.16. What would be the change in electric potential energy of a particle traveling along a path from location A to location B? (Assume that there is a tiny hole in the middle plate, so a particle can pass through it.)

Figure 17.16 The path from location A to location B passes through two regions in which the field is different. In order to calculate ΔV in this situation, we need to divide the path into two pieces. Each displacement vector must be small enough that the electric field is uniform in the region through which it passes. Essentially, we have to invent a new point, which we can call C, on the boundary between the two regions, as shown in Figure 17.17.

Figure 17.17 We divide the path into two segments by inventing a point C. Each segment of the path now passes through a region of uniform electric field.

Now we can calculate the difference in potential along the two segments of the path: From location A to location C: From location C to location B: From location A to location B:

In general, we need to add up all components (x, y, and z), so we can write the general equation as a sum:

POTENTIAL DIFFERENCE ACROSS SEVERAL REGIONS

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EXAMPLE Potential Difference Across Two Regions Suppose that from x = 0 to x = 3 the electric field is uniform and given by to x = 5 the electric field is uniform and given by

, and that from x = 3 (see Figure 17.18). What is the potential

difference ΔV = V C − V A ?

Figure 17.18 The electric field differs in the two regions.

Solution Split the path into two parts, A to B and B to C. In each part the electric field is constant in magnitude and direction. We have the following:

17.X.10 In the earlier Figure 17.17, location A is at 0.5, 0, 0 m, location C is 1.3, 0, 0 m, and location B is 1.7, 0, 0 m. and . Calculate the following quantities: (a) ΔV along a path going from A to B, and (b) ΔV along a path going from B to A.

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Answer

In Chapter 15 we proved that inside a conductor, such as a metal object, in equilibrium, the net electric field is zero (because otherwise mobile charges would shift until they contribute a field large enough to cancel the applied field). Therefore it must be the case that in equilibrium the net electric field is zero at all locations along any path through the conductor. This means that the potential difference is zero between two locations inside the metal (Figure 17.19).

Figure 17.19 Potential difference is zero inside a metal in equilibrium.

In a conductor in equilibrium the difference in potential between any two locations is zero, because E = 0 inside the conductor.

We'll use this fact, in combination with the approach developed in this section, to examine a situation involving the potential difference across multiple regions, one of which is inside a metal object. Suppose that a capacitor with large plates and a small gap of 3 mm initially has a potential difference of 6 volts from one plate to the other (Figure 17.20).

Figure 17.20 A capacitor with large plates and small gap.

QUESTION What are the direction and magnitude of the electric field in the gap?

The direction of the electric field in the gap is toward the left, away from the positive plate and toward the negative plate. The magnitude can be found from noting that the potential difference ΔV = Es = 6 volts, so that E = (6 volts)/(0.003 m) = 2000 volts/meter. Does Inserting a Metal Slab Change ΔV? Next we insert into the center of the gap a 1-mm-thick metal slab with the same area as the capacitor plates, as shown in Figure 17.21. We are careful not to touch the charged capacitor plates as we insert the metal slab. We want to calculate the new potential difference between the two outer plates.

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Figure 17.21 Insert a metal slab in the middle of the capacitor gap. First we need to understand the new pattern of electric field that comes about when the metal slab is inserted. The charges on the outer plates are −Q 1 and +Q 1, both distributed approximately uniformly over a large plate area A. The metal slab of course polarizes and has charges of +Q 2 and −Q 2 on its surfaces, as indicated in Figure 17.21. The electric field inside a capacitor is approximately E = (Q/A)/ε 0, where Q is the charge on one plate and A is the area of the plate, if the plate separation s is small compared to the size of the plates.

QUESTION Since the electric field inside the metal slab must be zero, we can conclude that Q 2 is equal to Q 1. Why?

The plates and the surfaces of the slab have the same area A. The outer charges −Q 1 and +Q 1 produce an electric field in the metal slab E 1 = (Q 1/A)/ε 0 to the left. The inner charges +Q 2 and −Q 2 are arranged like the charges on a capacitor, so they produce an electric field in the metal slab E2 = (Q 2/A)/ε 0 to the right. The sum of these two contributions must be zero, because the electric field inside a metal in equilibrium must be zero. Hence Q 2 is equal to Q 1. As a result, the left pair of charges produces a field like that of a capacitor, and the right pair of charges also produces a field like that of a capacitor. The effect is that after inserting the metal slab, the electric field is still approximately 2000 V/m in the air gaps but is now zero inside the metal slab. (The fringe fields are small if the gap is small.)

QUESTION Now that we know the electric field everywhere, what are the potential differences across each of the three regions between the plates (air gap, metal slab, air gap)?

The electric field in the air gap is essentially unchanged, so ΔV inside the metal slab must be zero, because E = 0 inside the slab. The potential difference between the plates of the capacitor is now: not 6 V as it was originally. (Note that there is no such thing as “conservation of potential”; the potential difference changed when we inserted the slab.) A Powerful Reasoning Tool. When you know that the electric field inside a metal must be zero, you know that the sum of all the contributions to that electric field must be zero. This is a powerful tool for reasoning about fields and charges in and on metals in equilibrium.

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17.X.11 Explain qualitatively why the new potential difference calculated above is less than the original 6 volts. 17.X.12 In Figure 17.22, location A is inside a charged metal block, and location B is outside the block. The metal block sits on an insulating surface and is not in contact with any other object. The electric field outside the block is .

Answer

Answer

Figure 17.22 Location A is inside a charged metal block. Location B is outside the block. (a) Calculate ΔV along a path going from A to B. (b) Calculate ΔV along a path going from B to A.

A Region of Varying Electric Field Requires an Integral In some situations, the electric field in a region varies continuously. For example, the electric field due to a single point charge is different in magnitude and direction at every location in space. To find a change in electric potential between two locations in such a region, we need to take into account the varying electric field. One approach would be to divide the path between the two locations into a finite number of steps, and use an average value of the electric field along each step to compute an approximate potential difference along that step. We could add these approximate values to get an approximate value for the potential difference along the entire path. As we make the length of each step smaller and smaller, the sum turns into an integral:

POTENTIAL DIFFERENCE IN A REGION OF VARYING FIELD

which can be evaluated as the sum of three separate integrals:

A Complete Procedure for Calculating Potential Difference LibraryPirate

To calculate the potential difference between points a and b, always begin by considering the equation

(a) Decide whether varies in the region of interest. Draw the pattern of electric field throughout this region, making sure that you are considering the electric field not only at points a and b, but at locations in space around those points as well. (b) Specify the origin and a set of axes for this situation. (c) Write

as a function of position using the coordinates specified by your chosen axes.

(d) Draw a path that connects points a and b. Be sure that your path starts at the initial location and ends at the final location. Based on this path and the electric field in the region, predict the sign for the potential difference associated with moving along the path. (e) Convert axes. (f) Evaluate

to a quantity that represents a displacement along the path you have chosen in the coordinates given by the

and integrate (or sum numerically). Check the sign that you obtain after performing your integration and compare it against your prediction.

Hints and Rules of Thumb Remember that you are free to choose any path that connects the two points, since the electric potential difference between two locations in space is independent of path. Picking a path that takes into consideration the dot product simplify the problem.

and the field pattern you've already drawn could significantly

Potential Difference Near a Point Charge We'll apply this procedure in a familiar situation: a region near a single point charge, as shown in Figure 17.23.

Figure 17.23 The electric field varies continuously along a path near a point charge. (a) In our drawing of the region around the point charge (Figure 17.23) we see that

varies, so we need to use an integral.

(b) We pick the location of the charge as the origin, with the x axis in the direction from location A to location B. (c) We write the field between A and B as a function of x:

(d) We choose as our path a straight line along our x axis, from A to B. Since the direction of the path from A to B is the same as the direction of the electric field in the region, the potential difference should be negative. (e) We divide the path up into a very large number of infinitesimal steps, each of the same magnitude and direction. Instead of , we will call each in-finitesimal displacement vector (f)

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.

The value of x at location A is x a, and the value of x at location B is x b, so these are the limits of integration:

We can determine the sign of this quantity by looking at both the sign of Q and the difference

. In this case Q > 0 and

is negative, and ΔV is negative, as we expected in step (d).

x b > x a, so

If the point charge were negative, the potential difference would be positive; this also makes sense, since the electric field would be opposite to the direction of the path.

EXAMPLE Potential Difference Due to a Proton A proton is located at the origin. Location C is 1 × 10−10 m from the proton, and location D is 2 × 10−8 m from the proton, along a line radially outward. (a) What is the potential difference V D − V C ? (b) How much work would be required to move an electron from location C to location D?

Solution (a) Potential difference:

(b) Work to move an electron: The least amount of work will be required if we don't change the kinetic energy of the proton.

General Definition of Potential Difference The most general definition of potential difference is the integral expression, which can be used in any situation.

GENERAL DEFINITION OF POTENTIAL DIFFERENCE

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This general expression is correct for regions in which the field is uniform, as well as for regions in which the field varies. For example, in the situation shown in Figure 17.24, where the field is uniform,

The result of the integration is the same as we got in the example involving uniform field in Section 17.3.

Figure 17.24 A region of uniform electric field.

Limit on Mathematical Complexity in This Text Using

to calculate the potential difference along a path that is not straight, such as that shown in

Figure 17.25, along which the electric field is varying in magnitude and direction, can be mathematically challenging. In many cases the best way to do this is to do it numerically, using a computer. In this text we will not ask you to do arbitrarily complex integrations of this kind analytically.

Figure 17.25 The potential difference for each step along this path is

. The potential difference

is

the sum of all these contributions.

To establish the basic principles we will typically consider relatively simple situations. Often the electric field is uniform (same magnitude and direction) all along a straight path, which makes the calculation very simple. Another relatively simple situation is one in which the magnitude varies but not the direction along a straight path. A third simple situation that we will encounter is one in which we move along a curved path, but the electric field is always in the same direction as this curved path, and often constant in magnitude, which occurs in electric circuits. In each of these simple cases it can be quite easy to evaluate the potential difference, as we saw in the preceding exercises.

EXAMPLE

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A Disk and a Spherical Shell A thin spherical shell made of plastic carries a uniformly distributed negative charge −Q 1. A thin circular disk made of glass carries a uniformly distributed positive charge +Q 2. The radius R 1 of the plastic spherical shell is very small compared to the large radius R2 of the glass disk. The distance from the surface of the spherical shell to the glass disk is d, and d is much smaller than R2. Find the potential difference V 2 − V 1. Location 1 is at the center of the plastic sphere, and location 2 is just outside the glass disk. State what approximations or simplifying assumptions you make.

Solution Alternatively, we could add the two electric field contributions, then integrate along a path from 1 to 2. Initial location: 1 Final location: 2 Path: Straight line from 1 to 2 (see Figure 17.27). so

according to the Superposition Principle,

Figure 17.26 A uniformly charged plastic shell and a uniformly charged glass disk.

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Figure 17.27 Electric field contributions of the shell and the disk, shown at locations along the path. This allows us to find ΔV due to the shell and ΔV due to the disk separately, then add them. Simplifying assumption: Neglect the polarization of the plastic and glass, because there is little matter in the thin shell and thin disk, so the field of the polarized molecules is negligible compared to the contributions of −Q 1 and +Q 2. ΔV due to shell: Put origin at center of shell. Inside the shell: V surface of shell − V 1 = 0 because E shell = 0 inside the shell. To the right of the shell:

Check sign: As we move toward the disk, we're moving opposite to the field, so the potential should increase. Result agrees, since +Q 1/R 1 is the larger term. ΔV due to disk: an approximation is that since

and

,

We could set up an integral, placing the origin at the center of the disk. However, since uniform, we recognize that the result of integrating will be

is approximately

Check sign: As we move toward the disk, we're moving opposite to the field, so the potential should increase. Result agrees.

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ΔV due to both shell and disk:

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PATH INDEPENDENCE In Chapter 7 we showed that potential energy differences depend only on the initial and final states of a system and are independent of path: different path, different process, but same change of state. Similarly, the potential difference ΔV = V B − V A between two locations A and B does not depend on the path taken between the locations. To calculate the potential difference between two locations based on the electric field in the region, it is necessary to choose a path. We are free, however, to choose a path that simplifies our calculations.

A Simple Example of Two Different Paths To see how the potential difference between two locations can be the same along different paths, we'll consider two different paths through a capacitor. Suppose you move from the positive plate of a capacitor to the negative plate, moving at an angle to the electric field (Figure 17.28).

Figure 17.28 Moving at an angle to the electric field.

QUESTION Calculate the potential difference in going from A to C: ΔV = V C − V A = ?

Since the electric field is the same in magnitude and direction all along the path, we can write

Suppose instead that we move along the two-part path shown in Figure 17.29.

Figure 17.29 Moving at right angles to the electric field.

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QUESTION Now ΔV = V C − V A = ?

Along the path from A to B,

Along the path from B to C,

Therefore we again find that which is what we found for the direct path from A to C.

QUESTION Does it make sense that V C − V B = 0?

Yes. Just as no work is required to move at right angles to a force, the electric potential does not change along a path perpendicular to the electric field. The potential difference doesn't depend on the angle of our path in Figure 17.28 because the sideways components of our steps don't affect ΔV.

Two Different Paths Near a Point Charge Along a path straight away from a stationary charge Q we know this:

It is instructive to work out the potential difference along a different path.

QUESTION In the situation shown in Figure 17.30, calculate the potential difference for each branch of the path from location i to A to B to C to location f, in terms of Q and the radii r 1, r 2, and r 3, then add these up to get the potential difference along this path. As a first step, draw the electric field at various locations along the path.

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Figure 17.30 What is the potential difference along this complicated path?

From i to A,

, so

From A to B, From B to C,

. , so

.

From C to f,

.

Adding up these ΔV's, we get:

This is exactly the same result that you would get by going directly in a straight line from i to f. We see here that it is only when has a radial component that there is a nonzero contribution to ΔV. Along the curved arcs the electric field is perpendicular to the path, so the arcs contribute zero to the potential difference. Going along the straight lines is equivalent to just going from r 1 to r 2. Evidently it doesn't matter what path we take. We say that potential difference is “path independent.”

An Arbitrary Path Near a Point Charge We can see more generally how this works out for any path near a stationary point charge. Follow an arbitrary path from initial location i to final location f (Figure 17.31). At each location along the path draw the component dlradial of in the direction of , pointing away from the charge Q.

Figure 17.31 Calculating the potential difference along an arbitrary path.

At every location along the path we have However, dl cos θ is the radial component of this arbitrary path we are simply integrating

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and is equal to dlradial , the change in distance from the charge. Therefore along , so

All that matters are the initial and final distances from the charge, not the path we happen to follow. This is consistent with our earlier result for potential difference along a straight path between location i and location f. What is didn't depend on what path we took. interesting is seeing how it worked out that the integral of

17.X.13 Calculate the potential difference along the closed path consisting of two radial segments and two circular segments centered on the charge q (Figure 17.32). Show that the four ΔV's add up to zero. It is helpful to draw electric field vectors at several locations on each path segment to help keep track of signs.

Answer

Figure 17.32 Exercise 17.X.13.

Round-Trip Potential Difference Is Zero An important consequence of path independence is that the integral of along a round-trip path gives a potential difference of zero, as we can now show (Figure 17.33). Move along path 1 (A-B-C), and then back to location A along path 2 (C-D-A).

Figure 17.33 Round-trip potential difference on a path from A back to A again is zero.

QUESTION Suppose that the potential difference along path 1 is −25 volts. What would be the potential difference in going backward along path 2 (A-D-C)? Forward along path 2 (C-D-A)?

Going backward along path 2 is the same as going forward along path 1, so the potential difference along the path (A-D-C) must be −25 volts. Going in the opposite direction, along path 2 (C-D-A), reverses the sign of the parallel component of the electric field, so

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the potential difference along path 2 must be +25 volts.

QUESTION If we walk all the way around the closed path, A −B −C −D −A, what is the potential difference?

The round-trip potential difference is zero (−25 volts plus +25 volts). Note that this result is the direct consequence of the path independence of electric potential. It is also not surprising, since we do expect VA − VA = 0!

Since potential difference can be calculated as an integral involving the electric field, and the electric field is the sum of all the contributions of many stationary point charges, potential difference is the sum of the contributions of all the point charges. The potential difference integral is independent of the path for one point charge, and since all charge distributions are made up of atomic point charges, the round-trip potential difference must be zero for any configuration of stationary charges whatsoever.

PATH INDEPENDENCE

Conservation of Energy The fact that ΔV = 0 for a round trip along any path is fundamentally related to the principle of Conservation of Energy. We can use a proof by contradiction to show this connection. Let's assume that it is possible to fasten down a collection of stationary charged particles in such a configuration that a round-trip along the path shown in Figure 17.33 gives a potential difference greater than zero. If this is possible, then an electron could travel around this path many times, and ΔU = −eΔ V of the system would be negative each time. Therefore the kinetic energy of the system would increase each time, despite the fact that no external energy inputs were made. We would have invented a perpetual motion machine! Because this outcome would violate the Energy Principle, we can conclude that our assumption (that it was possible for ΔV round trip to be nonzero) must have been incorrect. These are extremely important general results, and in the future we will often have occasion to refer to these properties of electric potential due to charges. Path independence is important because when we use potential we are reasoning about the relationship between initial and final states, independent of the intervening process. We are free to analyze the most convenient process that takes the system from the initial state to the final state.

Application: A Metal Not in Equilibrium During the (extremely) brief time when a metal is polarizing due to an external charge, there is a nonzero electric field inside the metal. Before equilibrium is reached, the electric field in the metal is not zero. As we will see in detail in a later chapter, in an electric circuit the battery maintains a nonzero electric field inside the wires, as a kind of continuing polarization process that doesn't lead to equilibrium. The electric field continually pushes the sea of mobile electrons through the wires. The electric field is not zero in this situation, which is not equilibrium.

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QUESTION In Figure 17.34 there is a nonzero electric field of uniform magnitude E throughout the interior of a wire of length L, and the direction of the electric field follows the direction of the wire. What is the potential difference VB − V A ?

Figure 17.34 Inside a wire in a circuit, there can be an electric field that has uniform magnitude E but follows the direction of the wire. The wire's length is L.

In this case, since for each step

, the potential difference is simply −EL, the sum of all the

terms

along the wire. Thus if a metal is not in equilibrium, there can be a nonzero potential difference inside the metal. However, in a circuit a thick copper wire may have such a small electric field in it that there is very little potential difference from one end to the other, in which case the potential is almost (but not quite) constant in the wire. 17.X.14 In a circuit there is a copper wire 40 cm long with a potential difference from one end to the other of 0.01 volts. What is the magnitude of the electric field inside the wire?

Answer

Impossible Patterns of Electric Field We can use the fact that the potential difference over a round-trip path must be zero to reason about particular patterns of electric field. Surprisingly, we can actually conclude that certain patterns of electric field can never be created by a collection of stationary charged particles. For example, consider the pattern of electric field shown in Figure 17.35, where the electric field is tangent to the circle at every point on the circle.

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Figure 17.35 Is it possible to produce this pattern of electric field (

tangent to a circle at every point on the circle) with some

arrangement of point charges?

QUESTION Try to think of an arrangement of stationary charges that could produce the pattern of electric field shown in Figure 17.35.

We can demonstrate very simply that it is impossible to produce this pattern of electric field by any arrangement of stationary point charges! Consider a path that starts at location A and goes clockwise around the circle, ending back at location A. Since at every location along the path is parallel to , the round-trip potential difference along this path is:

However, we showed previously that for a single point charge, and hence for any assemblage of point charges, ΔV = 0 for a round trip. Therefore this “curly” pattern of electric field must be impossible to produce by arranging any number of stationary point charges! Although we have shown that we cannot produce this pattern of electric field with any arrangement of stationary point charges, it might be possible if there were some other way to produce electric fields. We'll return to this issue in a later chapter.

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THE POTENTIAL AT ONE LOCATION Usually we are interested in the potential difference between two locations; for example, ΔV = V B − V A . However, it is sometimes useful to consider the potential at only one location. The potential at location A is defined as the potential difference between a location infinitely far away from all charged particles, and the location of interest:

POTENTIAL AT ONE LOCATION

The potential at location A, sometimes referred to as “the potential relative to infinity,” is the potential difference between location A and a location infinitely far away from all charged particles. Of course, this equation makes sense only if V∞ , the potential at infinity, is zero, and by convention it is defined to be so. This is consistent with the fact that the potential energy of a system of two charged particles that are infinitely far from each other must be zero. Since potential has a value at every location in space, potential (but not potential difference) is itself a field, but it is a scalar field, not a vector field.

Potential at a Location Near a Point Charge As an example, we will find the potential at a location a distance r from a single point particle of charge q. To do this, we will find the potential difference between this location and a location infinitely far away.

QUESTION How does the sign of this result depend on the sign of the charged particle? If q < 0, V r < 0, but if q > 0, Vr > 0. Figure 17.36 is a graph of potential as a function of distance from a positive or negative point charge.

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Figure 17.36 Potential as a function of distance from a positive (red) or negative (blue) point charge.

Potential at One Location and Potential Energy If we know the value of the potential at location A, we can use this to find what the potential energy of the system would be if we placed a charged object at that location: Using our result for V we get the familiar expression for the electric potential energy of two particles with charges q1 and q2 separated by a distance r:

EXAMPLE Potential at a Location Near a Sphere A metal sphere of radius 3 mm has a charge of −2 × 10−9 C. What is the potential at location C, which is 5 mm from the center of the sphere?

Solution

We can calculate the potential due to a charge distribution in two ways: either by dividing the distribution into pointlike pieces and adding the potential due to each piece, or by calculating

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along a path from infinity to the location of interest.

EXAMPLE Potential Along the Axis of a Ring Use superposition to calculate the potential at a location z from the center of a thin ring, along the axis, by adding the potential due to each piece of the ring. The ring carries a total charge Q. The total charge Q is made up of point charges, from the location of interest (Figure 17.37). each of charge q, and each of these point charges is a distance

Figure 17.37 The potential due to a uniformly charged ring.

Solution The potential contributed by any one of these point charges q is

for one point charge q. Adding up the potential contributed by all of the point charges, we have the following result:

Integrating This same result can be obtained by integrating the electric field of the ring along a path, using the expression for this field that we found in Chapter 16.

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Conversely, from the result for the potential of the ring we can determine the electric field of the ring by differentiation, since Ez = −∂V/∂z (see Problem 17.P.104). 17.X.15 Show that if you are very far from the ring (z >> R), the potential is approximately equal to that of a point charge. (This is to be expected, because if you are very far away, the ring appears to be nearly a point.)

Answer

Potential Inside a Conductor at Equilibrium In a previous section we noted that because the net electric field is zero inside a metal object in equilibrium, the potential difference between any two locations inside the metal must be zero. Since this is true, we can conclude that the potential at any location inside the metal must be the same as the potential at any other location (Figure 17.19). In a conductor in equilibrium the potential is exactly the same everywhere inside the conductor, because E = 0 inside the conductor. so V is constant.

QUESTION Does this mean that the potential is zero at every location inside a metal in equilibrium?

No. The potential inside a metal object in equilibrium is constant (the same at every location), but it need not be zero. Charges on the surface of the metal object or on other objects may contribute to a nonzero but uniform potential inside the metal.

Figure 17.38 The potential is constant inside a metal in equilibrium.

EXAMPLE The Potential Inside a Charged Metal Sphere A solid metal sphere of radius R has a charge of Q uniformly distributed over its surface. What is the potential (relative to infinity) throughout the sphere?

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Solution At any distance r > R, the potential relative to infinity due to the charged sphere is infinity just outside the surface of the sphere is

, so the potential relative to

. Inside the sphere the electric field is zero everywhere, so the

potential difference between any location inside the sphere and the surface is zero. Therefore the potential inside the sphere is

everywhere.

FURTHER DISCUSSION This situation is a good example of the indirect relationship between field and potential. Inside the sphere the electric field is zero everywhere but the electric potential is nonzero everywhere.

Avoiding a Common Confusion A common error is to assume that the electric field at a location determines the potential at a location. In fact, the electric field at location A has very little to do with the potential at location A ! Similarly, the electric field at A and the electric field at B have very little to do with the potential difference between locations A and B; it is the electric field in the intervening region that determines ΔV. As we have just seen, the electric field due to a charged spherical shell is zero inside the shell, but the potential inside the shell due to the charges on the sphere is nonzero.

Shifting the Zero of Potential As we saw in Chapter 6, electric potential energy (and therefore potential) must go to zero at large distances, to satisfy the requirements of relativity. Nevertheless, we are often interested only in differences of potential or potential energy, in which case we can choose a different zero location as a matter of convenience. This is exactly analogous to approximating change in gravitational potential energy as Δ(−GMm/r) ≈ Δ(mg y) near the surface of the Earth, and then choosing the zero of potential energy to be at or near the surface, rather than at infinity. The electric field is given by the gradient of the potential, not by the value of the potential. This means that only potential difference ΔV matters in determining the electric field, not the value V of the potential.

QUESTION Find the magnitude of the electric field in the 0.2 mm gap of a narrow-gap capacitor whose negative plate is at a potential of 35 volts and whose positive plate is at a potential of 135 volts (Figure 17.39).

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Figure 17.39 What is the magnitude of the electric field?

The magnitude of the electric field in this case is

The direction is from the positive plate toward the negative plate (direction of decreasing potential).

QUESTION Suppose instead that the potential of the negative plate is 75,000 volts and the potential of the positive plate is 75,100 volts (Figure 17.40). Now what is the electric field in the 0.2 mm gap?

Figure 17.40 What is the magnitude of the electric field?

This makes no difference in the calculation, because it is only the 100-volt difference in the potential that counts, not the particular values of the potential.

QUESTION Now suppose that the potential of the negative plate is −500 volts and the potential of the positive plate is −400 volts (Figure 17.41). What is the electric field in the 0.2 mm gap?

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Figure 17.41 What is the magnitude of the electric field?

The magnitude of the electric field is again 5 × 105 volts/m. These calculations show that what counts in calculating the electric field is the gradient of potential or potential difference per unit distance, not the value of the potential. This is exactly analogous to the fact that it is no harder to walk up from the 75th floor to the 76th floor of a building than it is to walk up from the 1st floor to the 2nd floor. To put it another way, adding a constant V 0 to all the potential values changes nothing as far as potential difference and electric field are concerned:

Reflection: Potential and Potential Difference We have seen that there are two different ways to find the potential at a particular location: Add up the contributions of all point charges at all other locations:

Travel along a path from a point very far away to the location of interest, adding up

at each step:

One way to think of the meaning of the potential at location A is to consider how much work per unit charge you would have to do to move a charged particle from a location very far away to location A. This work would of course depend on the electric field in the region through which you moved the charge. We have also seen that there are two ways to find the potential difference between two locations A and B: Subtract the potential at the initial location A from the potential at the final location B:

Travel along a path from A to B, adding up

at each step:

Again, note that the potential difference between two locations depends on the electric field in the region between the two locations. If you were to move a charge from location A to location B, the electric field in the intervening region would determine how much work per unit charge you would have to do.

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POTENTIAL DIFFERENCE IN AN INSULATOR Reasoning from the definition of equilibrium and the existence of a sea of mobile electrons in a metal, we were able to conclude that in equilibrium the net electric field everywhere inside a metal is zero. Given this, we were able to conclude that the potential difference between any two locations inside a metal object in equilibrium must be zero. The situation inside a polarized insulator is more complex. An applied field, such as the uniform field inside a a capacitor (Figure 17.42), polarizes the molecules in the insulator. These polarized molecules themselves contribute to the net field inside the material. Presumably, however, the electric field inside the plastic due to the polarized molecules varies depending on the observation location.

Figure 17.42 Polarization of molecules in a piece of plastic inside a capacitor.

For example, consider locations A and B inside the polarized plastic shown in Figure 17.43, where we show electric field contributed solely by the induced dipoles in the polarized plastic. If we consider columns of polarized molecules to be similar to capacitors (consisting of two vertical sheets of charge), then location A is inside “capacitor” 1, while location B is between “capacitors” 2 and 3. At location A the dominant contribution by the dipoles will be from the molecules in “capacitor” 1, and the electric field due to the dipoles will be large and to the left.

Figure 17.43 Electric field contributed solely by the induced dipoles in the plastic.

At location B, the fringe fields of the nearest two “capacitors” represent the dominant contributions by the dipoles, and the field due to the dipoles will point to the right and be smaller than the field due to the dipoles at A. If we follow a path from the left side of the plastic to the right side, due to the dipoles will sometimes point in the direction of and sometimes opposite to the . direction of

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It would be useful to consider an average electric field due to the polarized molecules inside the plastic, but it isn't clear how we would calculate such a field. It is not even clear from simply inspecting the situation what the direction of this average field will be. We can construct a surprisingly simple argument based on potential to help us answer this question.

Round-Trip Potential Difference Although the pattern of E dipoles is complex inside the material, it is much less complex at locations outside the plastic. Intuitively, you may be able to see that at locations outside the plastic the electric field due to the polarized plastic will have a similar pattern to the electric field near a single dipole. In Figure 17.44 the electric field due just to the polarized plastic is shown at locations along a path outside the plastic. If we travel is always a positive quantity: there is never a component of clockwise around the path, it is clear from the diagram that opposite to

. Therefore

along the path outside the plastic.

Figure 17.44 Electric field due solely to the induced dipoles in the plastic, along a path outside the plastic.

Since the molecular dipoles consist of stationary point charges, the round-trip path integral of the electric field (the round-trip potential difference) due to the molecular dipoles must sum to zero.

QUESTION Explain why this implies that the average field inside the plastic must point to the left and cannot point to the right.

If the average field inside the plastic pointed to the right, the round-trip path integral of electric field in Figure 17.44 would be nonzero, which is impossible. A very general argument concerning the nature of potential lets us deduce the direction of the average field inside the plastic.

Dielectric Constant and Net Field The net electric field inside the plastic is the sum of the field due to the capacitor plates and the field due to the induced dipoles in the plastic, which, as we have just demonstrated, points in a direction opposite to the capacitor's field (Figure 17.45). The net field inside the plastic is therefore smaller than the field due to the capacitor alone. We can say informally that the electric field is “weakened” inside an insulator. Note carefully that the net field is in the same direction as the field made by the plates, but smaller

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in magnitude.

Figure 17.45 The net field inside the polarized insulator is smaller in magnitude than the applied field.

We define a constant K, called the “dielectric constant” (dielectric is another word for insulator), as the factor by which the net electric field is weakened inside an insulator:

For a capacitor, the applied field is and the net field inside a capacitor that is filled with an insulator is (Q/A)/ε 0/K. The dielectric constant is related to the atomic polarizability, but the details are beyond the scope of this introductory textbook. (The difficulty in making the connection is in accounting for the electric fields produced by the other polarized molecules.) Note that the dielectric constant K is always bigger than 1, because polarization always weakens the net electric field inside the insulator. The easier it is to polarize a molecule, the bigger is the field-weakening effect and the bigger the value of K. Figure 17.46 gives representative values of the dielectric constant K for various insulators.

Figure 17.46 Values of the dielectric constant of several materials. If there is no insulator in the gap, the potential difference across the capacitor is |ΔV | = Es (since is uniform, and parallel to If the insulator fills the gap and we maintain the same charges +Q and −Q on the metal plates, both the electric field E and the potential difference ΔV are reduced by the same factor K:

Since we can't actually get inside the insulator and measure the field there, we determine K for a particular insulator by measuring the effect on the potential difference ΔV between the plates for a fixed capacitor charge Q. In summary, placing an insulator between the plates of a capacitor

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).

decreases electric field inside the insulator decreases potential difference across the insulator If the insulator doesn't fill the gap, the electric field inside the insulator is reduced by the factor K, but the electric field at other places in the gap is hardly affected, because the electric field of the insulator in those places turns out to be small compared to the field made by the plates. (In our model of the insulator, consisting of capacitor-like sheets of plus and minus charges, the field made by the dipoles outside the insulator is a fringe field.) Outside the capacitor plates, however, the fringe field of the plates is small, and the small field contributed by the insulator significantly reduces the net field. In Section 17.5 we examined the electric field and potential difference between the two metal plates of a capacitor, then inserted a metal slab. In the following exercise we will start with the same device, but insert a glass slab. 17.X.16 A capacitor with a 3 mm gap has a potential difference of 6 volts (Figure 17.47). A disk of glass 1 mm Answer thick, with area the same as the area of the metal plates, has a dielectric constant of 2.5. It is inserted in the middle of the gap between the metal plates. Now what is the potential difference of the two metal plates? (It helps to make a diagram showing the electric field along a path.)

Figure 17.47 Capacitor before inserting a glass disk 1 mm in thickness between the plates.

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ENERGY DENSITY AND ELECTRIC FIELD Until now we have thought of energy (that is, electric potential energy) as associated with interacting charged particles. There is an alternative view, however, that considers energy to be stored in electric fields themselves. To see how this view works out quantitatively, we'll consider moving one plate of a capacitor. The force that one capacitor plate exerts on the other (Figure 17.48) is equal to the charge Q on one plate times the field made by the other plate, which is half the total field in the gap:

Figure 17.48 To move a capacitor plate you must exert a force equal to the force on the plate by the other plate.

so the force on one plate is

Suppose that you pull the positive plate of the capacitor slowly away from the negative plate, exerting a force only infinitesimally larger than the force exerted by the other plate. The work that you do in moving the plate a distance Δs goes into increasing the electric potential energy U:

We can rearrange this expression in the following way:

The expression in parentheses is the electric field E inside the capacitor. The quantity A Δs is the change in the volume occupied by electric field inside the capacitor (length times width times increase in the plate separation). Therefore we can write

We ascribe an energy density (joules per m3) to the electric field. By pulling the capacitor plates apart we increased the volume of space in which there is a sizable electric field. We say that the energy expended by us was converted into energy stored in the electric field. This view is numerically equivalent to our earlier view of electric potential energy, but it turns out to be a more fundamental view. For example, the energy carried by electromagnetic radiation propagating through space far from any charges can best be expressed in terms of the energy density associated with the field.

FIELD ENERGY DENSITY

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The energy density (joules per cubic meter) is stored in a region where there is an electric field of magnitude E.

Although we derived the field energy density for the particular case of a capacitor, the result is general. Anywhere there is electric field, there is energy density given by this formula. 17.X.17 The energy density inside a certain capacitor is 10 J/m 3. What is the magnitude of the electric field inside the capacitor? 17.X.18 What is the energy density associated with an electric field of 3 × 106 V/m (large enough to initiate a spark)?

Answer

Answer

*An Electron and a Positron In Chapter 6 we stated the principle of conservation of energy in this way: The change in energy of a system plus the change in energy of its surroundings must be zero. In the following example we will see that in order to understand energy conservation in quite a simple situation it is necessary to invoke the idea that energy is stored in fields. Consider an electron and a positron that are released from rest some distance from each other. We will take the electron to be the system under consideration, so therefore the positron and everything else in the Universe are the “surroundings.” Because of the attractive electric force between the particles, the electron accelerates toward the positron, gaining kinetic energy. By the principle of conservation of energy, the energy of the surroundings must therefore decrease. (Remember that the system can have no potential energy, since it consists of a single particle, and potential energy is a property of pairs of particles.)

Figure 17.49 An electron and a positron are released from rest, some distance apart. We choose the electron as the system and everything else (including the positron) as the surroundings.

QUESTION Does the energy of the positron decrease?

No, the energy of the positron also increases, since it accelerates toward the electron, gaining kinetic energy.

QUESTION Where is there a decrease of energy in the surroundings?

