Instructor’s Solution Manual Introduction to Electrodynamics Fourth Edition David J. Griffiths 2012
2
Contents 1 Vector Analysis
4
2 Electrostatics
26
3 Potential
53
4 Electric Fields in Matter
92
5 Magnetostatics
110
6 Magnetic Fields in Matter
133
7 Electrodynamics
145
8 Conservation Laws
168
9 Electromagnetic Waves
185
10 Potentials and Fields
210
11 Radiation
231
12 Electrodynamics and Relativity
261
c
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3
Preface Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I would like to thank him particularly for all his help. If you find errors, please let me know (
[email protected]).
David Griffiths
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3
CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 4
3 CHAPTER 1. VECTOR ANALYSIS 3
CHAPTER 1. VECTOR ANALYSIS
Chapter 1 Chapter 1 Chapter 1
1 VectorChapter Analysis
Vector Analysis Vector Analysis Vector Analysis Problem 1.1 " $
Problem 1.1
}} }
" $
C sin θ2 C
C
+
C
C
" $
B
(a) From the diagram, B1.1 + C cos θ3 = B cos θ1 + C cos θ2 . Multiply by A. Problem Problem 1.1 AB + C cos θ(a) AB θ1 + AC cos θ2θ.3 = B cos θ1 + C cos θ2 . thecos diagram, B + C cos 3 =From From the diagram, B + C cos θ= B cos θ1θ ++C cos θ2cos . Multiply by A. 3 = AB + C cos θ AB cos θ2θ.2 . So:(a)A·(B + C) = A·B + A·C. (Dot product (a) From the diagram, B B cosAC θ1 + C cos 3 + C cos θ3 = is 1 distributive)
C sin
B
B
C
+
C
+
AB + C cos θ3 + =A·(B AB θ=1 A·B + AC cos θ(Dot 2 . cosproduct C sin θ2 AB C cos+ θ3cos = AB cos+θ1A·C. + AC θ2 . So: C) is distributive) θ2 So: A·(B C) = A·B + A·C. (Dot product is Similarly: B ++C sin θ = B sin θ + C sin θ Mulitply by A n ˆ. 2 . distributive) So: A·(B3 + C) = A·B 1+ A·C. (Dot product is distributive) θ3 # θ2 θ3 = Bsin sinθθ1n + C sin θ2 . Mulitply by A n ˆB . AB + C sin θ3 n ˆSimilarly: = ABB sin+θ1Cn ˆ sin + AC θ3 # B sin θ1 2 ˆ. θ2 Similarly: B + C sin θ = B sin θ + C sin θ . Mulitply by A n ˆ . Similarly: B +AB C sin+ θC = B sin θ + C sin θ . Mulitply by A n ˆ . 3 1 2 3 sin θ3 n ˆ of =1 the AB sin θ2it ˆfollows + AC sin θ2 n ˆ. θ1 θ3 # B 1n B sin ! If n ˆ isAB the unit vector pointing out $B sin AB Cthe sin θunit ˆsin =vector AB sin θpage, ˆ +sin AC n ˆthat . θ1 3n 1n θ1 + C sin θ3 n ˆ+is = AB θ1 n ˆ +pointing AC θ2ofn ˆ .sin If ˆn out theθ2page, it follows that! "#θ1B $ !! "# A "# θ2 $ !! "# $ ! A×(B C)the = unit (A×B) (A×C). (Cross product is distributive) B cos θ1 $ ! C"# cos If n ˆ vector is the+unit vector out pointing out of theit page, it follows that A ! "# $ If + n ˆ is pointing of the page, follows that A×(B + C) = (A×B) + (A×C). (Cross product is distributive) B cos θ1 C A cos θ2 A×(B + C) = (A×B) + (A×C). (Cross product is distributive) B cos θ1 C cos θ2 A×(B + C)case, = (A×B) + E. (A×C). product is distributive) (b) For the general see G. Hay’s (Cross Vector and Tensor Analysis, Chapter 1, Section 71,(dot product) and general E. Hay’s Vector and Analysis, Tensor Analysis, Section 7 (dot (b)(b) ForFor the the general case,case, see G.see E. G. Hay’s Vector and Tensor Chapter 1,Chapter Section 7 (dot product) and product) Section (cross product) Section 8 (cross (b) For8 the general case, see G. E. product) Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product) Section 8 (cross product) Problem 1.2 Problem Problem 1.2 1.2 Problem 1.2 C CC % TheThe triple crossproduct is not isinassociative. general associative. For example, triple crossproduct not in general associative. For example, The triple crossproduct is not in general For example, % % The triple crossproduct is not in general associative. For example, suppose A = B and C is perpendicular to A, as in the diagram. suppose = B and C istoperpendicular to diagram. A, as in the diagram. suppose A = B and C isAperpendicular A, as in the ! outofthepage, andasA×(B×C) points down, suppose A Then = Then B (B×C) and(B×C) C points is perpendicular to A, in theA×(B×C) diagram. !AA==BB ! A = B points outofthepage, and points down, Then (B×C) points A×(B×C) down, andpoints hasoutofthepage, magnitude ABC. and But (A×B) = 0, sopoints (A×B)×C = 0 != & Then (B×C) outofthepage, and A×(B×C) points down, and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C 0 != & and has ABC. But (A×B) A×(B×C). & ' A×(B×C) and magnitude has magnitude ABC. But (A×B)==0,0,so so (A×B)×C (A×B)×C = 00 6=!= B×C ' A×(B×C). A×(B×C) B×C ' A×(B×C). A×(B×C) A×(B×C).
}
}
}
z
Problem 1.3 z%% Problem 1.3 1.3 Problem 1.3 Problem z% √ √ A = +1 x ˆ + 1√ y ˆ −√1 ˆ z; A = 3; B√= 1 x ˆ +1y ˆ + 1ˆ z√ ; A =x ˆy ˆ1Aˆ − ˆ z;=A 3; B ˆ y ˆ+√ + B =1y 3. !B 3; ˆB =1 = 1;x ˆ1Ax +=+1 1y ˆ3; ˆ zˆz;;1B 3.+ 1 ˆz; A = +1 ˆ +1 + 1x ˆ+−1 y z= ; 1A+1 ˆ1= + ˆ z B131√ = x ˆ= ˆcos 3 cos θ+⇒ θ. A·B = +1 +x − 11 y =− 1= AB cos√θ = √ B √ √ 11 √ √ ! θ ! !By & AB A·B + 11A·B − 11−1 == 1%+1 = cos θ=◦=1 3=3AB ..θ ⇒ cos θ. 3cos cos θθ⇒ ⇒ cos θ 3= =cos 3 1 − 1 cos = A·B = +1=++1 1− = = AB θ = 3 θ cos θ 1 +cos 3 θ θ θ = cos 3 3 ≈ 70.5288 !y " !y % & % & 1 θ= A ( cos◦−1◦ 31 ≈ 70.5288◦ θ = 70.5288 −1cos1−1 3 ≈ x " θ = cos " 3 ≈ 70.5288 A ( A ( x Problem 1.4 x Problem 1.4 The crossproduct of any two vectors in the plane will give a vector perpendicular to the plane. For example, Problem 1.4 Problem 1.4 we might pick the base (A) and the left side (B): The crossproduct of any two vectors in the plane will give a vector perpendicular to the plane. For example,
The crossproduct The of any two vectorsofin the plane willingive a vector to the plane. example, crossproduct any two vectors the plane willperpendicular give a vector perpendicular to For the plane. For exam we might pick the base (A) and the left side (B): c base we Pearson might pick base (A) and the NJ. leftAllside (B): #2005 Upper Saddle River, rights reserved. This material is we might pick the (A)Education, and the theInc., left side (B): protected under all copyright laws as they currently exist. No portion of this material may be
A = −1 x ˆreproduced, + 2y ˆ +in0 any ˆ z; B =or−1 ˆany+means, 0y ˆ +without 3ˆ z. permission in writing from the publisher. form by x
A = −1 x ˆ +2y ˆ#2005 + 0ˆ zPearson ; B = Education, −1 ˆ x + 0Inc., y ˆ +Upper 3ˆ z. Saddle River, NJ. All rights reserved. c
This material is protected under all copyright laws as they currently exist. No portion of this material may be c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is reproduced, in any form or by any means, without permission in writing from the publisher. protected under copyright laws as they currently exist. c #2005 Pearson Education, Inc., Upper Saddle River, NJ.all All rights reserved. This material is No portion of this material may be in anyNo form or by any means, without permission in writing from the publisher. protected under all copyright laws as theyreproduced, currently exist. portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 1. VECTOR ANALYSIS
5
x ˆˆ z ˆ y ˆ + 3y ˆ + 2ˆ z. A×B = −1 2 0 = 6 x −1 0 3 This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: √ A×B 6 A×B = 36 + 9 + 4 = 7. n ˆ = A ˆ + 73 y ˆ + 27 ˆ z. ×B = 7 x Problem 1.5 x ˆ y ˆ ˆ z Ax Ay Az A×(B×C) = (By Cz − Bz Cy ) (Bz Cx − Bx Cz ) (Bx Cy − By Cx ) =x ˆ[Ay (Bx Cy − By Cx ) − Az (Bz Cx − Bx Cz )] + y ˆ() + ˆ z() (I’ll just check the xcomponent; the others go the same way) =x ˆ(Ay Bx Cy − Ay By Cx − Az Bz Cx + Az Bx Cz ) + y ˆ() + ˆ z(). B(A·C) − C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) − Cx (Ax Bx + Ay By + Az Bz )] x ˆ + () y ˆ + () ˆ z =x ˆ(Ay Bx Cy + Az Bx Cz − Ay By Cx − Az Bz Cx ) + y ˆ() + ˆ z(). They agree. Problem 1.6 A×(B×C)+B×(C×A)+C×(A×B) = B(A·C)−C(A·B)+C(A·B)−A(C·B)+A(B·C)−B(C·A) = 0. So: A×(B×C) − (A×B)×C = −B×(C×A) = A(B·C) − C(A·B). If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.) Conclusion: A×(B×C) = (A×B)×C ⇐⇒ either A is parallel to C, or B is perpendicular to A and C. Problem 1.7
r
r
= (4 x ˆ + 6y ˆ + 8ˆ z) − (2 x ˆ + 8y ˆ + 7ˆ z) = 2 x ˆ − 2y ˆ+ ˆ z √ = 4+4+1= 3
rˆ
=
r r
=
2 ˆ 3x
− 23 y ˆ + 13 ˆ z
Problem 1.8 ¯y + A¯z B ¯z = (cos φAy + sin φAz )(cos φBy + sin φBz ) + (− sin φAy + cos φAz )(− sin φBy + cos φBz ) (a) A¯y B = cos2 φAy By + sin φ cos φ(Ay Bz + Az By ) + sin2 φAz Bz + sin2 φAy By − sin φ cos φ(Ay Bz + Az By ) + cos2 φAz Bz = (cos2 φ + sin2 φ)Ay By + (sin2 φ + cos2 φ)Az Bz = Ay By + Az Bz . X (b) (Ax )2 + (Ay )2 + (Az )2 = Σ3i=1 Ai Ai = Σ3i=1 Σ3j=1 Rij Aj Σ3k=1 Rik Ak = Σj,k (Σi Rij Rik ) Aj Ak . 1 if j = k This equals A2x + A2y + A2z provided Σ3i=1 Rij Rik = 0 if j 6= k Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also necessary. For suppose A = (1, 0, 0). Then Σj,k (Σi Rij Rik ) Aj Ak = Σi Ri1 Ri1 , and this must equal 1 (since we 2 2 2 want Ax +Ay +Az = 1). Likewise, Σ3i=1 Ri2 Ri2 = Σ3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0). Then we want 2 = Σj,k (Σi Rij Rik ) Aj Ak = Σi Ri1 Ri1 + Σi Ri2 Ri2 + Σi Ri1 Ri2 + Σi Ri2 Ri1 . But we already know that the first two sums are both 1; the third and fourth are equal, so Σi Ri1 Ri2 = Σi Ri2 Ri1 = 0, and so ˜ = 1, where R ˜ is the transpose of R. on for other unequal combinations of j, k. X In matrix notation: RR c
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6
CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 5 CHAPTER 1. VECTOR ANALYSIS 5 Problem 1.9 y" y z! " " y" y $ z! " " % % $ down the axis: ! x LookingLooking down the axis: * ( !x Looking down the axis: * ( ! ) &y ! )z ' & # y ' & z ! ' z x # x &x '! z x
A 120◦ rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az , Ay = Ax , Az = Ay . 0 01 R = 1 0 0 0 10 Problem 1.10 (a) No change. (Ax = Ax , Ay = Ay , Az = Az ) (b) A −→ −A, in the sense (Ax = −Ax , Ay = −Ay , Az = −Az ) (c) (A×B) −→ (−A)×(−B) = (A×B). That is, if C = A×B, C −→ C . No minus sign, in contrast to behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A×B) −→ (A)×(B) = (A×B). So the crossproduct of two pseudovectors is again a pseudovector. In the crossproduct of a vector and a pseudovector, one changes sign, the other doesn’t, and therefore the crossproduct is itself a vector. Angular momentum (L = r×p) and torque (N = r×F) are pseudovectors. (d) A·(B×C) −→ (−A)·((−B)×(−C)) = −A·(B×C). So, if a = A·(B×C), then a −→ −a; a pseudoscalar changes sign under inversion of coordinates. Problem 1.11 (a)∇f = 2x x ˆ + 3y 2 y ˆ + 4z 3 ˆ z (b)∇f = 2xy 3 z 4 x ˆ + 3x2 y 2 z 4 y ˆ + 4x2 y 3 z 3 ˆ z (c)∇f = ex sin y ln z x ˆ + ex cos y ln z y ˆ + ex sin y(1/z) ˆ z Problem 1.12 (a) ∇h = 10[(2y − 6x − 18) x ˆ + (2x − 8y + 28) ˆ]. ∇h = 0 at summit, so y 2y − 6x − 18 = 0 2y − 18 − 24y + 84 = 0. 2x − 8y + 28 = 0 =⇒ 6x − 24y + 84 = 0 22y = 66 =⇒ y = 3 =⇒ 2x − 24 + 28 = 0 =⇒ x = −2. Top is 3 miles north, 2 miles west, of South Hadley. (b) Putting in x = −2, y = 3: h = 10(−12 − 12 − 36 + 36 + 84 + 12) = 720 ft. (c) Putting in x = 1, y = 1: ∇h = 10[(2 − 6 − 18) x ˆ + (2 − 8 + 28) y ˆ] = 10(−22 x ˆ + 22 y ˆ) = 220(− x ˆ+y ˆ). √ ∇h = 220 2 ≈ 311 ft/mile; direction: northwest.
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be in any form or All by rights any means, without c "2005 Pearson Education, Inc.,reproduced, Upper Saddle River, NJ. reserved. Thispermission material isin writing from the publisher. c "2005 Pearson Education, Inc., Upper Saddle NJ. All rights reserved. This material is protected under all copyright laws River, as they currently exist. No portion of this material may be protected under all copyright laws as they exist. No portion of this may the be publisher. reproduced, in any form or bycurrently any means, without permission in material writing from reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 1. VECTOR ANALYSIS
7
Problem 1.13
r
r
= (x − x0 ) x ˆ + (y − y 0 ) y ˆ + (z − z 0 ) ˆ z;
=
p (x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 .
(a) ∇( r 2 ) =
∂ ∂ ∂ 0 2 0 2 0 2 ˆ + ∂y () y ˆ + ∂z () ˆ z ∂x [(x−x ) +(y −y ) +(z −z ) ] x
(b) ∇( r1 ) =
∂ ∂x [(x
3 − 21 ()− 2 2(x 3 −2
∂ ∂x (
r
1 − 32 2(y 2 () 0
0
1 − 23 2(z 2 ()
0
∂ − 21 ∂y () 0
y ˆ+
∂ − 12 ∂z ()
r
.
ˆ z
− x )x ˆ− − y )y ˆ− − z )ˆ z 0 0 3 [(x − x ) x ˆ + (y − y ) y ˆ + (z − z ) ˆ z] = −(1/ r ) r = −(1/ r 2 ) rˆ .
= = −() (c)
1
ˆ+ − x0 )2 + (y − y 0 )2 + (z − z 0 )2 ]− 2 x
= 2(x−x0 ) x ˆ +2(y −y 0 ) y ˆ +2(z −z 0 ) ˆ z=2
n
)=n
r
r
n−1 ∂ ∂x
=n
r
r 2r
n−1 1 1 (2
x)
=n
r
n−1
rˆ x , so
∇( r n ) = n
r
n−1
rˆ
Problem 1.14 y = +y cos φ + z sin φ; multiply by sin φ: y sin φ = +y sin φ cos φ + z sin2 φ. z = −y sin φ + z cos φ; multiply by cos φ: z cos φ = −y sin φ cos φ + z cos2 φ. Add: y sin φ + z cos φ = z(sin2 φ + cos2 φ) = z. Likewise, y cos φ − z sin φ = y. ∂y ∂z ∂z So ∂y = cos φ; ∂y ∂z = − sin φ; ∂y = sin φ; ∂z = cos φ. Therefore ) ∂f ∂y ∂f ∂z (∇f )y = ∂f ∂y = ∂y ∂y + ∂z ∂y = + cos φ(∇f )y + sin φ(∇f )z So ∇f transforms as a vector. ∂f ∂y ∂f ∂z (∇f )z = ∂f ∂z = ∂y ∂z + ∂z ∂z = − sin φ(∇f )y + cos φ(∇f )z
qed
Problem 1.15 (a)∇·va =
∂ 2 ∂x (x )
+
∂ 2 ∂y (3xz )
(b)∇·vb =
∂ ∂x (xy)
+
∂ ∂y (2yz)
(c)∇·vc =
∂ 2 ∂x (y )
+
∂ ∂y (2xy
+
+
∂ ∂z (−2xz)
∂ ∂z (3xz)
+ z2) +
= 2x + 0 − 2x = 0.
= y + 2z + 3x.
∂ ∂z (2yz)
= 0 + (2x) + (2y) = 2(x + y)
Problem 1.16 i h 3 ∂ ∂ ∂ ∂ ∇·v = ∂x ( rx3 ) + ∂y ( ry3 ) + ∂z ( rz3 ) = ∂x x(x2 + y 2 + z 2 )− 2 i i h h 3 3 ∂ ∂ + ∂y y(x2 + y 2 + z 2 )− 2 + ∂z z(x2 + y 2 + z 2 )− 2 3
5
3
5
3
= ()− 2 + x(−3/2)()− 2 2x + ()− 2 + y(−3/2)()− 2 2y + ()− 2 5 + z(−3/2)()− 2 2z = 3r−3 − 3r−5 (x2 + y 2 + z 2 ) = 3r−3 − 3r−3 = 0. This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin. How, then, can ∇·v = 0? The answer is that ∇·v = 0 everywhere except at the origin, but at the origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, ∇·v is infinite at that one point, and zero elsewhere, as we shall see in Sect. 1.5. Problem 1.17 v y = cos φ vy + sin φ vz ; v z = −sin φ vy + cos φ vz . ∂v y ∂vy ∂vy ∂y ∂vy ∂z ∂vz ∂vz ∂z z ∂y cos φ + ∂v sin φ. Use result in Prob. 1.14: ∂y = ∂y cos φ + ∂y sin φ = ∂y ∂y + ∂z ∂y ∂y ∂y + ∂z ∂y ∂vy ∂vy ∂vz ∂vz = ∂y cos φ + ∂z sin φ cos φ + ∂y cos φ + ∂z sin φ sin φ. ∂vy ∂vy ∂y ∂vy ∂z ∂v z ∂vz ∂vz ∂y ∂vz ∂z = − sin φ + cos φ = − + sin φ + + cos φ ∂z ∂z ∂y ∂z ∂z ∂z ∂z ∂y ∂z ∂z ∂z ∂v ∂v ∂vz z = − − ∂yy sin φ + ∂zy cos φ sin φ + − ∂v ∂y sin φ + ∂z cos φ cos φ. So c
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8 ∂v y ∂y
+
∂v z ∂z
=
∂vy ∂y
cos2 φ + =
Problem 1.18 x ˆ ∂ (a) ∇×va = ∂x x2 x ˆ ∂ (b) ∇×vb = ∂x xy x ˆ ∂ (c) ∇×vc = ∂x y2
∂vy ∂z
CHAPTER 1. VECTOR ANALYSIS
sin φ cos φ +
∂vz ∂y
sin φ cos φ + ∂vz ∂z
z − ∂v ∂y 2
∂vz ∂z
sin2 φ +
sin2 φ −
∂vy ∂z
sin φ cos φ
2
sin φ cos φ + cos φ ∂vz 2 cos φ + sin φ + ∂z sin2 φ + cos2 φ =
∂vy ∂y
∂vy ∂y
∂vy ∂y
+
∂vz ∂z .