Evidently the energy stored in the fields surrounding the two particles must decrease. Clearly, the electric field at any location in space does change as the positions of the particles change. The electric field in the region between the particles gets larger, but the electric field everywhere else in space decreases (since E dipole is proportional to s, the distance between the particles). It would be a somewhat daunting task to integrate E 2 over the volume of the Universe, with the additional complication that close to a charged particle E approaches infinity. However, we do not actually need to do this integral to figure out the change in energy of the electric field throughout space. Since then

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In this example, the principle of conservation of energy leads us directly to the idea that energy must be stored in electric fields, since there is no other way to account for the decrease of energy in the surroundings. If we had chosen the electron plus the positron as our system, we would have found that ΔU el is equal to −2(ΔK electron). The change in potential energy for the two-particle system is the same as the change in the field energy. Evidently in a multiparticle system we can either consider a change in potential energy or a change in field energy (but not both); the quantities are equal. The idea of energy stored in fields is a general one. It is not only electric fields that carry energy, but magnetic fields and gravitational fields as well.

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*POTENTIAL OF DISTRIBUTED CHARGES *Potential Along the Axis of a Uniformly Charged Disk Consider a disk of radius R (area A = π R 2) with charge Q uniformly distributed over its surface. To calculate the potential due to this disk we follow a procedure similar to the one we used in finding the electric field of a charge distribution. We divide the disk into rings as we did in finding the electric field of a disk in the Chapter 16 (Figure 17.50).

Figure 17.50 A ring of radius r and width Δr makes a contribution Vring to the potential of the disk. At location z along the axis of a disk, the potential contributed by one ring is given by the result found above:

We have used the result from the previous chapter for the charge on one of the rings, Δq = (Q/A)2πrΔr. Add up the contributions of all the rings:

The units for this result check: meters in the bracket, and m2 in the A term in the denominator, so the units are those of Q/ (ε0R), which are indeed units of potential. We can find the electric field from the negative gradient of the potential. We want the z component of electric field, and taking the derivative of the potential we find the following:

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This is the same result obtained in Chapter 16 by summing the contributions to the electric field. 17.X.19 What is the potential at the center of a spherical shell of radius R that has a charge Q uniformly distributed over its surface? Don't do an integral of the electric field; just add up the contributions to the potential by the charges. (Afterward, compare with the alternative analysis in the next section.)

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Answer

*INTEGRATING THE SPHERICAL SHELL In this section, we'll use potential to prove that a uniformly charged spherical shell looks from the outside like a point charge but on the inside has a zero electric field. We did this by integrating the electric field directly in chapter 16, and a very different kind of proof using Gauss's law is given in a later chapter. Divide the spherical shell into rings of charge, each delimited by the angle θ and the angle θ + Δθ, and carrying an amount of charge ΔQ (Figure 17.51). The angle θ will be the integration variable used in summing up the contributions of the various rings.

Figure 17.51 The sphere may be divided into ring-shaped segments.

Each ring contributes V ring at an observation point a distance r from the center of the spherical shell. The center of each ring is a distance (r − R cos θ) from the observation location (we will eventually choose Δθ so small that it makes no difference whether we measure to the edge of the ring or to the center of the ring). Therefore the distance d from the observation location to each charge : on the ring is

The total amount of charge on each ring is

where ΔA is the surface area of the ring, since the total charge Q is spread uniformly all over the spherical surface, whose total area is 4π R 2. To calculate the surface area ΔA of the ring, as before we lay the ring out flat (Figure 17.52), noting that the radius of the ring is R sin θ and its width is R Δθ (since arc length is radius times angle, with angle measured in radians).

Figure 17.52 We can calculate the area of the ring of charge by “unrolling” it.

Putting these elements together, we have for the potential due to the ring

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We can add up all these contributions in the form of a definite integral:

The limits on the integral are determined by the fact that if we let θ range between 0 and π radians (180 degrees), we add up rings that account for the entire surface of the spherical shell. A change of variables lets us evaluate this integral. Let

However, (R cos θ)2 + (R sin θ)2 = R 2, from the Pythagorean theorem. Therefore

This gives two results, depending on which sign is taken for the square root. Take the − sign, which turns out to correspond to a location outside the shell:

This says that the potential outside a uniformly charged spherical shell is exactly the same as the potential due to a point charge located at the center of the shell, as though the shell were collapsed to a point. Since the electric field is the negative gradient of the potential, the electric field outside the shell is the same as if the shell were collapsed to its center. If we take the + sign, we have

This says that the potential inside the shell is constant, and equal to the potential just outside the shell. A constant potential means that the electric field inside the shell is zero. Note that for locations inside the spherical shell, some of the rings give a field to the left, and some give a field to the right. It turns out that these contributions exactly cancel each other. It is the 1/r 2 behavior of the electric field of point charges that leads to these unusual results. If the electric field of a point charge were not exactly proportional to 1/r 2, the electric field outside a uniform spherical shell would not look like the field of a point charge at the center, and the electric field would not be zero inside the shell.

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SUMMARY Potential difference

along a path from i to f

Units of electric potential are joules/coulomb, or volts. Special case: Regions of uniform electric field:

Change in electric potential energy for a particle of charge q moved through a potential difference ΔV: Sign of ΔV and direction of path relative to

:

Path going in the direction of Path going opposite to Path perpendicular to Point charge:

Path independence Change in potential between two locations is independent of path. For a round-trip path, ΔV = 0. Conductor in equilibrium along any path, so potential is uniform (constant everywhere) throughout a metal in equilibrium. Field is gradient of potential

Dielectric constant Inside an insulator

; K is “dielectric constant”;

In summary, placing an insulator between the plates of a capacitor

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, so

decreases the electric field inside the insulator decreases the potential difference across the insulator Field energy density

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EXERCISES AND PROBLEMS Sections 17.1, 17.2 17.X.20 What is the kinetic energy of a proton that is traveling at a speed of 3725 m/s? 17.X.21 If the kinetic energy of an electron is 4.4 × 10−18 J, what is the speed of the electron? (You can use the approximate (nonrelativistic) formula here.) 17.X.22 At a particular instant an electron is traveling with speed 6000 m/s. (a) What is the kinetic energy of the electron? (b) If a proton were traveling at the same speed (6000 m/s), what would be the kinetic energy of the proton? 17.X.23 The electric potential energy of a system of two particles is given by the equation

(a) What is the electric potential energy of two protons separated by a distance of 9 nanometers? (b) What is the electric potential energy of a proton and an electron separated by the same distance? 17.X.24 Which statements about potential energy are correct? Select all that apply. (A) Potential energy can be either positive or negative. (B) Potential energy is a property of a pair of objects. (C) When two interacting charged objects get very far away from each other, their potential energy approaches zero. (D) A single particle does not have potential energy. (E) A single isolated object can have potential energy. (F) Potential energy is always a positive number. (G) The potential energy of a pair of interacting charged objects increases as the distance between them increases. 17.X.25 A proton that initially is traveling at a speed of 300 m/s enters a region where there is an electric field. Under the influence of the electric field the proton slows down and comes to a stop. What is the change in kinetic energy of the proton? 17.X.26 The graph in Figure 17.53 shows the electric potential energy for a system of two interacting objects, as a function of the distance between the objects. What system(s) might this graph represent?

Figure 17.53 (A) Two protons (B) Two sodium ions (C) Two neutrons

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(D) Two chloride ions (E) Two electrons (F) A proton and an electron (G) A sodium ion and a chloride ion 17.X.27 The graph in Figure 17.54 shows the electric potential energy for a system of two interacting objects, as a function of the distance between the objects. What system(s) might this graph represent?

Figure 17.54 (A) Two protons (B) Two sodium ions (C) Two neutrons (D) Two chloride ions (E) Two electrons (F) A proton and an electron (G) A sodium ion and a chloride ion

Section 17.3 17.X.28 What is the difference between electric potential energy and electric potential? 17.X.29 What are the units of electric potential energy, of electric potential, and of electric field? 17.X.30 Locations A, B, and C are in a region of uniform electric field, as shown in the diagram in Figure 17.55. Location A is at −0.5, 0,0 m. Location B is at 0.5, 0, 0 m. In the region the electric field has the value 750, 0, 0 N/C.

Figure 17.55

For a path starting at B and ending at A, calculate: (a) The displacement vector (b) The change in electric potential (c) The potential energy change for the system when a proton moves from B to A

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(d) The potential energy change for the system when an electron moves from B to A 17.X.31 Locations A, B, and C are in a region of uniform electric field, as shown in Figure 17.55. Location A is at −0.5, 0, 0 m. Location B is at 0.5, 0, 0 m. In the region the electric field has the value 750, 0, 0 N/C. (a) For a path starting at B and ending at C, calculate: (1) The displacement vector (2) The change in electric potential (3) The potential energy change for the system when a proton moves from B to C (4) The potential energy change for the system when an electron moves from B to C (b) Which of the following statements are true in this situation? Choose all that are correct. (1) The potential difference cannot be zero because the electric field is not zero along this path. (2) When a proton moves along this path, the electric force does zero net work on the proton. (3)

is perpendicular to

.

17.X.32 Locations A, B, and C are in a region of uniform electric field, as shown in Figure 17.56. Location A is at −0.3, 0, 0 m. Location B is at 0.4, 0, 0 m. In the region the electric field has the value −850, 400, 0 N/C. For a path starting at A and ending at B, calculate:

Figure 17.56 1. The displacement vector 2. The change in electric potential 3. The potential energy change for the system when a proton moves from A to B 4. The potential energy change for the system when an electron moves from A to B 17.X.33 Locations A, B, and C are in a region of uniform electric field, as shown in Figure 17.56. Location A is at −0.3, 0, 0 m. Location B is at 0.4, 0, 0 m. In the region the electric field has the value −850, 400, 0 N/C. For a path starting at C and ending at A, calculate: 1. The displacement vector 2. The change in electric potential 3. The potential energy change for the system when a proton moves from C to A 4. The potential energy change for the system when an electron moves from C to A 17.X.34 You move from location i at 2, 7, 5 m to location f at 5, 6, 12 m. All along this path there is a nearly uniform electric field of 1000, 200, −510 N/C. Calculate V f − V i , including sign and units. 17.X.35 A capacitor with a gap of 1 mm has a potential difference from one plate to the other of 36 volts. What is the magnitude of

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the electric field between the plates? 17.X.36 An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential difference of 44 volts. (a) What is the kinetic energy of the electron in electron volts (eV)? (b) Which of the following statements would be true if the particle were a proton? Choose all that are correct. (1) The kinetic energy of the proton would be negative. (2) The proton would move in the opposite direction from the electron.

17.X.37 If the electric field exceeds about 3 × 106 N/C in air, a spark occurs. Approximately, what is the absolute value of the maximum possible potential difference between the plates of a capacitor whose gap is 3 mm, without causing a spark in the air between them? 17.X.38 A capacitor with a gap of 2 mm has a potential difference from one plate to the other of 30 volts. What is the magnitude of the electric field between the plates? 17.X.39 Figure 17.57 shows a region of uniform electric field. For a path starting at location and going to location, calculate the following quantities: (a) the difference in electric potential, (b) the potential energy change for the system when a proton moves along this path, and (c) the potential energy change for the system when an electron moves along this path.

Figure 17.57 17.X.40 For a path starting at location C and going to location A in Figure 17.57, calculate the following quantities: (a) the difference in electric potential, (b) the potential energy change for the system when a proton moves along this path, and (c) the potential energy change for the system when an electron moves along this path. 17.X.41 You move from location i at 2, 5, 4 m to location f at m. All along this path there is a nearly uniform electric field whose value is 1000, 200, −500 N/C. Calculate ΔV = V f − V i , including sign and units. 17.X.42 An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential difference of 35 volts. (a) What is the kinetic energy of the electron in electron volts (eV)? (b) What would happen if the particle were a proton? 17.X.43 In a particular region there is a uniform electric field of −760, 380, 0 V/m. Location A is 0.2, 0.1, 0 m, location B is 0.7, 0.1, 0 , and location C is 0.7, 0.4, 0 m. (a) What is the change in potential along a path from B to A? (b) What is the change in potential along a path from A to C? (c) An alpha particle (two protons and two neutrons) moves from A to C. What is the change in potential energy of the system (alpha + source charges)? 17.P.44 The potential difference from one end of a 1-cm-long wire to the other in a circuit is ΔV = V B − V A = 1.5 volts, as shown in Figure 17.58. Which end of the wire is at the higher potential? What are the magnitude and direction of the electric field E inside the wire?

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Figure 17.58 17.P.45 In a television picture tube, electrons are boiled out of a very hot metal filament placed near a negative metal plate. These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen, as shown in the diagram in Figure 17.59. If the high-voltage supply in the television set maintains a potential difference of 15,000 volts between the two plates, what speed do the electrons reach?

Figure 17.59 17.P.46 Two very large disks of radius R are carrying uniformly distributed charges Q A and Q B . The plates are parallel and 0.1 millimeters apart, as shown in Figure 17.60. The potential difference between the plates is V B − V A = −10 volts.

Figure 17.60 (a) What is the direction of the electric field between the disks? (b) Invent values of Q A , Q B , and R that would make V B − V A = −10 volts. 17.P.47 A particular oscilloscope, shown in Figure 17.61, has an 18,000-volt power supply for accelerating electrons to a speed adequate to make the front phosphor-coated screen glow when the electrons hit it. Once the electron has emerged from the accelerating region, it coasts through a vacuum at nearly constant speed.

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Figure 17.61

One can apply a potential difference of plus or minus volts across the deflection plates to steer the electron beam up or down on the screen to paint a display (other deflection plates not shown in the diagram are used to steer the beam horizontally). Each of the two deflection plates is a thin metal plate of length L = 8 cm and width (into the diagram) 4 cm. The distance between the deflection plates is s = 3 mm. The distance from the deflection plates to the screen is d = 30 cm. When there is a 40-volt potential difference between the deflection plates, what is the deflection y of the electron beam where it hits the screen? An approximate treatment is fine, but state your assumptions. As is usually the case, it pays to carry out all of your calculations algebraically and only evaluate the final algebraic result numerically. Note the exaggerated vertical scale: the deflection is actually small compared to the distance to the screen.

Section 17.4 17.X.48 In Figure 17.62, what is the direction of the electric field? Is ΔV = V f − V i positive or negative?

Figure 17.62 17.X.49 Locations A and B are in a region of uniform electric field, as shown in Figure 17.63. Along a path from B to A, the change in potential is −2200 V. The distance from A to B is 0.28 m. What is the magnitude of the electric field in this region?

Figure 17.63 17.X.50 You travel along a path from location A to location B, moving in the same direction as the direction of the net electric field

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in that region. What is true of the potential difference V B − V A ? (A) V B − V A > 0 (B) V B − V A < 0 (C) V B − V A = 0 17.X.51 You travel along a path from location A to location B, moving in a direction opposite to the direction of the net electric field in that region. What is true of the potential difference V B − V A ? (A) V B − V A > 0 (B) V B − V A < 0 (C) V B − V A = 0 17.X.52 You travel along a path from location A to location B, moving in a direction perpendicular to the direction of the net electric field in that region. What is true of the potential difference V B − V A ? (A) V B − V A > 0 (B) V B − V A < 0 (C) V B − V A = 0 17.X.53 Figure 17.64 shows several locations inside a capacitor. You need to calculate the potential difference VD − V C .

Figure 17.64 (a) What is the direction of the path (+x or −x)? (b) If the charge on the right plate is negative and the charge on the left plate is positive, what is the sign of VD − V C . 17.X.54 Figure 17.65 shows a portion of a long, negatively charged rod. You need to calculate the potential difference VA − V B .

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Figure 17.65 (a) What is the direction of the path (+y or −y)? (b) What is the sign of V A − V B ? 17.X.55 Location A is a distance d from a charged particle. Location B is a distance 2d from the particle. Which of the following statements are true? It may help to draw a diagram. (A) If the charge of the particle is negative, VB − V A is negative. (B) If the charge of the particle is positive, V A < V B . (C) If V B < V A , we know that the particle must be positive. (D) V B < V A , regardless of the sign of the charge of the particle. (E) The sign of (V B − V A ) does not give us any information about the sign of the charge of the particle.

Section 17.5 17.X.56 Location C is 0.02 m from a small sphere that has a charge of 4 nanocoulombs uniformly distributed on its surface. Location D is 0.06 m from the sphere. What is the change in potential along a path from C to D? 17.X.57 What is the change in potential ΔV in going from a location 3 × 10−10 m from a proton to a location 4 × 10−10 m from the proton? 17.X.58 An electron is initially at rest. It is moved from a location 4 × 10−10 m from a proton to a location 6 × 10−10 m from the proton. What is the change in electric potential energy of the system of proton and electron? 17.X.59 Locations A = a, 0, 0 and B = a, 0, 0 are on the +x axis, as shown in Figure 17.66. Four possible expressions for the electric field along the axis are given below. For each expression for the electric field, select the correct expression (1–8) for the potential difference V A − V B . In each case K is a numerical constant with appropriate units.

Figure 17.66 (a)

(b)

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(c)

(d)

1. V A − V B = 0 2. V A − V B = K (a − b) 3. 4.

VA − VB = K (

)

5. 6. V A − V B = K ln ( ) 7. V A − V B = K (a2 − b2) 8.

17.P.60 As shown in Figure 17.67, three large, thin, uniformly charged plates are arranged so that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate. Location A is −0.4, 0, 0 m, and location B is 0.2, 0, 0 m. The electric field has the value 725, 0, 0 V/m, and is −425, 0, 0 V/m.

Figure 17.67 (a) Consider a path from A to B. Along this path, what is the change in electric potential? (b) What is the change in electric potential along a path from B to A? (c) There is a tiny hole in the central plate, so a moving particle can pass through the hole. If an electron moved from A to B along the path shown, what would be the change in its kinetic energy? (d) What is the minimum kinetic energy the electron must have at location in order to ensure that it reaches location B? 17.P.61 If throughout a particular region of space the potential can be expressed as V = 4xz + 2y − 5z, what are the vector components of the electric field at location x, y, z ? 17.P.62 A dipole is oriented along the x axis. The dipole moment is p (=qs). (a) Calculate exactly the potential V (relative to infinity) at a location x, 0, 0 on the x axis and at a location 0, y, 0 on the y axis, by superposition of the individual 1/r contributions to the potential. (b) What are the approximate values of V at the locations in part (a) if these locations are far from the dipole? (c) Using the approximate results of part (b), calculate the gradient of the potential along the x axis, and show that the negative gradient is equal to the x component E x of the electric field.

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(d) Along the y axis, dV/dy = 0. Why isn't this equal to the magnitude of the electric field E along the y axis? 17.P.63 A capacitor consists of two large metal disks placed a distance s apart. The radius of each disk is R (R >> s), and the thickness of each disk is t, as shown in Figure 17.68. The disk on the left has a net charge of +Q, and the disk on the right has a net charge of −Q. Calculate the potential difference V2 − V 1, where location 1 is inside the left disk at its center, and location 2 is in the center of the air gap between the disks. Explain briefly.

Figure 17.68 17.P.64 The diagram in Figure 17.69 shows three very large metal disks (seen edgewise), carrying charges as indicated. On each surface the charges are distributed approximately uniformly. Each disk has a very large radius R and a small thickness t. The distances between the disks are a and b, as shown; they also are small compared to R. Calculate V 2 − V 1, and explain your calculation briefly.

Figure 17.69 17.P.65 Three charged metal disks are arranged as shown in Figure 17.70 (cutaway view). The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2.5 m2 (this is the area of one flat surface of the disk). The charge Q 1 = 5 × 10−8 C and the charge Q 2 = 4 × 10−7 C.

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Figure 17.70 (a) What is the electric field (magnitude and direction) in the region between disks 1 and 2? (b) Which of the following statements are true? Choose all that apply. (1) Along a path from A to B, (2) V B − V A = 0 (3)

volts

(c) To calculate V C − V B , where should the path start and where should it end? (d) Should V C − V B be positive or negative? Why? (1) Positive, because

is opposite to the direction of

(2) Negative, because

is in the same direction as

(3) Zero, because (e) What is the potential difference V C − V B ? (f) What is the potential difference V D − V C ? (g) What is the potential difference V F − V D? (h) What is the potential difference V G − V F ? (i) What is the potential difference V G − V A ? (j) The charged disks have tiny holes that allow a particle to pass through them. An electron that is traveling at a fast speed approaches the plates from the left side. It travels along a path from A to G. Since no external work is done on the system of plates + electron, ΔK + ΔU = We xt = 0. Consider the following states: initial, electron at location A; final, electron at location G. (1) What is the change in potential energy of the system? (2) What is the change in kinetic energy of the electron? 17.P.66 The long rod shown in Figure 17.71 has length L and carries a uniform charge −Q. Calculate the potential difference VA −

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V C . All of the distances are small compared to L. Explain your work carefully.

Figure 17.71 17.P.67 Three very large charged metal plates are arranged as shown in Figure 17.72. The radius of each plate is 4 meters, and each plate is w = 0.05 mm thick. The separation d1 is 6 mm, and the separation d2 is 2 mm. Each plate has a tiny hole in it, so it is possible for a small charged particle to pass through all the plates.

Figure 17.72

You are able to adjust the apparatus by varying the electric field in the region between location D and location F. You need to adjust this setting so that a fast-moving electron moving to the right, entering at location A, will have lost exactly 5.2 × 10−18 joules of kinetic energy by the time it reaches location. Using a voltmeter, you find that the potential difference volts. Based on this measurement, you adjust the electric field between D and F to the appropriate value. (a) Consider the system of (electron + plates). Neglecting the small amount of work done by the gravitational force on the electron, during this process (electron going from A to G), what is ? (b) What is the change in potential energy for the system during this process? (c) What is V G − V A ? (d) What is V F − V D? (e) What is the electric field (magnitude and direction) in the region between locations D and F?

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17.P.68 In the Van de Graaff generator shown in Figure 17.73, a rubber belt carries electrons up through a small hole in a large hollow spherical metal shell. The electrons come off the upper part of the belt and drift through a wire to the outer surface of the metal shell, so that the metal shell acquires a sizable negative charge, approximately uniformly distributed over the sphere. At a time when the sphere has acquired a sizable charge −Q, approximately how much work must be done by the motor to move one more electron from the base (a distance h below the sphere) to the upper pulley (located a distance R /2 from the center of the hollow sphere)? Explain your work, and state explicitly what approximations you had to make.

Figure 17.73 17.P.69 A thin spherical shell made of plastic carries a uniformly distributed negative charge −Q 1. As shown in Figure 17.74, two large thin disks made of glass carry uniformly distributed positive and negative charges +Q 2 and −Q 2. The radius R 1 of the plastic spherical shell is very small compared to the radius R2 of the glass disks. The distance from the center of the spherical shell to the positive disk is d, and d is much smaller than R2.

Figure 17.74 (a) Find the potential difference V 2 − V 1 in terms of the given quantities (Q 1, Q 2, R 1, R 2, and d). Point 1 is at the center of the plastic sphere, and point 2 is just outside the sphere. (b) Find the potential difference V 3 − V 2. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk. (c) Suppose that the plastic shell is replaced by a solid metal sphere with radius R1 carrying charge −Q 1. State whether the absolute magnitudes of the potential differences would be greater than, less than, or the same as they were with the plastic shell in place. Explain briefly, including an appropriate diagram. 17.P.70 A thin spherical shell of radius R 1 made of plastic carries a uniformly distributed negative charge −Q 1. A thin spherical

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shell of radius R 2 made of glass carries a uniformly distributed positive charge +Q 2. The distance between centers is L, as shown in Figure 17.75.

Figure 17.75 (a) Find the potential difference V B − V A . Location Ais at the center of the glass sphere, and location B is just outside the glass sphere. (b) Find the potential difference V C − V B . Location B is just outside the glass sphere, and location C is a distance d to the right of B. (c) Suppose the glass shell is replaced by a solid metal sphere with radius R2 carrying charge +Q 2. Would the magnitude of the potential difference V B − V A be greater than, less than, or the same as it was with the glass shell in place? Explain briefly, including an appropriate physics diagram. 17.P.71 As shown in Figure 17.76, a solid metal sphere of radius r 1 has a charge +Q. It is surrounded by a concentric spherical metal shell with inner radius r 2 and outer radius r 3 that has a charge −Q on its inner surface and +Q on its outer surface. In the diagram, point A is located at a distance r 4 from the center of the spheres. Points B and C are inside the metal shell, very near the outer and inner surfaces, respectively. Point E is just inside the surface of the solid sphere. Point D is halfway between C and E. Point F is a distance r 1/2 from the center.

Figure 17.76 (a) Is each of the following potential differences greater than zero, equal to zero, or less than zero? Briefly explain why in terms of the directions of the electric field and of the path. a. V B − V A b. V C − V B c. V D − V C d. V F − V E

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(b) Calculate V F , the potential at location F. Explain your work. 17.P.72 A long thin metal wire with radius r and length L is surrounded by a concentric long narrow metal tube of radius R, where R > x) or, equivalently, the magnetic field very near a short wire (x 15°, the magnetic field at the ends of the needle may be significantly weaker than the magnetic field at the center of the needle.

Elevating the Wire By elevating the wire about 1 cm above the top of the compass, we minimize this effect. Now even if the needle deflects up to 40°, the ends and the center will still be approximately the same distance from the wire. An easy way to elevate the wire is to support it on two fingers, as shown in Figure 19.59.

Figure 19.59 One way to elevate the wire by a reproducible distance is to support it on two fingers. 2. Is Current Proportional to Deflection Angle? As can be seen from Figure 19.60, the tangent of the deflection angle is proportional to the magnetic field made by the . However, this approximation is not good current in the wire. For small deflection angles (measured in radians), for angles greater than about 0.3 radians (17°).

Figure 19.60 Bwire /BEarth = tan θ Tangent of Deflection Angle

For deflection angles less than 15°, we can assume that the magnetic field, and hence the current in the wire, is proportional to the deflection angle. For deflection angles greater than 15°, we need to use the tangent of the deflection angle to calculate magnetic field (and therefore current).

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19.EXP.16 Effect of twice the length Here is an experiment in which we compare the electron current in a Nichrome wire of length L to the current in a Nichrome wire of length 2L.

QUESTION What does our model for current in a circuit predict for the result of the experiment?

Around the circuit loop we have emf −EL = 0, so if we double the length L, we'll get half as big an electric field (since the emf doesn't change). With half the field we get half the drift speed ( ) and half the electron current ( ) and half the conventional current ( ). Therefore we expect half as large a magnetic field and half the compass deflection. Next make the measurement. Be careful when connecting Nichrome wire. If you run current through a short (< 10 cm) length of the wire it can get hot enough to burn you. (It may even get hot enough to glow.) Following the directions given above, use the compass to compare the amount of current in the two different circuits shown in Figure 19.61:

Figure 19.61 How does the current change when we double the length of the Nichrome wire? (a) One battery (to keep the compass deflection small) and half the length of the thin Nichrome wire (about 20 cm). Don't cut the Nichrome wire, just connect clips to a portion half the length of the wire. (b) One battery and the full length (about 40 cm) of the thin Nichrome wire.

Make careful, quantitative measurements of the angles, and record your measurements. Are you able to verify the prediction that the current is halved when you double the length of the Nichrome wire? Do you find that the current is unaffected if you bend the wire or add additional connecting wires? 19.EXP.17 Current in a two-bulb series circuit Two identical light bulbs in series (Figure 19.62) are essentially the same as one light bulb with twice as long a filament, since the copper connecting wires contribute little energy dissipation. From the preceding experiment with doubling the length of a Nichrome wire, we might expect to get half the current compared with just one bulb.

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Figure 19.62 Two light bulbs in series. (a) Following the directions given above, use a compass to measure the current in the circuit with two thick filament bulbs in series shown in Figure 19.62, and record the observed compass deflections. Compare your observations to your prediction. (b) Compare the compass deflection you observe in the two-bulb circuit with the compass deflection you observe in a one-bulb circuit. Record your measurements (by removing one bulb from the circuit).

You presumably found that the two-bulb circuit had more than half the current of the one-bulb circuit (if necessary, repeat this measurement carefully). The issue here is that the electron mobility in very hot tungsten (several thousand degrees!) is less than the electron mobility in cooler tungsten. The greater thermal agitation of the atomic cores leads to larger “friction.” Note that the bulbs in the two-bulb circuit emit “redder” light than the single bulb, indicating that their filaments are not as hot. 19.X.18 Explain why the current in the two-bulb series circuit is more than half the current in the one-bulb circuit. How is the electric field affected? How is the drift speed affected? 19.X.19 Why didn't we see this effect when we experimented with doubling the length of the Nichrome wire?

Answer

Answer

19.EXP.20 Effect of doubling the cross-sectional area Figure 19.63 illustrates an experiment in which we see what happens to the current in a Nichrome wire when we double its cross-sectional area.

Figure 19.63 How does the current change when we double the cross-sectional area of the Nichrome wire?

QUESTION Explain why the electron current in the wire should increase by a factor of two if the cross-sectional area of the wire doubles.

In Figure 19.63 it seems plausible that doubling A should double the conventional current

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. It is like having

two wires in parallel, doubling the drain on the battery. We can make this argument quantitative by considering energy. Since we have emf −EL = 0 around the loop, the electric field E in the wire does not change when we use a thicker wire, so the current doubles. (This neglects internal resistance in the battery, which has the effect of limiting the amount of current that the battery can supply, so that for large crosssectional areas and large currents, doubling a large cross-sectional area may not yield twice the current.) Use the compass to compare the amount of current in two different circuits (Figure 19.63): one battery (to keep the compass deflection small) and the same lengths of thin and thick Nichrome wires. Use only one battery. Connect the connecting wires about 30 cm apart on each Nichrome wire (don't cut the Nichrome wire). Following the directions given above, make careful, quantitative measurements of the angles. The thick Nichrome wire has a cross section about twice the cross section of the thin Nichrome wire. Your instructor may give you a more precise ratio of the areas. What do you measure for the ratio of the current in the thick wire to the current in the thin wire? Does this agree with predictions? Your own data are probably consistent with what more precise experiments confirm. These measurements imply that the drift speed of individual electrons is the same in the thick wire and in the thin wire (because the electric field is the same in the two circuits if the length of the wires is the same), and that the many-electron current is distributed uniformly over the cross section. If the drift speed were not the same everywhere across the cross section of the wire, we couldn't simply write for the electron current, using the same for all the electrons passing that point in the circuit. In Section 19.3 we proved that the current is indeed uniform across the cross section. 19.EXP.21 A parallel circuit Connect two thick filament bulbs in “parallel” (that is, one beside the other), as shown in Figure 19.64.

Figure 19.64 Two thick-filament bulbs connected in parallel. (a) How does the brightness of the thick filament bulbs in this parallel circuit compare to the brightness of two thick filament bulbs in series? (b) Unscrew one of the bulbs in the parallel circuit. What effect does this have on the brightness of the remaining bulb? (c) Based on these observations, what relationship would you expect to see between the compass deflections at locations A, B, C, and D, in this circuit? Record your prediction. (d) Measure the compass deflections at locations A, B, C, and D, and record your measurements. Do your observations agree with your predictions in (c)?

In the preceding experiment you should again have seen that the current entering a location is equal to the current leaving the same location in the steady state. In particular, the current at A should be equal to the sum of the currents at B and C. If you did not observe this, you should repeat the experiment. We can also think of the two bulbs in parallel as equivalent to increasing the cross-sectional area of one of the bulb filaments.

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We can analyze the situation quantitatively. If we follow the path through one of the bulbs, we have 2 emf −EL = 0, (2 emf for two batteries), and L is the length of the bulb filament (neglecting dissipation in the connecting wires). If we follow the path through the other thick filament bulb, we also have 2 emf −EL = 0. Thus the electric field E = 2 emf/L is the same in both bulb filaments, and it is the same as there would be with only one bulb present. Figure 19.65 shows an approximate surface-charge distribution on the circuit. Each bulb is flanked by positive and negative charges, which make a large electric field in the bulb filament. Removing one of the bulbs doesn't change the surface-charge distribution very much, and the brightness of the remaining bulb hardly changes. The batteries of course have to deliver more current ( ) to the two parallel bulbs than they do to one bulb.

Figure 19.65 Approximate surface-charge distribution for two thick-filament bulbs in parallel. 19.EXP.22 Effect of two batteries Use the compass to compare the amount of current in two different circuits: one battery and a length of thin Nichrome wire, and two batteries and the same length of thin Nichrome wire. Following the directions given with Experiment 19.EXP.21, make careful, quantitative measurements of the angles. Are you able to verify the prediction that the current doubles when you use two batteries? (If you wish, try three or more batteries with the same length of thin Nichrome wire—but don't use more than two batteries with your light bulbs, because it may burn them out.)

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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EXERCISES AND PROBLEMS Sections 19.1, 19.2 19.X.23 What is the most important general difference between a system in “steady state” and a system in “equilibrium”? 19.X.24 Describe the following attributes of a metal wire in steady state vs. equilibrium: Metal wire

In steady-state

In equilibrium

Location of excess charge Motion of mobile electrons E inside the metal wire 19.X.25 How can there be a nonzero electric field inside a wire in a circuit? Isn't the electric field inside a metal always zero? 19.X.26 In the circuit shown in Figure 19.66, what are the relationships among iA, iB , iC , and iD?

Figure 19.66 19.X.27 State your own theoretical and experimental objections to the following statement: In a circuit with two thick filament bulbs in series, the bulb farther from the negative terminal of the battery will be dimmer, because some of the electron current is used up in the first bulb. Cite relevant experiments. 19.X.28 Write the node equation for the circuit in Figure 19.67. What is the value of the outward-going current I2?

Figure 19.67 19.X.29 Which of the following statements about a metal wire in equilibrium are true? List all that apply. (A) There cannot be excess charges on the surface of the wire. (B) There is no net flow of mobile electrons inside the wire. (C) There are no excess charges in the interior of the wire. (D) There may be excess charges in the interior of the wire. (E) The net electric field everywhere inside the wire is zero. (F) There may be excess charges on the surface of the wire.

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(G) The interior of the metal wire is neutral. (H) There may be a constant flow of mobile electrons inside the wire. (I)

The electric field inside the wire may be nonzero but uniform.

19.X.30 Which of the following statements about a metal wire in the steady state are true? List all that apply. (A) There is a constant flow of mobile electrons inside the wire. (B) The net electric field everywhere inside the wire is zero. (C) There are no excess charges in the interior of the wire. (D) There cannot be excess charges on the surface of the wire. (E) There is no net flow of mobile electrons inside the wire. (F) There may be excess charges on the surface of the wire. (G) There may be excess charges in the interior of the wire. (H) There may be a nonzero, uniform electric field inside the wire. (I)

The interior of the metal wire is neutral.

Section 19.3 19.X.31 Electron current

:

(a) What are the units of electron current? (b) What is n? What are its units? (c) What is A? What are its units? (d) What is ? What are its units? (e) What is u? What are its units? 19.X.32 Since there is an electric field inside a wire in a circuit, why don't the mobile electrons in the wire accelerate continuously? 19.X.33 Why don't all mobile electrons in a metal have exactly the same speed? 19.X.34 There are very roughly the same number of iron atoms per m3 as there are copper atoms per m3, but copper is a much better conductor than iron. How does uiron compare with ucopper? 19.X.35 The drift speed in a copper wire is 7 × 10−5 m/s for a typical electron current. Calculate the magnitude of the electric field E inside the copper wire. The mobility of mobile electrons in copper is 4.5 × 10−3 (m/s)/(N/C). (Note that though the electric field in the wire is very small, it is adequate to push a sizable electron current through the copper wire.) 19.X.36 Suppose that a wire leads into another, thinner wire of the same material that has only a third the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the drift speed in the thick wire is 4 × 10−5 m/s, what is the drift speed in the thinner wire? 19.X.37 Suppose that a wire leads into another, thinner wire of the same material that has only half the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the electric field E1 in the thick wire is 1 × 10−2 N/C, what is the electric field E 2 in the thinner wire? 19.X.38 Suppose that wire A and wire B are made of different metals and are subjected to the same electric field in two different circuits. Wire B has 6 times the cross-sectional area, 1.3 times as many mobile electrons per cubic centimeter, and 4 times the mobility of wire A. In the steady state, 2 × 1018 electrons enter wire A every second. How many electrons enter wire B every second? 19.X.39 In the circuit in Figure 19.68 the narrow resistor is made of the same material as the thick connecting wires.