X
∂ ∂ ˆ(0 − 6xz) + y ˆ(0 + 2z) + ˆ z(3z 2 − 0) = −6xz x ˆ + 2z y ˆ + 3z 2 ˆ z. ∂y ∂z = x 2 3xz −2xz y ˆ ˆ z ∂ ∂ ˆ(0 − 2y) + y ˆ(0 − 3z) + ˆ z(0 − x) = −2y x ˆ − 3z y ˆ − xˆ z. ∂y ∂z = x 2yz 3xz y ˆ ˆ z ∂ ∂ ˆ(2z − 2z) + y ˆ(0 − 0) + ˆ z(2y − 2y) = 0. ∂y ∂z = x 2 (2xy + z ) 2yz y ˆ
ˆ z
Problem 1.19
y v v
v
B
A
x z
As we go from point A to point B (9 o’clock to 10 o’clock), x increases, y increases, vx increases, and vy decreases, so ∂vx /∂y > 0, while ∂vy /∂y < 0. On the circle, vz = 0, and there is no dependence on z, so Eq. 1.41 says ∂vy ∂vx ∇×v =ˆ z − ∂x ∂y points in the negative z direction (into the page), as the right hand rule would suggest. (Pick any other nearby points on the circle and you will come to the same conclusion.) [I’m sorry, but I cannot remember who suggested this cute illustration.]
v Problem 1.20
v = yx ˆ + xy ˆ; or v = yz x ˆ + xz y ˆ + xy ˆ z; or v = (3x2 z − z 3 ) x ˆ + 3y ˆ + (x3 − 3xz 2 ) ˆ z; or v = (sin x)(cosh y) x ˆ − (cos x)(sinh y) y ˆ; etc. Problem 1.21 (i) ∇(f g) =
∂(f g) ∂x
=f
x ˆ+
∂g ∂x
(iv) ∇·(A×B) = = =
x ˆ+ ∂ ∂x
∂g ∂g ∂f ∂g ∂f ˆ z = f ∂x + g ∂f x ˆ + f + g y ˆ + f + g ˆ z ∂x ∂y ∂z ∂z ∂y y ˆ + ∂g z + g ∂f ˆ + ∂f ˆ + ∂f z = f (∇g) + g(∇f ). qed ∂z ˆ ∂x x ∂y y ∂z ˆ
∂(f g) ∂y ∂g ∂y
∂(f g) ∂z
∂ ∂ ∂y (Az Bx − Ax Bz ) + ∂z (Ax By − Ay Bx ) ∂A ∂B ∂Az ∂Bx ∂Az ∂Bz ∂Ax y y z Ay ∂B ∂x + Bz ∂x − Az ∂x − By ∂x + Az ∂y + Bx ∂y − Ax ∂y − Bz ∂y ∂By ∂A ∂Bx y x +Ax ∂z + By ∂A ∂z − Ay ∂z − Bx ∂z ∂A ∂A ∂By ∂Az y x z z x Bx ∂A + By ∂A + Bz ∂xy − ∂A − Ax ∂B ∂y − ∂z ∂z − ∂x ∂y ∂y − ∂z
(Ay Bz − Az By ) +
∂B z x − ∂B − Az ∂xy − ∂B = B· (∇×A) − A· (∇×B). ∂x ∂y ∂(f A ) ∂(f Ay ) Az ) ∂(f Ax ) − ∂z y x ˆ + ∂(f∂zAx ) − ∂(f∂x y ˆ+ − ˆ z ∂x ∂y
−Ay (v) ∇× (f A) =
y ˆ+
∂(f Az ) ∂y
∂Bx ∂z
qed
c
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CHAPTER 1. VECTOR ANALYSIS
9
∂Ay ∂f ∂f ∂f ∂f ∂Az x z ˆ + f ∂A ˆ = f ∂A ∂y + Az ∂y − f ∂z − Ay ∂z x ∂z + Ax ∂z − f ∂x − Az ∂x y ∂Ay ∂f ∂f ∂Ax + f ∂x + Ay ∂x − f ∂y − Ax ∂y ˆ z h i ∂Ay ∂Ay ∂Ax ∂Az ∂Ax z = f ∂A x ˆ + ˆ z − − y ˆ + − ∂y ∂z ∂x h ∂z ∂x ∂y i ∂f ∂f ∂f ∂f ∂f − Ay ∂f − A x ˆ + A − A y ˆ + A − A z z x z x y ∂z ∂y ∂x ∂z ∂y ∂x ˆ = f (∇×A) − A× (∇f ). qed Problem 1.22 ∂By ∂By ∂By ∂Bx ∂Bx x x ˆ + A y ˆ (a) (A·∇) B = Ax ∂B + A + A + A + A x y z y z ∂x ∂y ∂z ∂y ∂z ∂x ∂Bz ∂Bz ∂Bz + Ax ∂x + Ay ∂y + Az ∂z ˆ z. xx ˆ+y y ˆ+z ˆ z =√ . Let’s just do the x component. x2 +y 2 +z 2 ∂ ∂ ∂ √ 2 x2 2 [(ˆ r·∇)ˆ r]x = √1 x ∂x + y ∂y + z ∂z x +y +z n h i h i h io = 1r x √1 + x(− 12 ) (√1 )3 2x + yx − 21 (√1 )3 2y + zx − 12 (√1 )3 2z = 1r xr − r13 x3 + xy 2 + xz 2 = 1r xr − rx3 x2 + y 2 + z 2 = 1r xr − xr = 0.
(b) ˆ r=
r r
Same goes for the other components. Hence: (ˆ r·∇) ˆ r=0. ∂ ∂ ∂ (c) (va ·∇) vb = x2 ∂x + 3xz 2 ∂y − 2xz ∂z (xy x ˆ + 2yz y ˆ + 3xz ˆ z) = x2 (y x ˆ + 0y ˆ + 3z ˆ z) + 3xz 2 (x x ˆ + 2z y ˆ + 0ˆ z) − 2xz (0 x ˆ + 2y y ˆ + 3x ˆ z) 2 2 2 3 2 2 = x y + 3x z x ˆ + 6xz − 4xyz y ˆ + 3x z − 6x z ˆ z ˆ + 2xz 3z 2 − 2y y ˆ − 3x2 z ˆ z = x2 y + 3z 2 x Problem 1.23 ∂A
∂ ∂x (Ax Bx
+ Ay By + Az Bz ) =
∂Ax ∂x Bx
∂B
y y x + Ax ∂B ∂x + ∂x By + Ay ∂x + ∂B ∂Bz x x − Az ∂B [A×(∇×B)]x = Ay (∇×B)z − Az (∇×B)y = Ay ∂xy − ∂B ∂y ∂z − ∂x ∂A x x z [B×(∇×A)]x = By ∂xy − ∂A − Bz ∂A − ∂A ∂y ∂z ∂x ∂Bx ∂Bx ∂ ∂ ∂ x [(A·∇)B]x = Ax ∂x + Ay ∂y + Az ∂z Bx = Ax ∂B ∂x + Ay ∂y + Az ∂z ∂Ax ∂Ax ∂Ax [(B·∇)A]x = Bx ∂x + By ∂y + Bz ∂z
(ii) [∇(A·B)]x =
∂Az ∂x Bz
z + Az ∂B ∂x
So [A×(∇×B) + B×(∇×A) + (A·∇)B + (B·∇)A]x ∂B ∂Ay ∂Bx ∂Bz ∂Ax ∂Ax ∂Az x = Ay ∂xy − Ay ∂B ∂y − Az ∂z + Az ∂x + By ∂x − By ∂y − Bz ∂z + Bz ∂x ∂Bx ∂Bx ∂Ax ∂Ax ∂Ax x +Ax ∂B ∂x + Ay ∂y + Az ∂z + Bx ∂x + By ∂y + Bz ∂z ∂Ay ∂B ∂Bx ∂Ax ∂Ax ∂Bx x x = Bx ∂A + Ay ∂xy − ∂B ∂x + Ax ∂x + By ∂x − ∂y + ∂y ∂y + ∂y ∂Az ∂Ax ∂Bz ∂Bx x x +Bz − ∂A + Az − ∂B ∂z + ∂x + ∂z ∂z + ∂x + ∂z = [∇(A·B)]x (same for y and z)
/
/
(vi) [∇×(A×B)]x = =
/
/ /
/
/
/
∂ ∂ ∂ ∂ ∂y (A×B)z − ∂z (A×B)y = ∂y (Ax By − Ay Bx ) − ∂z (Az Bx − Ax Bz ) ∂By ∂Ay ∂Ax ∂Bx ∂Az ∂Bx ∂Ax ∂Bz ∂y By + Ax ∂y − ∂y Bx − Ay ∂y − ∂z Bx − Az ∂z + ∂z Bz + Ax ∂z
[(B·∇)A − (A·∇)B + A(∇·B) − B(∇·A)]x ∂Ax ∂Ax ∂Bx ∂Bx ∂Bx x = Bx ∂A ∂x + By ∂y + Bz ∂z − Ax ∂x − Ay ∂y − Az ∂z + Ax
∂Bx ∂x
+
∂By ∂y
c
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z + ∂B − Bx ∂z
∂Ax ∂x
+
∂Ay ∂y
z + ∂A ∂z
10
/
CHAPTER 1. VECTOR ANALYSIS
/
∂By ∂Bx ∂Bz ∂Bx x = By ∂A + Bx ∂y + Ax − ∂x + ∂x + ∂y + ∂z ∂Bx ∂Bx ∂Ax + Ay − ∂y + Az − ∂z + Bz ∂z = [∇×(A×B)]x (same for y and z)
/
∂Ax ∂x
/
x − ∂A ∂x −
∂Ay ∂y
−
∂Az ∂z
Problem 1.24 ∇(f /g) = = =
∂ ∂ ∂ ˆ + ∂y (f /g) y ˆ + ∂z (f /g) ˆ z ∂x (f /g) x ∂f ∂g ∂f ∂g ∂g g ∂y −f ∂y g ∂x −f ∂x g ∂f −f x ˆ+ y ˆ + ∂z g2 ∂z ˆ z g2 hg2 ∂f ∂f ∂f ∂g ∂g 1 ˆ + ∂y y ˆ + ∂z ˆ z − f ∂x x ˆ + ∂y y ˆ g 2 g ∂x x
∇·(A/g) = = =
+
∂ ∂ ∂ ∂x (Ax /g) + ∂y (Ay /g) + ∂z (Az /g) ∂A y ∂g ∂g x g −Ay g ∂A g ∂Az −A ∂g ∂x −Ax ∂x + ∂y g2 ∂y + ∂z g2 z ∂x h g2 ∂Ay ∂g ∂g ∂Az ∂Ax 1 − Ax ∂x + Ay ∂y g2 g ∂x + ∂y + ∂z
[∇×(A/g)]x = = = =
∂ ∂ ∂y (Az /g) − ∂z (Ay /g) ∂Ay ∂Az ∂g g ∂y −Az ∂y g −A ∂g − ∂z g2 y ∂z h g2 ∂Ay ∂g ∂Az 1 − Az ∂y − g2 g ∂y − ∂z g(∇×A)x +(A×∇g)x (same for y g2
Ay ∂g ∂z
i
∂g z ∂z ˆ
=
+ Az ∂g ∂z
g∇f −f ∇g . g2
i
=
qed
g∇·A−A·∇g . g2
qed
i
and z). qed
Problem 1.25 x ˆ ˆ z ˆ y ˆ(6xz) + y ˆ(9zy) + ˆ z(−2x2 − 6y 2 ) (a) A×B = x 2y 3z = x 3y −2x 0 ∂ ∂ ∇·(A×B) = ∂x (6xz) + ∂y (9zy) + ∂ ∂ ∇×A = x ˆ ∂y (3z) − ∂z (2y) + y ˆ ∂ ∂ ∇×B = x ˆ ∂y (0) − ∂z (−2x) + y ˆ
∂ 2 2 ∂z (−2x − 6y ) = 6z+ 9z + 0 = 15z ∂ ∂ ∂ ∂ z ∂x (2y) − ∂y (x) ∂z (x) − ∂x (3z) + ˆ
∂ ∂ ∂ (3y) − (0) + ˆ z ∂z ∂x ∂x (−2x) −
= 0; B·(∇×A) = 0 ∂ (3y) = −5 ˆ z; A·(∇×B) = −15z ∂y
?
∇·(A×B) = B·(∇×A) − A·(∇×B) = 0 − (−15z) = 15z. X ∂ ∂ (b) A·B = 3xy − 4xy = −xy ; ∇(A·B) = ∇(−xy) = x ˆ ∂x (−xy) + y ˆ ∂y (−xy) = −y x ˆ − xy ˆ x ˆ ˆ z ˆ y A×(∇×B) = x 2y 3z = x ˆ(−10y) + y ˆ(5x); B×(∇×A) = 0 0 0 −5 ∂ ∂ ∂ (A·∇)B = x ∂x + 2y ∂y + 3z ∂z (3y x ˆ − 2x y ˆ) = x ˆ(6y) + y ˆ(−2x) ∂ ∂ (B·∇)A = 3y ∂x − 2x ∂y (x x ˆ + 2y y ˆ + 3z ˆ z) = x ˆ(3y) + y ˆ(−4x)
A×(∇×B) + B×(∇×A) + (A·∇)B + (B·∇)A = −10y x ˆ + 5x y ˆ + 6y x ˆ − 2x y ˆ + 3y x ˆ − 4x y ˆ = −y x ˆ − xy ˆ = ∇·(A·B). X ∂ ∂ ∂ ∂ ∂ (c) ∇×(A×B) = x ˆ ∂y (−2x2 − 6y 2 ) − ∂z (9zy) + y ˆ ∂z (6xz) − ∂x (−2x2 − 6y 2 ) + ˆ z ∂x (9zy) − =x ˆ(−12y − 9y) + y ˆ(6x + 4x) + ˆ z(0) = −21y x ˆ + 10x y ˆ ∇·A =
∂ ∂x (x)
+
∂ ∂y (2y)
+
∂ ∂z (3z)
= 1 + 2 + 3 = 6; ∇·B =
∂ ∂x (3y)
+
∂ ∂y (−2x)
∂ ∂y (6xz)
=0
c
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CHAPTER 1. VECTOR ANALYSIS
11
(B·∇)A − (A·∇)B + A(∇·B) − B(∇·A) = 3y x ˆ − 4x y ˆ − 6y x ˆ + 2x y ˆ − 18y x ˆ + 12x y ˆ = −21y x ˆ + 10x y ˆ = ∇×(A×B). X Problem 1.26 (a)
∂ 2 Ta ∂x2
= 2;
(b)
∂ 2 Tb ∂x2
=
∂ 2 Ta ∂y 2
∂ 2 Tb ∂y 2
∂ 2 Tc ∂x2
=
=
∂ 2 Ta ∂z 2
∂ 2 Tb ∂z 2
= 0 ⇒ ∇2 Ta = 2.
= −Tb ⇒ ∇2 Tb = −3Tb = −3 sin x sin y sin z.
2
2
= 25Tc ; ∂∂yT2c = −16Tc ; ∂∂zT2c = −9Tc ⇒ ∇2 Tc = 0. 2 2 2 (d) ∂∂xv2x = 2 ; ∂∂yv2x = ∂∂zv2x = 0 ⇒ ∇2 vx = 2 ∂ 2 vy ∂ 2 vy ∂ 2 vy 2 ∇2 v = 2 x ˆ + 6x y ˆ. ∂x2 = ∂y 2 = 0 ; ∂z 2 = 6x ⇒ ∇ vy = 6x ∂ 2 vz ∂ 2 vz ∂ 2 vz 2 ∂x2 = ∂y 2 = ∂z 2 = 0 ⇒ ∇ vz = 0
(c)
Problem 1.27 ∂vy ∂vy ∂vz ∂vx ∂vz ∂vx ∂ ∂ ∂ + − − + − ∇·(∇×v) = ∂x ∂y ∂y ∂z 2 ∂z 2 ∂x 2 ∂z 2∂x ∂y ∂ vy ∂ vy ∂ vz ∂ 2 vz ∂ vx ∂ 2 vx = ∂x ∂y − ∂y ∂x + ∂y ∂z − ∂z ∂y + ∂z ∂x − ∂x ∂z = 0, by equality of crossderivatives. From Prob. 1.18: ∇×va = −6xz x ˆ+2z y ˆ+3z 2 ˆ z ⇒ ∇·(∇×va ) =
∂ ∂ ∂ 2 ∂x (−6xz)+ ∂y (2z)+ ∂z (3z )
= −6z+6z = 0.
Problem 1.28 x ˆ y ˆ ˆ z ∂ ∂ ∂ ∇×(∇t) = ∂x ∂y ∂z = x ˆ ∂t ∂t ∂t
∂2t ∂y ∂z
−
∂2t ∂z ∂y
+y ˆ
∂2t ∂z ∂x
−
∂2t ∂x ∂z
+ˆ z
∂2t ∂x ∂y
−
∂2t ∂y ∂x
∂x ∂y ∂z
= 0, by equality of crossderivatives. In Prob. 1.11(b), ∇f = 2xy 3 z 4 x ˆ + 3x2 y 2 z 4 y ˆ + 4x2 y 3 z 3 ˆ z, so x ˆ y ˆ ˆ z ∂ ∂ ∂ ∇×(∇f ) = ∂x ∂y ∂z 2xy 3 z 4 3x2 y 2 z 4 4x2 y 3 z 3 =x ˆ(3 · 4x2 y 2 z 3 − 4 · 3x2 y 2 z 3 ) + y ˆ(4 · 2xy 3 z 3 − 2 · 4xy 3 z 3 ) + ˆ z(2 · 3xy 2 z 4 − 3 · 2xy 2 z 4 ) = 0. X Problem 1.29 R R1 (a) (0, 0, 0) −→ (1, 0, 0). x : 0 → 1, y = z = 0; dl = dx x ˆ; v · dl = x2 dx; v · dl R= 0 x2 dx = (x3 /3)10 = 1/3. (1, 0, 0) −→ (1, 1, 0). x = 1, y : 0 → 1, z = 0; dl = dy y ˆ; v · dl = 2yz dy = 0; v · dl = 0. R R1 (1, 1, 0) −→ (1, 1, 1). x = y = 1, z : 0 → 1; dl = dz ˆ z; v · dl = y 2 dz = dz; v · dl = 0 dz = z10 = 1. R Total: v · dl = (1/3) + 0 + 1 = 4/3. R (b) (0, 0, 0) −→ (0, 0, 1). x = y = 0, z : 0 → 1; dl = dz ˆ z; v · dl = y 2 dz = 0; v · dl = 0. R R1 (0, 0, 1) −→ (0, 1, 1). x = 0, y : 0 → 1, z = 1; dl = dy y ˆ; v · dl = 2yz dy = 2y dy; v · dl = 0 2y dy = y 2 10 = 1. R R1 (0, 1, 1) −→ (1, 1, 1). x : 0 → 1, y = z = 1; dl = dx x ˆ; v · dl = x2 dx; v · dl = 0 x2 dx = (x3 /3)10 = 1/3. R Total: v · dl = 0 + 1 + (1/3) = 4/3. (c) x = y = z : 0 → 1; dx = dy = dz; v · dl = x2 dx + 2yz dy + y 2 dz = x2 dx + 2x2 dx + x2 dx = 4x2 dx; R R1 v · dl = 0 4x2 dx = (4x3 /3)10 = 4/3. H (d) v · dl = (4/3) − (4/3) = 0.
c
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12
CHAPTER 1. VECTOR ANALYSIS
Problem 1.30 R R2 R2 x, y : 0 → 1, z = 0; da = dx dy ˆ z; v · da = y(z 2 − 3) dx dy = −3y dx dy; v · da = −3 0 dx 0 y dy = 2
−3(x20 )( y2 20 ) = −3(2)(2) = −12. In Ex. 1.7 we got 20, for the same boundary line (the square in the xyplane), so the answer is no: the surface integral does not depend only on the boundary line. The total flux for the cube is 20 + 12 = 32. Problem 1.31 R R T dτ = z 2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last: Z Z Z dx dy dz. z2 R (1−y−z) The sloping surface is x+y +z = 1, so the x integral is 0 dx = 1−y −z. For a given z, y ranges from 0 to R (1−z) (1−z) 1 − z, so the y integral is 0 (1 − y − z) dy = [(1 − z)y − (y 2 /2)]0 = (1 − z)2 − [(1 − z)2 /2] = (1 − z)2 /2 = R R 1 z2 2 4 3 4 5 1 2 1 z 2 (1/2) − z + (z /2). Finally, the z integral is 0 z ( 2 − z + 2 ) dz = 0 ( 2 − z 3 + z2 ) dz = ( z6 − z4 + z10 )10 = 1 6
−
1 4
+
1 10
= 1/60.
Problem 1.32 T (b) = 1 + 4 + 2 = 7; T (a) = 0. ⇒ T (b) − T (a) = 7. ∇T = (2x + 4y)ˆ x + (4x + 2z 3 )ˆ y + (6yz 2 )ˆ z; ∇T ·dl = (2x + 4y)dx + (4x + 2z 3 )dy + (6yz 2 )dz 1 R R1 (a) Segment 1: x : 0 → 1, y = z = dy = dz = 0. ∇T ·dl = 0 (2x) dx = x2 0 = 1. R R R1 b 1 ∇T ·dl = 7. X Segment 2: y : 0 → 1, x = 1, z = 0, dx = dz = 0. ∇T ·dl = 0 (4) dy = 4y0 = 4. a R R1 2 1 3 Segment 3: z : 0 → 1, x = y = 1, dx = dy = 0. ∇T ·dl = 0 (6z ) dz = 2z 0 = 2. R R1 (b) Segment 1: z : 0 → 1, x = y = dx = dy = 0. ∇T ·dl = 0 (0) dz = 0. R R1 1 Segment 2: y : 0 → 1, x = 0, z = 1, dx = dz = 0. ∇T ·dl = 0 (2) dy = 2y0 = 2. R b R R1 ∇T ·dl = 7. X a Segment 3: x : 0 → 1, y = z = 1, dy = dz = 0. ∇T ·dl = 0 (2x + 4) dx 1 = (x2 + 4x) 0 = 1 + 4 = 5. (c) x : 0 → 1, y = x, z = x2 , dy = dx, dz = 2x dx. ∇T ·dl = (2x + 4x)dx + (4x + 2x6 )dx + (6xx4 )2x dx = (10x + 14x6 )dx. 1 Rb R1 ∇T ·dl = 0 (10x + 14x6 )dx = (5x2 + 2x7 ) 0 = 5 + 2 = 7. X a Problem 1.33 ∇·v = y + 2z + 3x o R R RR nR 2 (y + 2z + 3x) dx dy dz (∇·v)dτ = (y + 2z + 3x) dx dy dz = 0 2 (y + 2z)x + 23 x2 0 = 2(y + 2z) + 6 n o R R2 = (2y + 4z + 6)dy dz 0 2 2 y + (4z + 6)y 0 = 4 + 2(4z + 6) = 8z + 16
,→
,→
=
R2 0
2 (8z + 16)dz = (4z 2 + 16z) 0 = 16 + 32 = 48.