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Figure 19.68

In the steady state, which graph in Figure 19.69 correctly shows the magnitude of the electric field at locations around the circuit?

Figure 19.69

In the steady state, which graph in Figure 19.70 correctly shows the drift speed of the electrons at locations around the circuit?

Figure 19.70 19.X.40 A steady-state current runs in the circuit shown in Figure 19.71. The narrow resistor and thick connecting wires are made of the same material.

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Figure 19.71

Which of the following quantities are greater in the thin resistor than in the thick wire? List all that apply: u, n, E, v, A, i, or none of these. 19.P.41 All of the wires in the circuit shown in Figure 19.72 are made of the same material, but one wire has a smaller radius than the other wires. Which of the following statements are true of this circuit in the steady state? List all that apply.

Figure 19.72 (A) The drift speed of electrons passing location D is greater than the drift speed of electrons passing location G. (B) The magnitude of the electric field is the same at each location labeled by a letter. (C) The electric field at location F points up. (D) The electric field at G is larger in magnitude than the electric field at C. (E) The number of electrons passing location B each second is the same as the number of electrons passing location D each second. The radius of the thin wire is 0.22 mm, and the radius of the thick wire is 0.55 mm. There are 4 × 1028 mobile electrons per cubic meter of this material, and the electron mobility is 6 × 10−4 (m/s)/(V/m). If 6 × 1018 electrons pass location D each second, how many electrons pass location B each second? What is the magnitude of the electric field at location B? 19.P.42 Figure 19.73 is a top view of a portion of a circuit containing three identical light bulbs (the rest of the circuit including the batteries is not shown). The connecting wires are made of copper.

Figure 19.73

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(a) The compass is placed on top of the wire, and it deflects 20 degrees away from north as shown (the wire is underneath the compass). In what direction are the electrons moving at location P1? How do you know? (b) In the steady state, 3 × 1018 electrons pass location P1 every second. How many electrons pass location P2 every second? Explain briefly. (c) Give the relative brightnesses of bulbs B 1, B 2, and B 3. Explain briefly. (d) There are 6.3 × 1028 mobile electrons per cubic meter in tungsten. The cross-sectional area of the tungsten filament in bulb B 1 is 0.01 mm (which is 1 × 10−8 m2. The electron mobility in hot tungsten is 1.2 × 10−4 (m/s)/(N/C). Calculate the electric field inside the tungsten filament in bulb B 1. Give both the direction and the magnitude of the electric field. 19.P.43 Inside a chemical battery it is not actually individual electrons that are transported from the + end to the − end. At the + end of the battery an “acceptor” molecule picks up an electron entering the battery, and at the − end a different “donor” molecule gives up an electron, which leaves the battery. Ions rather than electrons move between the two ends to transport the charge inside the battery. When the supplies of acceptor and donor molecules are used up in a chemical battery, the battery is dead, because it can no longer accept or release electrons. The electron current in electrons per second, times the number of seconds of battery life, is equal to the number of donor (or acceptor) molecules in the battery. A flashlight battery contains approximately half a mole of donor molecules. The electron current through a thick filament bulb powered by two flashlight batteries in series is about 0.3 ampere. About how many hours will the batteries keep this bulb lit? 19.P.44 Consider the circuit containing three identical light bulbs shown in Figure 19.74. North is indicated in the diagram. Compasses are placed under the wires at locations A and B.

Figure 19.74 (a) The magnitude of the deflection of compass A is 13 degrees away from north. In what direction does the needle point? Draw a sketch. (b) What is the magnitude of the deflection of the needle of compass B? In what direction does the needle point? Draw a sketch. Explain your reasoning. (c) In the steady state 1.5 × 1018 electrons per second enter bulb 1. There are 6.3 × 1028 mobile electrons per cubic meter in tungsten. The cross-sectional area of the tungsten filament in bulb 1 is 1 × 10−8 m. The electron mobility in

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hot tungsten is 1.2 × 10−4 (m/s)/(N/C). Calculate the electric field inside the tungsten filament in bulb 3. Give both the direction and the magnitude of the electric field.

Sections 19.4, 19.5, 19.6, 19.7 19.X.45 In the few nanoseconds before the steady state is established in a circuit consisting of a battery, copper wires, and a single bulb, is the current the same everywhere in the circuit? Explain. 19.X.46 During the initial transient leading to the steady state, the electron current going into a bulb may be greater than the electron current leaving the bulb. Explain why and how these two currents come to be equal in the steady state. 19.X.47 At a typical drift speed of 5 × 10−5 m/s, it takes an electron about 100 minutes to travel through one of your connecting wires. Why, then, does the bulb light immediately when the connecting wire is attached to the battery? 19.X.48

(a) The emf of a particular flashlight battery is 1.7 volts. If the battery is 4.5 cm long, and the radius of the cylindrical battery is 1 cm, estimate roughly the amount of charge on the positive end plate of the battery. (b) Is this amount of charge sufficient to repel noticeably a positively charged piece of invisible tape?

19.X.49 A steady-state current flows through the Nichrome wire in the circuit shown in Figure 19.75.

Figure 19.75

Before attempting to answer these questions, draw a copy of this diagram. All of the locations indicated by letters are inside the wire. (a) On your diagram, show the electric field at the locations indicated, paying attention to relative magnitude. (b) Carefully draw pluses and minuses on your diagram to show the approximate surface charge distribution that produces the electric field you drew. Make your drawing show clearly the differences between regions of high surfacecharge density and regions of low surface-charge density. Use your diagram to determine which of the following statements about this circuit are true. (1) There is some excess negative charge on the surface of the wire near location B. (2) Inside the metal wire the magnitude of the electric field is zero. (3) The magnitude of the electric field is the same at locations G and C. (4) The electric field points to the left at location G. (5) There is no excess charge on the surface of the wire. (6) There is excess charge on the surface of the wire near the batteries but nowhere else. (7) The magnitude of the electric field inside the wire is larger at location G than at location C. (8) The electric field at location D points to the left. (9) Because the current is not changing, the circuit is in static equilibrium.

19.X.50 In the circuit shown in Figure 19.76, all of the wire is made of Nichrome, but one segment has a much smaller crosssectional area.

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Figure 19.76

On a copy of this diagram, using the same scale for magnitude that you used in the previous question for Figure 19.75, show the steady-state electric field at the locations indicated, including in the thinner segment. Before attempting to answer these questions, draw a copy of this diagram. All of the locations indicated by letters are inside the wire. (a) On your diagram, show the electric field at the locations indicated, paying attention to relative magnitude. Use the same scale for magnitude as you did in the previous question. (b) Carefully draw pluses and minuses on your diagram to show the approximate surface charge distribution that produces the electric field you drew. Make your drawing show clearly the differences between regions of high surfacecharge density and regions of low surface-charge density. Use your diagram to determine which of the following statements about this circuit are true. (1) There is a large gradient of surface charge on the wire between locations C and E. (2) The electron current is the same at every location in this circuit. (3) Fewer electrons per second pass location E than location C. (4) The magnitude of the electric field at location G is smaller in this circuit than it was in the previous circuit (Figure 19.75). (5) The magnitude of the electric field is the same at every location in this circuit. (6) The magnitude of the electric field at location D is larger than the magnitude of the electric field at location G. (7) There is no surface charge at all on the wire near location G. The electron current in this circuit is less than the electron current in the previous circuit (Figure 19.75).

19.X.51 In the circuit in Figure 19.77 a mechanical battery keeps a steady-state current running in a wire that has rather low electron mobility.

Figure 19.77

Which of the following statements about the circuit are true? List all that apply. (A) The electric field inside the wire varies in magnitude, depending on location. (B) At location 3 inside the wire the electric field points to the right.

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(C) The electric field is zero at all locations inside the metal wire. (D) At location 3 the electric field points to the left. (E) The magnitude of the electric field inside the wire is the same at all locations inside the wire. (F) Mobile electrons inside the wire push each other through the wire. (G) The nonzero electric field inside the wire is created by the moving electrons in the wire. (H) At every location inside the wire the direction of the electric field is parallel to the wire. (I)

The nonzero electric field inside the wire is created by excess charges on the surface of the wire and in and on the mechanical battery.

19.P.52 Some students intended to run a light bulb off two batteries in series in the usual way, but they accidentally hooked up one of the batteries backwards, as shown in Figure 19.78 (the bulb is shown as a thin filament).

Figure 19.78 (a) Use +'s and −'s to show the approximate steady-state charge distribution along the wires and bulb. (b) Draw vectors for the electric field at the indicated locations inside the connecting wires and bulb. (c) Compare the brightness of the bulb in this circuit with the brightness the bulb would have had if one of the batteries hadn't been put in backwards. (d) Try the experiment to check your analysis. Does the bulb glow about as you predicted?

Sections 19.8, 19.9 Base your analyses in the following problems on the principles discussed in this chapter. If you wish to use formulas derived elsewhere, you must justify them in terms of the microscopic analysis in terms of field that was introduced in this chapter. 19.X.53 Criticize the statement below on theoretical and experimental grounds. Be specific and precise. Refer to your own experiments, or describe any new experiments you perform: “A flashlight battery always puts out the same amount of current, no matter what is connected to it.” 19.X.54 What is the difference between emf and electric potential difference? 19.X.55 Compare the direction of the average electric field inside a battery to the direction of the electric field in the wires and resistors of a circuit. 19.X.56 Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery? (A) Very little energy is dissipated in the thick connecting wires.

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(B) The electric field in connecting wires is very small, so

.

(C) Electric field in the connecting wires is zero, so

.

(D) Current in the connecting wires is smaller than current in the bulb. (E) All the current is used up in the bulb, so the connecting wires don't matter. 19.X.57 In Figure 19.79, suppose that

volts, and

volts.

Figure 19.79 (a) What is the potential difference V C − V D? (b) If the element between C and D is a battery, is the + end of the battery at C or at D? 19.X.58 A Nichrome wire 48 cm long and 0.25 mm in diameter is connected to a 1.6-volt flashlight battery. What is the electric field inside the wire? Next, the wire is replaced by a different Nichrome wire with the same length, but diameter 0.20 mm. Now what is the electric field inside the wire? 19.X.59 For each of the following experiments, state what effect you observed (how the current in the circuit was affected) and why, in terms of the relationships:

Effect on current (be quantitative)

Experiment

Circle the parameter(s) that changed and explain briefly

Double the length of a Nichrome wire

nAuE

Double the cross-sectional area of a Nichrome wire

nAuE

Two identical bulbs in series compared to a single bulb

nAuE

Two batteries in series compared to a single battery

nAuE

19.X.60 In a table like the one shown, write an inequality comparing each quantity in the steady state for a narrow resistor and thick connecting wires, which are made of the same material as the resistor. Electron current in resistor

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Electron current in thick wires

nR

nW

AR

AW

uR

uW

ER

EW

vR

vW

19.X.61 In a circuit with one battery, connecting wires, and a 12 cm length of Nichrome wire, a compass deflection of 6 degrees is observed. What compass deflection would you expect in a circuit containing two batteries in series, connecting wires, and a 36 cm length of thicker Nichrome wire (double the cross-sectional area of the thin piece)? Explain. 19.X.62 What would be the potential difference V C − V B across the thin resistor in Figure 19.80 if the battery emf is 3.5 volts? Assume that the electric field in the thick wires is very small (so that the potential differences along the thick wires are negligible). Do you have enough information to determine the current I in the circuit?

Figure 19.80 19.P.63 A circuit is constructed from two batteries and two wires, as shown in Figure 19.81. Each battery has an emf of 1.3 volts. Each wire is 26 cm long and has a diameter of 7 × 10−4 meters. The wires are made of a metal that has 7 × 1028 mobile electrons per cubic meter; the electron mobility is 5 × 10−5 (m/s)/(V/m). A steady current runs through the circuit. The locations marked by “×” and labeled by a letter are in the interior of the wire.

Figure 19.81 (a) Which of these statements about the electric field in the interior of the wires, at the locations marked by “×”s, are true? List all that apply. (1) The magnitude of the electric field at location G is larger than the magnitude of the electric field at location F. (2) At every marked location the magnitude of the electric field is the same. (3) At location B the electric field points to the left. (b) Write a correct energy conservation (round-trip potential difference) equation for this circuit, along a round-trip path starting at the negative end of battery 1 and traveling counterclockwise through the circuit (that is, traveling to the left through the battery, and continuing on around the circuit in the same direction). (c) What is the magnitude of the electric field at location B?

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(d) How many electrons per second enter the positive end of battery 2? (e) If the cross-sectional area of both wires were increased by a factor of 2, what would be the magnitude of the electric field at location B? (f) Which of the diagrams in Figure 19.82 best shows the approximate distribution of excess charge on the surface of the circuit?

Figure 19.82

19.P.64 Three identical light bulbs are connected to two batteries as shown in Figure 19.83.

Figure 19.83 (a) To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-trip path that begins and ends at the same location. Each segment of the path should go through a wire, a bulb, or a battery (not through the air). How many valid energy conservation (loop) equations is it possible to write for this circuit? (b) Which of the following equations are valid energy conservation (loop) equations for this circuit? E1 refers to the electric field in bulb 1; L refers to the length of a bulb filament. Assume that the electric field in the connecting wires is small enough to neglect. (1) (2) (3)

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(4) (5) (6) (7) (c) It is also necessary to write charge conservation equations (node) equations. Each such equation must relate electron current flowing into a node to electron current flowing out of a node. Which of the following are valid charge conservation equations for this circuit? (1) (2) (3)

Each battery has an emf of 1.5 volts. The length of the tungsten filament in each bulb is 0.008 m. The radius of the filament is 5 × 10−6 m (it is very thin!). The electron mobility of tungsten is 1.8 × 10−3 (m/s)/(V/m). Tungsten has 6 × 1028 mobile electrons per cubic meter. Since there are three unknown quantities, we need three equations relating these quantities. Use any two valid energy conservation equations and one valid charge conservation equation to solve for E1, E 2, i1, and i2. 19.P.65 The circuit shown in Figure 19.84 consists of a single battery, whose emf is 1.8 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross-sectional area 1.4 × 10−6 m2 and is 25 cm long. The thin wire has cross-sectional area 5.9 × 10−8 m2 and is 6.1 cm long.

Figure 19.84 In this metal, the electron mobility is 5 × 10−4 (m/s)/(V/m), and there are 4 × 1028 mobile electrons/m 3. (a) Which of the following statements about the circuit in the steady state are true? (1) At location B the electric field points toward the top of the page. (2) The magnitude of the electric field at locations F and C is the same. (3) The magnitude of the electric field at locations D and F is the same. (4) The electron current at location D is the same as the electron current at location F. (b) Write a correct energy conservation (loop) equation for this circuit, following a path that starts at the negative end of the battery and goes counterclockwise. (c) Write a correct charge conservation (node) equation for this circuit. (d) Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the magnitudes E D and E F of the electric field at locations D and F. (e) Use the appropriate equation(s) to calculate the electron current at location D in the steady state. 19.P.66 In the circuit shown in Figure 19.85, two thick copper wires connect a 1.5-volt battery to a Nichrome wire. Each thick connecting wire is 17 cm long and has a radius of 9 mm. Copper has 8.4 × 1028 mobile electrons per cubic meter and an

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electron mobility of 4.4 × 10−3 (m/s)/(V/m). The Nichrome wire is 8 cm long, and has a radius of 3 mm. Nichrome has 9 × 1028 mobile electrons per cubic meter and an electron mobility of 7 × 10−5 (m/s)/(V/m).

Figure 19.85 (a) What is the magnitude of the electric field in the thick copper wire? (b) What is the magnitude of the electric field in the thin Nichrome wire? 19.P.67 When a single thick-filament bulb of a particular kind and two batteries are connected in series, 3 × 1018 electrons pass through the bulb every second. When two batteries connected in series are connected to a single thin-filament bulb, with a filament made of the same material and the same length as that of the thick-filament bulb but smaller cross section, only 1.5 × 1018 electrons pass through the bulb every second. (a) In the circuit shown in Figure 19.86, how many electrons per second flow through the thin-filament bulb?

Figure 19.86 (b) What approximations or simplifying assumptions did you make? (c) Show approximately the surface charge on a diagram of the circuit. 19.P.68 In the circuit shown in Figure 19.87, the two thick wires and the thin wire are made of Nichrome.

Figure 19.87 (a) Show the steady-state electric field at the locations indicated, including in the thin wire. (b) Carefully draw pluses and minuses on your own diagram to show the approximate surface-charge distribution in the steady state. Make your drawing show clearly the differences between regions of high surface-charge density and regions of low surface-charge density.

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(c) The emf of the battery is 1.5 volts. In Nichrome there are n = 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is u = 7 × 10−5 (m/s)/(N/C). Each thick wire has length L1 = 20 cm = 0.2 m and cross-sectional area A 1 = 9 × 10−8 m2. The thin wire has length L2 = 5 cm = 0.05 m and cross-sectional area A 2 = 1.5 × 10−8 m2. (The total length of the three wires is 45 cm.) In the steady state, calculate the number of electrons entering the thin wire every second. Do not make any approximations, and do not use Ohm's “law” or series-resistance formulas. State briefly where each of your equations comes from. 19.P.69 Three identical thick-filament bulbs are in series as shown in Figure 19.88. Thick copper wires connect the bulbs. In the steady state, 3 × 1017 electrons leave the battery at location A every second.

Figure 19.88 (a) How many electrons enter the second bulb at location D every second? If there is insufficient information to give a numerical answer, state how it compares with 3 × 1017. Justify your answer carefully. (b) Next, the middle bulb (at DE) is replaced by a wire, as shown in Figure 19.89. Now how many electrons leave the batteries at location A every second? Explain clearly! If you have to make an approximation, state what it is. Do not use ohms or series-resistance formulas in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter.

Figure 19.89 (c) Check your analysis by trying the experiment with a partner. Try to find three thick-filament bulbs that glow equally brightly when in series with each other, because bulb construction varies slightly in manufacturing. Remember to arrange the circuit so that the largest compass deflection is no more than 15 degrees. Report the deflections that you observe. Does the experiment agree with your prediction? (If not, can you explain the discrepancy? Be specific. For example, if the current is larger than predicted, explain why it is larger than predicted.) (d) Finally, the last bulb (at FG) is replaced by a bulb identical in every way except that its filament has twice as large a cross-sectional area, as shown in Figure 19.90. Now how many electrons leave the batteries at location A every second? Explain clearly! If you have to make an approximation, state what it is.

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Figure 19.90

19.P.70 The following questions refer to the circuit shown in Figure 19.91, consisting of two flashlight batteries and two Nichrome wires of different lengths and different thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires).

Figure 19.91

The thin wire is 50 cm long, and its diameter is 0.25 mm. The thick wire is 15 cm long, and its diameter is 0.35 mm. (a) The emf of each flashlight battery is 1.5 volts. Determine the steady-state electric field inside each Nichrome wire. Remember that in the steady state you must satisfy both the current node rule and energy conservation. These two principles give you two equations for the two unknown fields. (b) The electron mobility in room-temperature Nichrome is about 7 × 10−5 (m/s)/(N/C). Show that it takes an electron 36 minutes to drift through the two Nichrome wires from location B to location A. (c) On the other hand, about how long did it take to establish the steady state when the circuit was first assembled? Give a very approximate numerical answer, not a precise one. (d) There are about 9 × 1028 mobile electrons per cubic meter in Nichrome. How many electrons cross the junction between the two wires every second? 19.P.71 In the circuit shown in Figure 19.92, bulbs 1 and 2 are identical in mechanical construction (the filaments have the same length and the same cross-sectional area), but the filaments are made of different metals. The electron mobility in the metal used in bulb 2 is three times as large as the electron mobility in the metal used in bulb 1, but both metals have the same number of mobile electrons per cubic meter. The two bulbs are connected in series to two batteries with thick copper wires (like your connecting wires).

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Figure 19.92 (a) In bulb 1, the electron current is i1 and the electric field is E 1. In terms of these quantities, determine the corresponding quantities i2 and E 2 for bulb 2, and explain your reasoning. (b) When bulb 2 is replaced by a wire, the electron current through bulb 1 is i0 and the electric field in bulb 1 is E0. How big is i1 in terms of i0? Explain your answer, including explicit mention of any approximations you must make. Do not use ohms or series-resistance formulas in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter. (c) Explain why the electric field inside the thick copper wires is very small. Also explain why this very small electric field is the same in all of the copper wires, if they all have the same cross-sectional area. (d) Figure 19.93 is a graph of the magnitude of the electric field at each location around the circuit when bulb 2 is replaced by a wire. Copy this graph and add to it, on the same scale, a graph of the magnitude of the electric field at each location around the circuit when both bulbs are in the circuit. The very small field in the copper wires has been shown much larger than it really is in order to give you room to show how that small field differs in the two circuits.

Figure 19.93

19.X.72 A Nichrome wire 75 cm long and 0.25 mm in diameter is connected to a 1.7 volt flashlight battery. (a) What is the electric field inside the wire? (b) Next, the Nichrome wire is replaced by a wire of the same length and diameter, and same mobile electron density but with electron mobility 4 times as large as that of Nichrome. Now what is the electric field inside the wire? (c) The electron current in the first circuit (Nichrome) is i1. The electron current in the second circuit (wire with higher mobility) is i2. Which of the following statements is true? (1) (2)

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(3) (4) Not enough information is given to compare the two currents.

19.P.73 Two circuits are assembled using 1.5-volt batteries, thick copper connecting wires, and thin-filament bulbs (Figure 19.94). Bulbs A, B, and C are identical thin-filament bulbs. Compasses are placed under the wires at the indicated locations.

Figure 19.94 (a) On sketches of the circuits, draw the directions the compass needles will point, and indicate the approximate magnitude of the compass deflections in degrees. Note that the deflection is given at one location. If you do not have enough information to give a number, then indicate whether it will be greater than, less than, or equal to 5 degrees. (Assume that the compasses are adequately far away from the steel-jacketed batteries.) (b) Briefly explain your reasoning about the magnitudes of the compass deflections. (c) On the same sketches of the circuits, show very approximately the distribution of surface charge. (d) The tungsten filament in each of the bulbs is 4 mm long with a radius of 6 × 10−6 meter. Calculate the electric field inside each of the three bulbs, EA , E B , and E C . 19.P.74 A circuit is assembled that contains a thin-filament bulb and a thick-filament bulb as shown in Figure 19.95, with four compasses placed underneath the wires (we're looking down on the circuit). When the thin-filament bulb is unscrewed from its socket, the compass on the left (next to the battery) deflects 15 degrees. When the thin-filament bulb is screwed back in and the thick-filament bulb is unscrewed, the compass on the left deflects 4 degrees.

Figure 19.95

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With both bulbs screwed into their sockets, draw the orientations of the needle on each compass, and write the compass deflection in degrees beside the compass. Explain briefly. 19.P.75 A solid metal sphere of radius R carries a uniform charge of + Q. Another solid metal sphere of radius r carries a uniform charge −q. The amount of charge is not enough to cause breakdown in air. The two spheres are very far apart (distance >> R and distance >> r). At t = 0 a very thin wire of length L is connected to the two spheres (Figure 19.96). The mobility u of mobile electrons in this wire is very small, and the wire conducts electrons so poorly that it takes about an hour for the system to reach equilibrium. In a short time Δt (a few seconds) how many electrons leave the sphere of radius r? There are n mobile electrons per cubic meter in the wire, and the wire has a constant cross-sectional area A. Explain your work, and any approximations you need to make.

Figure 19.96

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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Circuit Elements

KEY IDEAS In a circuit containing a capacitor, the transient leading to final equilibrium or a steady state can last many seconds. Charging a capacitor in a series circuit: current is initially large, then slowly decreases to zero as charge builds up on the capacitor plates. Discharging a capacitor in a series circuit: current is initially large, then slowly decreases to zero as charge leaves the capacitor plates. Macroscopic descriptions of circuit elements include: Capacitors: Capacitance Resistors: Conductivity, resistivity, and resistance Resistors: Ohmic and nonohmic behavior Batteries: Internal resistance All elements: Power supplied or used Loop and node equations may be written using macroscopic quantities.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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CAPACITORS Surface-charge rearrangements are extremely rapid, taking place in nanoseconds. In circuits containing only batteries and resistors, this extremely rapid rearrangements of surface charge leads to the establishment of the steady state in just a few nanoseconds. However, if there is a capacitor in the circuit this initial rapid rearrangement is followed by a “quasi-steady state,” in which the currents change slowly with time as the circuit slowly comes into equilibrium. Experiment 20.EXP.20 shows this time-dependent behavior. In the experiment you connect a battery in series with a capacitor and a light bulb (Figure 20.1), and what you observe in this “charging” process is that initially the bulb glows brightly, as brightly as if there were no capacitor in the circuit, despite the fact that no current can flow across the gap between the capacitor plates. As time goes on, however, the bulb gets dimmer and dimmer and eventually goes out after many seconds.

Figure 20.1 Charging a capacitor. Initially the bulb glows brightly, but with time the bulb dims and finally goes out.

After the bulb has stopped glowing, if you remove the battery and attach the bulb to the capacitor (“discharging,” Figure 20.2), the bulb again glows brightly but then gets dimmer and dimmer and eventually stops glowing after many seconds.

Figure 20.2 Discharging a capacitor. Initially the bulb glows brightly, but with time the bulb dims and finally goes out.

There are two different time scales for events in a circuit containing a capacitor. The bulb lights instantly when a connection is made (surface charge rearrangement takes only nanoseconds), but the current in the circuit declines slowly over a period of many seconds —the two time scales differ by a factor of .

Construction and Symbols A capacitor usually consists of a sandwich of two metal foils separated by insulating material and coiled up into a small package (Figure 20.3). Note that there is no conducting path through the capacitor, even though you are able to light a bulb in series with the capacitor. Charges cannot jump from one plate to the other.

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Figure 20.3 Typical capacitor construction—a coiled-up sandwich of metal foils separated by insulating material.

In symbolic circuit diagrams a capacitor is often shown as though we had unrolled the coiled-up capacitor and made external connections to the center of the metal plates (Figure 20.4). The insulating material between the plates is usually not indicated.

Figure 20.4 Symbols for capacitors.

Discharging a Capacitor It is possible to explain the charging and discharging of a capacitor in terms of the fundamental properties of charge and field. First we'll discuss the discharging process, which is easier to understand. In Figure 20.5 we show the pattern of electric field in the connecting wires 0.1 s after connecting the bulb to the charged capacitor, while the bulb is glowing brightly.

Figure 20.5 Charge and electric field 0.1 s after connecting the bulb to the charged capacitor. The bulb is bright.

Just outside the capacitor, we know that the electric field at a location inside the wire points away from the positive plate and toward the negative plate, due mainly to the fringe field of the capacitor. After the quick transient (a matter of nanoseconds), the electron current everywhere in the connecting wires must be the same, and surface charge will build up to make a uniform electric field everywhere in the wires. There is a big field inside the bulb filament (not shown), across which is a big surface charge gradient, going from + to −.

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QUESTION In Figure 20.5, electron current will move opposite to the electric field. In the next short time interval, given the direction of electron current, what will happen to the amount of charge on the negative plate of the capacitor? On the positive plate?

Since electron current moves away from the negative plate, the negative plate will lose some of its negative charge. Since electron current moves toward the positive plate, that plate will become less positively charged. Since the plates now have less charge on them, they contribute a smaller fringe field in the connecting wires, and the electric field everywhere in the circuit gets smaller, as seen in Figure 20.6. Smaller electric field means smaller electron current, so the bulb glows less brightly.

QUESTION As the process continues, what will be the final state?

Figure 20.6 Charge and electric field 1 s after connecting the bulb to the charged capacitor. The bulb is less bright than before.

With time, the capacitor plates will lose more and more of their charge, so the electric field will get smaller and smaller, and the electron current will get smaller and smaller. The bulb will get dimmer and dimmer and finally stop glowing visibly when the current is so small that the bulb filament isn't hot enough to glow. Eventually the capacitor plates lose all their charge and the current stops completely (Figure 20.7). The circuit is now in equilibrium.

Figure 20.7 After many seconds there is no more charge on the capacitor plates, and the current is zero (equilibrium).

The dimming of the light bulb is gradual because the less charge there is on the capacitor plates, the smaller the fringe field and the smaller the current draining charge off the plates. The amount of charge that leaves the plates in the first second is larger than the amount that leaves in the second second, which in turn is larger than the amount that leaves in the third second. The bulb glows brightly for a short time but dimly for a long time.

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Here is a plausible-sounding objection: “When we connect a light bulb to a charged capacitor, the bulb lights up for a while. But I'm really puzzled. I don't see how charge can come off the capacitor and go through the light bulb. The positive and negative charges on the two plates are very close to each other and attract each other strongly. The charge can't leak off!”

QUESTION Try to explain the source of this confusion. You might refer to the previous discussion of discharging.

You need to look at the electric field in the wire close to but outside the plates. Initially, right next to the negative plate, the electric field in the wire is not zero and points toward the negative plate. Electrons in the wire at this location feel a force and move away from the negative plate. Electrons in the wire near the positive plate likewise move toward the positive plate, under the influence of the fringe electric field in the wire.

Charging a Capacitor Explaining the discharging of a capacitor was relatively easy on the basis of charge and field. The fringe field of the capacitor plus the electric field of the charges on the surfaces of the wires drives current in such a way as to reduce the charge on the capacitor plates (and to reduce the charges on the surfaces of the wires). As a result of these reduced charges, the electric field inside the wires is reduced, and so the current is reduced. The process continues with ever-decreasing current until all the charges have been neutralized. The system reaches equilibrium, and there is no more current. Explaining the initial charging by the battery is a bit more complicated.

CONTRIBUTION OF CHARGE ON CAPACITOR TO THE FIELD IN THE WIRES It takes many seconds for the capacitor to get charged and for the bulb to stop glowing. During the first few nanoseconds while the surface charges are arranging themselves, the capacitor makes almost no contribution to the electric field in the wires because there is very little charge on its plates yet. It is as though the capacitor weren't there—as though there were a continuous wire with no break in it.

QUESTION Therefore, how bright will the bulb be initially, compared to a circuit containing just batteries and the bulb?

Evidently the initial current will be the same as it would be in a simple bulb circuit, and the bulb just as bright as usual (Figure 20.8). However, after current has been running in the wires for a second or more, there will be some significant amount of + and − charge on the capacitor plates.

QUESTION How will the fringe field of the capacitor affect the net electric field in the neighboring wire—will it increase or decrease the net field? What will happen to the current?

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Figure 20.8 Charge and electric field 0.1s after connecting the circuit. The bulb is as bright as in a noncapacitor circuit.

The fringe field of the capacitor acts in a direction opposite to the conventional current, thus decreasing the net field. As a result, the current will decrease (Figure 20.9).

Figure 20.9 Charge and electric field 1 s after connecting the circuit. The bulb is less bright than before.

ROLE OF THE FRINGE FIELD OF THE CAPACITOR Eventually the current stops completely when there is enough charge on the capacitor plates to create a large enough fringe field to counteract the field made by all the other charges. The light stops glowing (Figure 20.10). We see that for both charging and discharging, a key to explaining the qualitative behavior of the circuit is the varying fringe field of the capacitor.

Figure 20.10 After many seconds the current is zero (equilibrium).

The initial establishment of surface charges takes a few nanoseconds, whereas it takes several seconds to reach equilibrium. During the several seconds while the bulb is glowing, the current is not constant, but it changes so slowly that it makes sense to describe the situation as a “quasi-steady state.”

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The Effect of Different Resistors You might wonder what effect the light bulb in the circuit has on the charging process and on the final state of the capacitor. Suppose that you have two light bulbs, one with a thick filament and the other with a thin filament (Figure 20.11). If you charge a capacitor through a thin-filament bulb, the bulb is initially dim (low current) and takes a long time to go out. In contrast, if you charge the capacitor through a thick-filament bulb, the bulb is initially bright (high current) but glows for only a short time.

QUESTION Should the final amount of charge on the capacitor plates depend on what kind of bulb is used during the charging process?

Figure 20.11 When a thin-filament bulb is connected to a battery, the current is small and the bulb is dim. With a thick-filament bulb the current is high and the bulb is bright.

Since charging the capacitor through the thin-filament bulb goes on much longer than charging through the thick-filament bulb, one might argue that the capacitor gets charged up a lot more when the thin-filament bulb is used. Alternatively, one might argue that since the thick-filament bulb glows much more brightly, the capacitor would get charged up a lot more even though the bulb goes out much quicker.

Experiment 20.EXP.21 is a test of the effect of the kind of bulb on the final amount of charge on the capacitor plates.

QUESTION Which argument do you think is right? Why? (Hint: Think about the fields in the final state after the bulb stops glowing.)

In the final state there is no current, so the only potential differences are across the battery and the capacitor. Therefore the potential difference across the capacitor is equal to that across the battery, no matter which bulb is in the circuit, and the charge on the plates is the same no matter which bulb we use. With the thin-filament bulb a small current runs for a long time, and with the thickfilament bulb a large current runs for a short time, but the total amount of charge is the same in either case. In Section 20.6 we will see how to predict the current and the amount of charge on the capacitor quantitatively at any time during the charging or discharging process.

Effect of Capacitor Construction The design of a capacitor has an effect on the behavior of a capacitor circuit. By thinking about what happens in the first fraction of a second when the current decreases slightly, and how the state of the system at the end of this time interval will affect what happens next, we can predict qualitatively how capacitor circuits will behave. In Section 20.6 we will see how to make these predictions quantitative.

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20.X.1 Consider two capacitors whose only difference is that the plates of capacitor number 2 are larger than those of capacitor number 1 (Figure 20.12). Neither capacitor has an insulating layer between the plates. They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Answer

Figure 20.12 What is the effect of increasing the size of the capacitor plates?

Show that in the first fraction of a second the current stays more nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail. Hint: Show charges on the metal plates, and consider the electric fields they produce in the nearby wires. Remember that the fringe field near a plate outside a circular capacitor is approximately

A more extensive analysis shows that this trend holds true for the entire charging process: the larger capacitor ends up with more charge on its plates. The capacitor you have been using has a very large plate area, folded into a small package.

PARALLEL CAPACITORS Two capacitors in parallel as shown in Figure 20.13 can be thought of as one capacitor with plates that have twice the area (see Ex. 20.X.1). Experiment 20.EXP.22 shows that the bulb glows twice as long with this arrangement.

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Figure 20.13 Two capacitors in parallel; Experiment 20.EXP.22.

AN ISOLATED BULB In Experiment 20.EXP.23. shown in Figure 20.14, you can light a bulb despite the fact that there is no conducting path between the bulb and the battery! In the isolated section of the circuit, electrons flow from the upper plate of the capacitor on the left through the bulb onto the upper plate of the capacitor on the right, at the same time that there is current in the lower wires.

Figure 20.14 A glowing bulb completely isolated from the batteries; Experiment 20.EXP.23.

REVERSING A CAPACITOR In Experiment 20.EXP.24. you can see that reversing the capacitor in a circuit after charging it leads to an interesting effect. Temporarily, it's like putting an extra battery in the circuit.

USES OF CAPACITORS Experiment 20.EXP.25. shown in Figure 20.15 helps understand how these capacitors can be useful. What you find if you intermittently break the connection to the batteries for very brief fractions of a second is that the bulb stays lit despite being disconnected from the battery. The current is supplied from the capacitor, which had been charged by the batteries. We say that the

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capacitor is connected “in parallel” with the batteries, rather than in series.

Figure 20.15 What capacitors are used for; Experiment 20.EXP.25.

The large-capacity capacitor used in these experiments is designed to be used in this way, connected in parallel to a power supply in a computer, to supply the computer circuits with current during momentary power outages. For a given voltage, these capacitors hold almost a million times more charge than ordinary capacitors! More generally, capacitors placed in parallel are used in all kinds of electric circuits to even out rapid changes in voltage. For example, in a power supply, parallel capacitors are used to filter out high-frequency (AC) voltage changes that are riding on top of a constant (DC) voltage. You might say that a capacitor makes a circuit behave in a sluggish manner, being unable to change conditions rapidly.