Numbering the surfaces as in Fig. 1.29: c
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CHAPTER 1. VECTOR ANALYSIS
13
2 R RR (i) da = dy dz x ˆ, x = 2. v·da = 2y dyRdz. v·da = 2y dy dz = 2y 2 0 = 8. (ii) da = −dy dz x ˆ, x = 0. v·da = 0. v·daR= 0. RR (iii) da = dx dz y ˆ, y = 2. v·da = 4z dx R dz. v·da = 4z dx dz = 16. (iv) da = −dx dz y ˆ, y = 0. v·da = 0. v·da R = 0. (v) da = dx dy ˆ z, z = 2. v·da = 6x dxRdy. v·da = 24. (vi)R da = −dx dy ˆ z, z = 0. v·da = 0. v·da = 0. ⇒ v·da = 8 + 16 + 24 = 48 X Problem 1.34 ∇×v = x ˆ(0 − 2y) + y ˆ(0 − 3z) + ˆ z(0 − x) = −2y x ˆ − 3z y ˆ − xˆ z. da = dy dz x ˆ, if we agree that the path integral shall run counterclockwise. So (∇×v)·da = −2y dy dz. o R R nR 2−z (∇×v)·da = (−2y)dy dz 0 z6 2 2−z 2 @ y 0 = −(2 − z) @ 2 R2 3 @ = − 0 (4 − 4z + z 2 )dz = − 4z − 2z 2 + z3 0 @ = − 8 − 8 + 83 = − 83 @ @ y Meanwhile, v·dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments. y
,→
=
2
−
z
z6 @ (3)
@@ I (2) @@ @ ? @ @  y (1)
R (1) x = z = 0; dx = dz = 0. y : 0 → 2. v·dl = 0. (2) x = 0; z = 2 − y; dx = 0, dz = −dy, y : 2 → 0. v·dl = 2yz dy. 2 R R0 R2 v·dl = 2 2y(2 − y)dy = − 0 (4y − 2y 2 )dy = − 2y 2 − 32 y 3 0 = − 8 − 23 · 8 = − 83 . R H (3) x = y = 0; dx = dy = 0; z : 2 → 0. v·dl = 0. v·dl = 0. So v·dl = − 83 . X Problem 1.35 R By Corollary 1, (∇×v)·da should equal 43 . ∇×v = (4z 2 − 2x)ˆ x + 2z ˆ z. R R1 (i) da = dy dz x ˆ, x = 1; y, z : 0 → 1. (∇×v)·da = (4z 2 − 2)dy dz; (∇×v)·da = 0 (4z 2 − 2)dz 1 = ( 34 z 3 − 2z) 0 = 43 − 2 = − 23 . R (ii) da = −dx dy ˆ z, z = 0; x, y : 0 → 1. (∇×v)·da = 0; R (∇×v)·da = 0. (iii) da = dx dz y ˆ, y = 1; x, z : 0 → 1. (∇×v)·da = 0; (∇×v)·da = 0. R (iv) da = −dx dz y ˆ, y = 0; x, z : 0 → 1. (∇×v)·da = 0; (∇×v)·da = 0. R (v) da = dx dy ˆ z, z = 1; x, y : 0 → 1. (∇×v)·da = 2 dx dy; (∇×v)·da = 2. R ⇒ (∇×v)·da = − 32 + 2 = 43 . X
c
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14
CHAPTER 1. VECTOR ANALYSIS
Problem 1.36 (a) Use the product rule ∇×(f A) = f (∇×A) − A × (∇f ) : Z Z Z I Z f (∇×A) · da = ∇×(f A) · da + [A × (∇f )] · da = f A · dl + [A × (∇f )] · da. qed S
S
S
P
S
(I used Stokes’ theorem in the last step.) (b) Use the product rule ∇·(A × B) = B · (∇×A) − A · (∇×B) : Z Z Z I Z B · (∇×A)dτ = ∇·(A × B) dτ + A · (∇×B) dτ = (A × B) · da + A · (∇×B) dτ. qed V
V
V
S
V
(I used the divergence theorem in the last step.) Problem 1.37 r =
p x2 + y 2 + z 2 ;
θ = cos−1
√
z x2 +y 2 +z 2
;
φ = tan−1
y x
.
Problem 1.38 There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry from Fig. 1.36. The most systematic approach is to study the expression: r = xx ˆ+yy ˆ +zˆ z = r sin θ cos φ x ˆ + r sin θ sin φ y ˆ + r cos θ ˆ z. ∂ If I only vary r slightly, then dr = ∂r (r)dr is a short vector pointing in the direction of increase in r. To make it a unit vector, I must divide by its length. Thus: ∂r
∂r
∂r
∂r
∂θ
∂φ
ˆ = ∂φ . ; θˆ = ∂θ ; φ ˆ r = ∂r ∂r ∂r ∂r ∂r ∂r ∂r ∂θ ∂r ∂φ
∂r 2 = sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ = 1. = sin θ cos φ x ˆ + sin θ sin φ y ˆ + cos θ ˆ z; ∂r ∂r 2 = r2 cos2 θ cos2 φ + r2 cos2 θ sin2 φ + r2 sin2 θ = r2 . = r cos θ cos φ x ˆ + r cos θ sin φ y ˆ − r sin θ ˆ z; ∂θ ∂r 2 = −r sin θ sin φ x ˆ + r sin θ cos φ y ˆ; ∂φ = r2 sin2 θ sin2 φ + r2 sin2 θ cos2 φ = r2 sin2 θ.
ˆ r = sin θ cos φ x ˆ + sin θ sin φ y ˆ + cos θ ˆ z. ⇒ θˆ = cos θ cos φ x ˆ + cos θ sin φ y ˆ − sin θ ˆ z. ˆ = − sin φ x φ ˆ + cos φ y ˆ. Check: ˆ r·ˆ r = sin2 θ(cos2 φ + sin2 φ) + cos2 θ = sin2 θ + cos2 θ = 1, X ˆ ˆ = − cos θ sin φ cos φ + cos θ sin φ cos φ = 0, X etc. θ·φ sin θ ˆ r = sin2 θ cos φ x ˆ + sin2 θ sin φ y ˆ + sin θ cos θ ˆ z. 2 ˆ cos θ θ = cos θ cos φ x ˆ + cos2 θ sin φ y ˆ − sin θ cos θ ˆ z. Add these: (1) sin θ ˆ r + cos θ θˆ = + cos φ x ˆ + sin φ y ˆ; ˆ = − sin φ x (2) φ ˆ + cos φ y ˆ. Multiply (1) by cos φ, (2) by sin φ, and subtract: ˆ x ˆ = sin θ cos φ ˆ r + cos θ cos φ θˆ − sin φ φ. c
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CHAPTER 1. VECTOR ANALYSIS
15
Multiply (1) by sin φ, (2) by cos φ, and add: ˆ y ˆ = sin θ sin φ ˆ r + cos θ sin φ θˆ + cos φ φ. cos θ ˆ r = sin θ cos θ cos φ x ˆ + sin θ cos θ sin φ y ˆ + cos2 θ ˆ z. ˆ sin θ θ = sin θ cos θ cos φ x ˆ + sin θ cos θ sin φ y ˆ − sin2 θ ˆ z. Subtract these: ˆ ˆ z = cos θ ˆ r − sin θ θ. Problem 1.39 ∂ (a) ∇·v1 = r12 ∂r (r2 r2 ) = r12 4r3 = 4r 4 R R R RR Rπ R (∇·v1 )dτ = (4r)(r2 sin θ dr dθ dφ) = (4) 0 r3 dr 0 sin θ dθ 2π (2)(2π) = 4πR4 0 dφ = (4) 4 R R 2 R R π 2π v1 ·da = (r ˆ r)·(r2 sin θ dθ dφ ˆ r) = r4 0 sin θ dθ 0 dφ = 4πR4 X (Note: at surface of sphere r = R.) R ∂ (b) ∇·v2 = r12 ∂r r2 r12 = 0 ⇒ (∇·v2 )dτ = 0 R R 1 2 R v2 ·da = r (r sin θ dθ dφ ˆ r) = sin θ dθ dφ = 4π. r2 ˆ They don’t agree! The point is that this divergence is zero except at the origin, where it blows up, so our R calculation of (∇·v2 ) is incorrect. The right answer is 4π.
Problem 1.40 ∂ 1 ∂ 1 ∂ ∇·v = r12 ∂r (r2 r cos θ) + r sin θ ∂θ (sin θ r sin θ) + r sin θ ∂φ (r sin θ cos φ) 1 1 1 = r2 3r2 cos θ + r sin θ r 2 sin θ cos θ + r sin θ r sin θ(− sin φ) = 3 cos θ + 2 cos θ − sin φ = 5 cos θ − sin φ i R R RR R θ hR 2π (∇·v)dτ = (5 cos θ − sin φ) r2 sin θ dr dθ dφ = 0 r2 dr 02 0 (5 cos θ − sin φ) dφ dθ sin θ =
R3 3
(10π)
R
π 2
0
,→2π(5 cos θ)
sin θ cos θ dθ π sin2 θ 2 2 =
,→
=
0
1 2
5π 3 3 R .
Two surfaces—one the hemisphere: da = R2 sin θ dθ dφ ˆ r; r = R; φ : 0 → 2π, θ : 0 → π2 . π R R R R 2π v·da = (r cos θ)R2 sin θ dθ dφ = R3 02 sin θ cos θ dθ 0 dφ = R3 12 (2π) = πR3 . ˆ = r dr dφ θˆ (here θ = π ). r : 0 → R, φ : 0 → 2π. other the flat bottom: da = (dr)(r sin θ dφ)(+θ) 2 R R R R 2 R 2π R3 v·da = (r sin θ)(r dr dφ) = r dr dφ = 2π . 3 0 0 R Total: v·da = πR3 + 23 πR3 = 53 πR3 . X Problem 1.41 ∇t = (cos θ + sin θ cos φ)ˆ r + (− sin θ + cos θ cos φ)θˆ +
1 sin /θ
(− sin / θ sin φ)φˆ
∇2 t = ∇·(∇t) ∂ 1 ∂ 1 ∂ r2 (cos θ + sin θ cos φ) + r sin = r12 ∂r θ ∂θ (sin θ(− sin θ + cos θ cos φ)) + r sin θ ∂φ (− sin φ) 2 1 1 1 2 = r2 2r(cos θ + sin θ cos φ) + r sin θ (−2 sin θ cos θ + cos θ cos φ − sin θ cos φ) − r sin θ cos φ 2 2 1 2 = r sin θ [2 sin θ cos θ + 2 sin θ cos φ − 2 sin θ cos θ + cos θ cos φ − sin θ cos φ − cos φ] 2 1 2 = r sin θ (sin θ + cos θ) cos φ − cos φ = 0. c
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16
CHAPTER 1. VECTOR ANALYSIS
⇒ ∇2 t = 0 Check: r cos θ = z, r sin θ cos φ = x ⇒ in Cartesian coordinates t = x + z. Obviously Laplacian is zero. Rb Gradient Theorem: a ∇t·dl = t(b) − t(a) Segment 1: θ = π2 , φ = 0, r : 0 → 2. dl = dr ˆ r; ∇t·dl = (cos θ + sin θ cos φ)dr = (0 + 1)dr = dr. R R2 ∇t·dl = 0 dr = 2. ˆ = 2 dφ φ. ˆ dl = r sin θ dφ φ π R R π2 ∇t·dl = (− sin φ)(2 dφ) = −2 sin φ dφ. ∇t·dl = − 0 2 sin φ dφ = 2 cos φ02 = −2.
Segment 2: θ =
π 2,
r = 2, φ : 0 →
π 2.
Segment 3: r = 2, φ = π2 ; θ : π2 → 0. ˆ ∇t·dl = (− sin θ + cos θ cos φ)(2 dθ) = −2 sin θ dθ. dl = r dθ θˆ = 2 dθ θ; R R0 0 ∇t·dl = − π 2 sin θ dθ = 2 cos θ π = 2. 2 2 Rb Total: a ∇t·dl = 2 − 2 + 2 = 2 . Meanwhile, t(b) − t(a) = [2(1 + 0)] − [0( )] = 2. X ˆ = − sin φ x Problem 1.42 From Fig. 1.42, ˆ s = cos φ x ˆ + sin φ y ˆ; φ ˆ + cos φ y ˆ; ˆ z=ˆ z Multiply first by cos φ, second by sin φ, and subtract: ˆ sin φ = cos2 φ x ˆ s cos φ − φ ˆ + cos φ sin φ y ˆ + sin2 φ x ˆ − sin φ cos φ y ˆ=x ˆ(sin2 φ + cos2 φ) = x ˆ. ˆ So x ˆ = cos φ ˆ s − sin φ φ. Multiply first by sin φ, second by cos φ, and add: ˆ cos φ = sin φ cos φ x ˆ s sin φ + φ ˆ + sin2 φ y ˆ − sin φ cos φ x ˆ + cos2 φ y ˆ=y ˆ(sin2 φ + cos2 φ) = y ˆ. ˆ So y ˆ = sin φ ˆ s + cos φ φ. ˆ z=ˆ z. Problem 1.43 ∂ ∂ ∂ (a) ∇·v = 1s ∂s s s(2 + sin2 φ) + 1s ∂φ (s sin φ cos φ) + ∂z (3z) 2 2 1 1 2 = s 2s(2 + sin φ) + s s(cos φ − sin φ) + 3 = 4 + 2 sin2 φ + cos2 φ − sin2 φ + 3 = 4 + sin2 φ + cos2 φ + 3 = 8. R5 R R R2 Rπ (b) (∇·v)dτ = (8)s ds dφ dz = 8 0 s ds 02 dφ 0 dz = 8(2) π2 (5) = 40π. Meanwhile, the surface integral has five parts: R R2 Rπ top: z = 5, da = s ds dφ ˆ z; v·da = 3z s ds dφ = 15s ds dφ. R v·da = 15 0 s ds 02 dφ = 15π. bottom: z = 0, da = −s ds dφ ˆ z; v·da = −3z s ds dφ = 0. v·da = 0. R π ˆ back: φ = 2 , da = ds dz φ; v·da = s sin φ cos φ ds dz = 0. v·da = 0. R ˆ v·da = −s sin φ cos φ ds dz = 0. v·da = 0. left: φ = 0, da = −ds dz φ; front: s = 2, da = s dφ dz ˆ s; v·da = s(2 + sin2 φ)s dφ dz = 4(2 + sin2 φ)dφ dz. R5 R R π2 v·da = 4 0 (2 + sin2 φ)dφ 0 dz = (4)(π + π4 )(5) = 25π. H So v·da = 15π + 25π = 40π. X ∂ ∂ ∂ ∂ ˆ (c) ∇×v = 1s ∂φ (3z) − ∂z (s sin φ cos φ) ˆ s + ∂z s(2 + sin2 φ) − ∂s (3z) φ ∂ ∂ (s2 sin φ cos φ) − ∂φ s(2 + sin2 φ) ˆ z + 1s ∂s =
1 s (2s sin φ cos φ
− s 2 sin φ cos φ) ˆ z = 0.
c
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CHAPTER 1. VECTOR ANALYSIS
17
Problem 1.44 (a) 3(32 ) − 2(3) − 1 = 27 − 6 − 1 = 20. (b) cos π = 1. (c) zero. (d) ln(−2 + 3) = ln 1 = zero. Problem 1.45 R2 (a) −2 (2x + 3) 31 δ(x) dx = 13 (0 + 3) = 1. (b) By Eq. 1.94, δ(1 − x) = δ(x − 1), so 1 + 3 + 2 = 6. (c)
R1 −1
9x2 13 δ(x + 13 ) dx = 9 − 13
2
1 3
=
1 3.
(d) 1 (if a > b), 0 (if a < b). Problem 1.46 d R∞ R∞ ∞ (a) −∞ f (x) x dx δ(x) dx = x f (x)δ(x)−∞ − −∞
d dx
(x f (x)) δ(x) dx.
df df dx + dx f = x dx + f. The first term is zero, since δ(x) = 0 at ±∞; (x f (x)) = x dx R ∞ df R∞ So the integral is − −∞ x dx + f δ(x) dx = 0 − f (0) = −f (0) = − −∞ f (x)δ(x) dx. d dx
d δ(x) = −δ(x). qed So, x dx R∞ R ∞ df R∞ ∞ dθ (b) −∞ f (x) dx dx = f (x)θ(x)−∞ − −∞ dx θ(x)dx = f (∞) − 0 R∞ dθ = f (0) = −∞ f (x)δ(x) dx. So dx = δ(x). qed
df dx dx
= f (∞) − (f (∞) − f (0))
Problem 1.47 (a) ρ(r) = qδ 3 (r − r0 ). Check:
R
R ρ(r)dτ = q δ 3 (r − r0 ) dτ = q.
X
(b) ρ(r) = qδ 3 (r − a) − qδ 3 (r). (c) Evidently ρ(r) = Aδ(r − R). To determine the constant A, we require R R Q Q = ρ dτ = Aδ(r − R)4πr2 dr = A 4πR2 . So A = 4πR ρ(r) = 2.
Q 4πR2 δ(r
− R).
Problem 1.48 (a) a2 + a·a + a2 = 3a2 . R (b) (r − b)2 513 δ 3 (r) dτ =
1 2 125 b
=
1 2 125 (4
+ 32 ) =
1 5.
(c) c2 = 25 + 9 + 4 = 38 > 36 = 62 , so c is outside V, so the integral is zero. 2
2
(d) (e − (2 x ˆ + 2y ˆ + 2ˆ z)) = (1 x ˆ + 0y ˆ + (−1) ˆ z) = 1 + 1 = 2 < (1.5)2 = 2.25, so e is inside V, and hence the integral is e·(d − e) = (3, 2, 1)·(−2, 0, 2) = −6 + 0 + 2 = 4. Problem 1.49 R First method: use Eq. 1.99 to write J = e−r 4πδ 3(r) dτ = 4πe−0 = 4π. Second method: integrating by parts (use Eq. 1.59). c
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18
Z J =−
ˆ r · ∇(e−r ) dτ + r2
V
Z =
CHAPTER 1. VECTOR ANALYSIS
I
−r
e
ˆ r · da. But ∇ e−r = 2 r
∂ −r e ˆ r = −e−rˆ r. ∂r
S
1 −r e 4πr2 dr + r2
Z
−r
e
ˆ r · r2 sin θ dθ dφ ˆ r = 4π r2
ZR
e−r dr + e−R
Z sin θ dθ dφ
0 −r
= 4π −e
R + 4πe−R = 4π −e−R + e−0 + 4πe−R = 4π.X 0
Problem 1.50 (a) ∇·F1 =
∂ ∂x (0)
+
∂ ∂y (0)
+
∂ ∂z
x2 = 0 ;
∇·F2 =
x y ˆ ˆ z ˆ ∂ ∂ ∂ ∂ y y; ∇×F1 = ∂x x2 = −2xˆ ∂y ∂z = −ˆ ∂x 0 0 x2
Here R = ∞, so e−R = 0.
∂x ∂x
+
∂y ∂y
+
∂z ∂z
=1+1+1= 3
x y ˆ ˆ z ˆ ∂ ∂ ∂ ∇×F2 = ∂x ∂y ∂z = 0 x y z
F2 is a gradient; F1 is a curl U2 = 12 x3 + y 2 + z 2 would do (F2 = ∇U2 ). ∂Ay ∂A ∂Az ∂Ax x3 2 z x = 0; For A1 , we want ∂zy − ∂A = ∂A ∂y ∂z − ∂x ∂x − ∂y = x . Ay = 3 , Ax = Az = 0 would do it. ˆ A1 = 13 x2 y
(F1 = ∇×A1 ) . (But these are not unique.) x ˆ y ˆ ˆ z ∂ ∂ ∂ ∂ ∂ ∂ (yz) + ∂y (xz) + ∂z (xy) = 0; ∇×F3 = ∂x ˆ (x − x) + y ˆ (y − y) + ˆ z (z − z) = 0. (b) ∇·F3 = ∂x ∂y ∂z = x yz xz xy So F3 can be written as the gradient of a scalar (F3 = ∇U3 ) and as the curl of a vector (F3 = ∇×A3 ). In fact, U3 = xyz does the job. For the vector potential, we have
∂Az ∂y ∂Ax ∂z ∂Ay ∂x
− − −
∂Ay ∂z ∂Az ∂x ∂Ax ∂y
= yz, which suggests Az = 14 y 2 z + f (x, z); Ay = − 41 yz 2 + g(x, y) = xz, suggesting Ax = 14 z 2 x + h(x, y); Az = − 41 zx2 + j(y, z) = xy, so Ay = 14 x2 y + k(y, z); Ax = − 14 xy 2 + l(x, z)
Putting this all together: A3 =
1 4
x z2 − y2 x ˆ + y x2 − z 2 y ˆ + z y 2 − x2 ˆ z (again, not unique).