Energy Stored in a Capacitor Let's consider briefly the energy aspects of charging and discharging a capacitor through a light bulb. 20.X.2 In the charging phase, from a time just before the circuit is completed to the time when there is essentially no more current, state whether there is energy gain or loss in the battery, in the bulb, in the capacitor, and in the surroundings. (Hint: Think especially carefully about the initial and final states of the bulb.)

Answer

20.X.3 You may have noticed that during the discharging phase, the light bulb glows just about as brightly, and for Answer just about as long, as it does during the charging phase. Let E stand for the energy emitted by the light bulb (as light and heat) in the discharging phase, from just before the bulb is connected to the capacitor until the time when there is essentially no more current. In terms of +E or −E, what was the energy change of the battery, capacitor, bulb, and surroundings during the charging phase, and during the discharging phase? One answer is already given in the following table: Charging Battery: Bulb: Capacitor:

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Discharging

Surroundings:

+E

It is somewhat surprising that we can get this much information out of one simple observation.

The Current Node Rule in a Capacitor Circuit The current node rule in circuits results from charge conservation in the steady state. Figure 20.16 shows a simple application of the node rule to a node in a circuit where the incoming current I1 supplies the outgoing currents I2 and I3, so that .

Figure 20.16 In the steady state it must be that

.

Figure 20.17 shows a more complex situation. Consider locations 1 and 2 as nodes. While the capacitor is being charged, there is current running onto one plate of the capacitor without any current coming off that plate (region 1 in Figure 20.17), and as a result the charge on the plate is changing. This is clearly not a steady state. Similarly, current comes off of the other plate (region 2 in Figure 20.17), which changes the charge on that plate.

Figure 20.17 In a non-steady-state situation, the current node rule does not apply.

However, if you look at the capacitor as a whole (region 3 in Figure 20.17), just as much current leaves one plate as enters the other plate. Otherwise the two plates together would acquire a nonzero charge that would quickly repel or attract electrons in such a way as to restore the net charge of the capacitor to be zero. So during charging or discharging of a capacitor the current node rule applies to the capacitor as a whole, even though it doesn't apply to just one plate of the capacitor.

Capacitance The capacitors you used in your circuit experiments may have had something like “0.47 F” or “1 F” marked on them, which would mean that they have a “capacitance” of about 0.5 or 1 “farad” (a unit honoring Michael Faraday). Let's see what is meant by this “capacitance” rating. The more charge on the capacitor plates (+Q and −Q), the bigger the electric field in the gap, and the greater the potential difference ΔV across the gap (Figure 20.18). Thus the absolute magnitude of the potential difference |ΔV| is proportional to the amount of

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positive charge Q on the positive plate, and conversely Q is proportional to |ΔV|. We define “capacitance” C as this proportionality constant:

Figure 20.18 The more charge there is on the plates, the bigger the electric field and the bigger the potential difference.

DEFINITION OF CAPACITANCE

The units of capacitance are coulombs per volt, or “farads.”

QUESTION By using what you know about E and ΔV, show that separation s.

in a parallel-plate capacitor with plate area A and plate

Since the magnitude of the electric field inside the gap is we can calculate the potential difference as which shows that the potential difference is proportional to the charge Q on one of the plates, as we expect. The capacitance is therefore Note that the capacitance depends on the geometry of the capacitor: the area of the plates and the separation between them. The concept of capacitance is useful in the detailed analysis of circuits, and we will use it later to study capacitor circuits quantitatively. Bear in mind, however, that capacitance is just a derived quantity relating more fundamental quantities: charge and potential difference (and ultimately charge and field). 20.X.4 A certain capacitor has rectangular plates 50 cm by 30 cm, and the gap width is 0.25 mm. What is its capacitance? If the gap were filled by an insulating material of dielectric constant 5.0, what would the capacitance be?

Answer

We see that typical capacitances are very small when measured in farads. A one-farad capacitor is quite extraordinary! Apparently it has a very large area A (all wrapped up in a small package), and a very small gap s. 20.X.5 Suppose that you charge a one-farad capacitor in a circuit containing two 1.5-volt batteries, so the final potential difference across the plates is 3 volts. How much charge is on each plate? How many excess electrons are on the negative plate?

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Answer

Reflection: Capacitors in Circuits We have seen a kind of circuit behavior that may be called a “quasi-steady state.” The current in the capacitor circuit is approximately steady over a time of a second or so, but only approximately: the current slowly decreases and finally goes to zero— equilibrium is attained. Actually, even an ordinary battery-and-bulb circuit is not in a true steady state, because when the chemicals in the battery are used up, the system comes into equilibrium (no current). There is an important difference, however. In a capacitor circuit the current continually decreases. In a battery-powered circuit, however, the current may stay nearly constant for many hours, then drop to zero in a much shorter time. We were able to understand the behavior of a capacitor circuit in terms of the fringe field of the capacitor competing with the electric field of the other charges (battery charges and surface charges). The more charge on the capacitor, the larger the fringe field. When charging the capacitor, equilibrium is finally reached when there is so much charge on the capacitor that the fringe field is large enough to cancel the other contributions to the electric field. We emphasized a qualitative analysis of capacitor circuits to get at the fundamental issues, and to provide additional practice in reasoning with charge and field. To analyze such circuits quantitatively we need to apply the energy principle in our analysis. Later in this chapter we do this, and we will find that we can quantitatively predict the rate of decrease of the current.

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RESISTORS Our study of electric circuits has been based on a microscopic picture of the motion of electrons inside the wires, due to electric fields produced by charges on and in the battery and on the surfaces of the wires. This microscopic picture gives us insight into the fundamental physical mechanisms of circuit behavior. However, it is not easy to measure electric field, surface charge, electron drift speed, or mobility directly. On the other hand, it is easy to use relatively inexpensive commercially available meters to measure conventional current (rather than electron current), potential difference (rather than electric field), and “resistance” (rather than mobility). For practical purposes, it is useful to describe and analyze circuits in terms of these macroscopic quantities. We will link what we already know about circuits at the atomic level to a macroscopic description in terms of quantities that can be measured easily with standard meters. To analyze the behavior of circuits in detail at the microscopic level, we have found it necessary to take into account both properties of particular materials and the geometry of particular circuit elements, such as thin and thick wires. It is sometimes useful to use a single number to summarize the properties of a particular object.

Conductivity: Combining the Properties of a Material Conventional current is , where is the absolute value of the charge of one of the mobile “charge carriers” (electrons or holes), n is the number density of the charge carriers, A is the cross-sectional area of the wire, u is the mobility of the charge carriers, and E is the electric field inside the wire that drives the mobile charges. This formula for conventional current mixes together three quite different kinds of quantities: , n, and u are properties of the material, A describes the geometry of the material, and E is the electric field due to charges outside the material (and/or on its surface). It is helpful to group the material properties together: arrive at the following way of describing the situation:

. Then if we divide both sides of the equation by the area A, we

where J = current density (A/m 2) and σ = conductivity = (|q|nu). By dividing the current by the cross-sectional area, we obtain a “current density” J in amperes per square meter, which is independent of the geometry of the situation. The current density is proportional to the applied electric field, and the proportionality constant σ is called the “conductivity,” which lumps together all the relevant properties of the material, . (σ is a lowercase Greek sigma.) to be a vector pointing in the direction of the conventional current flow, which is in the direction of the applied We can consider (Figure 20.19): electric field

Figure 20.19

is in the direction of the electric field, and in the direction of motion of a fictitious positive charge.

CURRENT DENSITY AND CONDUCTIVITY

The current density

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ampere/m2.

The conductivity

depends on

, the absolute value of the charge on each carrier; n, the number of charge carriers per m3; and u, the mobility of the charge carriers.

A higher conductivity means that a given applied electric field gives larger current densities. The conductivity is a property of the material and has nothing to do with what shape the material has or with how large an electric field is applied to the material. 20.X.6 In copper at room temperature, the mobility of mobile electrons is about

(m/s)/(volt/meter), and

Answer

mobile electrons per m3. Calculate the conductivity σ and include the correct units. In actual practice, it is usually easier to measure the conductivity σ and deduce the mobility u from this measurement.

there are about

20.X.7 Consider a copper wire with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a thick-filament bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Answer

Conductivity with Two Kinds of Charge Carrier When electric forces are applied to salt water, both the sodium ions (Na+) and the chloride ions (Cl−) move and constitute an electric current. If the electric field is to the right in the water, the force on the Na + ions is to the right, but the force on the Cl − ions is to the left (Figure 20.20).

Figure 20.20 Both positive and negative charges move in salt water. The flow of Na + ions to the right constitutes a conventional current to the right, and the flow of Cl − ions to the left also constitutes a conventional current to the right, because the motion of these negative ions has the effect of tending to make the region on the right less negative and hence more positive, with an effect similar to the flow of Na + ions to the right. If the charge on the positive ion is q1 (e in the case of Na + ions), the positive-ion contribution to the conventional current in . If the charge on the negative ion is q2 (the charge on a Cl − ion is , the negative-ion coulombs per second is . We have to take the absolute value of the charge contribution to the conventional current in coulombs per second is because the negative ions contribute conventional current in the same direction as do the positive ions.

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Note that the mobilities (and therefore the drift speeds) may differ for the two kinds of ions. Larger ions usually have lower mobilities because it is more difficult to move them through the water. The conventional current in amperes (coulombs per second) is the sum of the conventional current due to the positive charges and the conventional current in the same direction due to the negative charges: Note that both ionic currents contribute to the total current, and that the ionic charge q1 or q2 may be a multiple of e. For example, the charge of Zn ++ is The current density charge carrier is

. , so the conductivity in this situation where there are two kinds of

CONDUCTIVITY WITH TWO KINDS OF CHARGE CARRIERS

Resistance Combines Conductivity and Geometry Often when we have compared two different electric circuits, we have said such things as, “The cross-sectional area goes up, the length decreases, and the mobility increases.” It is practical to have a single quantity to describe a resistor that encapsulates both the properties of the material (mobility, mobile-carrier number density) and the geometrical properties of the particular object (crosssectional area, length). This quantity is called “resistance.” Consider the potential difference ΔV from one end to the other of a wire of constant cross-sectional area A, length L, and uniform composition (Figure 20.21). In the steady state, the electric field inside the wire has the same magnitude everywhere (otherwise the drift speed and current would vary) and is everywhere parallel to the wire (otherwise additional charge would be driven to the surface). Since potential difference is defined as

we have . The conventional current density is the same everywhere on the cross-sectional area A of the wire, as was proven in the previous chapter (if E varied across the cross section, one could construct a path with a nonzero round-trip potential difference, and uniform E implies uniform v and therefore uniform J).

QUESTION For some materials

. For such materials, write a formula for I in terms of σ and |ΔV|.

Figure 20.21 A resistor—for example, a Nichrome wire. Combining the two equations for |ΔV| and J, we find a relation between two macroscopic quantities, conventional current I and potential difference ΔV:

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It is useful to rearrange this relation between current and potential difference in the following way:

where the “resistance” R is measured in volts per ampere or “ohms” (abbreviated Ω and named for Georg Ohm, the German physicist who first studied resistance quantitatively in the early 1800s). We have this definition:

DEFINITION OF RESISTANCE

Note that resistance depends on the properties of the material: the geometry of the resistor: L, A These aspects are lumped into one quantity—resistance. We can make a large-resistance resistor in a variety of ways. The resistor could have a large length L, or a small cross-sectional area A, or a low conductivity σ. Sometimes resistance is expressed in terms of the “resistivity” . In that case, . We can of course rearrange and write . Usually, however, it is appropriate to think of the potential difference as causing a current, not of the current causing a potential difference, and the form reminds us that the current is the result of applying an electric field, with an associated potential difference. Note that we haven't developed any new physical principles. We have simply rewritten . potential difference, so that we have

in terms of

Reflection: Connecting Macroscopic and Microscopic Viewpoints We can relate the analysis we've just done to our earlier microscopic analysis of a current in a material with a single charge carrier, such as electrons in a metal:

.

. You should derive this relationship yourself without referring to notes. ) is more For some kinds of scientific work the microscopic picture in terms of electric field, mobility, and drift speed ( useful. However, when you need to make contact with the technological world of voltmeters and ammeters and ohmmeters, you also need to be able to describe a situation in terms of macroscopic quantities such as potential difference (voltage), resistance, and conventional current.

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Here is a side-by-side comparison of the two views:

In other books you may see this equation written as . This can lead to confusion between potential at one location and the potential difference between two locations, so we always write .

20.X.8 A carbon resistor is 5 mm long and has a constant cross section of 0.2 mm2. The conductivity of carbon at room temperature is per ohm · m. In a circuit its potential at one end of the resistor is 12 volts

Answer

relative to ground, and at the other end the potential is 15 volts. Calculate the resistance R and the current I. 20.X.9 A thin copper wire in the same circuit is 5 mm long and has a constant cross section of 0.2 mm2. The Answer −1 −1 ohm m . The copper wire is in series with the conductivity of copper at room temperature is carbon resistor in the same circuit you just studied, with one end connected to the 15-volt end of the carbon resistor, and the current you calculated runs through the carbon resistor wire. Calculate the resistance R of the copper wire and the potential V at end at the other end of the wire. You can see that for most purposes a thick copper wire in a circuit would have practically a uniform potential. This is because the small drift speed in a thick, high-conductivity copper wire requires only a very small electric field, and the integral of this very small field creates a very small potential difference along the wire.

Constant and Varying Conductivity As you may have seen in experiments with light bulbs, the mobility u of the tungsten in a light bulb filament varies with the temperature. For example, the current through one bulb is less than twice the current through two bulbs in series, because the mobility is lower at higher temperatures. Like the mobility, the conductivity and therefore the resistance may change for different amounts of current.

Ohmic Resistors In some materials the conductivity σ is nearly a constant, independent of the amount of current flowing through the resistor. We call such constant conductivity materials “ohmic.” An ohmic material has the property that , where the resistance is constant, independent of the amount of current. Because conductivity depends somewhat on temperature, and higher currents tend to raise the temperature of the material, no material is truly ohmic, but many materials can be considered to be approximately ohmic as long as the temperature doesn't change very much. Resistors made of ohmic materials are called “ohmic” resistors. Doubling the length L of an ohmic resistor doubles the resistance R and cuts the current in half if the potential difference ΔV across the resistor is kept constant. Doubling the cross-sectional area A cuts the resistance in half and doubles the current if the potential difference ΔV across the resistor is kept constant. Nichrome wires, and the carbon resistors used in electronic circuits, are nearly ohmic resistors, as long as not enough current runs through them to raise the temperature significantly. In metals the conductivity decreases somewhat at higher temperatures, and metals are only approximately ohmic. You can think of the electrons as having more frequent collisions with atoms that have increased

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random motion at higher temperatures. You may have seen experimental evidence for nonohmic behavior in the previous chapter, where the current through two bulbs in series was more than half the current through one bulb. 20.X.10 By actual measurement, the current through a thin-filament bulb when connected to two batteries in series (3 Answer volts) is about 100 milliamperes (mA); connected to one battery (1.5 volts) the current is about 80 mA; and connected to a small voltage of only 50 millivolts the current is about 6 mA. (Different thin-filament bulbs may differ from these values somewhat.) Using the formula , what is R for each of these cases? Is R a constant? Is a thin-filament bulb an ohmic resistor over this whole range of currents? So from 0 to 3 volts a thin-filament bulb is not ohmic, although in the range from 1.5 to 3 volts the resistance changes by only about 50% and the bulb can be said to be very roughly ohmic.

Semiconductors In a metal the current is zero when the potential difference is zero, but the slightest potential difference produces some current, as shown in Figure 20.22. Such behavior implies that at least some of the electrons in a metal are truly free, not bound to the atoms. If they were all bound, we would have to apply some field just to free some up, and then more field to increase the current, as shown in Figure 20.23.

Figure 20.22 Mobile electrons, not bound to atoms—ohmic behavior.

Figure 20.23 If electrons in a metal were bound to the atoms, the current might behave like this.

Pure silicon and germanium are nonohmic in a spectacular and spectacularly useful way. In these “semiconductor” materials, the density of mobile electrons n and therefore the conductivity (where u is the mobile electron mobility) depend exponentially on the voltage. Applying a slight voltage produces almost no current. After applying enough voltage to get a small current, doubling this voltage may make the current go up by a factor of a hundred rather than by a factor of two! (See Figure 20.24.)

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Figure 20.24 Behavior of a pure semiconductor such as silicon or germanium.

The key difference between a metal and a semiconductor lies in n, the number of available charge carriers per unit volume. In a metal, n is a fixed number (one mobile electron or hole per atomic core). In a semiconductor, at very low temperatures there are no mobile electrons or holes and the material acts nearly like an insulator for small applied fields. However, the electrons are not tightly bound to the atomic cores and it is possible to free up some electrons by applying a large enough electric field, so n is variable and increases with increasing potential difference. Raising the temperature can also free up some electrons (or holes), and the effect of increasing n more than compensates for the higher rate of collisions with the atomic cores, so that for a semiconductor the resistance decreases with increasing temperature. This is also true for carbon: the resistance of a carbon resistor decreases somewhat with increasing temperature.

Other Nonohmic Circuit Elements Not all resistors are ohmic, and we have seen that even those called “ohmic” are only approximately so if the temperature changes with current. Moreover, circuit elements other than resistors are definitely not ohmic. Capacitors aren't ohmic: If you double the current flowing onto and off of the capacitor plates, the capacitor gets charged faster, but it is not the case that |ΔV| is proportional to I for a capacitor. The voltage is proportional to the charge on one of the plates rather than being proportional to the current: . Therefore capacitors are not ohmic devices. Batteries aren't ohmic: If you double the current through a battery (by changing the circuit elements attached to the battery), the battery voltage hardly changes at all. The battery voltage certainly doesn't double and will actually decrease slightly. It definitely is not the case that |ΔV| is proportional to I for a battery.

HAS LIMITED VALIDITY It is extremely important to apply only to resistors. This formula is not a fundamental physics principle with broad validity. It is just an expression of the fact that some materials are (approximately) ohmic.

Conventional Symbols In circuit diagrams we will often use conventional symbols for resistors (jagged lines) and batteries (two parallel lines, the longer one representing the positive end of the battery); see Figure 20.25.

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Figure 20.25 Conventional symbols for resistors and batteries.

20.X.11 When a single thin-filament bulb is connected to a 1.5-volt battery, the current through the battery is about 80 milliampers. If you add another thin-filament bulb in parallel, the battery current of course increases to 160 milliamperes. Is the battery ohmic? That is, is the current through the battery proportional to the potential difference across the battery?

Answer

Series and Parallel Resistance SERIES RESISTANCE We can show that several ohmic resistors in series are equivalent to one ohmic resistor with a resistance that is the sum of the individual resistances. Consider the circuit of Figure 20.26, with three ohmic resistors in series with an ideal battery (one with negligible internal resistance).

Figure 20.26 Three resistors in series.

The current I is the same everywhere in this series circuit.

QUESTION Write the energy-conservation loop equation for this circuit, using the fact that

We can rearrange the loop equation,

for an ohmic resistor.

, in this form:

Evidently a series of ohmic resistors acts like a single ohmic resistor, with a resistance equal to the sum of all the resistances: This is not an entirely obvious result! The proof depends on the following: the definition of resistance

;

the assumption that the conductivity σ is independent of the current for the material in these resistors (ohmic resistors); the fact that the magnitude of the sum of all the resistor potential drops is equal to the potential rise across the battery, which is numerically equal to the emf of the battery; and the fact that in a series circuit the steady-state current I is the same in every element, due to charge conservation, and an

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unchanging charge distribution in the steady state. Because , if the various resistors are all made of the same material and have the same cross-sectional area, . note that the series-resistance formula is equivalent to just adding the various lengths:

QUESTION For example, show that two identical resistors in series are like a single resistor that is twice as long.

20.X.12 A certain ohmic resistor has a resistance of 40 ohms. A second resistor is made of the same material but is three times as long and has half the cross-sectional area. What is the resistance of the second resistor? What is the effective resistance of the two resistors in series?

Answer

PARALLEL RESISTANCE We can show that several ohmic resistors in parallel are equivalent to one ohmic resistor with a resistance that is the reciprocal of the sum of reciprocals of the individual resistances. Consider the circuit of Figure 20.27, with three different ohmic resistors in parallel.

Figure 20.27 Three resistors in parallel.

The potential difference is the same across each resistor in this parallel circuit and is numerically equal to the emf of the battery.

QUESTION Write the current node equation for this circuit, using the fact that

We can rearrange the current equation,

for an ohmic resistor:

, in this form:

Evidently ohmic resistors in parallel act like a single ohmic resistor, with a resistance whose reciprocal is equal to the sum of the reciprocals of the individual resistances:

Because , if the various resistors are all made of the same material and have the same length, note that the parallelresistance formula is equivalent to just adding the various cross-sectional areas: .

QUESTION

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For example, show that two identical resistors in parallel are like a single resistor that has twice the cross-sectional area.

This corresponds to what you see if you connect two bulbs in parallel to a battery: there is twice as much current through the battery, as though you had one bulb with twice the cross-sectional area. 20.X.13 When glowing, a thin-filament bulb has a resistance of about 30 ohms and a thick-filament bulb has a Answer resistance of about 10 ohms. If they are in parallel, what is their equivalent resistance? How much current goes through two flashlight batteries in series if a thin-filament bulb and a thick-filament bulb are connected in parallel to the batteries?

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WORK AND POWER IN A CIRCUIT If you move a small amount of charge Δq from one location to another, the small amount of electric potential energy change is , since electric potential is electric potential energy equal to the charge Δq times the potential difference per unit charge (joules per coulomb, or volts). If this work is done in a time Δt, we can calculate the power, which is energy per unit time:

However, is the amount of charge moved per unit time, which is the current I. Since important equation:

, we have the following

POWER FOR ANY KIND OF CIRCUIT COMPONENT

You should be able to repeat this argument without referring to your notes. This is a very general result, applying to any kind of circuit component. Here are some examples of the application of this equation to batteries and resistors: 20.X.14 Power in a battery: In a certain circuit, a battery with an emf of 1.5 volts generates a current of 3 amperes. What is the output power of the battery? Include appropriate units. (If on the other hand you were charging a rechargeable battery with a charging current of 3 amperes, this would be the input power to the battery.) 20.X.15 Power in a resistor: A resistor R conducts a current I. Show that the power dissipated in the resistor is Alternatively, show that the power can be written as

.

Answer

Answer

.

EXAMPLE Two Different Bulbs in Series The circuit in Figure 20.28 consists of two flashlight batteries, a thick-filament bulb, a thin-filament bulb, and very lowresistance copper wires. (a) Calculate the number of electrons that flow out of the battery every second at location A, in the steady state. Assume that in these steady-state conditions the resistance of the thick-filament bulb is 10 ohms and the resistance of the thin-filament bulb is 40 ohms. (b) Show the approximate surface charge on the steady-state circuit. (c) Draw an accurate graph of potential around the circuit, with the x axis running from A to H. Label the y axis with numerical values of the potential. (d) The tungsten filament in the thin-filament bulb is 8 mm long and has a cross-sectional area of 2 × 10−10 m2. How big is the electric field inside this metal filament? (e) What is the power output of the battery?

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Figure 20.28 A circuit with a thin-filament bulb and a thick-filament bulb.

Solution (a) Loop rule:

(b) See Figure 20.29. A bigger electric field is needed in the thin-filament bulb (because it has higher resistance), so the surface charge gradient across the thin-filament bulb is much bigger than the surface charge gradient across the thick-filament bulb. The surface charge gradient across the copper wires is extremely small, since only a very small electric field is needed in the wires. This gradient is too small to show up on this diagram, so the surface charge is shown as approximately uniform on the wires.

Figure 20.29 Approximate surface charge distribution on circuit in steady state. (c) See Figure 20.30. Distances on the x axis are not to scale.

Figure 20.30 Potential (volts) vs. location in circuit. Distances (x axis) are not to scale. (d) The drop in potential across the thin-filament bulb is given by

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. The electric field in the filament of

the thin-filament bulb can be found from the potential difference:

(e) For the battery:

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BATTERIES We have been treating batteries as if they were ideal devices: that is, as if they always managed to maintain the same charge separation regardless of what they are connected to. Real batteries, however, don't quite succeed in maintaining a potential difference equal to the battery emf, due to “internal resistance”—the resistance of the battery itself to the passage of current. We will show that a real battery can be modeled as an ideal, resistance-less battery in series with the internal resistance of the battery. Consider again our mechanical “battery” consisting of two large charged plates whose charge is continually replenished by transporting electrons on a conveyor belt from the “+” plate to the “−” plate. The electrons on the motor-driven belt are acted on both by the non-Coulomb force exerted by the motor, , and by the Coulomb force exerted by the charges on the plates, . These two forces oppose each other (Figure 20.31).

Figure 20.31 A mechanical battery with some friction in the mechanism.

In general, the non-Coulomb force has to be somewhat larger than the Coulomb force because of finite mobility of the charges moving through the battery. In a chemical battery, this takes the familiar form of collisions, leading to an average drift speed. The drift speed of ions through the battery is proportional to the net force ( ). If σ is the conductivity for moving is given not by the usual ions through the battery, and A is the cross-sectional area of the battery, the current density but by

Rearranging this formula, we have

If the (short) length of our mechanical battery is s, the electric field is nearly uniform, and the potential difference across the battery is , so

Note, however, that is the work per unit charge or emf of the battery, and the “internal resistance” of the battery, and finally we have

has the units of resistance. We call

This says: If there is very little internal resistance to movement of ions through the battery (or very little current I), the potential difference ΔV across the battery is numerically nearly equal to the emf of the battery. Up until now we have made this assumption to simplify the discussion. However, if the current I through the battery is large enough, the term may be sizable, and the potential difference across the battery (and therefore across the rest of the circuit) may be significantly less than the battery emf. Since

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, a real (nonideal) battery can be modeled as an ideal battery with the internal resistance in series, so

that the potential difference across the battery is less than the emf by an amount (Figure 20.32).

, the potential drop across the internal resistance

Figure 20.32 A real battery can be modeled as an ideal battery in series with an internal resistance.

If a resistor R is connected to a real, nonideal battery (Figure 20.33), energy conservation (round-trip potential difference = 0) gives Solving for the current I, we have

Let's see how this works out in practice. A fresh 1.5-volt “alkaline” flashlight battery has an internal resistance ohm.

of about 0.25

QUESTION What current would we get when we connect various resistors to the battery? Calculate the various values to get an idea of the effect: Ideal battery r int = 0

Real battery r int = 0.25 Ω

100 Ω:

0.015 A

100 Ω:

10 Ω:

0.15 A

10 Ω:

1 Ω:

1.5 A

1 Ω:

0 Ω (short circuit): infinite!

0 Ω (short circuit):

Figure 20.33 A nonideal battery connected to a resistor.

The currents with the real battery are 0.01496 A, 0.146 A, 1.2 A, and 6 A. There are several things to notice in this table: With a real battery, with nonzero internal resistance, you don't get 10 times the current through a 1-ohm resistor that you get through a 10-ohm resistor. Also, the internal resistance determines the maximum current that you can get out of a battery—this battery cannot deliver more than 6 amperes, no matter how small a resistor you attach to it. However, if the current is small, the behavior is nearly “ideal.” An example of the role of a battery's internal resistance is seen in the requirements for a car battery, which must deliver a very large current to the starting motor. The battery must be built in such a way as to have a particularly low internal resistance. There are very large metal electrodes inside a car battery (large cross-sectional area A), which helps reduce the internal resistance to the transport of

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ions. We should point out that the internal resistance of a chemical battery is only approximately constant. As the battery is used more and more, the internal resistance increases. However, at any particular moment, the formula does relate the potential difference across the battery to the current through the battery (using the values of emf and internal resistance valid at that moment). Experiments 20.EXP.26, 20.EXP.27, 20.EXP.28 let you explore internal resistance.

20.X.16 What is the potential difference ΔV across a 1.5-volt battery when it is short-circuited, if the internal Answer resistance is 0.25 ohm and the resistance of the connecting wire is much less than 0.25 ohm? What is the average net electric field E inside the battery? For a mechanical “battery,” how much charge is on the battery plates? This again shows the difference between the non-Coulomb emf of the battery, which does not change no matter what you connect to the battery, and the potential difference ΔV across the battery, which depends on how much charge the battery is able to maintain on its ends. Only for an ideal battery with no internal resistance is ΔV exactly equal to the emf. For real batteries that are discharging, ΔV is always somewhat less than the emf (and if you are charging a real rechargeable battery, you have to apply a ΔV that is somewhat greater than the emf).

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AMMETERS, VOLTMETERS, AND OHMMETERS An important application of circuit theory is to explain the operation of ammeters, voltmeters, and ohmmeters used for measuring conventional currents, potential differences, and resistances. Modern digital “multimeters” combine these functions into one compact instrument, with a switch that lets you choose between measuring current, voltage, or resistance, and with a convenient digital readout, complete with sign. Here are the most important things to know about ammeters and voltmeters: An ammeter must be inserted into the circuit in series with the circuit element whose current you want to measure. An ammeter must have a very small resistance, so as not to alter significantly the circuit into which it has been inserted (Figure 20.34).

Figure 20.34 An ammeter must be inserted in series with the circuit element whose current you want to measure. The ammeter must have low resistance so as not to alter the circuit significantly.

A voltmeter must be placed in parallel with the circuit element along which you want to measure the potential difference. A voltmeter must have a very high resistance, so as not to create a significant alternative pathway for current, which would alter the circuit (Figure 20.35).

Figure 20.35 A voltmeter is placed in parallel with the circuit element along which you want to measure the potential difference. The voltmeter must have high resistance so as not to alter the circuit significantly by providing an alternative current pathway.

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Using an Ammeter An ammeter must be inserted into a circuit, and the current to be measured is brought into the “+” socket and out the “−” socket. If conventional current flows into the “+” socket, the ammeter displays a positive number. When conventional current flows into the “−” socket, the ammeter indicates a negative current reading.

AMMETER SIGN CONVENTION If conventional current flows into the socket marked “+” an ammeter indicates a positive current.

(This is the opposite of the convention for batteries, since conventional current flows out of the “+” terminal of a battery.)

A Simple Ammeter If you have been doing experiments with currents and compasses, you have been using a simple ammeter, in the form of a compass whose needle is deflected by a passing current. If you could run known amounts of current over the compass, you could calibrate this ammeter by determining the relationship between the known current and the observed deflection angle. This is a “noninvasive” ammeter, because you don't have to alter the circuit in order to be able to detect and measure the current in a wire. A serious practical problem in using standard ammeters is that if you want to measure the current somewhere in an existing circuit, you must break the circuit at the place of interest and insert your ammeter. In order that this make as little change as possible in the current to be measured, ammeters are designed to have very low resistance, so that if you insert the ammeter it's as though you're just lengthening the wire a bit.

AN AMMETER SHOULD HAVE LOW RESISTANCE An ammeter should have as small a resistance as possible, so as not to change the current that you are trying to measure.

QUESTION Suppose that you are trying to measure the current through the bulb in the circuit shown in Figure 20.36. What is wrong with the placement of the ammeter?

Connecting the ammeter in parallel to the bulb means that there is a very low-resistance path for the current (a “short circuit”). Most current will go through the ammeter, with very little current going through the bulb. The bulb no longer glows, and the large current measured by the ammeter has nothing to do with the tiny current that goes through the bulb but rather is a measure of the maximum current these batteries can deliver. The fact that an ammeter must be inserted into a circuit may not seem like a problem with simple circuits, but it can be a big problem if the circuit of interest is built in such a way that you can't break in to insert your ammeter without damaging the circuit. Moreover, a commercial ammeter may blow a fuse or even be permanently damaged if used incorrectly as shown in Figure 20.36, because a current of several amperes will run through the very low-resistance ammeter, and many commercial ammeters cannot handle such large currents.

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Figure 20.36 This ammeter is placed incorrectly in the circuit.

Voltmeters Measure Potential Difference If we add a series resistor to an ammeter, in principle we can make a voltmeter to measure potential differences. Suppose that we attach an ammeter across a circuit element, through a resistor whose resistance R is very large (Figure 20.37).

QUESTION If the measured current through the ammeter is I, what is the potential difference

in this circuit, in terms of R and I?

Figure 20.37 An ammeter with a resistor in series forms a voltmeter.

The potential difference along the resistor is . A small current I running through the resistor is measured by the ammeter, and the potential difference is RI. If the ammeter is relabeled in volts (RI), the device reads potential differences directly. A voltmeter can be thought of as an ammeter with a series resistance, all in one box (shown inside dashed lines in Figure 20.37). Notice that a voltmeter has two leads and measures potential difference, not potential! Commercial voltmeters have their “+” and “−”

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sockets arranged as follows:

VOLTMETER SIGN CONVENTION If the potential is higher at the socket marked “+”a voltmeter indicates a positive potential difference.

If the “−” socket is at the higher potential, the voltmeter displays a negative potential difference. The meter in Figure 20.37 will read positive. An ammeter should have very small resistance so that inserting the ammeter into a circuit should disturb the circuit as little as possible. A voltmeter, however, should have as large a resistance as possible, so that placing the voltmeter in parallel with a circuit element will disturb the circuit as little as possible, with only a tiny current being deflected through the voltmeter.

A VOLTMETER SHOULD HAVE HIGH RESISTANCE A voltmeter (used in parallel) should have very large resistance. An ammeter (used in series) should have very small resistance.

There is a practical limit to the voltmeter resistance. The larger the resistance of the voltmeter, the more sensitive must be the ammeter part of the voltmeter. Commercial voltmeters often have resistances as high as 10 megohms (10 × 106 ohms), and the ammeter portion of the voltmeter is correspondingly sensitive. Even a commercial voltmeter with a resistance of 10 megohms is useless for measuring potential differences in a circuit consisting of 100-megohm resistors, because placing a 10-megohm voltmeter in parallel with a 100-megohm resistor drastically alters the circuit. 20.X.17 Two lab partners measured the current in a series circuit by placing a digital multimeter (set to read current Answer in milliamperes) in series between the two thick-filament bulbs (Figure 20.38). Then they wanted to measure the voltage across a bulb, so they simply switched the multimeter to the V setting (volts). They were surprised that the bulbs stopped glowing. What did they do wrong? What did their voltmeter read?

Figure 20.38 This voltmeter is placed incorrectly in the circuit.

Ohmmeters

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Commercial multimeters can act not only as ammeters and voltmeters but also as “ohmmeters” to measure the resistance of an element that has been removed from a circuit. The ohmmeter section of the multimeter consists of an ammeter in series with a small voltage source (for example, 50 millivolts). When you connect an unknown resistor to the ohmmeter, the small voltage drives a small current through the resistor and ammeter. The ohmmeter is calibrated to display the applied voltage divided by the observed current, which is the resistance (assuming that the resistance of the ammeter is small compared to the resistance you are trying to measure). You can see that you have to remove the resistor of interest from its circuit before you can use the ohmmeter, because an ohmmeter is an active device with its own voltage source.

How Commercial Meters Are Really Constructed We have seen that one could construct a voltmeter from a sensitive ammeter with a large resistor in series. Modern digital voltmeters read voltage directly, by timing how long it takes a known current to discharge a capacitor! A digital ammeter is basically a voltmeter with a very small resistance in parallel, and the voltmeter reads the very small potential difference across the small resistance. A digital ohmmeter drives a known current through the unknown resistor and measures the resulting potential difference. Independent of the construction details, it is important to understand that an ammeter has very low resistance, and a voltmeter has very high resistance.

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QUANTITATIVE ANALYSIS OF AN RC CIRCUIT We are now in a position to analyze quantitatively a series circuit containing an ideal battery with known emf, a resistor R, and a capacitor C —a so-called “RC” circuit. Recall that the potential difference across a capacitor is , where Q is the charge , where A is the area of on the positive plate and C is the “capacitance” (Section 20.1; for a parallel-plate capacitor, one of the plates, s is the gap distance between the plates, and K is the dielectric constant of the material filling the gap). The energy equation for the RC circuit in Figure 20.39 is

Figure 20.39 An RC circuit.

The Final State The simplest situation to analyze is the final equilibrium state of the circuit, in which the current has dropped to zero, and the capacitor is fully charged. Then the term RI is zero, and the loop equation gives this:

We see that the final charge on the capacitor depends only on the emf and the capacitance. The resistor in the circuit determines how long it takes to reach equilibrium, but it has no effect on the final charge on the capacitor.

The Initial Situation Recall what happens when a battery and thick-filament bulb are connected in series to a (discharged) capacitor. Immediately after making the final connection the light bulb is very bright (and then it gets dimmer). The initial brightness looks about the same as you get in a circuit containing only a thick-filament bulb without a capacitor. Let's see how we can understand this quantitatively. Rearrange the energy equation to solve for I in terms of the other quantities:

QUESTION Use this formula for I to explain why the initial brightness is the same as though the capacitor weren't there. (Hint: What is Q initially?)

The capacitor is initially uncharged, so initially Q = 0, and plugging this value for Q into the equation for I gives us . This is the same current we would have in a simple circuit without the capacitor. Having understood the initial situation, let's see what happens next. The rate at which charge Q accumulates on the positive capacitor plate is equal to

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. To put it another way, in a time dt

the plate receives an additional amount of charge

. We rewrite the energy equation:

This is in a form suitable for doing a numerical integration. At each instant, we know Q (initially zero), so we can calculate the change dQ from this equation. Then is the new value of Q, and we can use this new Q to calculate a new value of dQ, and so on.

Numerical Integration Let's take a couple of steps in this numerical integration to see how things go. In the first time interval dt, the initial value, Q 0, is zero. The new value, Q 1, is

, since

Now the capacitor has a small nonzero charge Q 1. In the second time interval, we have

Because Q 1 is nonzero, the increase in charge dQ is smaller this time. As time goes on, and Q gets larger and larger, dQ for each additional time step will be smaller and smaller. Therefore a graph of Q vs. time will look like the upper graph of Figure 20.40. The graph is the result of a numerical integration using R = 10 ohms (approximately the resistance of a glowing thick-filament bulb) and C = 1.0 farad, with two 1.5-volt batteries in series.

Figure 20.40 Charge and current as a function of time.

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The lower graph of Figure 20.40 is the current flowing into the resistor, calculated as . As Q increases, I decreases, because the fringe field of the capacitor increasingly opposes the current flow. In terms of potential, as Q increases, the potential difference across the capacitor increases, which means that the potential difference across the resistor must decrease, and therefore decreases.

Analytical Solution The graph in Figure 20.40 looks like some kind of exponential function. Let's make a guess that the current is in fact given by

where I0 is the initial current (which we found earlier to be reasoning:

This last equation is true for all time if

, and a is an unknown constant. Consider this line of

, so we have a solution:

CURRENT IN A SERIES RC CIRCUIT

This solution has all the right properties. When t = 0, the exponential is equal to 1, and , so the initial brightness is indeed the same as if there were no capacitor. As t becomes very large, the exponential goes to 0, and the current goes to zero, which is what is observed in an RC circuit. We used a very common and powerful method for finding an analytical solution for a differential equation (an equation involving derivatives): Guess the form of the solution (containing unknown constants such as a), plug this form and its derivatives into the differential equation, and determine what values the constants must have to satisfy the equation ( ). ,

Since

, and it is easy to show this:

CHARGE IN A SERIES RC CIRCUIT

This solution has all the right properties. When t = 0, the exponential is equal to 1, and Q = 0 (no charge yet on the positive plate of the capacitor). As t becomes very large, the exponential goes to 0, and Q approaches . This is correct: . When the current stops, the potential difference across the capacitor becomes equal to the potential difference across the battery.

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QUESTION What is the numerical value of the final charge on one plate of the capacitor in this circuit, in terms of coulombs?

We find that coulombs. That's a huge amount of charge! Compare, for example, with coulomb. the amount of charge on a strip of invisible tape, which is of the order of Actually, our analytical solution of the differential equation is physically only an approximation, because as the current changes the resistor changes temperature and changes resistance. This is particularly significant when the resistor is a light bulb, since its resistance drops a great deal as the current decreases and it cools off.

The RC Time Constant A rough measure of how long it takes to reach final equilibrium is the “time constant” RC. When the time has fallen from the value to the value

, the factor

QUESTION Calculate the “time constant” for a circuit with R = 10 ohms (approximately the resistance of a glowing thick-filament bulb) and C = 1.0 farad. This is the time when the current has decreased from the original 0.30 ampere to 0.11 ampere (0.37 times 0.30 ampere).

The time constant RC = (10 ohms)(1.0 farad) = 10 s. The bulb will be very dim with this current. This is roughly consistent with what is observed with a thick-filament bulb and a one-farad capacitor, which loses most of its brightness in this amount of time. Figure 20.41 shows graphs of Q vs. t and I vs. t for the RC circuit with the thick-filament bulb, with the 10-second RC time constant indicated.

QUESTION Show that the power dissipated in the bulb at t = RC is only 14% of the original power.

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Figure 20.41 Charge and current as functions of time.

The power dissipated in the bulb is

, so a reduction in the current by a factor of 0.37 gives a reduction in the power by a

factor of (0.37)2 = 0.136.

QUESTION Calculate the “time constant” for a circuit in which you replace the thick-filament bulb with a thin-filament bulb (whose resistance R when glowing is typically about 30 ohms, which may vary from one bulb to another).

The time constant RC = (30 ohms)(1 farad) = 30 s, which agrees roughly with observations with a one-farad capacitor. We now have a complete quantitative description of the behavior of an RC circuit. In experiments where one times how long it takes for the current to decrease to a point where the bulb no longer glows, this is not the same as timing how long it takes for the current to get smaller by a factor of e −1.

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REFLECTION: THE MACRO-MICRO CONNECTION In this chapter we made connections between a microscopic and macroscopic view of resistance. In our original microscopic view, individual electrons acquire a drift speed proportional to the electric field, with . This electric field is made by all the charges in and on the battery and on the surface of the conducting circuit elements. Electron current i is measured in number of electrons per second. In the macroscopic view, the electric field made by all the charges produces a conventional current density and a conventional current measured in coulombs per second (amperes). The integral of the electric field is potential difference, and any round-trip integral of the electric field due to point charges must equal zero, which represents energy conservation per unit charge. We also used a conventional-current version of the steady-state current node rule.

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*WHAT ARE AC AND DC? The steady constant current in a circuit consisting of a battery and a resistor is called “direct current” or DC. Home appliances such as a floor lamp use “alternating current” or AC. Two slots of the wall socket supply a sinusoidally varying voltage (“AC voltage”) that drives a sinusoidally varying current in the lamp (Figure 20.42). (A third wall socket connection goes to ground for safety reasons.) In many countries including the United States the frequency is 60 Hz (60 hertz means 60 cycles of a sine wave every second, with a period T = 1/60s). Many other countries, including those in Europe, use 50 Hz AC. For reasons discussed below, what is called “110 volts AC” is actually a voltage that varies sinusoidally between −155 volts and +155 volts.

Figure 20.42 The voltage between two slots in a wall socket varies sinusoidally.

Although current in an incandescent floor lamp rises and falls sinusoidally, the light doesn't seem to flicker for two reasons. First, humans have difficulty perceiving flickering that occurs 50 or 60 times per second. For example, movies are presented in the form of 24 still pictures per second, but you perceive a continuous image. Second, the temperature of an incandescent bulb filament cannot change rapidly; there isn't time in a fraction of a 60 Hz cycle for the filament to cool down. However, with a fluorescent lamp, where the light is produced nonthermally in sync with the applied voltage, you may see strobelike images of your fingers if you wave your fingers rapidly back and forth while looking through your fingers at the lamp. Very high voltages (100,000 volts or more) are used in power transmission lines to reduce energy losses between a generating station and a remote city. If R is the resistance of the transmission line, you want the voltage drop RI along the line to be small, so you want the current I to be small. The power delivered to the line is IΔV, so for I to be small you need ΔV to be large. However, very high voltages would be extremely dangerous in the home. It is relatively cheap and efficient to convert high AC voltages to low AC voltages using “transformers” whose operation is based on Faraday's law, which we will study in a later chapter. At least until quite recently, converting high DC voltages to low DC voltages was difficult and inefficient (though this is changing with modern semiconductor devices) so historically it made sense to use AC in distributing electrical power. Electronic devices such as computers need low DC voltages. The small box between a wall socket and a laptop uses semiconductor elements to convert the AC voltage to 12 volts DC; similar AC to DC converters are built into other devices such as desktop computers and radios.

*AC Power The reason a sinusoid varying between −155 volts and +155 volts is called “110 volts AC” is that it delivers the same energy in one second as would a constant 110 volts. Consider a circuit containing an AC power supply and a resistor, such as a floor lamp. With , we'll integrate the instantaneous power over one cycle (with period T and ω = 2π/T) and see how much energy is delivered to a resistor R in one cycle:

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The average power is the energy delivered in time T divided by the time T:

The factor of 1/2 comes ultimately from the fact that the average value of the square of a cosine is 1/2 (Figure 20.43).

Figure 20.43 The average value of the square of a cosine is 1/2.

The “root mean square” (rms) voltage is defined as

, in which case the average power can be written in a form

that looks like the equation for DC power:

The “110 volts AC” available at the wall socket means

volts, which corresponds to

volts.

*Phase Shift An unusual aspect of AC circuits is that the current need not be in phase with the voltage. For example, if an AC voltage is maintained across an RC circuit, this applied voltage peaks later than the current (Figure 20.44). We say that the current in the circuit “leads” the applied voltage. The energy equation (loop rule) is

where

, or

; the rate at which charge increases on a plate is equal to the charge per second entering the plate,

which is the current I. Differentiating the energy equation with respect to t we have

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Figure 20.44 An AC voltage is applied to an RC circuit. The current in the circuit peaks earlier than does the applied ΔV. By substituting into the energy equation it is possible to show that the current is where φ is called the “phase angle” and expresses the result that the current I peaks at a different time than the applied voltage. The values for φ and Imax can be obtained from the following relations, which emerge from the solution process:

The frequency-dependent factor

is called the “impedance” and plays the role of resistance in this circuit. If the

capacitance C or the angular frequency ω is very large, the impedance is approximately equal to R, and , just as in a and , so the current is in phase with the applied voltage as it is for a DC simple resistor circuit. Also, in this case circuit.

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*ELECTRONS IN METALS It is possible to show experimentally that the mobile charged particles in a metal are indeed electrons. For example, magnetic forces can lead to a transverse polarization of a current-carrying wire, and the sign of this polarization is different for positive and negative moving charges (this is called the “Hall effect,” and we will study this in Chapter 21). A different kind of experiment was carried out in 1916–1917 by Tolman and Stewart, which showed that the mobile charge carriers in a metal are indeed electrons, which was not known for certain before their experiment. We will describe a simpler experiment that is conceptually equivalent to the experiment they actually carried out. Suppose that we accelerate a metal bar, which means that we accelerate the atomic cores, which are bound to each other (Figure 20.45).

Figure 20.45 In an accelerated metal bar the mobile electrons are initially left behind, polarizing the bar.

On the average, the mobile electrons normally feel no force; the mutual repulsion of the mobile electrons is effectively canceled by the attraction to the atomic cores. Therefore as the bar and atomic cores accelerate, the mobile electrons are left behind. As a result, they pile up at the trailing end of the bar (they can't easily leave the bar). This polarizes the bar, with negative charge on the trailing surface and positive charge (deficiency of electrons) on the leading surface, as shown in Figure 20.45. The surface charges produce an electric field E to the left, which means that there is a force e E to the right on electrons inside the metal (Figure 20.45). As the polarization increases, E increases until the electric force e E is equal to the mass of the electron times the acceleration of the metal bar, at which point the acceleration of the mobile electrons matches the acceleration of the bar, and then the polarization no longer increases. The observed direction of the polarization (negative at the left end) led the experimenters to conclude that the mobile particles are indeed negatively charged. You should run through the argument to see that if the movable charges had been positive, the trailing end of the bar would have been positive instead of negative. Although this experiment does not give separate values for the charge and the mass of the mobile particles, it does provide a measurement of the ratio q/m. There had already been measurements of the ratio e/m for electrons moving in a vacuum, in a device similar to a television tube. The modern value for this ratio for the electron is about

Tolman and Stewart measured the acceleration a, and they used a voltmeter to measure the potential difference ΔV from one end of the metal bar to the other. The potential was lower at the left end of the bar, which shows that the mobile charge carriers are negative. Moreover, the experimental value of q/m had the same value as that obtained for electrons in other experiments. This was very strong evidence that currents in metals consist of moving electrons. 20.X.18 If the length of the bar is L, express q/m in terms of the quantities measured by Tolman and Stewart.

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Answer

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*A COMPLICATED RESISTIVE CIRCUIT We can use energy conservation (loop equations) and the current node rules to determine how much current would flow in each element of a complicated “multiloop” circuit made up of batteries and ohmic resistors (resistors for which the current is proportional to the potential difference). Analyzing complex circuits like the one discussed in this section is more a matter of technology than fundamental physics, so we won't analyze many such circuits. However, it is useful to analyze at least one such circuit as another illustration of the application of energy conservation and the current node rule. We will use the loop and node rules to generate as many independent equations as we have unknown currents, and then solve the system of equations. Assume that the emf's, battery internal resistances, and R 's are known for the circuit shown in Figure 20.46, but the currents are unknown. We will determine the current in each part of this circuit. The first step in the solution is to assign names and directions to the unknown currents (I1 through resistor R 1, I2 through resistor R 2, etc.). We don't know what the actual directions of these unknown conventional currents are, so we guess, and we record our guesses on a diagram, as shown in Figure 20.47. We also draw possible paths to follow for writing loop equations for energy conservation.

Figure 20.46 A circuit with unknown currents, which we will determine.

Figure 20.47 Name the unknown currents, draw paths along which to apply energy conservation.

We don't have to assign a different current variable to every branch. Consider conventional current I1. On the circuit diagram we have assumed that it flows from A toward B. This also means that we have assumed the same amount of current I1 goes from B to C through resistor R 1, and from H to A through resistor R 7. Similar comments apply to the other currents. These simplifications are trivial applications of the current node rule.

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First let's use energy conservation for loop 1 (the path ABCHA). We can start anywhere around the loop; let's start at location A and walk around the loop, adding up the potential differences we encounter on our walk (Figure 20.48). The potential rises across the battery (the electric field inside the battery is opposite to the direction we're walking) and the potential falls along the resistors, since we are walking in the assumed direction of the conventional current, which runs in the direction of the electric field from higher potential to lower potential.

Figure 20.48 Apply energy conservation for a round-trip path, loop 1 (ABCHA).

When we get back to our starting location at A, the net potential change along our walk should be zero, since potential difference in a round trip is zero (and ): Note that in writing this loop rule we implicitly used the current node rule to show that the current is the same I1 through the battery, r 1, R 1, and R 7. We also used the property of ohmic resistors to be able to relate the current through R1 to the potential . difference along R 1 (as If I1 turns out to be −1.3 amperes when we solve for the unknown currents, that will merely mean that we guessed the wrong direction and that I1 actually runs from C to B (and that C is at a higher potential than B). Thus there is no real penalty for guessing the wrong current directions. Next consider loop 2 (path CDEFC), starting at location C (Figure 20.49). When we get back to our starting location at C, the net potential change along our walk should be zero: Comments on loop 2: From D to E there is a potential drop (not a rise) across the battery numerically equal to ( , because we are traversing the battery from its high-potential + end to its low-potential − end, in the direction of the electric field inside the battery. (If we've guessed right about the direction of I2, this battery is being charged rather than discharged.) From F to C there is a potential rise + R 3I3 (not a drop) because we are heading upstream against the current I3. That is, the assumed direction of I3 (from C to F) implies that the potential at C is higher than the potential at F. Again, if we have made the wrong assumption, all that will happen is that the final numerical answer for I3 will have a negative sign.

QUESTION In the same manner, use energy conservation for loop 3 (path HCFGH) and for loop 4 (path ABCDEFGHA).

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Figure 20.49 Apply energy conservation for a round-trip path, loop 2 (CDEFC).

The equation for loop 4 around the outside of the circuit is the algebraic sum of the equations for the other three loops and is not an independent equation. Only three of these four loop equations are actually significant in solving for the unknown currents, although it doesn't matter which three we choose to use in the solution.

QUESTION Now write current node rule equations for nodes C, H, and F, paying attention to signs (+ for incoming current, − for outgoing current).

Adding the equations for node C and node F gives an equation that is equivalent to the equation for node H. Only two of these three node equations are actually independent, and we can use any two of them in solving for the unknown currents. Node C: Node F: Node H: We have a total of five independent equations (three independent loop equations and two independent node equations). It can be shown that in any resistive circuit there are as many independent loop equations as there are minimal-sized loops (that is, ignoring combined loops such as loop 4). The number of independent node equations can be determined by subtracting the number of independent loop equations from the number of independent currents. Our five equations are sufficient to be able to solve for the five unknown currents in terms of the known emf's and known resistances. This solution step can involve extremely tedious algebraic manipulation and should usually be turned over to a symbolic manipulation package such as Maple or Mathematica. Of course in some very simple cases it may be easy to solve the equations, but in general the algebra can be quite forbidding. Once the unknown currents have been determined (typically by using a computer program to solve the equations), we can know essentially everything there is to know about the circuit. For example, we can calculate the potential drop across each resistor, since we know R and I for each resistor. We can therefore also calculate the potential difference between any two locations in the circuit. Refer to Figure 20.46, and suppose that emf1 = 12 volts and R 1 = 20 ohms, and that after we solved the equations we found that I1 was 0.4 amperes.

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20.X.19 What is the potential difference ? How much power is dissipated in R1? What is the power expenditure of battery 1 (some of which is dissipated in the battery itself due to internal resistance r 1)?

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SUMMARY New concepts In a capacitor circuit there is a slow transient leading to a final equilibrium state. Current can run (for awhile) even if there is a gap in a circuit. The fringe field of the capacitor in the neighboring wires is critical for understanding the behavior of a capacitor circuit in terms of charge and electric field. Conventional current I: is in the direction of motion of positive carriers is measured in coulombs per second Conductivity

is a property of a material;

Ohms and nonohmic resistors Results Charging and discharging times are larger with larger plates, smaller gap, or with an insulator in the gap. Resistance

incorporates the geometry of a resistor. is the current through a resistor. is the potential difference across a resistor.

Power = IΔV for any circuit element. Potential difference across a real, nonideal battery (Figure 20.50):

Figure 20.50 Equivalent nonideal battery.

Capacitors in circuits can be analyzed in terms of exponentially with time, with a time constant RC.

, where C is the capacitance, and the charge Q and current I vary

depends on the geometry of the capacitor (Section 20.1; for a parallel-plate capacitor, , Capacitance where A is the area of one of the plates, s is the gap distance between the plates, and K is the dielectric constant of the material filling the gap).

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EXPERIMENTS Capacitors in Circuits 20.EXP.20 Charging and discharging a capacitor Find the capacitor in your experiment kit or lab (your instructor will tell you what it looks like). Caution: Do not exceed the voltage rating marked on the capacitor—typically a few volts. (If the rating is 2.5 volts, it is normally okay to use two ordinary “1.5 volt” batteries.) To make sure that the capacitor is initially uncharged, start by connecting a wire across the capacitor for a couple of seconds in order to discharge the capacitor. Figure 20.51: Connect the discharged capacitor in series with a thick-filament bulb (#14), and run the wire over a compass. The process is called “charging the capacitor.” What do you observe?

Figure 20.51 Charging a capacitor. After the bulb stopped glowing, was there still current flowing? (You may need to place the wire under the compass in order to see small deflections easily.) Figure 20.52: Remove the batteries and connect the bulb directly to the capacitor. The process is called “discharging the capacitor.” What do you observe?

Figure 20.52 Discharging a capacitor. Repeat both stages of the experiment (charging and discharging), but this time use a thin-filament bulb (#48) with the two batteries and the capacitor. What difference does the choice of bulb make? Sketch graphs of current through the bulb versus time for both kinds of bulbs. On your graphs label the time axis, and note which curve refers to which bulb. Pay special attention to the magnitude of the initial current (you may need to repeat the experiment to observe this). You see that the time can be varied by varying the resistance. If you like, you can experiment with different lengths or

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thicknesses of your Nichrome wires, which provide a wide range of charging and discharging times. 20.EXP.21 The effect of different bulbs on final charge Design an experiment that will show how the final amount of charge on each plate of the capacitor depends on which bulb is involved in charging the capacitor. You might discuss your ideas with other students before carrying out your experiments. (If you wish, you can conduct your study using different lengths or thicknesses of Nichrome wire rather than two kinds of light bulbs.) Be careful in your experiments. Before charging a capacitor be sure that it doesn't already have some charge on its plates: discharge it fully by connecting a wire across it for a couple seconds. Repeat any numerical measurements several times to make sure you know how reproducible the measurements are: 15.43 seconds and 15.67 seconds may not be significantly different! Describe your experiments, results, and conclusions. 20.EXP.22 Two capacitors in parallel Work with another group so that you can try lighting a thick-filament bulb with two capacitors in parallel as shown in Figure 20.53. What do you predict for the charging time compared to a circuit with one capacitor? (Again, consider what happens in the first fraction of a second, and what effect this will have on the process in the next time interval.) What do you observe?

Figure 20.53 Two capacitors in parallel. 20.EXP.23 Lighting an isolated bulb Work with another group so that you can make a circuit with two capacitors as shown in Figure 20.54. Note that the thin-filament bulb is in a conducting “island” completely isolated from the batteries.

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Figure 20.54 A bulb completely isolated from the batteries.

Before running the circuit, predict in detail what you expect will happen, and explain your reasoning. Try to work out agreement among the students in the two groups. Remember that during the few-nanosecond transient, the uncharged capacitors have no effect on the electric field. Now construct the two-capacitor circuit and observe what happens. Discuss any discrepancies between the prediction you made and what you see. You might use compasses to verify your theory about the currents in the various parts of the circuit. 20.EXP.24 Reversing the capacitor Follow the following instructions carefully—otherwise you might damage the capacitor or the bulb. Charge the capacitor using one battery and a thick-filament bulb (use only one battery!). Now that the capacitor is charged, what do you predict would happen if you reversed the connections to the capacitor, leaving all other connections intact, including the battery? Explain your reasoning, including an appropriate diagram. Now try it. What do you observe? If the observation is different from your prediction, explain what was wrong with the prediction. Why is it important to use only one battery? 20.EXP.25 The usefulness of a capacitor Connect the capacitor and a thin filament bulb in parallel rather than in series (Figure 20.55). Intermittently break the connection to the batteries for very brief fractions of a second. What happens to the brightness of the bulb during the very brief time that the batteries are disconnected, compared to what happens without a capacitor in the circuit? Explain.

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Figure 20.55 What these capacitors are used for.

Internal Resistance of a Battery 20.EXP.26 Two bulbs in parallel Connect two thick-filament bulbs in parallel to two batteries in series. Unscrew one of the two bulbs. When you screw it back in, you may be able to see a slight change in the brightness of the other bulb. Why is this? 20.EXP.27 Two batteries in parallel Connect two batteries in parallel to a thin-filament bulb. You will see that the bulb has about the same brightness (dimness) as it has with just one battery, because the potential difference is essentially unchanged. What would you expect concerning how long the batteries would last, compared with just one battery? Why? 20.EXP.28 The short-circuit current Hold a copper connecting wire high enough above a compass that when the wire is connected by other connecting wires to short-circuit a single battery, you get about a 20-degree compass deflection. Try a different battery, trying not to move the wire. What compass deflections do you get for each battery? For the two batteries in series? For the two batteries in parallel? Are your measurements consistent with your predictions in the exercises above for the maximum current from batteries in series and in parallel?

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EXERCISES AND PROBLEMS Section 20.1 20.X.29 The capacitor in Figure 20.56 is initially charged, then the circuit is connected. Which graph in Figure 20.57 best describes the current through the bulb as a function of time?

Figure 20.56

Figure 20.57 20.X.30 The capacitor in Figure 20.56 is initially charged, with the left plate positive, before the circuit is connected. What happens when the circuit is connected? (A) The charge on the left plate stays constant. (B) The left plate becomes less positive. (C) The left plate becomes more positive. 20.X.31 Which of the following statements about the discharging of a capacitor through a light bulb are correct? Choose all that are true. (A) The fringe field of the capacitor decreases as the charge on the capacitor plates decreases. (B) Electrons flow across the gap between the plates of the capacitor, thus reducing the charge on the capacitor. (C) The electric field at a location inside the wire is due to charge on the surface of the wires and charge on the plates of the capacitor. (D) Electrons in the wires flow away from the negative plate toward the positive plate, reducing the charge on the plates. 20.X.32 A particular capacitor is initially charged. Then a high-resistance Nichrome wire is connected between the plates of the capacitor, as shown in Figure 20.58. The needle of a compass placed under the wire deflects 20° to the east as soon as the

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connection is made. After 60 seconds the compass needle no longer deflects.

Figure 20.59 (a) Which of the diagrams in Figure 20.58 best indicates the electron current at three locations in this circuit: (1) 0.01 seconds after the circuit is connected? (2) 15 seconds after the circuit is connected? (3) 120 seconds after the circuit is connected? (b) Which of the diagrams in Figure 20.59 best indicates the net electric field inside the wire at three locations in this circuit: (1) 0.01 seconds after the circuit is connected? (2) 15 seconds after the circuit is connected? (3) 120 seconds after the circuit is connected?

Figure 20.58 20.X.33 The capacitor in Figure 20.60 is initially uncharged, then the circuit is connected. Which graph in Figure 20.57 best describes the current through the bulb as a function of time?

Figure 20.60 20.X.34 The capacitor in Figure 20.60 is initially uncharged. What happens when the circuit is connected? (A) Electrons jump across the gap between the plates. (B) Electrons accumulate on the left plate.

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(C) Electrons accumulate on the right plate. 20.X.35 The capacitor in Figure 20.60 is initially uncharged, then the circuit is connected. Which graph in Figure 20.57 best describes the absolute value of the charge on the left plate as a function of time? 20.X.36 The capacitor in Figure 20.61 is initially uncharged, then the circuit is connected. Which graph in Figure 20.57 best describes the magnitude of the fringe field of the capacitor at location A (inside the connecting wire) as a function of time?

Figure 20.61 20.X.37 The capacitor in Figure 20.61 is initially uncharged, then the circuit is connected. Which graph in Figure 20.57 best describes the magnitude of the net electric field at location A (inside the connecting wire) as a function of time? 20.X.38 How is the initial current through a bulb affected by putting a capacitor in series in the circuit? Explain briefly. 20.X.39 How is the charging time for your capacitor correlated with the initial current? That is, if the initial current is bigger, is the charging time longer, shorter, or the same? Explain briefly. 20.X.40 When a particular capacitor, which is initially uncharged, is connected to a battery and a small light bulb, the light bulb is initially bright but gradually gets dimmer, and after 45 seconds it goes out. The diagrams in Figure 20.62 show the electric field in the circuit and the surface charge distribution on the wires at three different times (0.01 s, 2 s, and 240 s) after the connection to the bulb is made. The diagrams in Figure 20.62 show the pattern of electric field in the wires and charge on the surface of the wires at three different times (0.01 s, 8 s, and 240 s) after the connection to the battery is made. Which of the diagrams best represents the state of the circuit at each time specified? (a) 0.01 s after the connection is made (b) 8 s after the connection is made (c) 240 s after the connection is made

Figure 20.62 20.X.41 How does the final (equilibrium) charge on the capacitor plates depend on the kind of bulb or the length of Nichrome wire in the circuit during charging? Very briefly state why this is. 20.X.42 How does the final (equilibrium) charge on the capacitor plates depend on the size of the capacitor plates? On the spacing between the capacitor plates? On the presence of a plastic slab between the plates? 20.X.43 Researchers have developed an experimental capacitor that takes about eight hours (!) to discharge through a thin-filament bulb. They propose using this in electric cars to provide a burst of energy for emergency situations (it would be charged at a slow rate during normal driving). Describe in general terms the key design elements of this extraordinary capacitor. 20.X.44 Consider two capacitors whose only difference is that the plates of capacitor number 2 are closer together than those of capacitor number 1 (Figure 20.63). Neither capacitor has an insulating layer between the plates. They are placed in two

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different circuits having similar batteries and bulbs in series with the capacitor.

Figure 20.63

Show that in the first fraction of a second the current stays more nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail. Hint: Show charges on the metal plates, and consider the electric fields they produce in the nearby wires. Remember that the fringe field near a plate outside a circular capacitor is approximately

A more extensive analysis shows that this trend holds true for the entire charging process: the capacitor with the narrower gap ends up with more charge on its plates. 20.X.45 The insulating layer between the plates of a capacitor not only holds the plates apart to prevent conducting contact but also has a big effect on charging. Consider two capacitors whose only difference is that capacitor number 1 has nothing between the plates, while capacitor number 2 has a layer of plastic in the gap (Figure 20.64). They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Figure 20.64

Show that in the first fraction of a second the current stays more nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail. Hint: Consider the electric fields produced in the nearby wires by this plastic-filled capacitor. Suppose that the plastic is replaced by a different plastic that polarizes more easily. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor 2?

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A more extensive analysis shows that this trend holds true for the entire charging process: the capacitor containing an easily polarized insulator ends up with more charge on its plates. The capacitor you have been using is filled with an insulator that polarizes extremely easily. 20.X.46 Suppose that instead of placing an insulating layer between the plates you inserted a metal slab of the same thickness, just barely not touching the plates. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor 2 in Exercise 20.X.45? Note that this is essentially equivalent to making a capacitor with a shorter distance between the plates, and this result is consistent with what we found for a narrower capacitor. 20.X.47 A certain capacitor has rectangular plates 56 cm by 24 cm, and the gap width is 0.20 mm. What is its capacitance? We see that typical capacitances are very small when measured in farads. A one-farad capacitor is quite extraordinary! Apparently it has a very large area A (all wrapped up in a small package), and a very small gap s. 20.X.48 A certain capacitor has rectangular plates 59 cm by 33 cm, and the gap width is 0.27 mm. If the gap is filled with a material whose dielectric constant is 2.9, what is the capacitance of this capacitor? 20.X.49 Suppose that you charged a 2.5 F capacitor with two 1.5-volt batteries. How much charge would be on each plate in the final state? How many excess electrons would be on the negative plate? 20.P.50 Give a complete but brief explanation for the behavior of the current during the discharging of a capacitor in a circuit consisting of a capacitor and light bulb. Include detailed diagrams. Explain, don't just describe! 20.P.51 Give a complete but brief explanation for the behavior of the current during the charging of a capacitor in a circuit consisting of batteries, bulb, and capacitor in series. Include detailed diagrams. Explain, don't just describe! 20.P.52 Figure 20.65 shows three circuits labeled A, B, and C. All the thin-filament bulbs, capacitors, and batteries are identical and are like the equipment you used in class. The capacitors are initially uncharged. In each circuit the batteries are connected for a short time T and then disconnected. The time T is only 10% of the total charging time through a single thin-filament bulb, so that the bulb brightness doesn't change much during the time T. (a) In which circuit (A, B, or C) does the capacitor now have the most charge? Explain. (b) In which circuit (A, B, or C) does the capacitor now have the least charge? Explain. (c) Design and carry out experiments to check your answers. Describe your experiments and the numerical results. Before each experiment, connect a wire across the capacitor for a few seconds to fully discharge the capacitor. One way to compare the amount of charge stored in the capacitor during the time T is to finish charging it through a single thin-filament bulb, and see how much less time is required than when you start with a discharged capacitor.

Figure 20.65 20.P.53 The two circuits shown in Figure 20.66 have different capacitors but the same batteries and thin-filament bulbs. The capacitors in circuit 1 and circuit 2 are identical except that the capacitor in circuit 2 was constructed with its plates closer together. Both capacitors have air between their plates. The capacitors are initially uncharged. In each circuit the batteries

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are connected for a time that is short compared to the time required to reach equilibrium, and then they are disconnected.

Figure 20.66

In which circuit (1 or 2) does the capacitor now have more charge? Explain your reasoning in detail. 20.P.54 The circuit shown in Figure 20.67 is composed of two batteries, two identical thin-filament bulbs, and an initially uncharged capacitor. Three compasses are placed on the desktop underneath the wires, all pointing north before the circuit is closed. Then the gap in the wire is closed, and the compass on the left immediately shows about a 20° deflection. (a) On a diagram, draw the new needle positions on all three compasses at this time, and write the approximate angle of deflection beside the compasses labeled A and B. Explain carefully. (b) When this circuit was connected with one of the bulbs removed from its socket, the single bulb glowed for T seconds. In the two-bulb circuit shown in Figure 20.67, predict how long the two bulbs would glow (in terms of T). Explain carefully. (c) If possible, carry out the experiment, observing initial compass deflections and length of time of glow with one or two thin-filament bulbs. State your numerical results. Do your observations agree with your predictions?

Figure 20.67 20.P.55 The circuit shown in Figure 20.68 consists of two flashlight batteries, a large air-gap capacitor, and Nichrome wire. The circuit is allowed to run long enough that the capacitor is fully charged with + Q and −Q on the plates.

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Figure 20.68

Next you push the two plates closer together (but the plates don't touch each other). Describe what happens, and explain why in terms of the fundamental concepts of charge and field. Include diagrams showing charge and field at several times. 20.P.56 An isolated capacitor consisting of two large metal disks separated by a very thin sheet of plastic initially has equal and opposite charges +Q and −Q on the two disks. Then we connect a long thin Nichrome wire to the capacitor as shown in Figure 20.69.

Figure 20.69

Describe and explain in detail what happens, including the first few nanoseconds, the approach to the final state, and the final state. Base your explanation on the fundamental concepts of charge and field; do not use the concept of potential in your explanation. To allow room for drawing charges and fields, make your diagram like Figure 20.69, with the thickness of the capacitor and the wire greatly exaggerated. 20.P.57 The charge on an isolated capacitor does not change when a sheet of glass is inserted between the capacitor plates, and we find that the potential difference decreases (because the electric field inside the insulator is reduced by a factor of 1/K). Suppose instead that the capacitor is connected to a battery, so that the battery tries to maintain a fixed potential difference across the capacitor. (A) A light bulb and an air-gap capacitor of capacitance C are connected in series to a battery with known emf. What is the final charge Q on the positive plate of the capacitor? (B) After fully charging the capacitor, a sheet of plastic whose dielectric constant is K is inserted into the capacitor and fills the gap. Does any current run through the light bulb? Why? What is the final charge on the positive plate of the capacitor? 20.P.58 A capacitor with a slab of glass between the plates is connected to a battery by Nichrome wires and allowed to charge completely. Then the slab of glass is removed. Describe and explain what happens. Include diagrams. If you give a direction for a current, state whether you are describing electron current or conventional current. 20.P.59 As shown in Figure 20.70, a spherical metal shell of radius r 1 has a charge Q (on its outer surface) and is surrounded by a concentric spherical metal shell of radius r 2 which has a charge −Q (on its inner surface).

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(a) Use the definition of capacitance:

to find the capacitance of this spherical capacitor.

(b) If the radii of the spherical shells r 1 and r 2 are large and nearly equal to each other, show that C can be written as ε0A/s (which is also the formula for the capacitance of a parallel-plate capacitor) where A = 4πr 2 is the surface area of one of the spheres, and s is the small gap distance between them (r 2 = r 1 + s).