Problem 1.51 (d) ⇒ (a): H∇×F = ∇×(−∇U ) = 0 (Eq. 1.44 – curl of gradient is always zero). R (a) ⇒ (c): F · dl = (∇×F) · da = 0 (Eq. 1.57–Stokes’ theorem). Ra H Rb Rb Rb (c) ⇒ (b): a I F · dl − a II F · dl = a I F · dl + b II F · dl = F · dl = 0, so Z
b
Z
b
F · dl = a I
F · dl. a II
(b) ⇒ (c): same as (c) ⇒ (b), only in reverse; (c) ⇒ (a): same as (a)⇒ (c). Problem 1.52 (d) ⇒ (a): H∇·F = ∇·(∇×W) = 0 (Eq 1.46—divergence of curl is always zero). R (a) ⇒ (c): F · da = (∇·F) dτ = 0 (Eq. 1.56—divergence theorem). c
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CHAPTER 1. VECTOR ANALYSIS (c) ⇒ (b):
R I
F · da −
R II
F · da =
H
19
F · da = 0, so Z Z F · da = I
F · da.
II
H (Note: sign change because for F · da, da is outward, whereas for surface II it is inward.) (b) ⇒ (c): same as (c) ⇒ (b), in reverse; (c)⇒ (a): same as (a)⇒ (c) . Problem 1.53 In Prob. 1.15 we found that ∇·va = 0; in Prob. 1.18 we found that ∇×vc = 0. So vc can be written as the gradient of a scalar; va can be written as the curl of a vector. (a) To find t:
(2)
∂t ∂x ∂t ∂y
= y 2 ⇒ t = y 2 x + f (y, z) = 2xy + z 2
(3)
∂t ∂z
= 2yz
(1)
From (1) & (3) we get 2
2xy + z (from (2)) ⇒ ∂Wz ∂y
(b) To find W:
−
∂f 2 2 ∂z = 2yz ⇒ f = yz + g(y) ⇒ t = y x ∂g ∂y = 0. We may as well pick g = 0; then
∂Wy ∂z
= x2 ;
∂Wx ∂z
−
∂Wz ∂x
= 3z 2 x;
∂Wy ∂x
−
+ yz 2 + g(y), so 2
∂t ∂y
= 2xy + z 2 +
∂g ∂y
=
2
t = xy + yz .
∂Wx ∂y
= −2xz.
Pick Wx = 0; then ∂Wz 3 = −3xz 2 ⇒ Wz = − x2 z 2 + f (y, z) ∂x 2 ∂Wy = −2xz ⇒ Wy = −x2 z + g(y, z). ∂x ∂Wz ∂y
−
∂Wy ∂z
=
∂f ∂y
+ x2 −
∂g ∂z
= x2 ⇒
∂f ∂y
−
∂g ∂z
= 0. May as well pick f = g = 0.
W = −x2 z y ˆ − 32 x2 z 2 ˆ z. x y ˆ ˆ z ˆ ∂ ∂ ∂ 2 =x ˆ 3xz 2 + ˆ z (−2xz).X Check: ∇×W = ∂x ∂y ∂z ˆ x +y 0 −x2 z − 3 x2 z 2 2 You can add any gradient (∇t) to W without changing its curl, so this answer is far from unique. Some other solutions: W = xz 3 x ˆ − x2 z y ˆ; W = 2xyz + xz 3 x ˆ + x2 y ˆ z; W = xyz x ˆ − 12 x2 z y ˆ + 12 x2 y − 3z 2 ˆ z.
c
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20
CHAPTER 1. VECTOR ANALYSIS
Problem 1.54
1 ∂ 1 ∂ 1 ∂ r2 r2 cos θ + sin θ r2 cos φ + −r2 cos θ sin φ 2 r ∂r r sin θ ∂θ r sin θ ∂φ 1 1 3 1 2 2 = 2 4r cos θ + cos θ r cos φ + −r cos θ cos φ r r sin θ r sin θ r cos θ = [4 sin θ + cos φ − cos φ] = 4r cos θ. sin θ
∇·v =
Z
Z (∇·v) dτ =
ZR
2
(4r cos θ)r sin θ dr dθ dφ = 4
Zπ/2 Zπ/2 r dr cos θ sin θ dθ dφ 3
0
= R
4
1 π = 2 2
πR 4
4
0
0
.
Surface consists of four parts: (1) Curved: da = R2 sin θ dθ dφ ˆ r; r = R. v · da = R2 cos θ R2 sin θ dθ dφ . Z
Zπ/2 Zπ/2 π πR4 1 v · da = R cos θ sin θ dθ dφ = R4 = . 2 2 4 4
0
0
R ˆ φ = 0. v · da = r2 cos θ sin φ (r dr dθ) = 0. (2) Left: da = −r dr dθ φ; v · da = 0. 2 ˆ (3) Back: da = r dr dθ φ; φ = π/2. v · da = −r cos θ sin φ (r dr dθ) = −r3 cos θ dr dθ. ZR
Z v · da =
Zπ/2 1 4 1 r dr cos θ dθ = − R (+1) = − R4 . 4 4 3
0
0
ˆ θ = π/2. v · da = r2 cos φ (r dr dφ) . (4) Bottom: da = r sin θ dr dφ θ; ZR
Z v · da =
Zπ/2 1 r dr cos φ dφ = R4 . 4 3
0
Total:
H
v · da = πR4 /4 + 0 − 41 R4 + 14 R4 =
πR4 4 .
0
X
Problem 1.55 x y ˆ ˆ z ˆ R ∂ ∂ ∂ ∇×v = ∂x z (b − a). So (∇×v) · da = (b − a)πR2 . ∂y ∂z = ˆ ay bx 0 v · dl = (ay x ˆ + bx y ˆ) · (dx x ˆ + dy y ˆ + dz ˆ z) = ay dx + bx dy; x2 + y 2 = R2 ⇒ 2x dx + 2y dy = 0, so dy = −(x/y) dx. So v · dl = ay dx + bx(−x/y) dx = y1 ay 2 − bx2 dx. c
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CHAPTER 1. VECTOR ANALYSIS
For the “upper” semicircle, y = Z
√
21
R2 − x2 , so v · dl =
a(R2 −x2 )−bx2 √ R2 −x2
dx.
Z−R
−R aR2 − (a + b)x2 R2 xp 2 −1 x −1 x 2 2 √ dx = aR sin v · dl = R −x + − (a + b) − sin R 2 2 R R2 − x2 +R R −R π π 1 1 1 = R2 (a − b) sin−1 (−1) − sin−1 (+1) = R2 (a − b) − − = R2 (a − b) sin−1 (x/R) 2 2 2 2 2 +R 1 2 = πR (b − a). 2
HAnd the same2 for the lower semicircle (y changes sign, but the limits on the integral are reversed) so v · dl = πR (b − a). X Problem 1.56 R (1) x = z = 0; dx = dz = 0; y : 0 → 1. v · dl = (yz 2 ) dy = 0; v · dl = 0. (2) x = 0; z = 2 − 2y; dz = −2 dy; y : 1 → 0. v · dl = (yz 2 ) dy + (3y + z) dz = y(2 − 2y)2 dy − (3y + 2 − 2y)2 dy; Z0
Z v · dl = 2
(2y 3 − 4y 2 + y − 2) dy = 2
1
0 14 4y 3 y2 y4 − + − 2y = . 2 3 2 3 1
(3) x = y = 0; dx = dy = 0; z : 2 → 0. v · dl = (3y + z) dz = z dz; Z0
Z v · dl =
z dz =
z 2 0 = −2. 2 2
2
Total:
H
v · dl = 0 +
14 3
−2=
8 3.
H R Meanwhile, Stokes’ thereom says v · dl = (∇×v) · da. Here da = dy dz x ˆ, so all we need is ∂ ∂ (∇×v)x = ∂y (3y + z) − ∂z (yz 2 ) = 3 − 2yz. Therefore Z
Z Z (∇×v) · da =
1
Z
Z
2−2y
(3 − 2yz) dy dz =
(3 − 2yz) dz dy 0
Z = 0
0
1
3(2 − 2y) − 2y 12 (2 − 2y)2 dy =
1 = −y 4 + 38 y 3 − 5y 2 + 6y = −1 + 0
8 3
1
Z
(−4y 3 + 8y 2 − 10y + 6) dy
0
− 5 + 6 = 83 . X
Problem 1.57 Start at the origin. (1) θ =
π 2,
φ = 0; r : 0 → 1. v · dl = r cos2 θ (dr) = 0.
(2) r = 1, θ =
π 2;
R
v · dl = 0.
φ : 0 → π/2. v · dl = (3r)(r sin θ dφ) = 3 dφ.
R
v · dl = 3
c
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π/2 R 0
dφ =
3π 2 .
22 (3) φ =
π 2;
CHAPTER 1. VECTOR ANALYSIS
r sin θ = y = 1, so r =
1 sin θ ,
dr =
−1 sin2 θ
cos θ dθ, θ :
π 2
→ θ0 ≡ tan−1 (1/2).
cos2 θ cos θ cos θ sin θ dθ v · dl = r cos2 θ (dr) − (r cos θ sin θ)(r dθ) = − 2 dθ − sin θ sin θ sin2 θ 3 cos θ cos θ cos θ cos2 θ + sin2 θ cos θ =− + dθ = − dθ = − 3 dθ. sin θ sin θ sin3 θ sin2 θ sin θ Therefore Zθ0
Z v · dl = −
π/2
(4) θ = θ0 , φ =
π 2;
r:
√
θ cos θ 1 0 1 1 5 1 = − = − = 2. 3 dθ = 2 2 · (1/5) 2 · (1) 2 2 sin θ 2 sin θ π/2
5 → 0. v · dl = r cos2 θ (dr) = 45 r dr. Z
0 Z0 4 4 r2 4 5 v · dl = r dr = = − · = −2. 5√ 5 2 √5 5 2 5
Total: I v · dl = 0 +
3π +2−2= 2
3π 2 .
R Stokes’ theorem says this should equal (∇×v) · da 1 ∂ ∂ 1 1 ∂ ∂ 2 ∇×v = (sin θ 3r) − (−r sin θ cos θ) ˆ r+ r cos θ − (r3r) θˆ r sin θ ∂θ ∂φ r sin θ ∂φ ∂r 1 ∂ ∂ 2 ˆ + (−rr cos θ sin θ) − r cos θ φ r ∂r ∂θ 1 1 1 ˆ [3r cos θ] ˆ r + [−6r] θˆ + [−2r cos θ sin θ + 2r cos θ sin θ] φ = r sin θ r r ˆ = 3 cot θ ˆ r − 6 θ. ˆ (∇×v) · da = 0. (1) Back face: da = −r dr dθ φ;
R
(∇×v) · da = 0.
ˆ (∇×v) · da = 6r sin θ dr dφ. θ = (2) Bottom: da = −r sin θ dr dφ θ; Z1
Z (∇×v) · da =
0
π 2,
so (∇×v) · da = 6r dr dφ
Zπ/2 1 π 3π 6r dr dφ = 6 · · = . X 2 2 2 0
Problem 1.58 v · dl = y dz. (1) Left side: z = a − x; dz = −dx; y = 0. Therefore R (2) Bottom: dz = 0. Therefore v · dl = 0.
R
v · dl = 0.
c
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CHAPTER 1. VECTOR ANALYSIS
23
(3) Back: z = a − 12 y; dz = −1/2 dy; y : 2a → 0.
R
v · dl =
R0 2a
2 0 y − 21 dy = − 12 y2 = 2a
4a2 4
R Meanwhile, ∇×v = x ˆ, so (∇×v) · da is the projection of this surface on the x y plane =
1 2
= a2 . · a · 2a = a2 . X
Problem 1.59 1 ∂ 1 ∂ 1 ∂ r2 r2 sin θ + sin θ 4r2 cos θ + r2 tan θ r2 ∂r r sin θ ∂θ r sin θ ∂φ 1 1 4r sin2 θ + cos2 θ − sin2 θ = 2 4r3 sin θ + 4r2 cos2 θ − sin2 θ = r r sin θ sin θ cos2 θ = 4r . sin θ
∇·v =
Z
Z (∇·v) dτ =
= 2πR
cos2 θ 4r sin θ
4
2
π/6 Zπ/6 Z2π θ sin 2θ cos2 θ dθ dφ = R4 (2π) 4r dr + 2 4 0 0 0 0 √ ! √ 4 3 2π + 3 3 . π+3 = πR 12 2
r sin θ dr dθ dφ =
π sin 60◦ + 12 4
=
πR4 6
ZR
3
Surface coinsists of two parts: (1) The ice cream: r = R; φ : 0 → 2π; θ : 0 → π/6; da = R2 sin θ dθ dφ ˆ r; v·da = R2 sin θ 2 4 R sin θ dθ dφ. Z
R2 sin θ dθ dφ =
π/6 Zπ/6 Z2π 1 π 1 1 πR4 2 4 4 4 ◦ v·da = R sin θ dθ dφ = R (2π) θ − sin 2θ = 2πR − sin 60 = 2 4 12 4 6 0 0
(2) The cone: θ =
0
π 6;
φ : 0 → 2π; r : 0 → R; da = r sin θ dφ dr θˆ = Z
R
v · da =
πR4 2
√
3 ˆ 2 r dr dφ θ;
v · da =
√
3 r3 dr dφ
R 2π √ √ R4 √ Z 3 Z 3 4 v · da = 3 r dr dφ = 3 · · 2π = πR . 4 2 0
Therefore
√ ! 3 π−3 2
π 3
√
−
3 2
+
√ 3 =
0 πR4 12
√ 2π + 3 3 .
X.
Problem 1.60 H H R R (a) Corollary 2 says (∇T )·dl = 0. Stokes’ theorem says (∇T )·dl = [∇×(∇T )]·da. So [∇×(∇T )]·da = 0, and since this is true for any surface, the integrand must vanish: ∇×(∇T ) = 0, confirming Eq. 1.44. H H R R (b) Corollary 2 says (∇×v)·da = 0. Divergence theorem says (∇×v)·da = ∇·(∇×v) dτ. So ∇·(∇×v) dτ = 0, and since this is true for any volume, the integrand must vanish: ∇(∇×v) = 0, confirming Eq. 1.46. Problem 1.61 H R (a) Divergence theorem: v · da = (∇·v) dτ. Let v = cT , where c is a constant vector. Using product rule #5 in front cover: ∇·v = ∇·(cT ) = T (∇·c) + c · (∇T ). But c is constant so R∇·c = 0. Therefore we have: R R R c · (∇T ) dτ = T c · da. Since c is constant, take it outside the integrals: c · ∇T dτ = c · T da. But c c
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24
CHAPTER 1. VECTOR ANALYSIS
is any constant vector—in particular, it could be be x ˆ, or y ˆ, or ˆ z—so each component of the integral on left equals corresponding component on the right, and hence Z Z ∇T dτ = T da. qed R R (b) Let v → (v × c) in divergence theorem. Then ∇·(v × c)dτ = (v × c) · da. Product rule #6 ⇒ ∇·(v × c) = c · (∇×v) − v · (∇×c) = c · (∇×v). (Note: ∇×c = R0, since c is constant.) R Meanwhile vector indentity (1) says da · (v × c) = c · (da × v) = −c · (v × da). Thus c · (∇×v) dτ = − c · (v × da). Take c outside, and again let c be x ˆ, y ˆ, ˆ z then: Z Z (∇×v) dτ = − v × da. qed R R (c) Let v = T ∇U in divergence theorem: ∇·(T ∇U ) dτ = T ∇U · da. Product rule #(5) ⇒ ∇·(T ∇U ) = T ∇·(∇U ) + (∇U ) · (∇T ) = T ∇2 U + (∇U ) · (∇T ). Therefore Z Z 2 T ∇ U + (∇U ) · (∇T ) dτ = (T ∇U ) · da. qed R R (d) Rewrite (c) with T ↔ U : U ∇2 T + (∇T ) · (∇U ) dτ = (U ∇T ) · da. Subtract this from (c), noting that the (∇U ) · (∇T ) terms cancel: Z Z T ∇2 U − U ∇2 T dτ = (T ∇U − U ∇T ) · da. qed R H (e) Stokes’ theorem: (∇×v) · da = v · dl. Let v = cTR. By Product Rule H#(7): ∇×(cT ) = T (∇×c) − c × (∇T ) = −c × (∇T ) (since c is constant). Therefore, − (c × (∇T R )) · da = T c ·Hdl. Use vector indentity #1 to rewrite the first term (c × (∇T )) · da = c · (∇T × da). So − c · (∇T × da) = c · T dl. Pull c outside, and let c → x ˆ, y ˆ, and ˆ z to prove: Z I ∇T × da = − T dl. qed Problem 1.62 (a) da = R2 sin θ dθ dφ ˆ r. Let the surface be the northern hemisphere. The x ˆ and y ˆ components clearly integrate to zero, and the ˆ z component of ˆ r is cos θ, so Z Z π/2 sin2 θ π/2 2 2 a = R sin θ cos θ dθ dφ ˆ z = 2πR ˆ z sin θ cos θ dθ = 2πR2 ˆ z = πR2 ˆ z. 2 0 0 H (b) Let T = 1 in Prob. 1.61(a). Then ∇T = 0, so da = 0. qed (c) This follows from (b). For suppose a = 6 a ; then if you put them together to make a closed surface, 1 2 H da = a1 − a2 6= 0. (d) For one such triangle, da = 12 (r × dl) (since r × dl is the area ofH the parallelogram, and the direction is perpendicular to the surface), so for the entire conical surface, a = 21 r × dl. (e) Let T = c · r, and use product rule #4: ∇T = ∇(c · r) = c × (∇×r) + (c · ∇)r. But ∇×r = 0, and ∂ ∂ ∂ (c · ∇)r = (cx ∂x + cy ∂y + cz ∂z )(x x ˆ+yy ˆ +zˆ z) = cx x ˆ + cy y ˆ + cz ˆ z = c. So Prob. 1.61(e) says I I Z Z Z T dl = (c · r) dl = − (∇T ) × da = − c × da = −c × da = −c × a = a × c. qed
c
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CHAPTER 1. VECTOR ANALYSIS
25
Problem 1.63 (1) 1 ∂ ∇·v = 2 r ∂r
1 r · r 2
=
1 ∂ (r) = r2 ∂r
1 r2 .
For a sphere of radius R: R R 1 R v · da = r · R2 sin θ dθ dφ ˆ r = R sin Rˆ ! θ dθ dφ = 4πR. So divergence R R 1 2 RR R (∇·v) dτ = sin θ dθ dφ = 4πR. r sin θ dr dθ dφ = dr 2 r theorem checks. 0
Evidently there is no delta function at the origin. ∇× (rn ˆ r) =
1 ∂ 1 ∂ 1 r2 rn = 2 rn+2 = 2 (n + 2)rn+1 = (n + 2)rn−1 2 r ∂r r ∂r r
(except for n = −2, for which we already know (Eq. 1.99) that the divergence is 4πδ 3(r)). (2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero. To be certain there is no lurking delta function here, we integrate over a sphere of radius R, using R H ? Prob. 1.61(b): If ∇×(rn ˆ r) = 0, then (∇×v) dτ = 0 = − v × da. But v = rn ˆ r and da = R2 sin θ dθ dφ ˆ r are both in the ˆ r directions, so v × da = 0. X Problem 1.64 (a) Since the argument is not a function of angle, Eq. 1.73 says 2r r3 1 1 d 1 1 d 2 D =− r − = 4π r2 dr 2 (r2 + 2 )3/2 4πr2 dr (r2 + 2 )3/2 3 2 1 3 1 32 r 2r 3r2 3r 2 2 2 = − = r + − r = .X 4πr2 (r2 + 2 )3/2 2 (r2 + 2 )5/3 4πr2 (r2 + 2 )5/2 4π(r2 + 2 )5/2 (b) Setting r → 0: D(0, ) =
32 3 = , 5 4π 4π3
which goes to infinity as → 0. X (c) From (a) it is clear that D(r, 0) = 0 for r 6= 0. X (d) Z Z ∞ 1 r2 2 2 2 D(r, ) 4πr dr = 3 dr = 3 = 1. X 32 (r2 + 2 )5/2 0 (I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for δ 3 (r).
c
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26
CHAPTER 2. ELECTROSTATICS
Chapter 2
Electrostatics Problem 2.1 (a) Zero. 1 qQ , where r is the distance from center to each numeral. F points toward the missing q. 4π0 r2 Explanation: by superposition, this is equivalent to (a), with an extra −q at 6 o’clock—since the force of all twelve is zero, the net force is that of −q only.
(b) F =
(c) Zero. 1 qQ , pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as 4π0 r2 a cancellation in pairs of opposite charges (1 o’clock against 7 o’clock; 2 against 8, etc.), with one unpaired q doing the job, then you’ll need a different explanation for (d). (d)
Problem 2.2 This time the “vertical” components cancel, leaving E= E=
q 1 4π0 2
r
2
sin θ x ˆ, or
1 qd x ˆ. 4π0 z 2 + d 2 3/2 2
 E AAU r θ AA z A A s As x q −q
1 qd z, which, as we shall see, is the field of a dipole. (If we From far away, (z d), the field goes like E ≈ 4π 3 ˆ 0 z set d → 0, we get E = 0, as is appropriate; to the extent that this configuration looks like a single point charge from far away, the net charge is zero, so E → 0.)
c
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1 Problem 2.3 CHAPTER 2. ELECTROSTATICS
27 1
Problem 2.3 Problem 2.3 z z " !" ! θ η z dq = λdx #L !x "# $!x θ r z dq = λdx
% L λ dx 1% 2 Ez =1 4π" =2 z 2 +2 x2 ; cos θ z= ηz ) L 2 cos θ; 2(η R λ ηcos 0 dx 0 λ dx 2 2 θ = η) z EzE = =4π"01 0 L θ; (η = z 2 η r =+z 2x+; xcos ; cos θ = r ) z 4π0 0 r 2 cos θ; ( % L 1 % 1 =1 4π"0 λzLR L0 (z12 +x2 )3/2 dx 1λz 1 = =4π" dxdx 2 3/2 '( λz0 &(z2 +x 0 2 ) 2 )3/2 (L 4π x '(L 10 & h0 1(z√+x 1 λ√ L . =1 4π"0 λz1 √z2 x z2 +x(2i L ( =1 4π" L zL 2 +L2 1λ 0√ λz√ . . = =4π"01λzλz z%2 12 z√2 +xx2 ( =0 =4π" 2 +L2 0 z z% z L 2 2 2 2 4π0 1% z λ dx 4π 0 1 x dx z +x +L %0λR zx dx 0 − E = −1 4π"0LR L 2 sin θ = 3/2 λ0 dx 1 4π" ηsin (x2x+z #L dx2 ) 1 1λ 0 ExEx=x =− − θ= − 4π" 2λ dx 2 +z 2 )3/2 & '( & ' ! λ sin θ = − 4π" η 0 0 0 (x 2 L 2 2 )3/2 4π10 & 0 r 4π0 1 & (x +z ! "# $ ( '( 1 1 1' i L x h z−√ i L = −1 4π"0 λh √− √ = − . λ ( ( 1 1 1 1 2 2 2 2 x 4π" x1 +z = − z1 +L. 1λ 0 1λ − √ 1 √√ = =− − − 2 +z 2 ( 0 = − 2 +L2 λ − λ − . 4π" 4π" z 0 0 x z 2 2 2 2 4π0 z z +L )* + * 4π0 + x, +z 0 0 λ z + * L + , 1 )* −1 + √z z x ˆ + √L L ˆ z . E =1 1 λ λ √√ z 2 + L2x √√ z 2 + L2ˆ ++ ˆ+ zˆ . . EE = = 4π% 0 z −1 x ˆ + z −1 2 2 2 2 4π% z z+ L L2 z z+ L L2 0 z z 2+ 2+ 4π 0 1 λL For z # L you expect it to look like a point charge q = λL: E →1 4π" z. It checks, for with z # L the x ˆ λL z 2 ˆ 1 20λL For z# L Lyou expect it ittotolook like a point charge q = λL: E → ˆ z. ˆ It checks, for with z # L the x ˆˆ 4π" For z you expect look like a point charge q = λL: E → 1 λL 0 z 2 z. It checks, for with z L the x 4π z ˆ z. term → 0, and the ˆ z term →1 4π" 0 zL.zz. term →→ 0, 0, and the ˆ zˆ term →→4π"01λz 0Lzλzˆ term and the z term 4π0 z z ˆ Problem 2.4 Problem 2.4 q 2 . /. a /2 Problem 2.4 2.1, with L → a and z → z +a 222 (distance from center of edge to P ), field of one edge is: From Ex. 2 z→ from center of of edge toto P ), field of of one is:is: z 2z+ From Ex. 2.1, with LL →→a2 aand 2 +2 a (distance From Ex. 2.2, with and z → (distance from center edge P ), field oneedge edge 2 2 1 λa λa. E =1 1 λa 0q 2 .a2. E1E=1= 4π% a2q 2 a2 + z + z + 2 2 2 1 4π%0 a 4 2 a 24 a 24 4π0 z 2z+ 2 +4 a2 z z+ 2 +4 a+ 4a 4 4 + 4 z There are 4 sides, and we want vertical components only, so multiply by 4 cos θ =q4 q z 2 a2 : There are 4 sides, and we want vertical components only, so multiply by 4 cos θ = 4 z 2+ :4 z q There are 4 sides, and we want vertical components only, so multiply by 4 cos θ = 4 z2 + a a2 : 4 z2 + 4 4λaz 1 4λaz E =1 1 . ˆ z . / 4λaz 2 EE = = 4π% z2.ˆ . 0 z 2 a+2 /aq 2 2 aˆ z. 4π% 0 z2 + 4 2 z a+ 2 4π 0 z 2 +4 a2 z z+ 2 +2 a2 4
Problem 2.5 Problem 2.5 Problem 2.5 ! ! θ z
θ z r
r
2
0% 1 1θ o ˆ 10%nR λdl cos z. “Horizontal” components cancel, leaving: E =1 4π" 1 0 λdlλdl η2 θ ˆ z. ˆ “Horizontal” components cancel, leaving: = =4π"4π “Horizontal” components cancel, leaving:E E z. 2 cos θ η 2%rcos 0 0 z 2 2 2 % Rdl = 2πr. So Here,2 η2 =2 r2 +2 z2 , cos θ z= zη (both constants), while Here, z z, cos θ θ==η (both constants), while So So Here, ηr ==r r++ , cos constants), whiledl =dl2πr. = 2πr. r (both 1 λ(2πr)z λ(2πr)z E =1 1 λ(2πr)z z. . zˆ 3/2 EE = = 4π ˆ z. ˆ 4π%00(r(r22++zz3/2 22 3/2 2 + z 2 ) )) 4π%0 (r
r
Problem 2.6 Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the field of each ring. Total charge of a ring is σ · 2πr · dr = λ · 2πr, so λ = σdr is the “line charge” of each ring. Ering
1 (σdr)2πrz 1 = ; Edisk = 2πσz 3/2 2 2 4π0 (r + z ) 4π0 1
1
1
R
Z 0
r (r2
c "2005 Pearson Education, Inc., Upper Saddle River, rights reserved. Edisk = NJ. All2πσz − √ This material ˆ z. is 2 + z 2 may be protected under all copyright laws as they currently exist. 4π0No portionz of thisRmaterial reproduced, in any form or by any means, without permission in writing from the publisher.
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is c protected "2005 Pearson Education, Inc., laws Upper River, NJ. All No rights reserved. This material under all copyright as Saddle they currently exist. portion of this material mayis be protected underinall copyright as means, they currently No portion of this material may be reproduced, any form or laws by any withoutexist. permission in writing from the publisher. reproduced, in any form or by any means, without permission in writing from the publisher.
3/2
+ z2)
dr.
1 CHAPTER 2. ELECTROSTATICS
28
For R z the second term → 0, so Eplane =
and E
−1/2
2 √ 1 = z1 1 + R z2 R2 +z 2 1 2πR2 σ 1 Q = 4π 2z 2 = 4π0 z 2 , 0
For z R,
Contents
≈
1 z
1 z 4π0 2πσˆ
1−
1 R2 2 z2
where Q = πR2 σ.
=
σ ˆ z. 20
, so [ ] ≈
1 z
−
1 z
+
1 R2 2 z3
=
R2 2z 3 ,
X z #
Problem 2.7 E is clearly in the z direction. From the diagram, dq = σda = σR2 sin θ dθ dφ, r 2 = R2 + z 2 − 2Rz cos θ, cos ψ = z−Rrcos θ .
z
ψ
r
So θ
R
" y
φ Z R 1 σR2 sin θ dθ dφ(z − R cos θ) Ez = dφ = 2π. . 4π0 (R2 + z 2 − 2Rz cos θ)3/2 ! Z π x (z − R cos θ) sin θ 1 θ = 0 ⇒ u = +1 (2πR2 σ) dθ. Let u = cos θ; du = − sin θ dθ; . = 2 2 3/2 θ = π ⇒ u = −1 4π0 0 (R + z − 2Rz cos θ) Z 1 1 !2 σR21sin θ dθ dφ(z z − Ru " − R cos θ)du. Integral = can be done by partial fractions—or look it up. 2 + z 2 − 2Rzu)3/2 . dφ = 2π. Ez4π = 0 (2πR σ) −1 (R 2 2 3/2 4π%0 (R + z − 2Rz cos θ) 1 ! # $ 1 1 2 1 π zu(z−−RR cos θ) sin θ 1 2πR2 σ (z − R) (−z − R) θ = 0 ⇒ u = +1 2 √ = = = (2πR − . σ) dθ. z 2 Let zu − = R cos θ; du =R − sin θ dθ; . 2 −22Rzu 3/2 0 4π04π%0 (2πR σ) z 2 0R2(R z + +2z+ θ = π ⇒ u = −1 z − 2Rz−1 cos θ)4π ! 1 1 z − Ru = (2πR2 σ) Integral can 1 be q done by partial fractions—or look it up. q 1 du. 1 2 4πR2 σ 2 4π%0the sphere),−1E(R = 3/2 z. For z > R (outside z0 −z2Rzu) 2 z =+ 4π 4π0 z 2 , so E = 4π z 2 ˆ 0 &1 % # $ 2 1 2πR σ (z − R) (−z − R) 1 1 zu − R 2 For z < R (inside), E(2πR so E = 0. √ = = − . z = 0,σ) 4π%0 z 2 R2 + z 2 − 2Rzu −1 4π%0 z 2 z − R z + R Problem 2.8 According to Prob. 2.7, all shells interior to the point (i.e. at smaller 1r) contribute as though their charge q 1 q 1 4πR2 σ ˆ z. = 4π" = z > R (outside thecenter, sphere), Ez all = exterior 2 , so E wereFor concentrated at the while contribute nothing. Therefore: 4π"0 z 2 shells 2 0 z 4π% z 0
For z < R (inside), Ez = 0, so E = 0.
E(r) =
1 Qint ˆ r, 4π0 r2
Problem 2.8 to Prob. 2.7,interior all shells to Outside the pointthe (i.e. at smaller r) charge contribute as though where QAccording charge to interior the point. sphere, all the is interior, so their charge int is the total were concentrated at the center, while all exterior shells contribute nothing. Therefore: 1 Q E= r.Qint 1ˆ ˆ r, E(r)4π =0 r 2 4π%0 r2 Inside the Q sphere, only that fraction of the total which is interior to the point counts: where int is the total charge interior to the point. Outside the sphere, all the charge is interior, so
Qint =
4 3 3 πr Q 4 3 3 πR
=
r3 1 r3 1 1 Q Q, so EE== 1 Q ˆ r= r. r3.Q r2 ˆ 3 3 R 4π R 4π 2 0 0 R 4π% r 0
Inside 2.9 the sphere, only that fraction of the total which is interior to the point counts: Problem ∂ (a) ρ = 0 ∇·E = 0 r12 ∂r r2 · kr3 = 0 r12 k(5r4 ) = 50 kr2 . c "2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be cmeans,Pearson Upper from Saddle NJ. All rights reserved. This material is reproduced, in any form or by any 2012 withoutEducation, permissionInc., in writing theRiver, publisher. protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Gaussian surface r $ #
111
Contents 30
CHAPTER 2. ELECTROSTATICS
Problem 2.15 Qenc = 0, so E = 0. Problem(i)2.15 & & (i) Qenc =%0, so E = 0. 2 ) = !10 QencR= !10 ρ dτR= !10 r¯k2 r¯2 sin θ d¯ r dθ dφ H (ii) E·da = E(4πr 1 1 2 2 '1 k2 r¯( Problem 2.13 (ii) E·da = E(4πr ) = Q = ρ dτ = sin θ d¯ r dθ dφ enc 0 0 r¯ &r 4πk 2.16 k 0r − a ˆ Problem r. r = 4πk = !0 !0 (r − a) ∴ kE =r #− a r2 a d¯ R E r 4πk 0 ! d¯ r = (r − a) ∴ E = ˆ r . = 4πk 0 0 a 2 0 r & 4πk 4πk b d¯ r = (b − a), so (iii) E(4πr2 ) = R !0 a b !0 ' a d¯ (iii) E(4πr2 ) = 4πk r( = 4πk 0 0 (b − a), so k b−a ! E=b − a 2 ˆ r. Problem Problem 2.13 ! E·da = E · 2πs · l = !10 Qenc = !10 ρπs2 l; Problem2.13 2.13k(i) Gaussian surface #0 r. ˆ r E= 2 % Problem 2.16 Problem Problem 2.16 2.16 0 r ρs l r b ˆ s. a E= 2#0 Problem 2.16 2.16 Problem H ! E·da = E · 2πs · l = 1 Q = 10 ρπs2 l; Gaussian c !2005 Pearson Education, Inc., Upper NJ. All! reserved. This material is0 enc !rights !surface " Saddle River, 222 under all copyright! laws they currently exist. No portion of may (i) Q l; E·da E 2πs s asGaussian (i) Q = ρπs l;l; ! E·da = EE·!·ρs 2πs = (i) Qenc = !!110!010ρπs ρπs ! E·da= =this ·material 2πs···lll= = !!110!010be (i) protected(ii) 1 1 enc 2 enc= surface Gaussian surface Gaussian surface E·da =E · 2πs ·l = E= ˆ s.publisher. reproduced, in any form or by any means, without permission in writing from the !0 Qenc = !0 ρπa l; ρs ρs ρs 2 0 lll 2 sˆ E sˆ E = s... E= = ˆ 2# 2# 2#000 E = ρa ˆ s. 2#0 s l
Contents Contents
(ii) (ii) (ii) (ii)
" " " sss (iii) lll
! H ! ! "Gaussian Gaussian surface Gaussiansurface surface E·da = E · 2πs ·1l = 10 Qenc = 10 ρπa2 l; ! Gaussian!!!surface s 222 E·da ·!··2πs l; E·da = EEρa 2πs = Q = ρπa l;l; E·da= =E 2πs = !!110!010Q Qenc = !!110!010ρπa ρπa 2 ···lll= enc enc= 1 Q E·da = E · 2πs · l = E = 222 ˆ s. !0 enc = 0; ρa ρa ρa 20 s ˆ sˆ E sˆ E = s...E = 0. E= = 2# 2# 2#000sss l ! ! ! Gaussian Gaussian surface Gaussiansurface surface
" " " sss
!!!H 1 = E·da ···ll·l= E·da= ·2πs 2πs l==!!110!01Q Q = 0; E·da = EEE···2πs 2πs = QQ ==0; 0;0; E·da ==E enc enc enc enc 00 E = 0. E = 0. EE == 0.0.
(iii) (iii) (iii) (iii)
Contents
E "
lll
Problem 2.17 # E E Ea" On the x z plane E = 0 by symmetry. Set up Gaussian “pillbox” with and the a one face b in this plane s " " other at y. c2.17 Pearson Upper River, NJ. Allup rights reserved. This material with is Problem!2005 On theEducation, x z planeInc., E= 0 bySaddle symmetry. Set a Gaussian “pillbox” one face in this plane protected under all copyright laws as they currently exist. No portion of this material may be and the other at y. reproduced, in any form or by any means, without permission in writing from the publisher. Gaussian pillbox
# # 1 1 # E·da = aa E a · A = 0bbbQenc = 0ssAyρ; s ρ yy ˆisisis(for y < d). E= cc!2005 !2005 Pearson material !2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material c PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This This material 0maybe protected ! protected under all copyright laws as they currently exist. No portion of this material may be protectedunder underall allcopyright copyright lawsas asthey theycurrently currentlyexist. exist.No Noportion portionof ofthis thismaterial materialmay be y " laws
R
#
reproduced, reproduced, in any form or by any means, without permission in writing from the publisher. reproduced,in inany anyform formor orby byany anymeans, means,without withoutpermission permissionin inwriting writingfrom fromthe thepublisher. publisher.
Qenc =
1 !0 Adρ
⇒ E=
ρ > d). Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is dy ˆ 2012 c(for yPearson "0 protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E$ ρd "0
−d
d
! y
Problem 2.17 On the x z plane E = 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the other at y. !
Gaussian pillbox # CHAPTER 2. ELECTROSTATICS
E·da = E · A = !10 Qenc = !10 Ayρ; ρ E = yy ˆ (for y < d). "0 31
! y" ρ Qenc =1 !10 Adρ ⇒ E =ρ d y ˆ (for y > d). Qenc = 0 Adρ ⇒ E = "d0 y ˆ (for y > d). 0 E$ ρd "0
−d
d
! y
Problem 2.18 Problem 2.18 From Prob. 2.12, the field inside the positive sphere is E+ = 3ρ0ρr+ , where r+ is the vector from the positive From Prob. 2.12, the field inside the positive sphere is E+ = 0 r+ , where rρ+ is the vector from the positive center to the point in question. Likewise, the field of the negative3!sphere is − 30ρr− . So the total field is center to the point in question. Likewise, the field of the negative sphere is − 3!0 r− . So the total field is ρ EE== ρ(r(r r−r) ) + −− "r−r− r + − 33" *&−− 0 0 % r+r+ d ρρ d r E = d. But (see diagram) r − r = d. So + − But (see diagram) r+ − r− = d. So E =30 d. ++ 3"0 Problem Problem2.19 2.19 Z" Z" # $ ˆ %& rˆηˆ 11 11 ηˆr ∇×E = ρ dτ = ∇× ∇× (sinceρρdepends dependson onr"r,0 ,not notr)r) ∇× r 22 ρ dτ = ∇× r2 2 ρ ρdτdτ (since ∇×E = 4π 4π 0 0 4π" η 4π"00 η $ % rˆηˆ ==0 0 (since 1.63). (since∇× ∇× 22 = 0, from Prob. 1.62). rη Problem Problem2.20 2.20 ' ' x 'x y ˆy zˆ ˆ ˆ z '' ˆ '∂ ˆ ' ∂ ∂ ∂ ∂ ∂ = ' (1) ==k k x[ˆ (0 ˆ(0 ˆ z(0 (1)∇×E ∇×E x(0−−2y) 2y)++y y ˆ(0−−3z) 3z)++ ˆ z(0−−x)] x)]6=#=0,0, 1 1 ∂x ' ∂x ∂y∂y ∂z∂z ' =k k[ˆ xy ' ' xy2yz 2yz3zx 3zx sosoEE ananimpossible impossibleelectrostatic electrostaticfield. field. 1 1is is
' ' 'x x ˆ z '' y ˆy ˆ zˆ '∂ ˆ ˆ ∂ ∂∂ ∂ ∂ ' ' ∂x x(2z−−2z) 2z)++y y ˆ(0−−0)0)++ ˆ z(2y−−2y)] 2y)]==0,0, (2)∇×E ∇×E (2) ==k k [ˆ x[ˆ (2z ˆ(0 ˆ z(2y 2 2 ∂y∂y ∂z∂z = ' =k k ' ∂x 2 2 y'2y 2xy ' 2xy++z 2z 2yz 2yz
possibleelectrostatic electrostaticfield. field. sosoEE a apossible 2 2is is
z
6
Let’s go by the indicated path: E·dl = (y 2 dx + (2xy +
c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be any form or by any means, without permission in writing from the publisher. 2 r(x0 , y0 , z0 ) zreproduced, )dy + 2yzindz)k
6
Step I: y = z = 0; dy = dz = 0. E·dl = ky 2 dx = 0.
III
Step II: x = x0 , y : 0 → y0 , z = 0. dx = dz = 0. 2 = 2kx0 y dy. RE·dl = k(2xy + Rz y)dy 0 E·dl = 2kx y dy = kx0 y02 . 0 0 II Step III : x = x0 , y = y0 , z : 0 → z0 ; dx = dy = 0. E·dl = 2kyz dz = 2ky0 z dz.
x
I = = II
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

y
32 R III
E·dl = 2y0 k
R z0 0
CHAPTER 2. ELECTROSTATICS
z dz = ky0 z02 .
(x0 ,y R0 ,z0 )
E·dl = −k(x0 y02 + y0 z02 ), or V (x, y, z) = −k(xy 2 + yz 2 ).
V (x0 , y0 , z0 ) = −
0
Check :
∂ ∂ ∂ −∇V =k[ ∂x (xy 2 +yz 2 ) x ˆ+ ∂y (xy 2 +yz 2 ) y ˆ+ ∂z (xy 2 +yz 2 ) ˆ z]=k[y 2 x ˆ+(2xy+z 2 ) y ˆ+2yz ˆ z]=E. X
Problem 2.21 Rr V (r) = − ∞ E·dl.
So for r > R: V (r) = −
Outside the sphere (r > R) : E =
1 q r. 4π0 r 2 ˆ
Inside the sphere (r < R) :
q 1 r. 4π0 R3 rˆ
Rr ∞
1 q 4π0 r¯2
RR
1 q 4π0 r¯2
d¯ r=
1 4π0 q
Rr
d¯ r− ∞ q 1 r2 3− 2 . = 4π0 2R R
and for r < R: V (r) = −
When r > R, ∇V =
q ∂ 4π0 ∂r
When r < R, ∇V =
q 1 ∂ 4π0 2R ∂r
1 r
R
1 r¯
E=
r = ∞
q 1 ¯ 4π0 R3 r
q 1 , 4π0 r d¯ r=
q 4π0
h
1 R
−
1 R3
r 2 −R2 2
i
q q 1 1 ˆ r = − 4π r, so E = −∇V = 4π r. X 2ˆ 2ˆ 0 r 0 r 2 q q 1 2r r 3 − Rr 2 ˆ r = 4π −R r = − 4π r; so E = −∇V = 2 ˆ 3ˆ 0 2R 0 R
q 1 r.X 4π0 R3 rˆ
V(r) 1.6 1.4 1.2 1
(In the figure, r is in units of R, and V (r) is in units of q/4π0 R.)