Figure 20.70 20.P.60 A capacitor is connected to batteries by Nichrome wires and allowed to charge completely. Then the plates are suddenly moved farther apart. Describe what happens and explain in detail why it happens, based on fundamental physical principles. If you give a direction for a current, state whether you are describing electron current or conventional current. Include appropriate diagrams to support your explanation. 20.P.61 Two 1.5-volt batteries in series were connected to a capacitor consisting of two very large metal disks placed very close to each other, so that the disks became charged (left disk positive). The batteries were then removed. Next, thick copper wires and a thick-filament bulb were attached to the charged capacitor, as indicated in Figure 20.71. Two magnetic compasses lying on top of the copper wires initially pointed north. You find experimentally that when this thickfilament bulb is connected directly to the two batteries in series you observe a compass deflection of 15°. (a) Briefly describe in detail what you would observe about the bulb and both compasses during the next minute. You do not need to explain yet why this happens. Just describe in detail what you would see. (b) At a time that is 0.01 second after assembling this circuit, draw and label the electric field at the 6 locations marked × on the diagram. Pay attention to the relative magnitudes of your vectors. Make sure you have clearly labeled your electric field vectors (for example, E bulb , etc.). (c) Sketch roughly a possible charge distribution on the diagram. If necessary for clarity, add comments on the diagram about the charge distribution. (d) The length of the left copper wire is L, the length of the right copper wire is also L, the length of the bulb filament is Lbulb , and the gap between the capacitor plates is s. The diameter of the copper wires is d, and the diameter of the bulb filament is dbulb . The mobility in the copper is u, and the mobility in the hot tungsten is ubulb . The number of mobile electrons per unit volume is approximately the same in the copper and the tungsten. At a time that is 0.01 second after assembling this circuit, write two valid physics equations involving the magnitudes of the electric fields, using the labels that you gave for the electric fields in part (b). What are the principles underlying these physics equations? (e) In part (a) you described what you would see happening with the bulb and compasses over the course of a minute. Now explain qualitatively but in detail why this happens. Be brief but be complete in your explanation.

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Figure 20.71 20.P.62 In circuit 1 (Figure 20.72), an uncharged capacitor is connected in series with two batteries and one light bulb. Circuit 2 (Figure 20.72) contains two light bulbs identical to the bulb in circuit 1; in all other respects it is identical to circuit 1. In circuit 1, the light bulb stays lit for 25 seconds. The following questions refer to these circuits. You should draw diagrams representing the fields and charges in each circuit at the times mentioned, in order to answer the questions.

Figure 20.72 (a) One microsecond (1 × 10−6 s) after connecting both circuits, which of the following are true? Choose all that apply: (1) The net electric field at location A in circuit 1 is larger than the net electric field at location B in circuit 2. (2) At location A in circuit 1, electrons flow to the left. (3) At location A in circuit 1, the electric field due to charges on the surface of the wires and batteries points to the right. (4) In circuit 1 the potential difference across the capacitor plates is equal to the emf of the batteries. (5) The current in circuit 1 is larger than the current in circuit 2. (b) Two seconds after connecting both circuits, which of the following are true? Choose all that apply: (1) There is more charge on the plates of capacitor 1 than there is on the plates of capacitor 2. (2) There is negative charge on the right plate of the capacitor in circuit 1. (3) At location B in circuit 2 the net electric field points to the right. (4) At location B in circuit 2 the fringe field of the capacitor point to the right. (5) At location A in circuit 1 the fringe field of the capacitor points to the left. (c) Which of the graphs in Figure 20.73 represents the amount of charge on the positive plate of the capacitor in circuit 1 as a function of time? (d) Which of the graphs in Figure 20.73 represents the current in circuit 1 as a function of time?

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Figure 20.73

Section 20.2 20.X.63 What are the units of conductivity σ, resistivity ρ, resistance R, and current density J? 20.X.64 In gold at room temperature, the mobility of mobile electrons is about 4.3 × 10−3 (m/s)/(V/m), and there are about 5.9 × 1028 mobile electrons per cubic meter. Calculate the conductivity of gold, including correct units. 20.X.65 Consider a silver wire with a cross-sectional area of 1 mm2 carrying 0.3 ampere of current. The conductivity of silver is 6.3 × 107 (A/m 2)/(V/m). Calculate the magnitude of the electric field required to drive this current through the wire. 20.X.66 Which of the following are ohmic resistors? For those that aren't, briefly state why they aren't. Nichrome wire A thin filament light bulb A plastic rod Salt water Silicon (a semiconductor) 20.X.67 When a thin-filament light bulb is connected to two 1.5 V batteries in series, the current is 0.075 A. What is the resistance of the glowing thin-filament bulb? 20.X.68 In the circuit shown in Figure 20.74, the emf of the battery is 7.9 volts. Resistor R1 has a resistance of 23 ohms, and resistor R 2 has a resistance of 44 ohms. A steady current flows through the circuit. (a) What is the absolute value of the potential difference across R 1? (b) What is the conventional current through R 2?

Figure 20.74 20.X.69 When a thick-filament bulb is connected to one flashlight battery, the current is 0.20 ampere. When you use two batteries in series, the current is not 0.40 ampere but only 0.33 ampere. Briefly explain this behavior. 20.X.70 If the combination of resistors shown in Figure 20.75 were to be replaced by a single resistor with the equivalent resistance, what should that resistance be?

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Figure 20.75 20.X.71 In the circuit shown in Figure 20.76 the emf of the battery is 7.4 volts. Resistor R1 has a resistance of 31Ω, resistor R 2 has a resistance of 47Ω, and resistor R 3 has a resistance of 52Ω. A steady current flows through the circuit. (a) What is the equivalent resistance of R 1 and R 2? (b) What is the equivalent resistance of all three resistors? (c) What is the conventional current through R 3?

Figure 20.76 20.P.72 A long iron slab of width w and height h emerges from a furnace, as shown in Figure 20.77. Because the end of the slab near the furnace is hot and the other end is cold, the electron mobility increases significantly with the distance x.

Figure 20.77

The electron mobility is u = u0 + kx, where u0 is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of mobile electrons (we can neglect the small thermal expansion of the iron). A steady-state conventional current runs through the slab from the hot end toward the cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at the two locations marked with ×. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms of x and the given quantities w, h, d, u0, k, n, and I (and fundamental constants)? (d) What is the sign of the potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantities w, h, d, u0, k, n, and I (and fundamental constants), what is the magnitude of the

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voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab? 20.P.73 The conductivity of tungsten at room temperature, 1.8 × 107 (A/m 2)/(V/m), is significantly smaller than that of copper. At the very high temperature of a glowing light-bulb filament (nearly 3000 degrees Kelvin), the conductivity of tungsten is 18 times smaller than it is at room temperature. The tungsten filament of a thick-filament bulb has a radius of about 0.015 mm. Calculate the electric field required to drive 0.20 ampere of current through the glowing bulb and show that it is very large compared to the field in the connecting copper wires. 20.P.74 A circuit consists of a battery, whose emf is K, and five Nichrome wires, three thick and two thin as shown in Figure 20.78. The thicknesses of the wires have been exaggerated in order to give you room to draw inside the wires. The internal resistance of the battery is negligible compared to the resistance of the wires. The voltmeter is not attached until part (e) of the problem. (a) Draw and label appropriately the electric field at the locations marked × inside the wires, paying attention to appropriate relative magnitudes of the vectors that you draw. (b) Show the approximate distribution of charges for this circuit. Make the important aspects of the charge distribution very clear in your drawing, supplementing your diagram if necessary with very brief written descriptions on the diagram. Make sure that parts (a) and (b) of this problem are consistent with each other. (c) Assume that you know the mobile-electron density n and the electron mobility u at room temperature for Nichrome. The lengths (L1, L2, L3) and diameters (d1, d2) of the wires are given on the diagram. Calculate accurately the number of electrons that leave the negative end of the battery every second. Assume that no part of the circuit gets very hot. Express your result in terms of the given quantities (K, L1, L2, L3, d1, d2, n, and u). Explain your work and identify the principles you are using. (d) In the case that d2 r 3) to determine the electric field outside the cable. (Don't forget to consider the flux on the ends of your Gaussian cylinder.) 22.P.24 Figure 22.108 shows a close-up of the central region of a capacitor made of two large metal plates of area A, very close together and charged equally and oppositely. There are +Q and −Q on the inner surfaces of the plates and small amounts of charge +q and −q on the outer surfaces.

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Figure 22.108 (a) Knowing Q, determine E: Consider a Gaussian surface in the shape of a shoe box, with one end of area Abox in the interior of the left plate and the other end in the air gap (surface 1 in the diagram). Using only the fact that the electric field is expected to be horizontal everywhere in this region, use Gauss's law to determine the magnitude of the field in the air gap. Check that your result agrees with our earlier calculations in Chapter 16. (Be sure to consider the flux on all faces of your Gaussian box.) (b) Proof by contradiction: Consider Gaussian box 2, with both ends in the gap. Use Gauss's law to prove that the electric field is constant in magnitude throughout the gap. (c) Knowing Q, determine E: Consider Gaussian box 3, with its left end in the gap and its right end in the interior of the right plate. Show that Gauss's law applied to this Gaussian box gives the correct magnitude of the electric field in the gap. (d) Knowing E, determine q/A: Finally, consider Gaussian box 4, with its left end outside the capacitor and its right end in the interior of the left plate. In Chapter 16 we determined that the fringe field is approximately given by

where s is the gap width and R is the radius of the circular plates. Use this information and Gaussian box 4 to determine the approximate amount of charge q on the outer surface of the plate. 22.P.25 In Chapter 21 we showed that the nonuniform direction of the magnetic field of a magnet is responsible for the force F = I(2π R)B sin θ exerted by the magnet on a current loop. However, we didn't know the angle θ, which is a measure of the nonuniformity of the direction of the magnetic field (Figure 22.109). Gauss's law for magnetism can be used to determine this angle θ.

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Figure 22.109 (a) Apply Gauss's law for magnetism to a thin disk of radius R and thickness Δ x located where the current loop will be placed. Use Gauss's law to determine the component B3 sin θ of the magnetic field that is perpendicular to the axis of the magnet at a radius R from the axis. Assume that R is small enough that the x component of magnetic Field in Space field is approximately uniform everywhere on the left face of the disk and also on the right face of the disk (but with a smaller value). The magnet has a magnetic dipole moment μ. (b) Now imagine placing the current loop at this location, and show that F = I(2π R)B 3 sin θ can be rewritten as F = μ loop |dB/dx|, which is the result we obtained from potential-energy arguments in Chapter 21. 22.P.26 A solid chunk of metal carries a net positive charge and is also polarized by other charges that are not on the metal, as shown in Figure 22.110. The electric field very near the surface of a metal in equilibrium is perpendicular to the surface: any component of the electric field parallel to the surface would drive currents in the metal. (Along a rectangular path, part inside and part outside the metal, the round-trip potential difference is zero, so the parallel component of electric field is the same just inside and just outside the surface.) Let S be the “local” surface charge density—the surface charge per unit area (C/m2) at a particular location on the surface.

Figure 22.110 (a) Calculate the magnitude E of the electric field in the air very near a location on the surface of the metal, in terms of the surface charge density S at that location. Use a cylindrical Gaussian surface as shown, with one face just inside the metal and one face just outside. Don't skip steps in your analysis! (b) Show that nearby charges contribute half of E, and distant charges contribute the other half.

Section 22.5 22.X.27 In the region shown in Figure 22.111, the magnetic field is vertical and was measured to have the values shown on the surface of a cylinder. Why should you suspect something is wrong with these measurements?

Figure 22.111

Section 22.6 22.X.28 The magnetic field has been measured to be horizontal everywhere along a rectangular path 20 cm long and 4 cm high, as shown in Figure 22.112. Along the bottom the average magnetic field B = 1.5 × 10−4 tesla, along the sides the average

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1 −4 magnetic field B 2 = 1.0 × 10 tesla, and along the top the average magnetic field B3 = 0.6 × 10−4 tesla. What can you

conclude about the electric currents in the area that is surrounded by the rectangular path?

Figure 22.112 22.P.29 Figure 22.113 shows a measured pattern of magnetic field in space. How much current I passes through the shaded area? In what direction?

Figure 22.113 22.P.30 A D-shaped frame is made out of plastic of small square cross section and tightly wrapped uniformly with N turns of wire as shown in Figure 22.114, so that the magnetic field has essentially the same magnitude throughout the plastic. (R is much bigger than w.) With a current I flowing, what is the magnetic field inside the plastic? Show the direction of the magnetic field in the plastic at several locations.

Figure 22.114 22.P.31 Figure 22.115 shows a large number N of closely packed wires, each carrying a current I out of the page. The width of this sheet of wires is L.

Figure 22.115

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(a) Show (and explain) the direction of the magnetic field at the two indicated locations, a distance d from the middle wire, where d is much less than L(d > D Why do we see a reflection of visible light from a smooth surface at only one angle? When visible light instead of x-rays hits a crystal, it accelerates electrons in the material, and there is indeed re-radiation in all directions. However, the x-ray diffraction condition 2d sin θ = nλ for adjacent layers is not relevant for visible light, because the distance between atomic layers is extremely

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small compared to the wavelength of visible light (d ≈ 1 × 10−10 m, λ ≈ 5000 × 10−10 m). The difference in path length for reradiation from adjacent layers is negligible, and adjacent layers are nearly in phase. Because the interatomic spacing d is much smaller than λ for visible light, there will be constructive interference between two waves only if the difference in path length is zero, so we will see only one high-intensity maximum. As can be seen from Figure 25.21, for interactions with atoms in the surface layer Δl = 0 only if the reflection angle equals the incident angle. At any other angle, the difference in path length is nonzero (Figure 25.22), and there is destructive interference. The interaction of the incident radiation with deeper layers of atoms is essentially the same, since the layers are very close together compared to the wavelength of the radiation.

Figure 25.21 Only when incident angle = reflection angle is s 1 = s 2 , so Δl = 0.

Figure 25.22 At any other angle, s 1 ≠ s 2 , so Δl is nonzero.

Metal Surfaces Reflection of visible light off a metal surface is quite different from the reflection off transparent insulating materials such as water or glass. The mobile electrons in a metal are very easy to accelerate. The mobile electrons near the surface gain a lot of energy, so we conclude that little energy is available to transfer to electrons far from the surface. It must be the case that the intensity of the radiation (and therefore the intensity of the electric field) inside the metal is small. How can this happen? Presumably there is so much re-radiation by accelerated electrons near the surface of the metal that there is very little net field in the interior of the metal. There are no significant interference effects between layers at different depths from the surface. Of course there are the usual interference effects involving electrons in different locations on the surface, so that there is strong intensity only at the “reflection” angle, and metals make good mirrors.

Thin-Film Interference Despite what we said about the condition 2d sin θ = nλ for adjacent layers in connection with reflection of visible light, there can be interesting interference effects when visible light hits a thin film of transparent material only a few wavelengths thick, such as a soap bubble, or a thin oil slick floating on a puddle of water. These thin-film interference effects depend critically on the exact thickness of the thin film. Assume for simplicity that single-frequency incoming light hits the surface of a thin film nearly head-on (rays nearly perpendicular to

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the surface, as in Figure 25.23). Suppose that the thin film has a thickness of only half a wavelength (λ/2). Though this is a very thin film, it nevertheless contains a large number N of atomic layers: N is about (250 × 10−9 m/1 × 10−10 m) = 2500 atomic layers.

Figure 25.23 Visible light encounters a very thin film of matter. Consider pairs of layers such as layer 1 and layer (N/2 + 1), etc.

In Figure 25.23, consider pairing the first layer of atoms with atomic layer number (N/2 + 1 = 1251), pairing the second atomic layer with atomic layer number (N/2 + 2 = 1252), and so on. Atomic layer number (N/2 = 1250) is paired with atomic layer number N = 2500 (the last layer). The path difference for each of these pairs is λ/2, so each pair produces zero intensity, and the total re-radiated intensity is zero for a thin film whose thickness is λ/2. By similar arguments you can show that there will be zero re-radiation for film thicknesses of λ/2, 2λ/2, 3λ/2, 4λ/2, and so on—any integer multiple of λ/2. For film thicknesses halfway between these thicknesses (λ/4, 3λ/4, 5λ/4, etc.), it can be shown that there is maximum re-radiation, although a formal proof is beyond the scope of this discussion.

QUESTION A soap film or oil slick when illuminated by ordinary white light shows a rainbow of colors. Briefly explain why this is.

For a given film thickness, some wavelength might be the right length to give fully constructive interference and be bright, but other wavelengths wouldn't be the right length and would be dim. If the film thickness isn't uniform or you look at a different angle, the wavelength that gives fully constructive interference will be different. In a soap bubble that lasts a long time, more and more of the liquid flows to the bottom of the bubble. The top part gets very thin, thinner than a quarter-wavelength of light even for violet light, which has the shortest visible wavelength. This very thin film seems to re-radiate almost no light at all (it appears black), because there are only a small number of atomic layers compared to the several thousand that contribute when the thickness is a quarter-wavelength. A closely related thin-film interference phenomenon is responsible for the brilliant iridescent colors seen in some butterfly wings and bird feathers, which contain structures with spacing comparable to a wavelength of light. There is constructive interference for some colors and destructive interference for others.

Index of Refraction The relevant wavelength in thin-film interference is the wavelength inside the material. As we saw in Chapter 24, this turns out to be shorter than the wavelength in air. Inside a dense transparent material such as glass or water, individual atoms are significantly affected not only by the incoming electromagnetic radiation but also by the electric fields re-radiated by the other atoms. When you add up all the electric fields, you find that the pattern of the net field has a crest-to-crest distance that is shortened. The factor by which the wavelength is decreased is called the index of refraction and is around 1.5 for many kinds of glass; it is 1.33 for water.

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In a material with index of refraction n, the wavelength λ′ = λ/n, where λ is the wavelength of the radiation in vacuum. The frequency of the wave is not affected, and since v = f λ the apparent speed of light inside a solid is noticeably slower than 3 × 108 m/s. This complication does not affect our analysis of x-ray diffraction because the re-radiation by an atom exposed to x-rays is very small compared to the re-radiation by an atom exposed to visible light. In the case of x-rays, we can consider the electric field inside the material to be due almost entirely to the incoming radiation, with a negligible contribution from the tiny re-radiation from the other atoms. 25.X.8 A film of oil 200 nm thick floats on water. The index of refraction is 1.6. For what wavelength of light will there be complete destructive interference?

Answer

Coherence Length There is an effect that limits interference effects in films. Most sources of visible light produce sinusoidal waves with a fairly short total length L, which corresponds to the short length of time Δt = L/c during which an atom was emitting the light (c is the speed of light). Even if the total length L corresponds to a large number of wavelengths, it might be less than a millimeter. If the total length L is shorter than twice the thickness of the film, by the time the light re-radiated by the last atomic layer emerges from the film, the first atomic layer has stopped emitting, and there is nothing to interfere with. The total length L of the sinusoidal emission is called the “coherence length.” One of the important properties of lasers is that they emit light with a very long coherence length. This is particularly important in making holograms, which are based on interference effects.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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THE WAVE MODEL vs. THE PARTICLE MODEL OF LIGHT We have established that a variety of phenomena can be explained by the wave properties of electromagnetic radiation. Yet in Chapters 8 and 10 we explained a variety of phenomena by a particle (photon) model of light. As a first step in comparing the wave and particle models of light, let's consider the predictions of the two models of light in two different experimental situations: the photoelectric effect and the collision of a beam of light with free electrons (Compton scattering).

The Photoelectric Effect The mobile electrons in a metal, although quite free to move around inside the metal, are nevertheless bound to the metal as a whole: they don't drip out! It takes some work to move an electron from just inside to just outside the surface of the metal. At that point (electron just outside the surface), the metal would now be positively charged and would attract the electron, so you would need to do additional work to move the electron far away. The total amount of work required to remove an electron from a metal is often called the “work function” but could just as well have been called the “binding energy.” We will refer to it with the symbol W. Typical values for common metals are a few electron volts (eV).

Prediction of the Wave Model One way to supply enough energy to remove an electron from a metal is to shine light on the metal (Figure 25.24).

Figure 25.24 The photoelectric effect: light can eject an electron from a metal.

QUESTION According to the wave model of light, how could a beam of light have sufficient energy W to overcome the binding energy and remove an electron from a piece of metal?

Since the energy per square meter (intensity) of an electromagnetic wave is proportional to E2, we would need a high-intensity electromagnetic wave, in which the amplitude of the electric field is very large, to supply enough energy. Alternatively, if the energy from the electromagnetic wave were not quickly dissipated into thermal energy, we might be able to shine a low-intensity beam on the metal for a long time to supply enough energy.

QUESTION According to the wave model of light, would we need to use electromagnetic radiation of any particular wavelength, or would any wavelength be adequate?

According to the wave model, the energy density of an electromagnetic wave depends on E2, the square of its amplitude, not on its wavelength. All we need is an intense beam of light—it could have any wavelength (or color). We might therefore expect a plot of the kinetic energy of an ejected electron vs. intensity of light hitting the metal to be linear, as in Figure 25.25.

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Figure 25.25 The wave model of light predicts that the kinetic energy of an ejected electron should be proportional to the intensity of light hitting the metal.

Prediction of the Particle Model Recall from the previous volume that according to the particle model a beam of light is a collection of particles traveling at speed c, each with zero rest mass but with a fixed energy and momentum.

QUESTION What does the particle model of light predict would be required to eject an electron from a metal?

The particle model suggests that the energy of an individual photon is what is important, since the absorption of a single photon of the appropriate energy would be adequate to free an electron from the metal (and give it some kinetic energy). This model predicts that an extremely intense beam (many photons per square meter) will have no effect if the energy of the individual particles is too low (Figure 25.26).

Figure 25.26 The particle model predicts that above a certain threshold energy W, the kinetic energy of an ejected electron should be proportional to (Ephoton − W).

What Is Observed? In fact, what is observed experimentally is that the wavelength of the light that falls on the metal surface determines whether or not an electron will be liberated from the metal (Figure 25.27). If the wavelength of the light is too large (the frequency is too low), nothing happens, even if the beam is very intense. However, as the wavelength of the radiation is decreased (frequency increased), at one particular wavelength we begin to see electrons liberated from the metal. As the wavelength of the incident light is decreased even further, the kinetic energy of the electrons liberated from the metal increases.

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Figure 25.27 What is actually observed is that above a minimum frequency f0 , the kinetic energy of ejected electrons is proportional to (f − f0 ).

What happens if we use a very low intensity beam? At low intensity, longwavelength light does not eject electrons, no matter how long we wait. However, as predicted by the particle model, with a low intensity beam of short wavelength light, we sometimes see an electron ejected almost immediately. For light of short enough wavelength, we do however observe that the rate at which electrons are ejected from the metal is proportional to the intensity of the light.

Unifying the Two Models How do we reconcile our models with the experimental observations? First, it seems clear that this phenomenon is essentially a quantum one. A certain amount of energy must be delivered in one chunk, or packet, to the system, in order to raise the system to an energy state in which the electron is unbound. Thus, the particle model of light seems appropriate here. The surprising thing, however, is that the wavelength of the incident light is important. Wavelength is certainly a property of a wave —how can this be reconciled with a particle model of light? Evidently, we need to model light as something with both wavelike and particle-like properties. The wave nature of light explains the interference phenomena we have observed. The particle nature of light explains how a fixed amount of energy can be delivered to a system by a photon. It turns out to be possible to describe photons as particles that have precise values of momentum and energy, yet strangely enough with precise wavelength as well (Figure 25.28).

Figure 25.28 A photon can be described as a particle with momentum and energy, yet strangely enough with precise wavelength as well.

Energy and Wavelength From experimental observations of the photoelectric effect, it is found that the energy of a photon is proportional to the frequency and inversely proportional to the wavelength. Quantitatively, the relation is

where h = 6.6 × 10−34 joule-second is Planck's constant, which you may remember from our study of quantized harmonic oscillators in Volume I.

Intensity and Number of Photons/Second Since the number of electrons per second ejected is proportional to the intensity of light of sufficiently small wavelength, apparently in the particle model the intensity of light is proportional to the number of photons per second striking a surface.

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Applying the Model QUESTION If the work function (binding energy) W = 3 eV for a certain metal, what is the minimum energy a photon must have to eject an electron? Evidently the photon must have an energy of 3 eV or more. If the photon has less energy, it can't knock an electron out of this metal.

QUESTION What is the wavelength of a photon with this much energy?

This is near the violet end of the visible spectrum, which extends from about 400 nm (3.1 eV) to about 700 nm (1.8 eV). For a metal with W = 3 eV, light in most of the visible spectrum or in the infrared cannot eject an electron. UV (ultraviolet) light, with wavelength shorter than 400 nm (photon energy greater than 3.1 eV), can knock electrons out of this metal.

QUESTION What if you expose this metal to light whose photon energy is 4 eV (in the UV)? How much kinetic energy would you expect the ejected electron to have when it was far from the metal?

Since it takes 3 eV just to remove the electron, a 4 eV photon can eject an electron with 1 eV to spare, so the electron will have a maximum kinetic energy of 1 eV (it could have less if it loses some energy on its way out of the solid). Experiments have verified this simple particle model for the photoelectric effect. The main points are these: There is a minimum photon energy W required to eject an electron from a metal; photons of lower energy (longer wavelength) do not eject electrons. Photons with excess energy eject electrons whose maximum kinetic energy is equal to the photon energy minus W.

Historically, at the beginning of the 20th century the wave model of light was the accepted model. It was Einstein who first proposed the particle interpretation of the photoelectric effect, in 1905. (In the same year Einstein also published the theory of special relativity, and he also correctly analyzed Brownian motion, the random motion of small objects due to molecular collisions, which led to the first accurate determination of atomic sizes. A truly remarkable year!) 25.X.9 What is the energy in eV of a photon whose wavelength is 334 nm? 25.X.10 If a particular metal surface is hit by a photon of energy 4.3 eV, an electron is ejected, and the maximum kinetic energy of the electron is 0.9 eV. What is the work function W (binding energy) of this metal?

Compton Scattering

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Figure 25.29 is a diagram of a collision between a photon and a stationary electron. This process is called “Compton scattering” in honor of the physicist who showed experimentally that this process has both particle and wave aspects. If the electron recoils with some kinetic energy, the photon necessarily has lost energy. In the wave model of the interaction, loss of energy corresponds to decreased amplitude but the same frequency, whereas in the wave–particle model a loss of photon energy corresponds to a longer wavelength. Compton showed experimentally that the combined wave–particle model did correctly predict the change of wavelength as a function of the scattering angle of the electron. This was one of the most compelling early pieces of evidence for the combined wave–particle nature of light.

Figure 25.29 Compton scattering: a photon scatters off a stationary electron.

In Volume 1 you used the Momentum Principle and the Energy Principle to study particle collisions, and you treated photons as particles with zero rest mass. The general relationship between energy and momentum for a particle is E2 = (pc)2 + (mc 2)2. For zero-mass particles such as photons (and probably neutrinos), this reduces to E = pc. We will use this in an analysis of Compton scattering. Let's use the combined wave–particle model of light to predict the change in wavelength Δλ for a photon that collides with a free electron. As indicated in Figure 25.29, let the energy of the incoming photon be E1, the outgoing photon energy be E2, and the energy of the recoil electron be E3, where E 3 = mc 2 + K, K is the kinetic energy of the recoil electron, and m is the mass of the electron. Energy Principle Solve for recoil electron energy Momentum Principle Solve for recoil electron momentum To get the square of the magnitude of a vector, we take the dot product of the vector with itself: Dot product Expand Multiply by c 2 Substitute (pc)2 = E 2 − (mc 2)2 Substitute E 3

After expanding the squared parenthesis and simplifying, we obtain this:

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Here we have a quantitative prediction for the increase in wavelength of the scattered photon, as a function of the scattering angle θ. The quantity

is called the Compton wavelength. To make it feasible to measure the wavelength increase, one must use light whose wavelength is comparable to the Compton wavelength. Such light is in the x-ray region of the electromagnetic spectrum. To study this kind of collision, it is necessary to have a beam of monochromatic (single wavelength) x-rays. However, most x-ray beams start out with a range of wavelengths.

QUESTION Given what you know about x-ray diffraction, can you think of a way to produce a single-wavelength beam?

If a beam of x-rays interacts with a crystal, re-radiated x-rays of a given wavelength will be particularly intense at an angle satisfying the x-ray diffraction condition. Such a single-wavelength beam could be used in the experiment.

QUESTION How might one measure the wavelength of the x-rays after they interact with an electron?

Again, they can be measured by using x-ray diffraction from a single crystal. If we know the spacing of atoms in a particular crystal, measuring the angles of diffraction maxima will tell us the wavelength of the x-rays. Compton did the experiment, shooting x-rays of known, single wavelength at a block of metal (which contains lots of nearly free electrons) and measuring the wavelength of the scattered x-rays at various scattering angles. He found that the wavelength did indeed increase by the amount predicted by the analysis given above. To summarize, a particle view of light was used to predict the change in energy of the photon, and the wave–particle relation E = hc/ λ was used to convert the energy change to a change in wavelength. The experiment showed that the combination of the particle model and the wave model did indeed predict what actually happens in this process. Neither a pure wave model of the process nor a pure particle model would make the correct prediction. 25.X.11 In a Compton scattering experiment, if the incoming x-rays have wavelength 5 × 10−11 m, what is the wavelength for x-rays observed to scatter through 90°?

Answer

Two-Slit Interference Revisited We saw previously that if a beam of monochromatic light passes through two slits, an interference pattern is formed on a distant surface. Using the wave model of light, and applying the superposition principle, we were able to predict accurately the angles at

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which maxima and minima would occur in the interference pattern. Suppose we shine a beam light through two slits, as before, but this time we decrease the intensity of the light until the beam is extremely weak. The amount of light hitting a distant screen is too small to be detected by human eyes, so we use a different kind of detector: a photomultiplier (Figure 25.30). The functioning of a photomultiplier tube is based on the photoelectric effect. If a single photon ejects an electron from a metal surface, the resulting oneelectron “current” is amplified electronically into a signal strong enough to detect. Photomultiplier tubes can detect individual photons. We will put an array of photomultipliers at the location of the screen where we previously saw an interference pattern with a strong beam of light, and make a plot of number of photons detected vs. location.

Figure 25.30 A two-slit experiment with a very low intensity beam, using photomultipliers as photon detectors.

QUESTION What would you expect the appearance of the resulting plot to be?

Although individual photons are detected at particular locations, over time an interference pattern builds up, like the one observed with an intense beam (Figure 25.31). This pattern, however, builds up statistically, with photons frequently detected at locations of maxima and never detected at locations of minima. Despite the fact that we are detecting individual particles, we still observe a wavelike phenomenon.

Figure 25.31 Number of photons detected vs. angle for a low-intensity beam passing through two slits.

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It is clear from this and other experiments that a complete model of light must have both a wavelike and a particle-like character, as we concluded earlier.

Electron Diffraction If we shoot a beam of x-rays at a single atom, re-radiation of x-rays by accelerated electrons in the atom goes in nearly all directions. However, if we shoot a beam of x-rays at a crystal made up a huge number of atoms arranged in a three-dimensional array, there is constructive interference of the re-radiation only in certain special directions for which the condition 2d sin θ = nλ is true. This is very much a wave phenomenon. Earlier in this chapter we discussed x-ray diffraction by polycrystalline powders, in which there is strong constructive interference for those tiny crystals that happen to be oriented at the appropriate angle for the condition 2d sin θ = nλ to be true. The result is rings of re-radiated x-rays surrounding the incoming x-ray beam. The larger the wavelength λ, the larger the angle θ, and the larger the rings seen on the x-ray film. We would expect very different behavior if we used particles instead of waves. Suppose that instead of using x-rays we shoot electrons at a single atom. There is a complicated electric interaction between an incoming electron and the atom, including polarization of the atom by the electron. The electron is deflected by the collision with the atom and can go in almost any direction, depending on how close to center it hits the atom.

QUESTION If electrons were shot at a crystal made up of a huge number of atoms arranged in a three-dimensional array, what would you expect to see?

We certainly don't expect to see any interference effects, since we assume that an electron is a particle, not a wave. Nevertheless, if we shoot a beam of electrons into a polycrystalline powder, we get rings of electrons, just as with x-ray diffraction, with the angles of the rings described by the usual x-ray diffraction equation! Experimentally, it appears that electrons also can act like waves. Moreover, an electron not only exhibits wavelike interference phenomena but even appears to interfere with itself! If you send in electrons one at a time (a very low-current beam of electrons), and detect the electrons one at a time with an array of detectors (or a photographic film), you get an interference pattern, but built up in a statistical way. Figure 25.32 shows what you observe with longer and longer exposures of film with a low-current beam of electrons hitting a polycrystalline powder. Each dot represents a location where an electron hit the film.

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Figure 25.32 Longer and longer exposures to electrons passing through a polycrystalline powder.

This is intriguing! You detect individual electrons, like particles. Yet if you wait long enough, you find that the pattern of detection looks like what you get with x-rays, which are waves. We say that electrons have properties that are both particle-like and wavelike. In these electron experiments the condition for (statistical) constructive interference is 2d sin θ = nλ, just the same as for x-ray diffraction. What is the wavelength λ of the electrons? It was predicted theoretically by de Broglie and verified experimentally by Davisson and Germer that

where p is the momentum of the electron, and h is Planck's constant, an extremely small quantity measured to be h = 6.6 × 10−34 joule · second. Note that since for a photon E = pc, this relation is valid for photons too:

This relationship is easily demonstrated with an electron-diffraction apparatus in which you can control the momentum of the incoming electrons by varying the accelerating voltage ΔV. We calculate the momentum corresponding to a given kinetic energy K:

so

, where K = eΔV if the electrons start from rest. Therefore we predict this wavelength for the electron:

QUESTION Based on this relation and on the diffraction relation nλ = 2d sin θ, how would you expect the spacing of electron diffraction rings to change if you increase the accelerating voltage?

Experimentally we find that if the accelerating voltage ΔV is quadrupled, the electron wavelength is decreased by a factor of 2, and the diffraction rings shrink due to sin θ being reduced by a factor of 2. Electron diffraction is used extensively to study the structure of materials. It has an advantage compared with x-ray diffraction in that electron beams can be focused easily to a very small spot using electric and magnetic fields, whereas this is difficult to do with xrays. For this reason, electron diffraction is particularly useful in studying small regions of a sample of material. In actual practice, rather high-energy electrons are used in order to achieve good penetration of the material, with typical accelerating potentials of several hundred thousand volts. That means that the wavelength λ is much smaller than a typical interatomic spacing d. Consequently θ is quite small, and the electrons are sent in nearly parallel to the atomic planes of interest.

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Neutron Diffraction Perhaps electrons and photons are very special in having both particle and wave aspects? Not really. Similar diffraction patterns have been observed when entire helium atoms are shot at a crystal, and the wavelength of the helium atom is again λ = h/p (but with a much larger mass m). A particularly important case is neutron diffraction. Neutrons have no charge and only a very small magnetic moment, so neutrons have no electric interactions with electrons and a small magnetic interaction. However, neutrons do have a strong nuclear interaction with protons and other neutrons, as is evident by the presence of neutrons in the nuclei of atoms. If you shoot a single neutron at a single atom, the neutron may be hardly affected by the electrons but can be deflected by the nucleus through just about any angle. When you shoot neutrons at a crystal, the usual formula 2d sin θ = nλ for constructive interference applies.

What Is It that Is “Waving”? Electromagnetic radiation consists of waves of electric field. Interference of electromagnetic radiation is due to superposition of electric fields, and the intensity is proportional to the square of the field amplitude. In the case of electron or neutron diffraction, what is it that is “waving”? It turns out that particles can be described by abstract “wave functions” whose amplitude squared at a location in space is the probability of finding the particle there. The same formula 2d sin θ = nλ that predicts the locations of rings of intensity for x-ray diffraction also predicts the locations of rings of probability for electron or neutron diffraction. Classical interference of light is therefore a good foundation for understanding quantum interference of particles—but the interpretation changes from classical intensity to quantum probability. 25.X.12 Roughly, what is the minimum accelerating potential ΔV needed in order that electrons exhibit diffraction effects in a crystal? 25.X.13 Roughly, with what minimum speed should neutrons should be moving in order to exhibit diffraction effects in a crystal?

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FURTHER APPLICATIONS OF THE WAVE MODEL Having explored the combined wave–particle model of light, we return in this section to the wave model of light. There are a large number of important phenomena that can be most easily analyzed in terms of the wave aspects of light, and we examine some of these in the following sections.

Scattering of Light In Chapter 24 we discussed why the sky is blue, which involves frequencydependent re-radiation by air molecules of light from the Sun, with blue light re-radiated more strongly than red light. This process, in which light passing through a nearly transparent medium is partially re-radiated to the side, is called “scattering” (the light is “scattered” into directions different from the original direction). We will describe an experiment that brings out some interference aspects that affect scattering phenomena. Perhaps your instructor will demonstrate this for you. A bright white light is directed through a long aquarium tank full of water and onto a screen, where it shows up as a white spot (Figure 25.33). If the water is quite clean, very little light is scattered to the side.

Figure 25.33 Light passes through a tank of water and appears on a screen.

A puzzle is why there is almost no light scattered off to the side. After all, there are huge numbers of molecules in the water, and light striking those molecules accelerates electrons, leading to re-radiation. We'll consider path differences for different molecules and see where we should expect lowintensity and high-intensity light. For any molecule in the water, find a second molecule a short distance λ/2 downstream (Figure 25.34). Off to the side, the path difference for the re-radiation from these two molecules is half a wavelength, so the intensity is zero. If we can pair up all the molecules in this way, we can expect to see very little intensity off to the side.

Figure 25.34 Viewed from above: re-radiation (scattering) from water molecules a distance λ/2 apart. Can we always find a suitable partner for every molecule? The answer for water (or glass) is yes, because in a liquid or solid the molecules are right next to each other, filling the entire space. Go downstream a distance λ/2, a distance of about 250 nm = 2500 atomic diameters, and you can be sure that you will find a molecule at that location (unless you are within λ/2 of the end of the water tank, but that short region contains very little water). It is true that molecules in a liquid have some limited freedom to move about, unlike molecules in a solid, but they do fill the space at all times. Note that in the downstream direction, straight through the water toward the screen, the re-radiation from these two molecules is in phase, so we can expect to see light coming out the end of the water tank, and we do.

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(It is reasonable to ask what would happen if we paired up molecules that were a distance λ apart. Wouldn't that give constructive interference? Yes, but we would also have to calculate how that re-radiated light would interfere with the light from all the other pairs that were one wavelength apart. This calculation is beyond the scope of this discussion, but the result is that we get zero intensity when we add up all the contributions. The simplicity of the argument in terms of pairs of molecules that are λ/2 apart comes from the fact that it is easy to add up all the zero intensities from all such pairs and get zero.)

Add Some Soap to the Water Next we drop some material such as soap into the water to provide a low concentration of specks of material. Now you see some light scattered off to the side, and it is distinctly bluish. At the same time, the spot on the screen becomes somewhat reddish, because the blue portion of the white light has been scattered more than the red portion, leaving light that has been depleted of the blue component. As explained in Chapter 24, the scattered bluish light is strongly polarized, as you can see by turning a piece of polarizing material in front of your eye as you look toward the tank from the side. (You also get a big effect by turning a piece of Polaroid between the light source and the water tank.) The visible scattering is due to there being a relatively small number of specks of material in the water that re-radiate a different amount than the water molecules they displace. Moreover, the specks are far away from each other in random locations and do not destructively interfere with each other. That is, if you go a distance λ/2 downstream from a soap particle, you won't necessarily find another, similar soap particle at that location.

Scattering from a Gas There is a subtle point concerning why the sky is blue. Why is there significant scattering off to the side of clear air, when we see little scattering from water? A gas is different from a liquid or solid in that the molecules are not in contact with each other but are roaming around quite freely, with lots of space between molecules. Since the air molecules are not in relatively fixed positions, the argument for destructive interference that we used for water is not valid. Consider a cube of air whose edge is only 0.1λ long, at standard temperature and pressure. Such a cube contains a surprisingly large number of molecules, though many fewer than in a comparable volume of a liquid or solid:

If there were exactly 3000 molecules in each tiny cube of air, there would be little scattering, due to destructive interference by radiation from pairs of cubes located a distance of λ/2 apart. However, the number of molecules in each cube fluctuates randomly as air molecules roam in and out of a cube. The scattering from the sky that we see and enjoy is largely due to the fluctuations in the number of molecules per unit volume. This effect is much less pronounced for liquids. Although the molecules do slide past each other in a liquid, the number of molecules per unit volume hardly changes, since the molecules remain in contact with each other. Of course there is no fluctuation at all in a solid, and if there were no impurities or spatial imperfections in a block of glass it would not scatter light passing through it. The typical statistical fluctuation of N molecules in a given volume of air can be shown to be equal to the square root of N. The electric fields re-radiated by these molecules are nearly in phase with each other (because they are in a volume that is small compared to λ). Thus, the intensity is proportional to

. The scattering is what one would naively expect—N times the

scattering from one molecule!

Diffraction Gratings A compact disc (CD) looks pretty much like a mirror, and you can see a lamp reflected in the shiny surface. However, if you hold the disc in a fixed position and move your eye away from the direct reflection you see a rainbow of colors. This is an interference

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effect. The compact disc has concentric rings of tiny pits etched into the plastic, and these rings of pits separate undisturbed rings of plastic from each other. You observe interference between reflections from adjacent undisturbed sections of the plastic. The distance d between adjacent rings is comparable to the wavelength of visible light (otherwise there would be no effect). There are a very large number of rings on a compact disc, and the effect is called “diffraction,” the term applied to interference phenomena when there are a large number of sources whose waves interfere with each other. A similar device is the “diffraction grating,” a piece of glass or plastic that has been accurately scratched with a large number of closely spaced straight parallel lines a very small distance d apart (Figure 25.35). As with a compact disc, you can see an ordinary reflection in a diffraction grating, but off at an angle you see a rainbow of colors. “Transmission” diffraction gratings are transparent between the scratches and act like a large number of parallel slits through which light passes. As with two-slit interference, each slit can be treated as though it were a source of light (see the derivation at the end of this chapter).

Figure 25.35 A diffraction grating (not to scale) has many fine parallel scratches on its surface.

The diffraction grating is an important scientific tool because it can be used to measure very accurately the wavelengths (and therefore the frequencies) of light emitted by atoms. Different atoms emit light of different frequencies, and diffraction gratings have not only played an important role in the study of atomic structure but have also provided a convenient way to identify the atoms present in a sample of material. A diffraction grating may have as many as 1000 parallel scratches per millimeter! In this case d = 1 × 10−3 mm = 1 × 10−6 m; compare with a typical wavelength of light of about 500 nm = 0.5 × 10−6 meter.

QUESTION Before we treat this phenomenon quantitatively, see whether you can explain very briefly and qualitatively why there are bands of colors off at an angle to a compact disc (or reflective diffraction grating), and why the ordinary direct reflection does not show a rainbow.

The angle θ at which you get another maximum depends on the wavelength (because the phase difference between adjacent strips depends on the wavelength). Therefore, different wavelengths in the white light show up at different angles. The ordinary reflection is normal, because in this case the path lengths are all equal, so there is no dependence on wavelength. To analyze a diffraction grating quantitatively, we'll consider a particularly simple geometry but one that illustrates the most important aspects of the device. Suppose that a beam of red light strikes a transmission diffraction grating nearly perpendicular to the grating (Figure 25.36). There will be a central peak at θ = 0 since all the slits are in phase with each other, but there can be one or more additional peaks in other directions, where the path difference between adjacent slits is one wavelength.

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Figure 25.36 A transmission diffraction grating.

QUESTION If the spacing between adjacent slits is d and the wavelength of the red light is λ, what is the first nonzero angle θ at which you see high intensity? (Hint: Think through what θ is required to get a path difference of one wavelength for adjacent slits of the grating; draw a diagram for adjacent slits.)

This maximum occurs at an angle where d sin θ = λ, so the angle at which you will see high intensity for red light (λ ≈ about 700 nm) is calculable from sin θ = λ/d.

QUESTION If you replace the source of red light with a source of violet light (wavelength of about 400 nm), will the high intensity be seen at a larger or smaller angle θ? Why?

Since sin θ = λ/d, a shorter wavelength λ will give a maximum at a smaller angle θ. Since you see these various colors coming at different angles from the diffraction grating (or compact disc), this implies that white light from the Sun or from an incandescent lamp must be a mixture of many colors, from red to yellow to green to violet. Diffraction gratings are used heavily in science and technology to determine what wavelengths are present in various kinds of light, which is an indicator of what atoms are producing that light. If you look through a diffraction grating at a slit illuminated by a neon lamp, you can see individual lines of color that are characteristic of the neon atoms inside the lamp that are producing the light. This “line spectrum” is different for different atoms, and quite different from the continuous spectrum produced by hot glowing objects such as the Sun or an incandescent lamp.

What's Important about Having Lots of Slits? In the next section we will show formally that the beams of light at the maxima of a diffraction grating are extremely narrow compared to those made by just two slits. It is easy to see from an energy argument that this must be true. With just two slits, the maximum intensity is 22 = 4 times the intensity due to one slit. However, in a typical diffraction grating 2 cm wide, with spacing between slits d = 1 × 10−3 mm, there are 2 × 104 slits, and the maximum intensity is (2 × 104)2 = 4 × 108 times the intensity of a single slit! For there to be such a large intensity in a few special directions, there must be wide regions where the intensity is extremely small. The beams of light from a diffraction grating are extremely narrow and sharply defined, which is why diffraction gratings can be used to make very precise measurements of wavelengths.

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25.X.14 A certain diffraction grating is known to have a line spacing of d = 1 × 10−3 mm. A source of singlewavelength light is observed to produce a high-intensity beam at an angle θ = 36° to the normal. If this is the first maximum at a nonzero angle, what is the wavelength of the light?

Answer

This wavelength is emitted by excited sodium, as in sodium vapor lamps. Detecting this wavelength in some light indicates the presence of excited sodium.

EXAMPLE Experimenting with a CD Hold a compact disc (CD) nearly perpendicular to a line from the CD to a distant light bulb. If you look nearly perpendicular into a region of the CD near the edge, you see an ordinary reflection of the light bulb (Figure 25.37). While holding the CD stationary in this position, move your head slowly away from the perpendicular (or equivalently, tip the CD) while continuing to look at this region of the CD, until you see the first color of a rainbow (Figure 25.38).

Figure 25.37 Looking straight at a CD, you see an ordinary reflection.

Figure 25.38 Looking at an angle to a CD, you see colors. (a) Is the first color you see red or violet? (Try it!) Why? (Violet ≈ 400 nm; red ≈ 700 nm.) (b) If you keep moving your head further away from the perpendicular (or tipping the CD) you see a complete spectrum. If you continue to even larger angles you may see additional spectra. How many complete spectra do you see? (c) Use your observations in parts (a) and (b) to estimate the distance between circular tracks on the CD. Report the data on which you base your estimate. (d) The CD is read with a relatively inexpensive infrared laser (wavelength longer than red). Why can't the tracks be placed closer together, to get more information on the CD? (Newer devices such as DVD players use shorterwavelength lasers.)

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Solution First, try the experiment for yourself! (a) Moving your head away from the perpendicular, the first color you see is violet, which has the shortest wavelength of the visible spectrum (Figure 25.39). This is what you would expect, because for the light reradiated by all the grooves to be in phase, sin θ = λ/d, so the smallest wavelength will have a maximum at the smallest angle (d is the spacing between grooves on the CD).

Figure 25.39 Violet and blue appear at the smallest angle. Yellow and red appear at larger angles. (b) Most people see three spectra, but this is a bit tricky, since you have to hold your head just right (c) We found the first maximum for red light to be at an angle θ1 of about 30 degrees, and we can obtain a value for d from this:

Alternatively, if we see three complete spectra, the red of the third spectra is an at angle θ2 ≈ 90°, so we can carry out an alternative, approximate calculation of the spacing d.

(d) If the spacing d between the tracks is shorter than the wavelength of the incident light, it is not possible to resolve the images of the two tracks (they overlap). If the tracks were closer together, the wavelength of light used to read the information would need to be shorter.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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ANGULAR RESOLUTION In this section we will examine a set of phenomena that at first may seem unrelated but can be explained by the same basic aspects of the wave model of light. We are interested in the following phenomena: The diameter of a telescope mirror (or lens) determines whether you see two distant stars as two separate bright spots or one large bright spot. In two-slit interference, not all the maxima have the same brightness. Light passing through one single slit shows maxima and minima on a screen placed in front of the slit. The width of a diffraction grating determines the angular separation of the maxima observed through the grating.

All of these phenomena involve what is called “angular resolution,” and can be explained in terms of the formula for angular spread

where W is the width of the device.

Angular Width of a Maximum An important property of a diffraction grating is that in the direction of a maximum, the re-radiated beam is extremely narrow. Just how wide (in angle) is one of the maxima observed when monochromatic light passes through a particular diffraction grating? The narrower this maximum, the easier it is to distinguish light of two slightly different wavelengths.

Width of a Maximum How do we define the “width” of a maximum? We could say that the width of the maximum is the angular distance from the brightest area of the maximum to the darkest area next to it—the adjacent minimum. Our task is to calculate the location of the adjacent minimum. Suppose that θ is the angle of the first maximum, with the path difference for adjacent slits being one wavelength, so that Since the maximum is at angle θ, we'll call the location of the adjacent minimum (θ + Δθ). Our task is to find the value of Δθ.

Conditions for a Minimum We note that there will be a minimum if light passing through the first slit and the (N/2)th slit (the middle slit) interferes destructively (Figure 25.40). This is reminiscent of the argument we made when considering sideways scattering of light from a liquid. Assume that there are an even number N of slits (or if N is odd, neglect the small amount of light from the last slit).

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Figure 25.40 The first slit and the middle slit of the diffraction grating interfere destructively if the path difference for adjacent slits is λ + λ/N.

For the first nontrivial maximum (not straight ahead), the path difference between adjacent slits is one wavelength, so the path difference between the first slit and the (N/2)th slit is (N/2)λ. For a minimum, this path difference has an additional λ/2:

Now consider the next pair of slits, slit number 2 and slit number (N/2 + 1). They also have a path difference of λ/2, so they too interfere destructively. We have succeeded in pairing up all the slits in such a way that they each contribute zero intensity, so the total intensity in the direction θ + Δθ is zero. Write the path difference between adjacent slits like this: The path difference between the first slit and the (N/2)th slit is this:

Solving for ε we find

Finding Δθ Now, to find Δθ we use trig identities to expand this relation for two adjacent slits:

Since Δθ is a small angle, we can make the approximations that so

Because d sin θ = λd

The total width of the diffraction grating W is N times the width of one slit:

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Unless θ is close to 90°, cos θ is on the order of 1, so we find that Δθ, the “half width” of the maximum, is approximately given by this:

ANGULAR HALF WIDTH OF A MAXIMUM

where W is the total width of the (illuminated portion) of the device.

Although we've proved this only for a diffraction grating, this result is quite general. Whenever a device contains a very large number of sources that interfere, the angular width of a maximum-intensity beam is approximately equal to the wavelength divided by the total width of the device. For visible light hitting a diffraction grating that is 2 cm wide, the angular spreading may be very small: Δθ ≈ λ/W ≈ (600 × 10−9 m)/(2 × 10−2 m) = 3 × 10−5 radians, which is less than 2 millidegrees.

An Alternative Geometric Argument Instead of using trig identities as we did above, we can find Δθ by a geometric argument involving two adjacent slits (Figure 25.41). In the dashed triangle, the side opposite the very small angle Δθ is ε = λ/N, because this is the additional path length that yields a minimum. The hypotenuse is approximately d cos θ. The angle Δθ in radians is approximately equal to the ratio of the opposite side to the hypotenuse:

where W is the total width of the grating (W = Nd). This is an extremely important result: The wider the grating, the narrower the beam. Unless θ is close to 90, cos θ is of order of 1, so we get, as before:

Figure 25.41 Two adjacent slits in a transmission diffraction grating, with a maximum at θ and a minimum at θ + Δθ.

What about Other Pairings of Slits? You might wonder whether we would not have gotten complete cancellation if we had chosen other ways of pairing up the slits, rather than pairing the first slit with the middle slit, and so on. For example, the first and last slits have a path difference of Nλ+λ, so they interfere constructively. However, this is the only pair that does this, and if we were to add up the effects of all the other slits we would find that their combined effects would cancel the re-radiation of the outer slits. Once we have shown that there is some pairing that gives a net intensity of zero, we don't actually have to consider other pairings.

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Single-Slit Diffraction What does a diffraction grating have to do with maxima and minima created when light passes through a single slit? Just as two narrow slits through which light passes are equivalent to two narrow sources of light, so we can model a single wide slit, of width W, to be equivalent to a very large number N of very narrow sources located a distance d = W/N apart. (Actually, the sum over N finite sources turns into an integral over an infinite number of infinitesimal slits.) There is of course a maximum straight ahead, but there is a minimum at a small angle Δθ given approximately by λ/W (Figure 25.42). The overall pattern of intensity looks like Figure 25.43.

Figure 25.42 The first minimum for a single slit is at Δθ ≈ λ/W.

Figure 25.43 The pattern of intensity vs. angle for diffraction by a single slit.

This also explains the variation in intensity of the maxima in two-slit interference. Each slit makes a pattern like Figure 25.43, and these two patterns interfere. The intensity pattern is the combined result of both single-slit diffraction and two-slit interference. 25.X.15 To get a feel for the size of these “diffraction” effects, consider a single slit W = 0.1 mm wide, illuminated Answer −9 by violet light. The wavelength λ of violet light is about 400 nm (400 × 10 m). Light from the slit hits a screen placed a distance L = 1 m = 1000 mm away from the slit, and the image of the slit is much larger than 0.1 mm due to significant amounts of light heading at angles as large as plus or minus λ/W (for a total angular spread of 2λ/W). Calculate this total width in millimeters.

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Telescope Resolution Limited by Diffraction When a telescope makes an image of a distant star on a photographic plate or array of electronic detectors, the image is broadened by diffraction, which spreads the light rays out into a cone with an approximate half-angle Δθ ≈ λ/W. The images of two stars that are close together in the sky may be broadened so much that they overlap each other and cannot be “resolved” into two separate images (Figure 25.44). The “resolving power” or “resolution” of a telescope or camera is limited by diffraction. The wider the lens, the sharper the image (bigger W, smaller Δθ). Alternatively, the shorter the wavelength, the sharper the image (smaller λ, smaller Δθ). One rule of thumb for resolution states that if φ, the angle between the stars, is greater than 2Δθ, the images will be resolvable.

Figure 25.44 Light from two distant stars passes through the aperture of a telescope. The half width of the diffractionbroadened image for each star determines whether or not it will be possible to distinguish the two images.

Radio telescopes used in making astronomical observations are sometimes linked electronically to other radio telescopes located thousands of miles away, in order to make a radio-frequency “lens” that is so large that the diffraction broadening is extremely small. This has made it possible to obtain exquisitely detailed pictures of extremely distant structures. It is much more difficult to do this with ordinary “optical” telescopes that capture visible light, because you have to bring the light from two different telescopes to the same place, where there can be interference. Nevertheless, there are now linked pairs of large optical telescopes with mirrors to bring light from the two telescopes to an observation location between the telescopes. This makes it possible to obtain pictures with much higher resolution than can be obtained with older optical telescopes.

Only One Large Diffraction Peak With a few sources we often find several large completely constructive peaks, corresponding to the path difference between neighboring sources being equal to 0, λ, 2λ, and so on. Even a diffraction grating with many sources may produce several maximumintensity beams. The situation is different with singleslit diffraction.

QUESTION Explain briefly why a single slit (or lens) produces just one large peak (there are smaller peaks, but only one large peak).

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With a single slit, the path difference “d” between adjacent “sources” is infinitesimal in size, so the path difference is an infinitesimal quantity and can never get as big as λ. There is no direction (except for θ = 0) where all the amplitudes are in phase. 25.X.16 Satellites used to map the Earth photographically are typically about 100 miles (160 kilometers) above the Answer surface of the Earth, where there is almost no atmosphere to affect the orbit. If the lens in a satellite camera has a diameter of 15 cm, roughly how far apart do two objects on the Earth's surface have to be in order that they can be resolved into two objects, if the resolution is limited mainly by diffraction? The key point is that the angle subtended by the two objects should be comparable to or bigger than the spread in angle 2Δθ introduced by diffraction of light going through the lens. Visible light has wavelengths from 400 to 700 nm; consider a wavelength somewhere in the middle of the spectrum. You should find that the numbers on a car license plate would be too blurred to read.

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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STANDING WAVES In this chapter and in Chapter 24 we have been studying “traveling waves”—waves whose crests travel through space at a speed that is characteristic of the type of wave (the speed of light for electromagnetic radiation, the speed of sound for sound waves, etc.). Next we will study waves that travel back and forth within a confined space. Examples include elastic waves traveling back and forth along a length of string, or sound waves traveling back and forth inside an organ pipe. The behavior of such waves in these classical systems provides hints for understanding some quantum aspects of atoms. Consider two sinusoidal traveling waves with exactly the same frequency that were initiated some time ago far to the left and far to the right and are heading toward each other (Figure 25.45). These waves might be electromagnetic waves in empty space, elastic waves propagating from two ends of a very long taut string, sound waves propagating from two ends of a long organ pipe, or water waves propagating from two ends of a lake. The description and analysis is practically the same in each of these very different physical situations. The wide applicability of the analysis we will develop is one of the reasons why we study waves.

Figure 25.45 Colliding waves add up to create a “standing wave.”

When the waves start to overlap, they of course interfere with each other constructively or destructively, depending on the relative phases of the two waves in space. This interference leads to a remarkable effect. Look at the sequence of snapshots in Figure 25.46, taken one-eighth of a period apart. In the overlap region, the “wave” doesn't propagate! It just stays in one place, with the crests waving up and down but not moving to the left or to the right. This is something new, called a “standing wave.”

Figure 25.46 Two traveling waves of the same frequency produce a standing wave.

There are even instants when the “wave” is completely zero throughout the entire region! At the instant that this occurs for a string, pieces of the string do have velocity (rate of change of displacement) and kinetic energy, so the waving does continue, even though there is momentarily no transverse displacement of the string anywhere in the region. In the case of electromagnetic radiation, the electric and magnetic fields are changing as they go through zero. One of the important properties of a standing wave is the existence of what are called “nodes”—locations where the wave is zero at all times. Positions of nodes of the standing wave are circled on the bottom snapshot in Figure 25.45. Halfway between the nodes the standing wave periodically gets to full amplitude. These locations are called “antinodes.”

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Standing Waves vs. Traveling Waves A single observer observing the wave as a function of time at one specific location that is not a node can't tell the difference between a traveling wave and a standing wave.

QUESTION Briefly explain why this is.

The single observer sees the wave going up and down at this location, which is just what it does when a traveling wave goes by. On the other hand, if the single observer is located at a node, the observer may conclude that nothing interesting is happening at all, or that this is a location where two traveling waves are destructively interfering with each other, as in two-slit interference, without forming a standing wave in space. Only if we look at the complete pattern of waves in space and time can we see that a standing wave is really very different from a traveling wave. Yet there are some properties that are the same for both kinds of wave. Look carefully at the time sequence of snapshots in Figure 25.45, and then answer the following important questions:

QUESTION If the wavelength of the two traveling waves is λ, what is the wavelength (crest-to-crest distance) of the standing wave? If the frequency of the two traveling waves is f, what is the frequency (number of complete cycles per second) of the standing wave?

The wavelength and frequency of the standing wave are the same as for the traveling waves. This means that although the standing wave isn't going anywhere (left or right), we can still use the important relationship v = f λ to relate frequency and wavelength for a standing wave through the speed v of a traveling wave.

Confined Waves: Standing Waves on a String One of the ways to create a standing wave is to initiate a single traveling sinusoidal wave in a confined space, so that it continually reflects off the boundaries of this confined space, and the reflected wave interferes with the original wave. The frequency doesn't change in reflecting off a stationary boundary, so the original wave and its reflection are guaranteed to have the same frequency, which is necessary in order to be able to create a standing wave. Examples of such confined waves include elastic waves on a string stretched between two supports, sound waves inside a pipe closed at both ends, and light waves reflecting back and forth between mirrors at each end of a laser. For concreteness we'll emphasize elastic waves on a taut string, but the analysis applies to many other quite different physical situations. Figure 25.47 shows two possible standing waves on a taut string of length L that is tied to two fixed supports. The wavelength in the upper case is λ = 2L, as can be verified by noting that the length L contains half a wavelength. The wavelength in the lower case is λ = L.

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Figure 25.47 Two of the possible standing waves on a taut string of length L.

QUESTION Why can't you have a standing wave on this string with wavelength 3L, or 1.1L, or 0.95L?

These wavelengths don't fit because the waves aren't zero at the ends of the string, where the string is tied to fixed supports. You have discovered that the possible standing waves on a string are “quantized.” Not just any old wavelength will do. The possible wavelengths of a standing wave on a taut string are completely determined by the distance between the supports. The same important principle applies to possible wavelengths of standing sound waves in an organ pipe or possible wavelengths of standing light waves in a laser.

STANDING WAVES ARE QUANTIZED Only certain wavelengths (and frequencies) are possible.

Enumerating the Possible Standing Waves Let's generate a more complete catalog of the possible wavelengths for standing waves on a taut string.

QUESTION In terms of the length L, write down the six longest possible wavelengths. You already know the first two: λ1 = 2L and λ2 = L.

Evidently we have λ3 = 2L/3, λ4 = 2L/4, λ5 = 2L/5, and λ6 = 2L/6. The waves are shown in Figure 25.48. A quick way to determine the possible wavelengths is to note that in the length L there must be an integer number n of half-wavelengths, so L = n(λ n/2), which means that λn = 2L/n. As long as the amplitude isn't too large, or the wavelength too small, the speed of propagation v of traveling waves on a taut string turns out to be independent of the wavelength.

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Figure 25.48 The next four possible standing waves.

QUESTION In terms of this speed v, what are the frequencies of the six lowest possible standing-wave frequencies? Solving for fn = v/λ n = nv/(2L), we have f1 = v/(2L), f2 = 2v/(2L), f3 = 3v/(2L), f4 = 4v/(2L), f5 = 5v/(2L), and f6 = 6v/(2L). These different standing waves are called “modes.” The technical use of the term is that a “mode” of oscillation of a system is one characterized by a standing wave that is sinusoidal in time with the same frequency at every location, and that necessarily satisfies the boundary conditions (in our case, that the wave is zero at the ends of the string). A mode may also be sinusoidal in space at a particular instant of time, as is true for a string, but it need not be spatially sinusoidal in more complicated systems, such as the twodimensional surface of a drum. We've seen that there is a longest possible wavelength 2L and a lowest possible frequency v/(2L). Is there a shortest possible wavelength, corresponding to a highest possible frequency? Yes. If the distance between atoms in the string is D, the shortest theoretically possible wavelength for a standing elastic wave is 2D, with adjacent atoms oscillating out of phase with each other (Figure 25.49). Any shorter wavelength has no physical significance.

Figure 25.49 The shortest possible wavelength (highest possible frequency).

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As is the case for many other systems, a vibrating string has a number of modes (number of different possible standing-wave patterns) that is not infinite but finite.

Building Up to the Steady State You may have had the experience of building up a standing wave in a string or rope. You tie one end of the rope to something, pull it taut, then shake your end of the rope up and down. If you shake at one of the mode frequencies, a large oscillation quickly builds up, because the traveling wave you send down the rope reflects off the other end in such a way as to produce a standing wave through constructive interference. However, if you try to shake the rope at a frequency that is not right for any mode, you don't build up a large oscillation, because you don't get constructive interference. As you continue to put energy into the rope by shaking it at an appropriate frequency, the standing wave grows and grows. What limits its growth? What determines the final amplitude of the standing wave? At first, almost all of your energy input goes into making the standing wave grow. As the amplitude increases, however, so does the rate of energy losses due to air resistance and “internal friction” in the rope itself associated with bending the material. The air resistance grows because air resistance is larger at higher speeds, and the rope has higher transverse speeds when the amplitude of the standing wave is larger. Internal friction inside the rope also grows as the bending angle of the rope gets larger with larger amplitude. At some particular amplitude the losses to air resistance and internal friction in each cycle have grown to be exactly equal to the energy input you make each cycle. During each cycle you simply make up for the losses, but no further growth of the wave occurs. You have reached a steady state. If the losses are relatively small, you don't have to do much to keep the wave going. In fact, you may have experienced keeping a rope oscillating with a sizable amplitude while your hand hardly moves. In a low-loss situation, your hand is very nearly a node, in that your hand moves a very tiny amount compared with the large motion at one of the antinodes (Figure 25.50).

Figure 25.50 With low losses, very little motion of the hand is sufficient to maintain the steady state.

25.X.17 Given the loss mechanisms we have outlined, which should be easier to build up and sustain: a longwavelength (low-frequency) mode or a short-wavelength (high-frequency) mode? If you have the opportunity, try this yourself with a rope.

Answer

Superposition of Modes We've established that standing waves (modes) are quantized for oscillating strings. Is that all a string can do—be in one of these modes? No! According to the superposition principle, it ought to be possible for a string to be oscillating in a manner described by the superposition of two or more modes. What would that look like? Would it be a standing wave? Figure 25.51 shows you a time sequence of the superposition of the two lowest-frequency modes (wavelengths 2L and L). Both modes have been given the same amplitude. The individual modes are shown with light lines, and their sum is shown with a colored line.

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Figure 25.51 A confined wave that is the superposition of the two lowest-frequency modes.

QUESTION Is this motion a standing wave? A traveling wave? Why or why not?

Neither. It is not a standing wave, because there are no nodes (places where the amplitude is always zero). It is not a simple traveling wave, because crests move to the left and to the right. Also, there isn't a “wavelength,” which is something that both standing and traveling waves have. Despite the complexity of this motion, it is at least periodic—the motion repeats after a time. To understand how the motion repeats, in Figure 25.51 it is instructive to follow mode 1 from top to bottom, then follow mode 2 from top to bottom.

QUESTION If the period of the longest-wavelength mode is T(= 2L/v), what is the period of this two-mode oscillation?

After T, the low-frequency mode has gone through one complete cycle, and the second mode has gone through two complete cycles, and we're back at the original configuration. Therefore the period is T. Another way to see this is that in Figure 25.51, mode 1 goes through a quarter cycle (from maximum positive to zero), while mode 2 goes through two quarter cycles (from maximum positive on the left to maximum negative on the left). Extending the set of diagrams to four times as much time would bring both modes back to the starting configuration. Combinations of large numbers of modes can represent pretty complicated motions. Figure 25.52 is a time sequence of the superposition of Particles particular amplitudes of ten different modes, chosen in such a way as to approximate a “square wave” initially.

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Figure 25.52 An approximation to an initial square wave, using ten modes.

How complicated can this get? It can be mathematically proven that any function f (x) that is zero at the ends of the string, f (0) = 0 and f (L) = 0, can be created by adding up carefully chosen amounts of a (possibly infinite) number of modes. In a section on Fourier analysis at the end of this chapter we show you how we determined the amplitudes of each of the ten modes we used to approximate the wave shown in Figure 25.52. This provides an extremely powerful way of analyzing the behavior of confined waves. First figure out what the modes are (the possible sinusoidal standing waves). Then figure out what combination of modes can make a particular shape f (x) at time t = 0. What will happen from then on can be calculated simply by following the sinusoidal behavior of the individual modes, each with their own characteristic frequencies, and adding up the individual contributions of the various modes.

Musical Tones When a violin string is bowed, it oscillates with a superposition of various modes, and these oscillations drive the body of the violin to oscillate, which alternately compresses and rarefies the air to produce the sound waves that you hear. Usually the largest-amplitude mode is the longest-wavelength (lowestfrequency) mode, which corresponds to what is called the “pitch” or “fundamental frequency.” A pure single-frequency tone sounds very boring. The rich musical tone you hear from a violin is due not just to the fundamental frequency but to the superposition of many higher-frequency modes that are also present. Mode frequencies that are simple integer multiples of the fundamental frequency are called “harmonics.” It is a peculiarity of the human ear and brain that we find the superposition of harmonics very pleasing. The superposition of frequencies that are not simple integer multiples of a fundamental frequency sounds much less pleasing.

More Complicated Systems The example of a string is highly representative of a very broad range of standing-wave phenomena. However, we should alert you to some issues that are not illustrated by the relatively simple behavior of a string.

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Nonharmonic Oscillations First, while the geometry of the system determines the possible wavelengths (in order to satisfy the boundary conditions), the different wavelengths need not be simply 1/n times the longest wavelength. A good example is a drum. The wavelengths of the twodimensional waves on a drumhead are determined by the requirement that the wave be zero at the drumhead boundary. The relationship among the various mode wavelengths is rather complicated, not simply 1/n times the longest wavelength. The fact that the wavelengths of drum modes are not simply related to each other (are not “harmonics”) corresponds to the fact that a drum does not produce a musical tone such as is produced by a violin.

Speed Dependent on Frequency Second, the speed of propagation in many systems is not constant but depends on the frequency (or wavelength). Since f = v/λ , in such systems the frequency of a mode need not be simply inversely proportional to the wavelength of that mode, as it is for a string. Such systems are called “dispersive” because traveling waves of different frequencies spread out (disperse) from each other due to differences in their speeds. The analysis in terms of modes for dispersive systems is still valid. It's just that when you calculate f = v/λ for a particular λ, you have to consider the fact that v isn't a constant.

Nonlinear Systems Third, we relied on the superposition principle to combine modes simply by adding them up. More fundamentally, we implicitly assumed that if we double the amplitude of a mode, the frequency won't change. Such systems are called “linear” systems (because the amplitude of the superposition of two waves of the same mode is just the same mode, with an amplitude that is the sum of the amplitudes of the two waves). However, many real systems are only approximately linear in this sense. In “nonlinear” systems, the superposition of two waves of the same frequency may lead to a wave with a different frequency. This can even happen with a string. The speed of propagation v for a string turns out to be proportional to the square root of the tension in the string. For small amplitudes, the string is not stretched very much more than it was at rest, and the tension is not changed very much. But for large amplitudes, the string may be stretched quite a bit more than it is when at rest, and that means a change in v, which means a change in frequency for the same wavelength. Hence two medium-sized waves might add up to a large wave with a different frequency.

Confined Waves and Quantum Mechanics We have been considering quantization in classical (non-quantum) systems. We found that standing waves in confined regions couldn't have just any old wavelength, because the possible wavelengths are quantized by the constraints of the boundaries. There is an analogous situation in the quantum world of atoms, due to the fact that particles such as electrons can have wavelike properties. For example, we found that electron diffraction demonstrates the startling fact that electrons have wave properties as well as particle properties. This implies that there could be three-dimensional “standing waves” of an electron in a hydrogen atom, because the electron is confined to a region near the proton, due to the electric attraction. A hydrogen atom does indeed have quantized states analogous to standing-wave modes. Each of these states has a specific energy—the energy is quantized. A hydrogen atom can drop from a higher-energy to a lower-energy state with the emission of energy in the form of light, and a study of the light emitted by atoms lets us determine the quantized energies of atomic states.

*Fourier Analysis How did we figure out what amount of each of the ten modes to superimpose to get the complicated behavior shown in Section 25.6? In principle we could have done it by trial and error—just try various combinations of modes until we get the shape we want. Though,

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you can easily imagine, it would be very tedious to try to get this right merely by trial and error. We'll outline the general scheme for doing this, then apply it to our specific case. We want to express the initial shape of the string, f (x), as a sum of appropriate amounts of many modes, which are sine functions with wavelengths 2L, 2L/2, 2L/3, and so on:

We need to adjust the coefficients A1, A 2, A 3, and so on in order to choose the appropriate superposition of modes that is equal to f (x). In order to figure out what the nth coefficient should be, consider the following integral:

It can be shown by using trig identities that the following is true:

Therefore the integral picks out just the An term:

Finally, this means that we can calculate the nth mode coefficient by the following formula:

This scheme is called “Fourier analysis.” It plays an enormously important role in many branches of science and engineering.