0.8 0.6 0.4 0.2 1
0.5
1.5
2
2.5
3
r
Problem 2.22 1 2λ E = 4π ˆ s (Prob. 2.13). In this case we cannot set the reference point at ∞, since the charge itself 0 s extends to ∞. Let’s set it at s = a. Then s R s 1 2λ 1 − V (s) = − a 4π d¯ s = 2λ ln . ¯ 0 s 4π0 a
(In this form it is clear why a = ∞ would be no good—likewise the other “natural” point, a = 0.) ∂ 1 1 2λ ∂s 2λ 1s ˆ ∇V = − 4π ln as ˆ s = − 4π s = −E. X 0 0 Problem 2.23 Rb R0 V (0) = − ∞ E·dl = − ∞ =
k 0
1−
a b
Ra R0 − b k0 (r−a) dr − a (0)dr = r2 k b a a − ln b − 1 + b = ln . 0 a k (b−a) dr 0 r 2
Problem 2.24 Using Eq. 2.22 and the fields from Prob. 2.16: Rb Ra Rb Ra V (b) − V (0) = − 0 E·dl = − 0 E·dl − a E·dl = − 2ρ0 0 s ds −
ρa2 20
k (b−a) 0 b
−
k 0
ln
a b
+a
1 a
−
Rb
1 ds a s
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1 b
Contents
CHAPTER 2. ELECTROSTATICS
=−
ρ 20
a
s2 2
+
0
ρa2 20
b
ln sa = −
33 ρa2 40
1 + 2 ln
b . a
Problem Problem2.25 2.25 11 2q 2q q .. ! 4π 4π"00 z 22+ "dd#22 z + 22 L √ R L 1 λ √z 2 + x2 ) %L $ L √ λ2λdx (b) V = 4π 10 −L dx2 = 4πλ0 ln(x + 2 + x2 )−L % z +x √ (b) V = 4π" z = ln(x + 4π"0 −L z 2 +x2 −L 0 ! " # √ √ ' & √ 2 2 λ z2 + L + z 2+ L 2 λ ( L+ ) L2 √ λ L + z + L √ . = ln = λ ln L+ z2 +L2 = 4π0 ln −L + √z 2 + L2 . = 2π 2π"00 ln z z 4π"0 −L + z 2 + L2 (a) (a) VV ==
) %R √ $ R σ 2πr dr σ (* 2 Z 1 2p 2 )% = R 2−z . R√ p (c) V = r + z R = 2πσ ( + z σ 2πr dr σ 1 2 2 4π"0 0 √r +z = 2πσ ( r2 +0z 2 ) 2" =0 R2 + z 2 − z . (c) V = 20 4π0 0 4π0 0 r2 + z 2 ∂V In each case, by symmetry ∂V ∴ E = − ∂V z. ∂y = ∂x = 0. ∂z ˆ ∂V In each case, by symmetry ∂V = = 0. ∴ E = − ∂V z. ∂y ∂x ∂z ˆ
z
r x
1 10 4π"
" # 2qz 1 1 2z ”3/2 ˆ z= z (agrees with Prob. 2.2a). (a) E = − 4π" 2q − 21 “ " #3/2 ˆ 0 d 2 2 4π"0 z12 + " d #22qz 1 1 z +( 2 ) 2z 2 ˆ z= (a) E = − ˆ z (agrees with Ex. 2.1). 2q − 3/2 4π0 + 2 2 4π0 z 2 + d 2 3/2 , d 2 z + 2 1 √ 21 λ √1 2z − (−L+√1z2 +L2 ) 12 √z21+L2 2z ˆ z (b) E = − 4π" 2 2 2 0 z 2 +L2 (L+ z +L ) λ 1 √ 1 1 1 1+ , √ 1 1 2Lλ 2 2 2 2 √ −L− z +L (b) E = − 2z − ˆ 2z with ˆ z Ex. 2.1). λ √ z √ −L+ z +L √ 2 ) 2 √zˆ z2 + =4π − 4π" z = + √z 2 + (agrees 2 2 2 + 2L2 (−L L2 0 0 (L z 2+ +L2 z + L )(z22 +Lz2 )−L 2 4π"0 z zL2 + L ) ( √ √ λ+ 2Lλ z −L + z 2 +L2 − L − z 2 + 1 . L2 , √ = −σ 1 √ 1 ˆ z= ˆ z (agrees with Ex. 2.2). σ2 z2 2 2 2 2 −L 4π0 zProb. z +2.6). L2 0 √ z 2 +2L z = (z +1L−)√ ˆ z (agrees with (c) E = − 2"4π 2 R +z 2z − 1 ˆ 0 2"0 R2 + z 2 σ 1 1 σ z the in − (a)1 is ˆ 0 , which, naively, suggests E = −∇V √ charge 2z (c) EIf = −righthand with Prob. 2.6). = 0, in contradiction z−q, = then V 1 −= √ ˆ z (agrees 2 + z2 2 20 2 toRProb. 20 is that R z 2 know V on the z axis, and from this we cannot with the answer 2.2b. The point we + only ∂V hope to compute Ex = − ∂V ∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that V = 0 ,sowhich, naively, E =is −∇V = 0, in If the righthand charge in (a) is −q, then Ex = Ey = 0. But now E points in the x direction, knowing V onsuggests the z axis insufficient tocontradiction determine E. 2 with the answer to Prob. 2.2. The point is that we only know V on the z axis, and from this we cannot hope ∂V to compute Ex = − ∂V ∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that Ex = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insufficient to determine E.
Problem 2.26 " c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b h
√
1 V (a) = 4π0
Z 0
2h
σ2πr
r
(where r =
r
2πσ 1 √ σh √ ( 2h) = dr = 4π0 2 20 √ / 2)
r¯ r
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
h
a
!
34
√
1 V (b) = 4π0
Z 0
2πσ 1 √ 4π0 2
=
σ √ 2 20
=
2h
√
Z
σ2πr r¯
CHAPTER 2. ELECTROSTATICS
dr
2h
0
q h2 +
p h2 +
r
2
−
r
√
(where r¯ =
r 2
−
2h r
√
2h r
q h2 +
r
2
−
√
2h r )
dr
q h + √ ln(2 h2 + 2
r
2
−
√
2h r + 2 r −
√
√ 2h 2h) 0
√ √ √ σ h h = √ h + √ ln(2h + 2 2h − 2h) − h − √ ln(2h − 2h) 2 20 2 2 √ ! h i √ √ σ h σh 2+ 2 σh √ ln(2h + 2h) − ln(2h − 2h) = √ = √ ln ln = 40 40 2 20 2 2− 2 √ √ i σh σh h = ln(1 + 2). ∴ V (a) − V (b) = 1 − ln(1 + 2) . 20 20
(2 +
√
2)2
!
2 3
Problem 2.27 Problem 2.27 Cut the Cut cylinder into slabs, shownasinshown the figure, the cylinder intoasslabs, in theand figure, and use result of result Prob. of 2.25c, z →with x and σ x→and ρ dx: use Prob.with 2.25c, z→ σ → ρ dx: V =
=
! "# $
√z+L/2 % &√ ' ρ R2 + x2 −2x dx2 R + x − x dx 2"0 z−L/2 # z−L/2 √ √ √ ( √ ρ 1 2 2z+L/2 2 )*z+L/2 21 + 2 +2R2 ln(x ρ 2 R2 + x2 ) − 2 2 * x R x + x ) − x z−L/2 R + x z−L/2 20 2 = 2"0 2 x R + x + R ln(x + 8 9 3 r 2 3 9 8 r 2 q q L 2 < = z+ L + R2 +(z+ q q L +2 )R2 + z+ L 2 = < 2 2 ρ z+ 2 L L 2 L 2 L 2 2 ( 4 2 5−2zL2 ) . 5 r 2 ρ z+ − z− +R ln R + z+ R + z− 2 L L L L ( ) ) ( ) ) ( ( 2 2 4 r2 40 : = 2 2 R +(z+ 2 ) −(z− ) 2 R +(z− ) +R ln −2zL . 4"0 :(z+ 2 ) 2 2 2 z− L + R2 + z− L 2 ; ( L +2 )R2 +(z− L; 2 z− )
ρ 20
=
z− L 2
z+L/2 R
V =
2
(Note:
2 2 L2 2' −(Note: z + L2− &z++zL−'2L2+ &z=−−z L 2− zL −2 4 = −z − 2 2
L2 2 + z− − L zL +2 4 = −2zL.) zL + z − zL + L4 = 4 2
2
L "#
!
$
!" "
$! x
dx
2
−2zL.)
s (s 2 2 2 , , L 2& + ' '2 z + . .z2 − L2 & ∂V ˆ zρ L L2 L 2 2 2 2 z + −2 R + z −2 z − L2 E = −∇V = −ˆ z = −∂V Rˆ + qL − qL zρ+ z +2 2 2 / / ∂z= −ˆ 2 z+ 2 + 2 z− 2 − R + E = −∇V z 40 = − L L '2 − R + '2 ∂z 4$0 2R + z +22 & 2R + z −2 2 & R + z + L2 R + z − L2 L L z− 2 z+ 2 ) " 1+ q 1 +0 q z− L # z+ L 2 L 2 1 2 2+ 1 (+z−qL2 )2 2 L 2 1 + R2 + z+q R ( ) 2 2 L 2 2 2 R +(z+ 2−) R +(z− 2 )− 2L q +R 2 q 2 2 − +R 2+ / 2+ / ' − 2L z + L2 + RL z +2L2 & z L−'2L2 + RL z −2L2 & L 2 z + 2 + R + z{z+ 2 z − 2 + R + }z − 2  $! 1 " # 1 q q − 1 1 2 / − / L 2& '22 + '2 R2 + z +2L2 & R z −22 R + z + L2 R + z − L2 s s 2 2 , Lˆ zρ .R22 + ,z − L  − 2L .2 2 E=− 2 ˆ R + z + − 2 zρ 4E0 2 z+ L 2 z − L − 2L =− − 2 R2 + 2 R2 + 4$0 2 2 River, NJ. All rights reserved. This material is c , 2012 , Inc.,Upper Saddle Education,  Pearson 2 copyright laws 2 protected under.all as they.currently exist. No portion of this material may be ρ L L 2 + z + in any + form R or2by without ˆ +anyzmeans, − = L − Rreproduced, z. permission in writing from the publisher. 2$0 2 2
Problem 2.28
z $
Orient axes so P is on z axis. P
V =
1
%
ρ
dτ.
=
2
Here√ρ is constant, dτ = r sin θ dr dθ dφ,
z
θ r
!
1 Hello
CHAPTER 2. ELECTROSTATICS
35
s 2 s 2 L L ρ ˆ = L − R2 + z + + R2 + z − z. 20 2 2 z #
Problem 2.28 Orient axes so P is on z axis. R ρ Here ρ√is constant, dτ = r2 sin θ dr dθ dφ, 1 V = 4π0 r dτ. r = z 2 + r2 − 2rz cos θ. ρ 4π0
Rπ
sin θ z 2 +r 2 −2rz cos θ
0
√
R
=
∴V =
ρ 4π0
· 2π · 2
( Rz 0
Contents
But ρ =
q 4 3, 3 πR
so V (z) =
R 2π
;
θ r
"y
0
dφ = 2π.
x!
√ √ π 1 r2 + z 2 − 2rz cos θ 0 = rz r2 + z 2 + 2rz − r2 + z 2 − 2rz 1 2/z , if r < z, (r + z − r − z) = 2/r , if r > z. √
1 rz
dθ =
z
φ
2 √r sin θ dr dθ dφ z 2 +r 2 −2rz cos θ
V =
r
P
1 rz
1 2 z r dr
+
RR 1
r r dr
z
3q 1 20 4πR3
) 2
R2 −
=
z2 3
n
ρ 0
=
1 z3 z 3
q 8π0 R
+
R2 −z 2 2
3−
z2 R2
o
=
ρ 20
; V (r) =
R2 −
z2 3
q 8π0 R
.
3−
r2 R2
. X
Problem 2.29 ∇2 V = =
1 4π0
1 2 4π ∇ R 00
R
ρ r
dτ =
1 4π0 0
R
r
ρ(r0 ) ∇2
1
dτ (since ρ is a function of r0 , not r)
ρ(r )[−4πδ 3 (r − r )] dτ = − 10 ρ(r). X
Problem 2.30. Problem 2.30.
(a) ˆn ˆn nnalways ˆn (a)Ex. Ex.2.5: 2.4: EEabove = 2σ2"σ00n ˆ; ;EEbelow =−−2σ2"σ00n ˆ(ˆ (ˆ alwayspointing pointingup); up);EEabove −EEbelow =σ0"σ0n ˆ. .X! above= below= above− below= Ex. Ex.2.6: 2.5: At Ateach eachsurface, surface,EE==00one oneside sideand andEE==σ0"σ0 other otherside, side,so so∆E ∆E==σ0"σ0. .X! 22
σR rˆ rˆ Prob. ˆ rr== σ0"σ0ˆ Prob.2.11: 2.11: EEout = σR r; ;EEinin==00; ;so so∆E ∆E== σ0"σ0ˆ r. .X! out= "0rr2 2 ˆ 0
! s
(b) (b)
H! Outside: E·da E·da==E(2πs)l E(2πs)l== 1"10QQenc = σ"10(2πR)l (2πR)l⇒ ⇒EE== σ"σ0RRsˆ s(at (atsurface). surface). ˆ s = σσsˆ enc= Outside: 0 0 0 s s = 0"0ˆ
! R
c σ !2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is s.this .X!material Inside: soEE== ∆E enc Inside: QQenc ==0,0, so 0.0. ∴∴ ∆E ==σ0"of0ˆ sˆ protected all copyright laws as they currently exist. No portion may be % " #$ under reproduced, in any form or by any means, without permission in writing from the publisher.
l
2
(c) Vout = RR0 r2σσ = Rσ (at surface); Vin = Rσ ; so Vout = Vin . X Rσ 0 0 (c) Vout = "0 r = Rσ "0 (at surface); Vin = "0 ; so Vout = Vin . !
∂Vin ∂Vin ∂Vout R2 σ σ out = 0 ; so ∂V = − σ0σ. X ∂rout = − 0Rr22σ = − 0σ (at surface); ∂r ∂rout − ∂r ∂Vin ∂V ∂Vin ∂V = − = − (at surface); = 0 ; so − ∂r "0 r 2 "0 ∂r ∂r ∂r = − "0 . ! c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be Problemin2.31 reproduced, any form or by any means, without permission in writing from the publisher.
1 4π"0
&
qi rij
1 4π"0
'
−q a
√q 2a
+ , + 2 q 1 . ∴ W4 = qV = −2 + √ 4π"0 a 2
(a) V =
=
1
)
−q2
*
+
1
−q a
(
)
=
q2
q 4π"0 a
q2
) −2 +
*
√1 2
*
. (1)
(4)
−
+
+
−
36
CHAPTER 2. ELECTROSTATICS
Problem 2.31 (a) V =
1 4π0
P
qi rij
1 4π0
n
−q a
√q 2a
+ 2 q 1 ∴ W4 = qV = −2 + √ . 4π0 a 2
(b) W1 = 0, W2 = Wtot =
2 1 q 4π0 a
=
1 4π0
−q 2 a
n −1 +
√1 2
+
; W3 =
−q a
1 4π0
o
q 4π0 a
=
−2 +
√1 2
. (1) r
2 √q 2a
−
q2 a
; W4 = (see (a)). o 1 2q 2 1 − 1 − 2 + √12 = . −2 + √ 4π0 a 2
(2)
r(4)
−
+
r+
−r
(3)
Problem 2.32 Conservation of energy (kinetic plus potential): 1 1 1 qA qB 2 2 mA vA + mB vB + = E. 2 2 4π0 r At release vA = vB = 0, r = a, so E=
1 qA qB . 4π0 a
When they are very far apart (r → ∞) the potential energy is zero, so 1 1 1 qA qB 2 2 mA vA + mB vB = . 2 2 4π0 a Meanwhile, conservation of momentum says mA vA = mB vB , or vB = (mA /mB )vA . So 1 1 2 mA vA + mB 2 2 s vA =
mA mB
2
1 qA qB 2π0 (mA + mB )a
2 vA =
1 2
mA ; mB
mA mB
2 (mA + mB )vA =
s vB =
1 qA qB . 4π0 a
1 qA qB 2π0 (mA + mB )a
mB . mA
Problem 2.33 From Eq. 2.42, the energy of one charge is W =
∞ ∞ X 1 1 1 (−1)n q 2 q 2 X (−1)n qV = (2) = . 2 2 4π0 na 4π0 a 1 n n=1
(The factor of 2 out front counts the charges to the left as well as to the right of q.) The sum is − ln 2 (you can get it from the Taylor expansion of ln(1 + x): 1 1 1 ln(1 + x) = x − x2 + x3 − x4 + · · · 2 3 4 with x = 1. Evidently α = ln 2 . c
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2
CHAPTER 2. ELECTROSTATICS
37
Problem 2.32 ! Problem 2.34 (a) W = 21 ρV dτ . From Prob. 2.21 (or P R 2 2 q 1 (a) W = 12 ρV dτ . From Prob. 2.21 (or Prob. 2.28): V = 2ρ0 R2 − r3 = 4π 3 − Rr 2 & $ R% 0 2R 1 1 q r2 W = ρ 3 − 2 4πr2 dr 3 Z R 2 R4π$ R 5 R 3 0 2R 0 r2 r qρ 1 1 q qρ r 1 3 = 3 − 2 4πr2 dr = W = ρ 3 − 2 R − % 2 2 4π0 2R 0 R 40 R 3 R 5 0 40 R qρ5 2 qR2 q 1 3q = = R = 4 2 3 5$ 5$ 4π$ 5 R 2 πR 0 0 3 0 qR q qρ 2 3q 1 R = = = . 50 50 43 πR3 4π0 5 R ! (b) W = "20 E 2 dτ . Outside (r > R) E = R q 1 q 1 *$ (b) W = 20 E 2 dτ . Outside (r > R) E = 4π r ; Inside (r < R) E = 4π r. 2ˆ 3 rˆ ∞ 0 r 0 R 1 1 2 $0 2 q (r 4π dr) + ∴ W = (Z ) 2 4 Z R 2 ∞ 2 (4π$ ) r 0 R 1 0 r 1 *% ∴W = q2 (r2 4π dr) + (4πr2 dr) &)∞ % 5& 4 2 (4π0 )2 R3 R r 0 1 q2 1 )) 1 r = − ) + 6 ) ( R ∞ r R R 5 1 r5 1 q2 1 1 1 3 q 2 4π$0 2 1 q2 1 = + = .X = − + 6 4π0 2 r R R 5 0 4π0 2 R 5R 4π0 5 R ! . ,V E·da + V E 2 dτ , whe (c) W = "20 S H R arbitrary. Let’s use a sphere of radius a > R. V E·da + V E 2 dτ , where V is large enough to enclose all the charge, but otherwise (c) W = 20 S 1 q *$ % arbitrary. Let’s use a sphere of radius a > R. Here V = 4π0 r . &% & 1 q 1 q $0 r2 sin W (Z ) 2 2= 2 Z R Z a 4π$ r 4π$ r 0 0 0 1 q 1 q 1 q 2r=a W = r2 sin θ dθ dφ + E 2 dτ + (4πr/ dr) 2 2 4π0 r 4π0 r2 4π r 0 0 R $0 q2 1 q2 4π r=a = 4π + + 2 2 5R a 2 (4π$ ) a (4π$ ) ( 2 2 0 0 q 1 q 4π 1 1 0 2 / 0 4π + + 4πq − = 2 1 q 1 1 1 1 2 (4π0 )2 a (4π0 )2 5R (4π0 )2 r R = = + − + 4π$ 2 a 5R a R 4π 0 1 q2 1 1 1 1 1 3 q2 = + − + = .X 4π0 2 a 5R a R 4π0 5 R As a → ∞, the contribution from the surf " # 2 1 q 1 q2 while As a → ∞, the contribution from the surface integral 4π0 2a goes to 4π" zero, the picks volume ( 6a − 1) up integral the slack. 0 2a 5R 2 1 q 6a Problem 2.33 4π0 2a ( 5R − 1) picks up the slack.
Problem 2.35
1 q¯ dW = d¯ q V = d¯ q , (¯ q = charge on sphere of radius r). 4π0 r 4 r3 q¯ = πr3 ρ = q 3 (q = total charge on sphere). 3 R 4πr2 3q d¯ q = 4πr2 dr ρ = 4 3 q dr = 3 r2 dr. R πR 3 3 1 qr 1 3q 2 1 3q 2 4 dW = r dr = r dr 3 3 4π0 R r R 4π0 R6 2 Z 1 3q 2 R 4 1 3q 2 R5 1 3q W = r dr = = .X 6 6 4π0 R 0 4π0 R 5 4π0 5 R c
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! r
"
d¯ q
q¯
c "2005 Pearson Ed protected under al reproduced, in any
38
Problem 2.36 R (a) W = 20 E 2 dτ. 2 R b q W = 20 4π a 0
E=
CHAPTER 2. ELECTROSTATICS
1 q 4π0 r 2
(a < r < b), zero elsewhere. Rb 1 q2 1 1 q2 1 2 2 4πr dr = = − . 2 2 r 8π0 a r 8π0 a b
2
2
1 q 1 q 1 q 1 −q (b) W1 = 8π , W2 = 8π , E1 = 4π r (r > a), E2 = 4π r (r > b). So 2 ˆ 2 ˆ 0 b 0 r 0 r 02 a 2 2 R R −q q2 1 1 2 ∞ 1 E1 · E2 = 4π0 E1 · E2 dτ = − 4π0 q b r4 4πr2 dr = − 4π . r 4 , (r > b), and hence 0b R 2 q 1 1 1 q 2 a1 + 1b − 2b = 8π Wtot = W1 + W2 + 0 E1 · E2 dτ = 8π a − b .X 0 0
Problem 2.37 z
r b r
q2
a q
x
E1 =
1 q1 ˆ r; 4π0 r2
y
q1
E2 =
1 q2 rˆ ; 4π0 r 2
Wi = 0
q1 q2 (4π0 )2
Z
1 r2
r
2
cos β r2 sin θ dr dθ dφ,
where (from the figure)
r
=
p
r2 + a2 − 2ra cos θ,
Therefore Wi =
q1 q2 2π (4π)2 0
cos β =
(r − a cos θ)
Z
r
3
r
It’s simplest to do the r integral first, changing variables to 2 r d r = (2r − 2a cos θ) dr As r : 0 → ∞,
r
: a → ∞, so Wi =
The
r
q1 q2 8π0
Z
π
⇒ ∞
Z
0
a
(r − a cos θ)
r
sin θ dr dθ.
:
(r − a cos θ) dr = 1
r
2
.
dr
r dr
.
sin θ dθ.
integral is 1/a, so Wi =
q1 q2 8π0 a
Z
π
sin θ dθ = 0
q1 q2 . 4π0 a
Of course, this is precisely the interaction energy of two point charges. Problem 2.38 (a) σR =
q −q q ; σa = ; σb = . 2 2 4πR 4πa 4πb2 c
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Problem 2.34 ! 1 q (a) W = !20 E 2 dτ. E = 4π! 2 (a < r < b), zero elsewhere. 0 r ' & " #2 ! $ % !b 1 q2 1 1 b 1 2 q q2 !0 2 4πr dr = 8π!0 a r2 = W = 2 4π!0 . − a r2 8π#0 a b 2
2
1 q 1 q 1 q 1 −q , W2 = 8π! , E1 = 4π! (b) W1 = 8π! r (r > a), E2 = 4π! r (r > b). So 2 ˆ 2 ˆ 0 b 0 r 0 r " " #02 a 2 # CHAPTER 2. ELECTROSTATICS 39 2 ! ! ∞ q2 −q 1 1 E1 · E2 = 4π! E1 · E2 dτ = − 4π! . q 2 b r14 4πr2 dr = − 4π! r 4 , (r > b), and hence 0 0 0b $ $ % % ! 2 q 1 1 2 1 1 WRtot = W1 + W − 1b .! = R8π! R 2b + #10 qE1 · E2 Rdτa = 8π!0 qR2 R a +1 b − 1 q q q 0 0 0 a q b (b) V (0) = − ∞ E·dl = − ∞ 4π . dr − (0)dr − dr − (0)dr = + − 2 2 b a 4π0 r R 0 r 4π0 b R a Problem 2.35
q −q q (a) σR = ; σa = ; σ R=a . RR (c) σb → 0 (the charge “drains (0)2 = −b ∞ (0)dr 4πR2 off”); V4πa 4πb2 − a
1 q q . − 4π0 R a !a !0 ! R$ 1 q % !0 !b $ 1 q % q q# 1 "q Problem 2.39 + − dr − dr − (0)dr − (0)dr = (b) V (0) = − ∞ E·dl = − ∞ 4π! 2 2 b R a 4π!0 r 0 r 4π#0 b R a qa qb qa + qb ; σb = − ; σR = . (a) σa = − !0 !a ! R$ 1 q % 1 "q q 4πa2 4πb2 4πR2 (c) σb → 0 (the charge “drains off”); V (0) = − ∞ (0)dr − a 4π! − 2 dr − R (0)dr = 0 r 4π#0 R a 1 qa + qb ˆ r, where r = vector from center of large sphere. (b) Eout = Problem2 2.36 4π0 r qa qb qa + qb σa = − ; qσb = − ; σR = . 2 2 2 1 q(a) 1 a b 4πa 4πb r (r ) is 4πR ˆ r , E = ˆ r , where the vector from center of cavity a (b). (c) Ea = a b b a b 4π0 ra2 4π0 rb2 1 qa + qb ˆ r, where r = vector from center of large sphere. (b) Eout = 4π# r2 0 (d) Zero. 1 q 4π0 r 2
dr −
R0
(0)dr = R
1 qb(but not Ea or Eb ); force on qa and qb still zero. (e) σR changes (but not σa 1or σqba); Eoutside changes ˆ ra , Eb = ˆ rb , where ra (rb ) is the vector from center of cavity a (b). (c) Ea = 2 4π#0 ra 4π#0 rb2 Problem 2.40 (a) No. For example, if it is very close to the wall, it will induce charge of the opposite sign on the wall, (d) Zero. and it will be attracted. (b) No. Typically itRwill be attractive, seeσbfootnote 12changes for a extraordinary (e) σ changes (but not but σa or ); Eoutside (but not Ea counterexample. or Eb ); force on qa and qb still zero.
Problem 2.41 Problem 2.37 Between plates, the E =plates 0; outside the 0plates = So σ/#0 = Q/#0 A. So Between the plates, E = the 0; outside E = σ/ = Q/E0 A. 2 2 Q2#0 Q = Q . 0 2 0 PQ=2 #0 E 2 = P = E = .2 2 = 2#0 A2 2 2 20 A2 2 20 A22 #0 A
Problem 2.42 Problem 2.38 1 Q Inside, E = 0; outside, E= = 0;4π r; soE = Inside, E outside, 2ˆ 0 r
1 Q r; 4π!0 r 2 ˆ
z!
so
"E θ
1 1 Q 1 1 EQ rave ; )fzz; aver= ; f2z4π! =0σ(E R2 ˆ 2 4π0 R2 ˆ
Q )z ; σ = Q 2 . = σ(Eave σ= 4πR 4πR2 . ! !$ Q % 1 $ 1 Q % R RF Q = f1z da1= Q 4πR cos θ R2 sin θ dθ dφ 2 Fz = fz da = z4πR cos2 θ 2R24π! sin0 θRdθ dφ 2 2 4π R2 0 $ $ Q %2 $ 1 %( $ % % ! Q2 2 (π/2 1 2 Q 2 1 π/2 2θ dθ R 1 Q 2 2π π/2 sin Q 2 sin θ = θ cos = = . Q 2 = π/2 Q Q 2 1 1 1 1 π!0 4R = 2 2π!0 4R. 0= = 20 4πR 2π 0 2!0sin4πR θ cos θ dθ0= π0 4R 32πR2 #0 2 sin θ 0 2π0 4R 32πR2 0
Eave =
Problem 2.43 #2005 c Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright as they No portion this material may be law: Say the charge on the inner cylinderlaws is Q, for acurrently length exist. L. The field isofgiven by Gauss’s R reproduced, in1any form or1 by any means, without Q 1 permission in writing from the publisher. E·da = E · 2πs · L = 0 Qenc = 0 Q ⇒ E = 2π0 L s ˆ s. Potential difference between the cylinders is Z V (b) − V (a) = − a
b
Q E·dl = − 2π0 L
Z a
b
1 Q ds = − ln s 2π0 L
As set up here, a is at the higher potential, so V = V (a) − V (b) =
Q 2π0 L
ln
c
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b a
.
b . a
cylinder is Q, for a length L. The field is given by Gauss’s law: !Say the charge on the inner s. Potential difference between the cylinders is E·da = E · 2πs · L = !10 Qenc = !10 Q ⇒ E = 2π!Q0 L 1s ˆ V (b) − V (a) = −
"
b
a
Q E·dl = − 2π"0 L
"
b
a
Q 1 ds = − ln s 2π"0 L
As set up here, a is at the higher potential, so V = V (a) − V (b) = C=
Q V
=
2π!0 L b , ln( a )
Q 2π!0 L
2π"0 % &. ln ab
40 so capacitance per unit length is
ln
# $ b . a
%b& a .
CHAPTER 2. ELECTROSTATICS
Problem 2.400 L 2π0 . C = VQ = 2π b , so capacitance per unit length is ln( a ) ln ab "0 2 E A". (a) W = (force)×(distance) = (pressure)×(area)×(distance) = 2 Problem 2.44 ( ' 2 (b) W = (energy per unit volume)×(decrease in volume) = "00 E22 (A"). Same as (a), confirming that the (a) W = (force)×(distance) = (pressure)×(area)×(distance) = E A. 2 energy lost is equal to the work done. 2 Problem 2.41 (b) W = (energy per unit volume)×(decrease in volume) = 0 E2 (A). Same as (a), confirming that the energyFrom lost Prob. is equal to the the field workatdone. 2.4, height z above the center of a square loop (side a) is Problem 2.45 1 4λaz ) E= ˆ z. % 2& 4π" a2 (side a) is a 0 2 From Prob. 2.4, the field at height z above the center loop z of + a4square z2 + 2 ! " da 2 ! " da 2 1 4λaz da over a from 0 to a ¯: ˆ Here λ → σ 2 (see figure), and we integrate q z. E= 2 2 4π0 z 2 + a " ! z2 + a 4
a
2
" a¯ 1 a2 a da Here λ → (see figure), and we&integrate over a from ¯:a da = 2 du. ) Eσ =da . Let u = 0 to 2σz , soa 2 % 2 2 4π"0 4 0 z 2 + a4 z 2 + a2 Z " +√ ,a¯2 /4 * 2 a ¯ 2 1 a da du a2 a ¯ /4 σz 2u + z2 q E = = 12σz . Let u = , so a da = 2 du. −1 √ a2 4σz0 = tan 4π a2 0 2 2 z2 + 2 4π" π"0 z4 z 2u + 0 0 z +(u4+ z ) 2 z 0 ) " !#a¯2 /4 + a¯2 , . Z a¯2 /4 / √ 2 12σ σz 2 2u + z 2 du 2 +z √ − tan−1 == = (1) ; tan−1 4σztan−1 4π π0 z z (u +zz 2 ) 2u + z 2 π"0 0 0 0 q ( ! * ) a ¯2 0 2 2σ 2 + z 2σ a2 π −1(1) ; tan−1 − tan−1 = ˆ z. 1+ 2 − π0 zE = π"0 tan 2z 4
"
a+da
!
1 % 2 & 2σ ! r tan−1 (∞) −# π = 2σ π − π = σ . ! a → ∞ (infinite plane):"E = π! 4 π!0 2 4 2!0 0 2 2σ a σ π a2 −1 −1 p ˆ z. E= tan 1 + −12 √− ˆ z = tan 2 + (a2 series: and as azTaylor z ' a (point charge): 1 4+ x − π4 , π π0 Let f (x) = tan2z /2) 0 expand4z
f (x) " 2σ +π1 x2πf "" (0) + f (0) π xf (0) σ ··· tan−1= (∞) −+ 4 = π0 22− 4 = 20 . X √ −1 z a (point charge): Let f (x) + x − π4 Inc., , andUpper expand asRiver, a Taylor series: c = tan #2005 Pearson 1Education, Saddle NJ. All rights reserved. a → ∞ (infinite plane): E =
2σ π0
This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any without permission in writing from the publisher. 1 means, 2 00 0
f (x) = f (0) + xf (0) + x f (0) + · · · 2
Here f (0) = tan−1 (1) −
π 4
=
π 4
−
π 4
= 0; f 0 (x) = f (x) =
Thus (since
a2 2z 2
= x 1), E ≈
2σ π0
1 a2 4 2z 2
=
1 1√1 1+(1+x) 2 1+x
=
1√ , 2(2+x) 1+x
so f 0 (0) = 41 , so
1 x + ( )x2 + ( )x3 + · · · 4 1 σa2 4π0 z 2
=
q 1 4π0 z 2 .
X
c
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CHAPTER 2. ELECTROSTATICS
43
The (double) integral is a pure number; Mathematica says it is 2. So V =
σR . π0
Problem 2.52 λ ln sa+ , where s+ is distance from λ+ (Prob. 2.22). 7 (a) Potential of +λ is V+ = − 2π 0 s− λ z Potential of −λ is V− = + 2π ln a , where s− is distance from λ− . " 0 (x, y, z) ! " s− λ s − ∴ Totalλ V = s− ln . ln 2π#0 . s+ ∴ Total V = 7 s+ 2π0 # s+ # z a a p s+ = (y − a)2 + z 2 , and p s− = (y + a)2 + z 2 , so ! Now " (x, y, z) 2 , and s = y Now s+ = (y − a)2!+ z" (y + a)2 + z 2 , so − −λ λ $ % ! " √ λ s − s 2 2 − (y+a)2 +z 2 ∴ Total V = ln √λ(y+a). 2 +z λ = (yλ + ln a)2 (y + z+2 a) + z . (x,2π# y, λ z) s+0 ln √2(y−a) 0ln= 2π# √ 2 +z 2 2 2 . = V (x, y, z)V = ln s+ 4π# 2π0 0 a)2 (y (y−a)2 +z 2 (y − + z−2 a) + z # #4π0 2 2 2 2 a a ! Now s+ = (y − a) + z , and s− = (y +2 a)2 + z , so y (y+a) +z (4π#0 V0 /λ) −λ λ % = k = constant. That is: ! √ are given " by (y−a)2 +z$2 = e 2 (b) Equipotentials 2 2 2 λ 2 2 (yV + (y+a)2 +z(y+a)2 +z (4π /λ)a) + z 2 constant. 2That is: 2 √ = e+ k= (b)V (x, Equipotentials are+given y, z) = ln . ln 2 +z y 2 2π# +λ2ay a2(y−a) + zby =(y−a) k(y'= − 22ay a20+0 z 2 ) = ⇒ 2 +z 2 2 y (k 2 − 1) + z (k − 1) + a (k − 1) − 2ay(k + 1) = 0, or 0 & 4π# (y − 0 2 2 2 2 2 2 2 2 2 a) + z 2 2 k+1 2 y + 2ayy+ + a z+ + z a= − k(y + a + z ) ⇒ y (k − 1) + z (k − 1) + a (k − 1) − 2ay(k + 1) = 0, or 2ay − 2ay k−1 = 0. The equation for a circle, with center at (y0 , 0) and radius R, is k+1 2 2 y 2 + z 2 +(ya− −y 2ay = 0. The a2circle, with center at (y0 , 0) and radius R, is 2 2 2 2equation (4π# 0 /λ) + zk−1 =byR2(y+a) , or 2y+z + z2 + (y002Vfor − R )− 0 ) given 0 =&0. 'That is: (b) Equipotentials are = k 2yy = constant. 2 = e 2 2 2 2 (y−a) 2 +z 2 2 (y − y ) +2z =2R the , orequipotentials −R ) − 2yy0 =0. k+1 2y + z + (y y 2 + 2ay0 +Evidently a + z &= k(y' − 2ay + a2 0+are z 2 )circles, ⇒ y 2 (kwith − 1)y+ z2a(k − 1) +and a2 (k − 1) − 2ay(k + 1) = 0, or 0 = k−1 k+1 Evidently the equipotentials are circles, with y = a and & ' 0 2 k+1 k−1 a circle, at (y20+2k−1) , 0) and radius y2 + z 2 + a a22 − (k2 +2k+1−k k+1 2 2 4kR, is k−1 a2 =a22 for = 2ay y02 − R ⇒= R20.=The y02 −equation −(ka22with = a2center 2 k−1 (k−1)2 2 4k= a (k−1)2 , or +2k+1−k +2k−1) k+1 2 2 2 2 2 2 2 2 2 2 − R2 ⇒√R 2 = y 2 − a2 = a 2 2 a = y − a = a = a , or 2 2 (y − y0 )0 + z = R , or y0 + z + (y0 −k−1 R ) − 2yy0 =&0. ' (k−1) (k−1) √R = 2a k ; or, in terms of V0 : k−1 2a kthe equipotentials are circles, with y = a k+1 and Evidently ; or, in terms of V0 : R = k−1 0 k−1 & '2 2 2 +2k−1) 2 2 k+1 ! = a2 (k−1) "4k 2 , or a2 = y02 − R2 ⇒ R2 4π# = y002V0− − a2 = a2 0(kV0+2k+1−k 2 2π#0 V0 /λ −2π# /λ (k−1) /λa = a k−1 2π#0 V0 e +2π 1 0 V0 /λe −2π0 V+0 /λ √ e4π0 V0 /λe+ 1 e =a + = 2π a coth y0 = a 4π# V /λ e . 0 V0 2a k /λ 0 a = 0aV0coth = a ; 4π . λ R =y0k−1 or, in terms of V− e 0= e2π#0 V0 /λ − e−2π# 0: 1 λ e 0 V0 /λ − 1 e2π0 V0 /λ − e−2π0 V0 /λ ! " 2π#0 V0 /λ 2π#0 V0 2 a 2π0 V0 /λe e = 2a 2 V /λ ! a "( 2π#0 V0 ) = 2π R a csch . = a −2π# = 0 V0 0 0 V0 /λ 0 /λ 0 V00 /λ . λ R = 2a =0 V0 /λ )2π# a csch −0 V10 /λ +(ee2π# − /λ e−2π# ea2π# V0 = e4π#4π +e14π#0 V= 2π0sinh 0 V0 λ λ e 0 V0 /λ − = 1 a (e2π0 V0 /λ − e−2π0 V0= ) a coth sinh y0 = a 4π# . λλ e 0 V0 /λ − 1 e2π#0 V0 /λ − e−2π#0 V0 /λ z ! " " e2π#0 V0 /λ 2π#0 V0 2 a ( ) = a csch R = 2a 4π# V /λ . = a 2π# V /λ = λ e 0 0 −1 (e 0 0 − e−2π#0 V0 /λ ) sinh 2π#λ0 V0 z
! R
"
! R −λ
! y
λ y0
−λ
λ y0
! y
Problem 2.48 c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently No 1 portion of this material may be d2 exist. V ρ reproduced, in any any means, without permission inρ. writing from the publisher. = − (Eq. 2.24), so (a) ∇2 Vform = or − #by 0 dx2 #0 * Problem 2.48 2qV 1 2 2 .1 (b) qV = 2 mv → v d= V ρ 2 =m− ρ. (a) ∇ V = − #0 (Eq. 2.24), so dx2 #0 dx (c) dq = Aρ dx*; dq dt = aρ dt = Aρv = I (constant). (Note: ρ, hence also I, is negative.) 2qV . (b) qV = 12 mv 2 → v = m Inc., Upper Saddle River, NJ. All rights reserved. This material is c #2005 Pearson Education,
44
CHAPTER 2. ELECTROSTATICS
Problem 2.53 (a) ∇2 V = − ρ0 (Eq. 2.24), so r (b) qV =
1 2 2 mv
→ v=
1 d2 V = − ρ. 2 dx 0
2qV 8 . m
! ! dq 2 d2 V (Note: −1/2 V Aρ dx 1; I m 1 dx I = Aρv I = I m(constant). ρ, hence also I, − is negative.) (c)(d) dq d= =−aρ , where β = = βV = − ρ = = − ⇒ dt dt 2 dx !0 !0 Av !0 A 2qV !0 A 2q . dx2 q q (Note: I is negative, so β isI positive; q isd2positive.) 2 V −1/2 I m I m (d) ddxV2 = − 10 ρ = − 10 Av = − 0 A 2qV ⇒ = βV , where β = − A 2q . 0 dx2 : (e) Multiply by V " = dV dx (Note: I is negative, so β is positive; q is positive.) " " " dV 1 ! 0 dV dV (e) Multiply by V V " = dx=:βV −1/2 ⇒ V " dV " = β V −1/2 dV ⇒ V 2 = 2βV 1/2 + constant. dx dx 2 Z Z dV 0 1 02 1/2 −1/2 dV + constant. V 0 dV 0 = V =is2βV β and V −1/2 dVat⇒cathode But V (0) = VV "0(0) ==0 βV (cathode is ⇒ at potential zero, field zero), so the constant is zero, and dx dx 2 # # ! dV 2 V 0 (0) = 1/4 But V (0) (cathode zero,dV and V = = 4βV 1/20⇒ β Vpotential ⇒ V −1/4 = field 2 βat dx;cathode is zero), so the constant is zero, and = 2is at dx" " # # p 41/43/4 −1/4 p 0 dV −1/4 1/2 x+ = V 2V = 4βV dV ⇒2 β = 2dx ⇒ βV V ⇒= V 2 βdV =constant. 2 β dx; 3 dxZ Z p p 4 −1/4 β dx ⇒is also dVso=this 2 constant V 3/4zero. = 2 β x + constant. But V V (0) = 0, 3 $ %4/3 $ %2/3 %1/3 $ 3 #constant is also zero. 3# 9 81I 2 m But V (0) = 0, 3/4so this 4/3 4/3 V = β x, so V (x) = β x , or V (x) = β x4/3 . x = 2 A2 q 2 2 4 32# 0 4/3 2/3 1/3 3p 3p 9 81I 2 m V 3/4 = β x, so V (x) = β& '4/3 x4/3 , or V (x) = β x4/3 = x4/3 . x 2 2 4 3220 A2 q (see graph). Interms of V0 (instead of I): V (x) = V0 d ( ) x 4/3 V" Without spacecharge, V would increase linearly: V (x) = V0 xd . (see graph). Interms of V0 (instead of I): V (x) = V0 d V0 4#0 V d2 V 4 1 −2/3 Without spacecharge, V would1 increase linearly: V (x) =0 V0 xd . ρ = −#0 2 = −#0 V0 4/3 · x = − . 2 2/3 dx 3 3 d 9(d x) without 1 4 1 −2/3 40 V0 *d2 V ' & ρ = −0 2q 2√ = −0# V0 4/3 · xx 2/3 = − . with 3 3 d 0 /m 9(d2 x)2/3 . v = dx V = 2qV d r m ! x 2/3 2q √ & p ' x d v= V =81I 2 m 2qV1/3 /m . 2 2 0 32!0 A q 3 81md4 2 2 (f) V (d) = m V0 = 32!2 A2 q d4/3d⇒ V03 = 32! I ; I = V ; 2 A2 q 81md4 0 0 0 * √ 1/3 √ 2 4 2q 3220 A2 q 3 4 2 ! A q 81I3/2 0A 3/2 81md4# (f) V I(d) V03 = K32 . == 9V√0m0=d2 32 =20 A2 q I22 ; I 2 = V020 Am2 q= KVd04/3 ,⇒ where 4 V0 ; 9d r m 81md √ √ 40 A 2q 4 2 A q 3/2 3/2 Problem I = 9√m02.49 V0 = KV0 , where K = . d2 9d2 m " & ' 1 η −η/λ ρˆ ηˆ (a) E = 2.54 e dτ. 1+ Problem 4π#0 η 2 λ Z r e− r /λ dτ. 1 ρ rˆ (a)(b)E Yes. = The field2 of 1a+point charge at the origin is radial and symmetric, so ∇×E = 0, and hence this is also r λ 4π0 true (by superposition) for any collection of charges. " r " r r ' −r/λ 1c 1 & Education, Pearson (c) V = − e Inc., drUpper Saddle River, NJ. All rights reserved. This material is 1+ q E·dl = − 2012 2 protected under all copyright laws as they currently exist. No portion of this material may be 4π# r λ 0 ∞ ∞ reproduced, in any form or+" by any means, without permission in writing ,from the publisher. " ∞ " ∞ 1 r ' −r/λ q 1 ∞ 1 −r/λ 1 & 1 −r/λ = 1+ e dr = q e dr + e dr . 4π#0 r r2 λ 4π#0 r2 λ r r r
c %2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2. ELECTROSTATICS
45