*A Specific Example In Section 25.6 we wanted to analyze a “square wave” as a sum of modes (Figure 25.53). This function f (x) has the value +1 for 0 < x < L/2, and 0 for L/2 < x < L, so we have

Figure 25.53 We will express this square wave in terms of a sum of sinusoids.

Dropping the overall factor of 2/π (which just sets the overall scale of how big the oscillation is), we get the following set of mode coefficients: A 1 = 1, A 2 = 1, A 3 = 1/3, A 4 = 0, A 5 = 1/5, A 6 = 1/3, A 7 = 1/7, A 8 = 0, A 9 = 1/9, A 10 = 1/5, and so on. Earlier in this section we used just these first ten modes, so we didn't mimic the desired square-wave function f (x) exactly right, but we were close. If we had used more modes, we would have come even closer to approximating the square wave. If we use an arbitrarily large number of modes we can come arbitrarily close to the desired function.

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This is actually a pretty extreme example, because this square wave has discontinuities that require very high-frequency modes in order to approximate the infinite slopes of the function. Functions that have less extreme behavior can often be approximated quite well with only a few modes.

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*DERIVATION: TWO SLITS ARE LIKE TWO SOURCES The proof that two illuminated slits act like two sources is an application of the superposition principle. Suppose that a laser illuminates an opaque sheet (Figure 25.54). According to the superposition principle, electric and magnetic fields produced by the laser must be present behind the sheet, despite the lack of light there. This means that the light from the laser accelerates electrons in the sheet in such a way that the accelerated electrons produce electric and magnetic fields behind the sheet that exactly cancel the fields produced by the laser.

Figure 25.54 Behind the opaque sheet, the net electric field is zero.

We can consider the sheet to consist of the material that will eventually be removed to make the slits (sections S1 and S2 of the sheet), plus the rest of the sheet (section R). The (zero) net electric and magnetic fields beyond the sheet can be considered to be produced by four sources: the laser and the accelerated electrons in sections S1, S2, and R of the sheet. The net electric field is

When we cut out the slits, the net electric and magnetic fields beyond the sheet are due just to the laser and to the accelerated electrons in section R of the sheet, (Figure 25.55).

Figure 25.55 With the slits cut out, the net field is due just to the laser and section R.

If we assume that the electron accelerations in section R are about the same in the two situations, whether or not the slits have been cut out, then the electric and magnetic fields produced by section R are about the same in both situations. Since , the electric field with the slits cut out is

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In calculating intensities we don't care about an overall sign, so instead of considering the difficult situation of a laser and a sheet with two slits cut out of it, we can consider the much simpler and already familiar situation of two sources, located where the slits actually are and producing a net electric field with no laser and no section R (Figure 25.56).

Figure 25.56 Beyond the slits, we can pretend that the field is due to just these two sources.

How good is the assumption that the electron accelerations in section R are about the same whether or not the slits have been cut out? This assumption may not work well very close to the slits, because the presence or absence of the material will affect the edge regions of section R, but we are interested in explaining the interference pattern on a distant screen, not close to the slits. The assumption also may not work very well if the slits are too small, because in that case their contributions are so small as to be comparable to the edge effects, but we typically deal with slits that are quite wide in the sense of having a width much larger than the wavelength of light. Predictions based on treating the two slits as though they were sources do give results in excellent agreement with observations. This derivation is due to Richard Feynman (The Feynman Lectures on Physics, by R. P. Feynman, R. B. Leighton, and M. Sands, Addison-Wesley 1964, page 31–10.

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SUMMARY Fundamental Principles Particles have wavelike properties. The momentum of a particle is related to its wavelength:

where h = 6.6 × 10−34 joule-second is Planck's constant. Light has both wavelike and particle-like properties. A photon is a particle of light with zero rest mass, with energy and momentum related to its wavelength:

Both photons and particles with nonzero rest mass such as electrons and neutrons can exhibit wavelike properties, such as interference. Properties of sinusoidal waves A wave propagating to the right: E is the amplitude T is the period f = 1/T is the frequency ω = 2π f is the angular frequency λ is the wavelength φ is a phase corresponding to choice of t = 0 and x = 0 v = f λ (v is the speed of propagation of the wave) Intensity I is proportional to (amplitude) 2 Interference Waves of the same frequency can interfere with each other. As a result of the superposition principle, some locations have unusually large amplitudes and intensities, whereas other locations have small or zero amplitudes and intensities. Two-slit interference may be observed even if photons go through the slits one at a time. The photoelectric effect A photon with energy equal to or greater than the binding energy of an electron and a metal (the “work function”) can eject an electron from a metal surface. Compton scattering A collision between a photon and a free electron results in a change of the wavelength of the photon, corresponding to the amount of energy gained by the electron in the collision.

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Standing waves Standing waves result from the interference of two traveling waves of the same frequency that are traveling in opposite directions. The standing waves have the same frequency and wavelength as the traveling waves. In a confined space standing waves are quantized—only certain standing-wave wavelengths are allowed. In the simple case of a taut string, the allowed wavelengths are 2L, 2L/2, 2L/3, and so on. The standing-wave motions are called “modes.” Arbitrarily complex confined waves can be built up in terms of superpositions of these simple modes. The steady state is reached when the energy losses are just equal to the input energy. Path difference and two-source interference Interference of two in-phase sources that are a distance d apart: Maximum intensity (4I0) where Δl = d sin θ is 0, λ, 2λ, 3λ, and so on. Minimum intensity (zero) where Δl = d sin θ is λ/2, 3λ/2, 5λ/2, and so on. No zero-intensity minimum is possible if d < λ/2. If the sources are not in phase, you need to take that phase difference into consideration in addition to the phase difference due to Δl. X-ray diffraction Condition for “reflection” is 2d sin θ = nλ, where n is an integer, d is the distance between adjacent crystal planes, and θ is the angle of the incoming x-rays to the surface. Polycrystalline powders produce rings around the beam. Compton scattering prediction

Thin films show interference effects. Examples are oil slicks and soap bubbles. Diffraction gratings Maxima at angles satisfying d sin θ = nλ, where d is the slit spacing. Angular resolution of devices

where W is the total width of the (illuminated portion) of the device. Single source of width W: angle to first diffraction minimum is θ ≈ λ/W. Scattering of light is affected by interference effects. From allowed wavelengths in a standing wave one can calculate the corresponding frequencies. In the simple case of a taut string, the allowed frequencies can be calculated from f = v/λ , and they are v/(2L), 2v/(2L), 3v/(2L), and so on. Fourier analysis A confined wave can be expressed as a sum of sinusoids:

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where

Derivation Two slits act like two sources.

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EXERCISES AND PROBLEMS Section 25.1 25.X.18 The height of a particular water wave is described by the function

Calculate the angular frequency ω, the frequency f, the period T, the wavelength λ, the speed and direction of propagation (the +x or the −x direction), and the amplitude of the wave. At x = 0 and t = 0, what is the height of the wave? 25.X.19 The relations between wavelength, frequency, period, and speed of propagation apply to all wave phenomena, not just electromagnetic radiation. Middle C on a piano has a fundamental frequency of about 256 hertz. What is the corresponding wavelength of the sound waves? (The speed of sound in air varies with temperature but is about 340 m/s at room temperature.) 25.X.20 Two audio speakers are side by side, 1 meter apart. They are connected to the same amplifier, which is producing a sine wave of 440 hertz (“concert A”). Calculate a direction in which you won't hear anything, and make a diagram showing the speakers and this direction. (The speed of sound in air is about 340 m/s.) 25.P.21 Coherent green light with a wavelength of 500 nm illuminates two narrow vertical slits a distance 0.12 mm apart. Bright green stripes are seen on a screen 2 meters away. (a) How far apart are these stripes, center-to-center? (b) Do they get farther apart or closer together if you move the slits closer together? (c) Do they get farther apart or closer together if you use violet light (wavelength = 400 nm)? 25.P.22 In Figure 25.57 an unpolarized sinusoidal electromagnetic wave with wavelength λ travels along the −x direction in a region where there are two short copper wires oriented along the z direction, a distance L= 2.5λ apart.

Figure 25.57 (a) You place a detector a long ways away, at an angle θ to the x axis. Despite being far outside the region of the unpolarized wave, you detect electromagnetic radiation, and it is polarized. Explain this phenomenon in detail, including the directions of the electric and magnetic fields that you detect. (b) Calculate an angle θ1 other than 0° where you will see maximum intensity, and calculate an angle θ2 where you will see zero intensity. Explain briefly.

Section 25.2 25.X.23 It is natural to say that “light bounces off mirrors,” but is there a physics principle that would account for such behavior? What really happens? Why isn't there light going in all directions, not just in the direction of the “reflection” angle?

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25.X.24 When x-rays with wavelength 0.4 × 10−10 m hit a crystal at an angle of 30° to the surface, a strong beam of x-rays is observed in the direction of the “reflection” angle. Assume that you know from other evidence that the spacing of the atomic layers parallel to this surface of the crystal is greater than 1 × 10−10 m and less than 1.8 × 10−10 m. What are possible values of the spacing between the layers? 25.P.25 Suppose that you have an x-ray source that produces a continuous spectrum of wavelengths, with the shortest wavelength being 0.3 × 10−10 m. You have a crystal whose atoms are known to be arranged in a cubic array with the distance between nearest neighbors equal to 1.2 × 10−10 m. (a) Design an arrangement of x-ray beam and crystal orientation that will give you a monochromatic beam of x-rays whose wavelength is 0.5 × 10−10 m. Explain your arrangement carefully and fully in a diagram. (b) If the shortest wavelength in the x-ray spectrum were 0.2 × 10−10 m instead of 0.3 × 10−10 m, explain why your beam would not be a pure single-wavelength beam. 25.P.26 The color magenta consists of a mixture of red (about 700 nm) and blue light (about 450 nm). If you see a magentacolored section of a soap bubble, about how thick is this section of the soap bubble? Assume that wavelengths in the material are shortened by a factor of 1.3.

Section 25.3 25.X.27 If the work function of a metal is 3.9 eV, what wavelength of light would you have to shine on the metal surface in order to eject an electron with negligible kinetic energy? 25.X.28 Summarize the different predictions of the wave and particle models of light regarding the photoelectric effect. What experimental observations support the particle model? 25.X.29 According to the wave model of light, what is the relationship between energy and intensity? According to the particle model of light? 25.X.30 In a collision between a photon and a stationary electron, why does the wavelength of the photon change as a result of the collision? Does it increase or decrease? 25.X.31 What is the energy of a photon whose wavelength is 690 nm? 25.X.32 If the work function of a metal is 3.4 eV, what would be the maximum wavelength of light required to eject an electron from the metal? 25.X.33 In electron diffraction, diffraction rings are produced when electrons go through polycrystalline material after being accelerated through an accelerating potential difference. What happens to the size of the diffraction rings when the accelerating potential difference is increased? 25.X.34 Electrons are accelerated through a potential difference of 1000 volts and strike a polycrystalline powder whose atomic layers are 1 × 10−10 m apart. Predict an angle at which you will see an electron-diffraction ring. 25.P.35 A 100-watt light bulb is placed in a fixture with a reflector that makes a spot of radius 20 cm. Calculate approximately the amplitude of the radiative electric field in the spot and the number of photons per second hitting the spot.

Section 25.4 25.X.36 A ray of violet light (wavelength 400 nm) hits perpendicular to a transmission diffraction grating that has 10,000 lines per centimeter. At what angles to the perpendicular are there bright violet rays, in addition to zero degrees? 25.X.37 At night you look through a transmission diffraction grating at a sodium-vapor lamp used for outdoor lighting, which emits nearly monochromatic light of wavelength 588 nanometers. The manufacturer of the grating states that it was ruled with 10,000 lines per centimeter. In addition to zero degrees, at what other angles will you observe bright light? 25.X.38 Explain why a beam of light can go straight through a rather long tank of clear water or a long rod of clear glass, with hardly any light emitted to the side despite the huge number of atoms whose electrons are accelerated by the incoming light. 25.P.39 At night you look down into an outdoor swimming pool that has a powerful underwater light whose horizontal beam heads to your right. The water is not completely clear, and there is significant scattered light. You have a polarizing film with a line drawn on it showing the direction of electric field that is passed through with little loss. To minimize the brightness of the scattered light, should you hold the film with the line in the direction of the beam (left/right) or perpendicular to the

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beam (up/down)? Explain briefly, including a diagram.

Section 25.5 25.X.40 A satellite in Earth orbit carries a camera. Explain why the diameter of the camera's lens determines how small an object can be resolved in a photo taken by the camera. 25.X.41 Red laser light with wavelength 630 nm goes through a single slit whose width is 0.05 mm. What is the width of the image of the slit seen on a screen 5 m fromthe slit? 25.P.42 A single vertical slit 0.01 mm wide is illuminated by red light of wavelength 700 nm. (a) About how wide is the bright stripe on a screen 2 meters away? (b) Does the stripe get wider or narrower if you make the slit narrower? (c) Does the stripe get wider or narrower if you use light? 25.P.43 Consider a communications satellite that orbits the Earth at a height of about 40,000 kilometers. At this height the orbit period is 24 hours, so that it seems to hang motionless above the Earth, which turns underneath the satellite once every 24 hours. This is called a “synchronous” satellite. The advantage of such an arrangement is that receiving antennas on the ground can point at a seemingly fixed location of the satellite in the sky and not have to be steered. You have probably seen such fixed receiving antennas pointed at the sky. If the broadcasting antenna on the satellite is 2 meters in diameter and broadcasts at a frequency of 10 gigahertz (f = 10 × 109 hertz), approximately what is the diameter of the region on the ground where the satellite transmission can be picked up?

Section 25.6 25.X.44 How do standing waves differ from traveling waves? How are they similar? 25.X.45 The lowest-frequency mode of the lowest (C) string on a cello is 64 hertz. The length of the string between its supports is 70 cm. (a) What is the speed of propagation of traveling waves on this string? (b) Calculate the frequencies and wavelengths of the next 4 modes (not including the 64 hertz), and make a sketch of the standing-wave shape for each mode. 25.X.46 When a cello string is bowed, many different modes are excited, not just the lowest-frequency mode, which contributes to the richness of the sound. (a) When the bow is taken away the string continues to vibrate. After some time, only the lowest-frequency mode remains in motion. Why? (b) When a cellist bows while lightly touching the midpoint of the string, without pressing the finger down on the string, the lowest-frequency mode and all modes with odd multiples of that frequency are absent. This changes the tone markedly and is used to achieve a special musical effect. Explain why the odd harmonics are missing.

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Matter & Interactions EXECUTIVE EDITOR Stuart Johnson ASSISTANT EDITOR Aly Rentrop ASSOCIATE MARKETING DIRECTOR Christine Kushner SENIOR PRODUCTION EDITOR Elizabeth Swain ART DIRECTOR Jeof Vita TEXT DESIGN Laura Ierardi SENIOR MEDIA EDITOR Thomas Kulesa SENIOR ILLUSTRATION EDITORS Sigmund Malinowski and Anna Melhorn COVER IMAGE Ruth Chabay Cover Description

This book was set in 10/12 Times Ten Roman in LaTex by Aptara ® , Inc. and printed and bound by Courier Kendallville. The cover was printed by Courier Kendallville. This book is printed on acid-free paper. Copyright © 2011, 2007, 2002 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. Library of Congress Cataloging-in-Publication Data Chabay, Ruth W.

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Matter & interactions / Ruth W. Chabay, Bruce A. Sherwood.—3rd ed. p. cm. Includes index. Volume 1 ISBN 978-0-470-50345-4 (pbk.) Volume 2 ISBN 978-0-470-50346-1 (pbk.) Complete ISBN 978-0-470-50347-8 (cloth) 1. Physics—Textbooks. 2. Mechanics—Textbooks. I. Sherwood, Bruce A. II. Title. III. Title: Matter and interactions. QC23.2.C43 2011 530—dc22

2009034010 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

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Preface This textbook emphasizes a 20th-century perspective on introductory physics with the goal of involving students in the contemporary physics enterprise: Apply fundamental principles to a wide range of systems (from nuclei to stars; unify mechanics and thermal physics, and electrostatics and circuits) Integrate contemporary physics throughout the curriculum (atomic models of matter, quantized energy, relativistic dynamics) Engage students in physical modeling (idealization, approximation, assumptions, estimation) Integrate computational physics (now a partner of theory and experiment)

Modern Mechanics (volume 1, Chapters 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13) focuses on the atomic structure of matter and interactions between material objects and emphasizes the wide applicability of a small number of fundamental principles: the Momentum Principle, the Energy Principle, and the Angular Momentum Principle. We can use these principles to explain and predict the behavior of systems as different as molecules and galaxies. Electric & Magnetic Interactions (volume 2, Chapters 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25) emphasizes the concepts of electric and magnetic fields and extends the study of the atomic structure of matter to include the role of electrons. The principles of electricity and magnetism are the foundation for much of today's technology, from cell phones to medical imaging.

Prerequisites This book is intended for introductory calculus-based college physics courses taken by science and engineering students. You will need a basic knowledge of derivatives and integrals, which can be obtained by studying calculus concurrently.

Modeling Matter & Interactions places a major emphasis on constructing and using physical models. A central aspect of science is the modeling of complex real-world phenomena. A physical model is based on what we believe to be fundamental principles; its intent is to predict or explain the most important aspects of an actual situation. Modeling necessarily involves making approximations and simplifying assumptions in order that the model can be analyzed in detail.

Computational Modeling Computational modeling is now as important as theory and experiment in contemporary science and engineering. We introduce you to serious computer modeling right away to help you build a strong foundation in the use of this important tool. Computational modeling allows us to analyze complex systems that would otherwise require very sophisticated mathematics or that

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could not be analyzed at all without a computer. Numerical calculations based on the Momentum Principle give us the opportunity to watch the dynamical evolution of the behavior of a system. Simple models frequently need to be refined and extended. This can be done straightforwardly with a computer model but is often impossible with a purely analytical (non-numerical) model.

Questions As you read the text, you will frequently come to a question that invites you to stop and think, to make a prediction, to carry out a step in a derivation or analysis, or to apply a principle. These questions are answered in the following paragraphs, but it is important that you make a serious effort to answer the questions on your own before reading further. Be honest in comparing your answers to those in the text. Paying attention to surprising or counterintuitive results can be a useful learning strategy.

Exercises Small exercises that require you to apply new concepts are found at the end of many sections of the text. These may involve qualitative reasoning or simple calculations. You should work these exercises when you come to them, to consolidate your understanding of the material you have just read. Answers to these exercises are found at the end of each chapter.

Conventions Used in Diagrams The conventions most commonly used to represent vectors and scalars in diagrams in this text are shown in the margin. In equations and text, a vector will be written with an arrow above it: .

TO THE INSTRUCTOR The approach to introductory physics in this textbook differs significantly from that in most textbooks. Key emphases of the

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approach are these: Starting from fundamental principles rather than secondary formulas Atomic-level description and analysis Modeling the real world through idealizations and approximations Computational modeling of physical systems Unification of mechanics and thermal physics Unification of electrostatics and circuits The use of 3D vectors throughout

Instructor Web Site Much useful information may be found on the Matter & Interactions public web site: http://www.matterandinteractions.org

This public web site includes information of a general character, including articles about M&I and useful educational software. Also at this public web site are links to the Wiley web site for instructors, where you will find useful resources that include the following: Lecture notes Videos of lectures, and videos that are useful to show during lectures Additional lecture demo software Figures from the textbook in jpeg format Sample course calendar Test questions Clicker questions Lab activities including experiments, computational modeling, and whiteboard problems for group problem solving

Articles Dealing with Matter & Interactions Five key articles dealing with this curriculum are the following: Chabay, R. & Sherwood, B. (1999). Bringing atoms into first-year physics. American Journal of Physics 67, 1045–1050. Chabay, R. W. & Sherwood, B. (2004). Modern mechanics. American Journal of Physics 72, 439–445. Chabay, R. W. & Sherwood, B. (2006). Restructuring Introductory E&M. American Journal of Physics 74, 329–336. Chabay, R. & Sherwood, B. (2008) Computational physics in the introductory calculus-based course. American Journal of Physics 76(4&5), 307–313. Chabay, R. & Sherwood, B. (2007) Matter & Interactions. A chapter in the on-line book Reviews in PER Volume 1: ResearchBased Reform of University Physics, E. F. Redish & P. J. Cooney, editors, http://www.compadre.org/per/per reviews/volume1.cfm.

These articles are available on the Matter & Interactions web site. The last article includes implementation aspects, evaluation

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information, and so on.

Computational Homework Problems Some important homework problems require the student to write a computer program. We strongly recommend VPython as the best tool for student use. An adequate subset of the underlying Python programming language can be taught in an hour or two, even to students who have never written a program before, and instructional materials are available to instructors. Real-time 3D animations are generated as a side effect of student computations, and these animations provide powerfully motivating and instructive visualizations of fields and motions. VPython supports true vector computations, which encourages students to begin thinking about vectors as much more than mere components. VPython can be obtained at no cost for Windows, Macintosh, and Linux at http://vpython.org.

Homework Problems In addition to the exercises and problems included at the end of a chapter, there are many exercises distributed throughout the chapter. We have prepared versions of the textbook exercises and problems for the WebAssign computer homework system (http://www.webassign.net).

Desktop Experiment Kit On the public web site indicated above is information on a desktop experiment kit for E&M that is distributed by PASCO that makes it possible for students to make key observations of electrostatic, circuit, and magnetic phenomena in the classroom, tightly integrated with the theory (http://www.pasco.com, search for EM-8675). Several chapters contain optional experiments that can be done with this kit. This does not preclude having other, more complex laboratory experiences associated with the curriculum. For example, one such lab that we use deals with Faraday's law and requires signal generators, large coils, and oscilloscopes. You may have lab experiments already in place that will go well with this textbook.

What's New in the 3rd Edition The 3rd edition of this text includes the following new features: Full color, which helps clarify many diagrams and allows examples to stand out more clearly. A larger set of exercises and problems at the end of each chapter. More numerical examples and problems to complement symbolic ones. Improvements in the treatments of computational modeling, angular momentum, the effect of electric fields on matter, and electromagnetic radiation. A section on geometric optics in Chapter 24.

Details of Changes from the 2nd Edition

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The iterative approach to predicting motion, which is the foundation for computational modeling, has been moved from Chapter 3 to Chapter 2 in recognition of its central importance in modern mechanics. This makes it possible to practice with the iterative approach in the context of simple force laws before applying it to more complicated 1/r 2 vector force laws in Chapter 3. Chapter 3 has been renamed “Fundamental Interactions” and includes examples of all four fundamental interactions: gravitational, electromagnetic, weak, and strong. What had been a very long Chapter 4 has been split into Chapter 4 on the ball-and-spring model of solids and spring-mass oscillations and Chapter 5 on statics and dynamics free-body problems, including dynamics problems involving curving motion. Included in the new Chapter 5 is extended and more consistent instruction on statics and dynamics free-body problems. The Momentum Principle in the form is still primary, but more consistent connections are made to acceleration. In the first chapter on energy (Chapter 6) potential energy is introduced in a more concrete way in the context of a particular force law (gravitation). The second energy chapter (Chapter 7) deals a bit more extensively with the full range of types of energy inputs, not just work and thermal energy transfer but also radiative transfer, mass transfer, sound, and electrical inputs. It also introduces explicitly the concept of a steady state, where there are several energy transfers but no change in the energy of the system. In Chapter 8 on quantized energy there is explicit mention of the existence of selection rules, which in this introductory treatment are not actually taken into account. Chapter 9 on multiparticle systems now introduces moment of inertia and

. There is an example of using integral

calculus to find the moment of inertia for a rod. Chapter 11 on angular momentum builds on the concept of moment of inertia and rotational kinetic energy which are now introduced in Chapter 9. There are many more numerical examples to make the concepts more concrete. There is a basic treatment of rotational kinematics. Chapter 12 on quantum statistical mechanics now uses k B instead of k for the Boltzmann constant. Chapter 15 on the effects of electric field on matter has been reorganized and strengthened, and v = uE is introduced to help students understand with a deeper sense of mechanism why E = 0 in a conductor in equilibrium. The case study on sparks has been moved to Chapter 21 on magnetic force and has been rewritten to use the concept of potential and to capitalize on the discussion of Jack and Jill and Einstein in discussing the extended muon lifetime. Chapter 17 on electric potential includes an explicit procedure for computing potential difference from a known electric field. Chapter 22 now contains the discussion of p-n junctions, formerly in a separate chapter, as a case study in the use of Gauss's law. Chapter 23 on Faraday's law is now careful to say that a curly electric field and a time-varying magnetic field accompany each other, rather than asserting that a time-varying magnetic field “causes” a curly electric field. Chapter 24 on electromagnetic radiation has been reorganized for greater clarity and includes sections on geometrical optics. There is now an optional section that derives the equation for the radiative field of an accelerated charge, based on a derivation by Purcell.

What Can Be Omitted? In a large course for engineering and science students with three 50-minute lectures and one 110-minute small-group studio lab per week, or in a studio format with five 50-minute sessions per week, it is possible to complete most but not all of the mechanics and E&M material in two 15-week semesters. In an honors course, or a course for physics majors, it is possible to do almost everything. You may be able to go further or deeper if your course has a weekly recitation session in addition to lecture and lab.

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What can be omitted if there is not enough time to do everything? In mechanics, the one thing we feel should not be omitted is the introduction to entropy in terms of quantum statistical mechanics (Chapter 12). This is a climax of the theme of integrating mechanics and thermal physics. One way to decide what mechanics topics can be omitted is to be guided by what foundation is required for Chapter 12. See detailed suggestions below. In E&M, one should not omit electromagnetic radiation and its effects on matter (Chapter 24). This is the climax of the whole E&M enterprise. One way to decide what E&M topics can be omitted is to be guided by what foundation is required for Chapter 24. See detailed suggestions below. Any starred section (*) can safely be omitted. Material in these sections is not referenced in later work. Chapter 1 (Interactions and Motion): Unless your students have a particularly strong background in vectors, all of this chapter is important for later work (except for starred sections). Chapter 2 (The Momentum Principle): None of the unstarred sections of this foundational chapter should be omitted. Chapter 3 (The Fundamental Interactions): The section on determinism may be omitted. Chapter 4 (Contact Interactions): Buoyancy and pressure may be omitted (one can return to these topics during Chapter 13 on gases). Chapter 5 (Rate of Change of Momentum): All of this chapter is important for later work. Chapter 6 (The Energy Principle): This is the foundation for the energy concept, and nothing should be omitted (other than starred sections, of course). Chapter 7 (Internal Energy): If you are pressed for time, you might choose to omit the second half of the chapter on energy dissipation, beginning with Section 7.10. Chapter 8 (Energy Quantization): None of the unstarred sections of this short chapter should be omitted. Chapter 9 (Multiparticle Systems): The formalism of finding the center of mass may be skipped, because the important applications have obvious locations of the center of mass. Chapter 10 (Collisions): A good candidate for omission is the analysis of collisions in the center-of-mass frame. Since there is a basic introduction to collisions in Chapter 3 (before energy is introduced), one could omit all of Chapter 10. On the other hand, the combined use of the Momentum Principle and the Energy Principle can illuminate both fundamental principles. Chapter 11 (Angular Momentum): The main content of this chapter should not be omitted, as it introduces the third fundamental principle of mechanics, the Angular Momentum Principle. One might choose to omit most applications involving nonzero torque. Chapter 12 (Entropy: Limits on the Possible): It is strongly recommend to do at least the first half of this chapter, on the Einstein solid. To make this work well, exposition and student computing should be interwoven. It might seem odd to ask the students simply to reproduce the results shown in the textbook, but in our experience and that of our students it is enormously educational to be required to make the ideas concrete in a computer program. You find yourself asking “What exactly is q1?” and answering very concretely by actually calculating with the initially unfamiliar quantities. It is the difference between writing an essay about entropy and calculating the actual value of the entropy. Despite its scientific importance, one might omit the second half of the chapter, on the Boltzmann distribution. However, if you cover Chapter 13 you will need a few key results from this analysis. Chapter 13 (Gases and Engines): This entire chapter may be omitted. If you include this chapter, you should not omit the initial sections, including the relationship of pressure to the results from Chapter 12. The important formula for particle flow and the concept of mean free path are both used in E&M, though they are rederived there. You might choose to omit the sections on macroscopic energy transfers (isothermal and adiabatic processes). The sections on heat engines represent additional applications of the second law of thermodynamics and may be omitted. Another possibility is to omit just the sections on nonzeropower engines. Chapter 14 (Electric Field): The field concept is the backbone of the treatment of E&M, so none of this brief chapter should be

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omitted. Electric dipoles, which are introduced here, are central to what follows. Chapter 15 (Electric Fields and Matter): Nothing in this foundational chapter should be omitted. Chapter 16 (Electric Field of Distributed Charges): It is important that students acquire a good working knowledge of the patterns of electric field around some standard charged objects (rod, ring, disk, capacitor, sphere). If however they themselves are to acquire significant expertise in setting up physical integrals, they need extensive practice, and you might decide that the amount of time necessary for acquiring this expertise is not an appropriate use of the available course time. Chapter 17 (Electric Potential): The section on dielectric constant can be omitted if necessary, as can the section on the potential of distributed charges. Chapter 18 (Magnetic Field): In the sections on the atomic structure of magnets, you might choose to discuss only the first part, in which one finds that the magnetic moment of a bar magnet is consistent with an atomic model. Omitting the remaining sections on spin and domains will not cause significant difficulties later. Chapter 19 (Electric Field and Circuits): It is best to do everything because it solidifies the field concept. However, you may choose not to do all of the final applications. Chapter 20 (Circuit Elements): The discussion of field in capacitor circuits helps significantly in solidifying the students' understanding of electric field, and offers an opportunity to think about transient phenomena on an ordinary time scale. The derivation of resistance, and the connection between macro and micro circuit analysis, is useful to most students of science and engineering. The sections on series and parallel resistors and on internal resistance, meters, quantitative analysis of RC circuits, and multiloop circuits can be omitted. Physics and engineering students who need to analyze complex multiloop circuits will later take specialized courses on the topic; in the introductory physics course the emphasis should be on giving all students a good grounding in the fundamental mechanisms underlying circuit behavior. Chapter 21 (Magnetic Force): We recommend discussing Jack and Jill and Einstein, but it is safe to omit the sections on relativistic field transformations. However, students often express high interest in the relationship between electric fields and magnetic fields, and here is an opportunity to satisfy some aspects of their curiosity. Motors and generators may be omitted or downplayed. The case study on sparks in air can be omitted, because nothing later depends critically on this topic, though it provides an introductory-level example of a phenomenon where an intuitively appealing model fails utterly, while a different model predicts several key features of the phenomenon. Another possibility is to discuss sparks near the end of the course, because it can be a useful review of many aspects of E&M. Chapter 22 (Patterns of Field in Space): Gauss's law and Ampere's law are important foundations for the following chapters on Faraday's law and electromagnetic radiation. It would in principle be possible to do Gauss's law just after Chapter 16, and Ampere's law just after Chapter 18, but it is not a good idea. Students need extensive experience with patterns of field in space before they can make sense of these topologically complex relations. Moreover, studying Gauss's law just before Faraday's law makes the concept of flux fresh and unproblematic in Chapter 23. Chapter 23 (Faraday's Law): Though it can safely be omitted, we recommend retaining the section on superconductors, because students are curious about this topic. The section on inductance can also be omitted. Chapter 24 (Electromagnetic Radiation): This is the climax toward which much of the textbook has been heading. We strongly recommend omitting nothing on radiation. However, you may wish to omit the treatment of geometrical optics. Chapter 25 (Waves and Particles): It may not be possible to get to this chapter at all, and its inclusion is optional. It contrasts the classical wave model of light developed in Chapter 24 with the particle model that was used in mechanics, and it attempts to show the relationship between these two very different models. It also develops key concepts in physical optics. Students who are interested in physics often find these sections interesting.

Acknowledgments

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We owe much to the unusual working environment provided by the Department of Physics and the former Center for Innovation in Learning at Carnegie Mellon, which made it possible during the 1990s to carry out the research and development leading to the first edition of this textbook in 2002. We are grateful for the open-minded attitude of our colleagues in the Carnegie Mellon physics department toward curriculum innovations. We are grateful to the support of our colleagues Robert Beichner and John Risley in the Physics Education Research and Development group at North Carolina State University, and to other colleagues in the NCSU physics department. We thank Fred Reif for emphasizing the role of the three fundamental principles of mechanics, and for his view on the reciprocity of electric and gravitational forces. We thank Robert Bauman, Gregg Franklin, and Curtis Meyer for helping us think deeply about energy. Much of Chapter 12 on quantum statistical mechanics is based on an article by Thomas A. Moore and Daniel V. Schroeder, “A different approach to introducing statistical mechanics,” American Journal of Physics vol. 65, pp. 26–36 (January 1997). We have benefited from many stimulating conversations with Thomas Moore, author of another introductory textbook that takes a contemporary view of physics, “Six Ideas that Shaped Physics.” Michael Weissman and Robert Swendsen provided particularly helpful critiques on some aspects of our implementation of Chapter 12. We thank Hermann Haertel for opening our eyes to the fundamental mechanisms of electric circuits. Robert Morse, Priscilla Laws, and Mel Steinberg stimulated our thinking about desktop experiments. Bat-Sheva Eylon offered important guidance at an early stage. Ray Sorensen provided deep analytical critiques that influenced our thinking in several important areas. Randall Feenstra taught us about semiconductor junctions. Thomas Moore showed us a useful way to present the differential form of Maxwell's equations. Fred Reif helped us devise an assessment of student learning of basic E&M concepts. Uri Ganiel suggested the high-voltage circuit used to demonstrate the reality of surface charge. The unusual light bulb circuits at the end of Chapter 23 are based on an article by P. C. Peters, “The role of induced emf's in simple circuits,” American Journal of Physics 52, 1984, 208-211. Thomas Ferguson gave us unusually detailed and useful feedback on the E&M chapters. Discussions with John Jewett about energy transfers were helpful. We thank David Andersen, David Scherer, and Jonathan Brandmeyer for the development of tools that enabled us and our students to write associated software. We thank our colleagues David Brown, Krishna Chowdary, Laura Clarke, Norman Derby, Thomas Foster, Chris Gould, Mark Haugan, Joe Heafner, Robert Hilborn, Eric Hill, Andrew Hirsch, Leonardo Hsu, Matthew Kohlmyer, Barry Luokkala, Sara Majetich, Jonathan Mitschele, Arjendu Pattanayak, Jeff Polak, Prabha Ramakrishnan, Vidhya Ramachandran, Michael Schatz, Robert Swendsen, Aaron Titus, Michael Weissman, and Hugh Young. We thank a group of reviewers assembled by the publisher, who gave us useful critiques on the second edition of this volume: Kelvin Chu, Michael Dubson, Tom Furtak, David Goldberg, Javed Iqbal, Shawn Jackson, Craig Ogilvie, Michael Politano, Norris Preyer, Rex Ramsier, Tycho Sleator, Robert Swendsen, Larry Weinstein, and Michael Weissman. We have benefited greatly from the support and advice of Stuart Johnson of John Wiley & Sons. Elizabeth Swain of John Wiley & Sons was exceptionally skilled in managing the 3rd edition project. Helen Walden did a superb job of careful copyediting; any remaining errors are ours. This project was supported, in part, by the National Science Foundation (grants MDR-8953367, USE-9156105, DUE-9554843, DUE9972420, DUE-0320608, DUE-0237132, and DUE-0618504). We are grateful to the National Science Foundation and its reviewers for the long-term support of this challenging twenty-year project. Opinions expressed are those of the authors, and not necessarily those of the Foundation.

Ruth Chabay

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Bruce Sherwood Department of Physics North Carolina State University July 2009

Copyright © 2011 John Wiley & Sons, Inc. All rights reserved.

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