(b) Yes. The field of a point charge at the origin is radial and symmetric, so ∇×E = 0, and hence this is also true (by superposition) for any collection of charges. Z r Z r r −r/λ 1 1 9 (c) V = − 1 + e E·dl = − q dr 4π0 ∞ r2 λ 9 ∞ Z Z ∞ Z ∞ ∞ 1 r q 1 1 1 1 ! ! 1= −r/λ q −r/λ −r/λ −r/λ −r/λ −r/λ ee dr! = e dr + e Therefore dr . − λλ1−r/λ ←− exactly right Now r2 e 4π! dr = −re2 r 1 + −r/λ r dr 4π r2 to kill theλlast term. r 0 ←−r exactly right to killr the last term. Therefore Now 0 r12 er−r/λ dr = − e r − λ1 e r dr R −r/λ " R −r/λ #∞ $right to kill−r/λ Now r12 e−r/λ dr = − e r − λ1 e r dr ←−−r/λ exactly the last term. Therefore q e " ## #∞q$ e −r/λ V (r) = = −(q # e ∞ ) ##4π"0 r q . e−r/λ r −r/λ V4π" (r)0 q= = − . −r/λ e r r # q e4π" 4π" 0 . r −0 =r V (r) = 4π0 r& r ' 4π0 r & ' % R & −R/λ' q 1 1 R −R/λ' (d) 2 % IE·da = e R 4π !R ! = . 1+ q & e q 2 1 1+1 (d) −R/λ 2 −R/λ 1 1 R q 4π" ! R ! λ " λ R1 + R 0 = 0R −R/λ 24π −R/λ 1 + e ! ! = . q (d) S E·da E·da = q4π" 1!2+ eλ 4π R = 1+ e λ e. ! 0 R ( S ( R S 4π (0 R R 2−r/λ λ 0 "0 λ) −r/λ +,R * q q e r ,R ) −−r/λ −r/λ ( Rdr = q e R * Z qR (−r/λ Z re ZV dτ (= −r/λ r2 e4π ! dr = − 1 Rq q−r/λ rR + 2 e λ r 2 "0 q −r/λ e q e q 4π" ! r " (1/λ) 0 0 −r/λ 2 V =0 ! dr0 = re redr = dr = − 2 −−1λ0 − 1 r r4π V dτ V= dτ 'rdr4π =. 2 ! 0 0& "0 (1/λ) V 4π 0 4π" r 0' 0 "0. 0 0 (1/λ) λ 0 0 V R& 0 2 q −R/λ q 1 R . =λ + −e 1+ " q 2 −R/λ λ R −e−R/λ = λ02 = λ −e 1 + 1 ++λ1 + . 1 . 0 "0 λ & ' & ' . % ( 1 q R R& q I Z & −R/λ' −R/λ' % ( − 1+ + 1 = .. E·da + 2 1 V dτ ∴ = q 1+ Re R Re R q qed 1 q −R/λ −R/λ −R/λ −R/λ λ+ V+ V dτ V "=0 dτ = 1 + λ 1 +e λ 1 +e "=01 .= qqed ∴S ∴ E·da E·da + e − 1− e+ 1 + . qed 2 2 λ Vλ V 0 "0 λ λ 0 "0 λ λ S S (e)(e) Does thethe result in (d) hold forfor a nonspherical surface? Suppose we we R Does result in (d) hold a nonspherical surface? Suppose 2 2 S R (e) Does the result in (d) hold for a nonspherical surface? Suppose we make a “dent” in the sphere—pushing a patch (area R sin θ dθ dφ) make a “dent” in the sphere—pushing a patch (area R sin θ2 dθ dφ) S q 2 2 make a “dent” in the sphere—pushing a patch (area R sin θ dθ dφ) from radius R out to radius S (area S sin θ dθ dφ). from radius R out to radius S (area S sin θ2 dθ dφ). q from radius R out to radius S (area S sin θ dθ dφ). & ' & ' . % 1 1 q + S & e + R & e . −S/λ ' 2 −R/λ' 2 % I (S sin θ dθ dφ) − (R sin θ dθ dφ) 1 1 ∆ E·da = 1λ R R S 1λ S q S 2q1 2 1 −S/λ 2 2 −S/λ 2 −R/λ −R/λ 2 4π" R e (S sin θ dθ dφ) − e (R sin θ dθ dφ) 1 + 1 + ∆ E·da = ∆ E·da = 0 )& 1 + e (S sin θ dθ dφ) − 1 + e (R sin θ dθ dφ) ' , λ& λ S02S 'S 2 λ R2 R 2 λ q 4π0 4π" R )& −S/λ ' −R/λ' = 1+ &Re sin θ dθ,dφ. S − −S/λ q 1 q+ λ Se −S/λ −R/λ −R/λ 4π" λ 1 +eR e − 1− e sin θ dθ sindφ. θ dθ dφ. = 0= + 1 + 1 +e λ λ 4π0 4π"0 λ λ ( SZ ( ( Z−r/λ S ( 1 1 Z 1 1q q ( 1 1q q e e−r/λ (r2 sin Sdr −r/λ ∆∆ θ dr dθ dφ = sin θ dθ dφ re−r/λ V dτ = 2 1 q e 1 q 1 2 2 2 V dτ = r sin θ dr dθ dφ = sin θ dθ dφ re−r/λ dr 2 λ λ2 ∆ λ dτ 4π" r r λ dφ 4π" 2= 2= 0 0 r sin θ dr dθ sin θRdθRdφ re−r/λ dr V λ 4π λ 4π 0 0 2 2 2 * * ++# λ λ 4π" r λ 4π" 0 0 S R q q r −r/λ * + *r ## Sr ++#S =− θ qdθθ dφ # dθ dφe e−r/λ1 −r/λ =4π" − 0 sinsin 1 λ+ R =)& − sin θ dθ dφ & e 4π λ1 + R # 0 0S ' R ' λ ,R q q 4π" ' &Re−R/λ ' sin θ dθ,dφ. )& −S/λ =− 1 q+1 + Se e−S/λ −R/λ S − −S/λ =4π" − 0 −1 +1 + eR sin θ dθ dφ. λ λ e − λ1 + e−R/λ sin θ dθ dφ. =− 4π λ1 + 0 4π"0 λ λ ! / H 1 1 1 1 R exactly compensates forfor thethe change in inE·da, we we getget thethe total using So So thethe change in in change V 1dτ compensates change E·da, q for total using ! exactly /andand λ2 λ2V dτ "0 q0for Vdid dτdid exactly compensates for Any theAny change in surface E·da, and we built getup"1up q for the total using So the change inasλ2as thethe dented sphere, justjust we we with thethe perfect sphere. closed surface cancan be be built by successive dented sphere, with perfect sphere. closed by successive 0 the of dented sphere, as we did with theall perfect sphere. Any closed surface be charges built upinside, byinside, successive distortions the sphere, sojust the result holds forfor all shapes. By superposition, if there arecan many distortions of the sphere, so the result holds shapes. By superposition, if there are many charges 1 distortions of .the sphere,outside so the do result for all(in shapes.argument By superposition, if therethat are many charges inside, total Qenc Charges not holds contribute 0 enc for this thethe total is is"10 Q do not contribute (in the the argument above above we wefound found that H. !1Charges1 outside R /the total is Q . Charges outside do not contribute (in the argument above we found that 1 E·da for this volume + V dτ = 0—and, again, the sum is not changed by distortions of the surface, as this enc 2 volume E·da +/ λ2 "0V dτ =!λ0—and, again, the sum is not changed by distortions of the surface, as long as q for 1 long as q remains outside). So the new “Gauss’s Law” holds for any charge configuration. E·da + volume V dτ = 0—and, again, the sum is not changed by distortions of the surface, as long as q remains outside. So the new λ2 “Gauss’s Law” holds for any charge configuration. remains outside. So the new “Gauss’s Law” holds for any charge configuration. 1 This material is 1 reserved. c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights V = (f) protected In differential “Gauss’s law”currently reads: exist. ∇·ENo+portion or,1putting under allform, copyright laws as they of this may be it all in terms of E: 1 ρ,material 2 λ " 0 Vthe = publisher. ρ, or, putting it all in terms of E: (f) Inindifferential form, “Gauss’s law” reads: in∇·E + from reproduced, any means, without permission writing (any form or by 2 λ "0 1 ( 1 ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇2 V + 1 V = 1 ρ. = ∇·E − 2 E·dl 1 1 λ λ22 V + "10 V = 1 ρ. 0 = ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇ ∇·E − 2 "E·dl λ "0 λ2 "0 c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected #2005 under all copyright laws as Inc., they Upper currently exist.River, No portion this material c Pearson Education, Saddle NJ. Allofrights reserved.may Thisbematerial is reproduced, in any form bycopyright any means, without in writing from the protected underorall laws as theypermission currently exist. No portion of publisher. this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
=−
q 4π$0
1+
S λ
e−S/λ − 1 +
R λ
e−R/λ sin θ dθ dφ.
) * So the change in λ12 V dτ exactly compensates for the change in E·da, and we get "10 q for the total using the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside, the total is "10 Qenc . Charges outside do not contribute (in the argument above we found that for this ) * volume E·da + λ12 V dτ = 0—and, again, the sum is not changed by distortions of the surface, as long as q 46new “Gauss’s Law” holds for any charge CHAPTER 2. ELECTROSTATICS remains outside. So the configuration. 1 1 (f) In differential form, “Gauss’s law” reads: ∇·E + 2 V = ρ, or, putting it all in terms of E: 1 1λ $0 (f) In differential form,!“Gauss’s law” reads: ∇·E + 2 V = ρ, or, putting it all in terms of E: λ 0 1 1 1 1 Z − ∇·E ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇2 V + 2 V = ρ. E·dl = 1 1λ 1 λ2 1 $0 $0 2 ∇·E − 2 E·dl = ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇ V + 2 V = ρ. λ 0 λ 0 ρ
−
dτ
λ
r/
# 0
)V
=−
ρ
R ρ r e−
−
(∇ 2
R
0 E=
×
V =−
#
E·dl
dτ
∇
/ rλ
ρ ; #0
1 λ2
λ
r )e
4π 1 # 0
ρ
= dl
E=−∇V
$
+ rˆ (1 r2
R E·
=
R
1 2 λ
V
1 0 # 4π
−
·E
" V
E=
∇
!%
& E
(g) ReferProblem to ”Gauss’s law” in differential form (f). Since E is zero, inside a conductor (otherwise charge would 2.50 move, and in such a direction as to cancel the field), V is constant (inside), and hence ρ is uniform, throughout ∂ = $0“extra” ∇·E = $charge everywhere). = $reside 0 ∂x (ax) 0 a (constant the volume. ρAny must on the surface. (The fraction at the surface depends on λ, and The same charge density would be compatible (as far as Gauss’s law is concerned) with E = ayˆ y, for on the shape of the conductor.) a instance, or E = ( )r, etc. The point is that Gauss’s law (and ∇×E = 0) by themselves do not determine 3 Problem 2.55 the field —like any differential equations, they must be supplemented by appropriate boundary conditions. ∂ ρ = Ordinarily, (ax)are = so 0 a“obvious” (constantthat everywhere). these we impose them almost subconsciously (“E must go to zero far from 0 ∇·E = 0 ∂x the source charges”)—or we appeal to symmetry ambiguity (“the with field E must beythe The same charge density would be compatible (as far to asresolve Gauss’sthe law is concerned) = ayˆ , forsame—in a magnitude—on both sides of an infinite plane of surface charge”). But in this case there are no natural instance, or E = ( 3 )r, etc. The point is that Gauss’s law (and ∇×E = 0) by themselves do not determine boundary conditions, and no persuasive symmetry conditions, to fix the answer. The question “What the field —like any differential equations, they must be supplemented by appropriate boundary conditions. is the electric produced by athat uniform chargethem density filling all of space?”(“E is simply it does Ordinarily, thesefield are so “obvious” we impose almost subconsciously must goillposed to zero: far from not give the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the field must be the same—in magnitude—on both sides of an infinite of surface charge”). inRiver, this NJ. case are no This natural c plane #2005 Pearson Education, Inc., Upper But Saddle All there rights reserved. material is underconditions, all copyright laws as they currently exist. No portion of this material boundary conditions, and no persuasiveprotected symmetry to fix the answer. The question “What is themay be reproduced, in any form or by any means, without permission in writing from the publisher. electric field produced by a uniform charge density filling all of space?” is simply illposed : it does not give us information to determine the answer. (Incidentally, it won’t help to appeal to Coulomb’s law sufficient R rˆ 1 E = 4π0 ρ r 2 dτ —the integral is hopelessly indefinite, in this case.) Problem 2.56 Compare Newton’s law of universal gravitation to Coulomb’s law: F = −G Evidently
1 4π0
m1 m2 ˆ r; r2
F=
1 q1 q2 ˆ r. 4π0 r2
→ G and q → m. The gravitational energy of a sphere (translating Prob. 2.34) is therefore Wgrav =
3 M2 G . 5 R
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2. ELECTROSTATICS
47
Now, G = 6.67 × 10−11 N m2 /kg2 , and for the sun M = 1.99 × 1030 kg, R = 6.96 × 108 m, so the sun’s gravitational energy is W = 2.28 × 1041 J. At the current rate this energy would be dissipated in a time
t=
W 2.28 × 1041 = 5.90 × 1014 s = 1.87 × 107 years. = P 3.86 × 1026
Problem 2.57 First eliminate z, using the formula for the ellipsoid:
σ(x, y) =
Q 1 p . 2 2 4 2 2 4 4πab c (x /a ) + c (y /b ) + 1 − (x2 /a2 ) − (y 2 /b2 )
Now (for parts (a) and (b)) set c → 0, “squashing” the ellipsoid down to an ellipse in the x y plane:
σ(x, y) =
Q 1 p . 2πab 1 − (x/a)2 − (y/b)2
(I multiplied by 2 to count both surfaces.) (a) For the circular disk, set a = b = R and let r ≡
p Q 1 √ x2 + y 2 . σ(r) = . 2πR R2 − r2
(b) For the ribbon, let Q/b ≡ Λ, and then take the limit b → ∞: σ(x) = (c) Let b = c, r ≡
Λ 1 √ . 2π a2 − x2
p y 2 + z 2 , making an ellipsoid of revolution: x2 r2 + = 1, a2 c2
with σ =
Q 1 p . 4πac2 x2 /a4 + r2 /c4
The charge on a ring of width dx is
dq = σ2πr ds,
where ds =
p p dx2 + dr2 = dx 1 + (dr/dx)2 .
r 2x dx 2r dr c2 x dr c4 x2 c2 p 2 4 Now = − 2 , so ds = dx 1 + 4 2 = dx + 2 =0⇒ x /a + r2 /c4 . Thus 2 a c dx a r a r r
λ(x) =
dq Q 1 c2 p 2 4 Q p = 2πr x /a + r2 /c4 = . (Constant!) dx 4πac2 x2 /a4 + r2 /c4 r 2a
c
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48
CHAPTER 2. ELECTROSTATICS
Problem 2.58
y ( _a2, ÷3 _a2 ) b
(0,a)
r
q
x
( _a2, ÷3 _a2) (a) One such point is on the x axis (see diagram) at x = r. Here the field is q cos θ 1 Ex = −2 2 = 0, 4π0 (a + r)2 b
or
2 cos θ 1 = . b2 (a + r)2
Now, (a/2) − r cos θ = ; b
2
b =
a 2
−r
2
√ +
!2 3 a = (a2 − ar + r2 ). 2
Therefore 2[(a/2) − r] 1 . To simplify, let = (a + r)2 (a2 − ar + r2 )3/2 (1 − 2u) 1 = , 2 3/2 (1 + u)2 (1 − u + u )
or
r ≡u: a
(1 − 2u)2 (1 + u)4 = (1 − u + u2 )3 .
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2. ELECTROSTATICS
49
Multiplying out each side: 1 − 6u2 − 4u3 + 9u4 + 12u5 + 4u6 = 1 − 3u + 6u2 − 7u3 + 6u4 − 3u5 + u6 , or 3u − 12u2 + 3u3 + 3u4 + 15u5 + 3u6 = 0. u = 0 is a solution (of course—the center of the triangle); factoring out 3u we are left with a quintic equation: 1 − 4u + u2 + u3 + 5u4 + u5 = 0. According to Mathematica, this has two complex roots, and one negative root. The two remaining solutions are u = 0.284718 and u = 0.626691. The latter is outside the triangle, and clearly spurious. So r = 0.284718 a. (The other two places where E = 0 are at the symmetrically located points, of course.)
y
a a
_,_ ( ÷2 ÷2 )
b+
b_
q+
q_
r
x
(b) For the square: q Ex = 4π0 where cos θ± =
cos θ− cos θ+ 2 2 −2 2 b+ b−
√ (a/ 2) ± r ; b±
Thus
b2± =
a √ 2
=0
2
+
⇒
a √ ±r 2
cos θ− cos θ+ = , b2+ b2− 2
= a2 ±
√
2 ar + r2 .
√ √ (a/ 2) + r (a/ 2) − r √ √ = . (a2 + 2 ar + r2 )3/2 (a2 − 2 ar + r2 )3/2
To simplify, let w ≡
√
2 r/a; then
1+w 1−w = , (2 + 2w + w2 )3/2 (2 − 2w + w2 )3/2
or
(1 + w)2 (2 − 2w + w2 )3 = (1 − w)2 (2 + 2w + w2 )3 .
Multiplying out the left side: 8 − 8w − 4w2 + 16w3 − 10w4 − 2w5 + 7w6 − 4w7 + w8 = (same thing with w → −w). c
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50
CHAPTER 2. ELECTROSTATICS
The even powers cancel, leaving 8w − 16w3 + 2w5 + 4w7 = 0,
or
4 − 8v + v 2 + 2v 3 = 0,
where v ≡ w2 . According to Mathematica, this cubic equation has one negative root, one root that is spurious (the point lies outside the square), and v = 0.598279, which yields r v r= a = 0.546936 a . 2
y
(a cos(2p/5), a sin(2p/5))
(a cos(p/5), a sin(p/5))
b q r
a
c f
x
For the pentagon: Ex = where cos θ =
q 4π0
cos θ cos φ 1 +2 2 −2 2 (a + r)2 b c
a cos(2π/5) + r , b
2
cos φ =
= 0,
a cos(π/5) − r ; c
2
b2 = [a cos(2π/5) + r] + [a sin(2π/5)] = a2 + r2 + 2ar cos(2π/5), 2
2
c2 = [a cos(π/5) − r] + [a sin(π/5)] = a2 + r2 − 2ar cos(π/5). 1 r − a cos(π/5) r + a cos(2π/5) +2 = 0. +2 3/2 3/2 2 2 2 (a + r)2 [a + r + 2ar cos(2π/5)] [a + r2 − 2ar cos(π/5)] Mathematica gives the solution r = 0.688917 a. For an nsided regular polygon there are evidently n such points, lying on the radial spokes that bisect the sides; their distance from the center appears to grow monotonically with n: r(3) = 0.285, r(4) = 0.547, r(5) = 0.689, . . . . As n → ∞ they fill out a circle that (in the limit) coincides with the ring of charge itself. Problem 2.59 The theorem is false. For example, suppose the conductor is a neutral sphere and the external field is due to a nearby positive point charge q. A negative charge will be induced on the near side of the sphere (and a positive charge on the far side), so the force will be attractive (toward q). If we now reverse the sign of q, the induced charges will also reverse, but the force will still be attractive. If the external field is uniform, then the net force on the induced charges is zero, and the total force on the conductor is QEe , which does switch signs if Ee is reversed. So the “theorem” is valid in this very special case. c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 2. ELECTROSTATICS
51
Problem 2.60 The initial configuration consists of a point charge q at the center, −q induced on the inner surface, and +q on the outer surface. What is the energy of this configuration? Imagine assembling it piecebypiece. First bring in q and place it at the origin—this takes no work. Now bring in −q and spread it over the surface at a—this takes −qV = −q(1/4π0 )(q/a). Finally, bring in +q and spread it over the surface at b—this costs nothing, since the net field of the other two charges is zero out there. Thus the energy of the initial configuration is q2 . Wi = − 4π0 a The final configuration is neutral shell and a distant point charge—the energy is zero. Thus the work necessary to go from the initial to the final state is W =
q2 . 4π0 a
Problem 2.61
y
R (2pj/n)
rj x
R
Suppose the n point charges are evenly spaced around the circle, with the jth particle at angle j(2π/n). According to Eq. 2.42, the energy of the configuration is 1 Wn = n qV, 2 where V is the potential due to the (n − 1) other charges, at charge # n (on the x axis). V =
n−1 X 1 1 q , 4π0 j=1 r j
r
j
= 2R sin
jπ n
(see the figure). So Wn =
n−1 q2 n X 1 q2 = Ωn . 4π0 R 4 j=1 sin(jπ/n) 4π0 R
c
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52
CHAPTER 2. ELECTROSTATICS
Mathematica says 9
Ω10 =
10 X 1 = 38.6245 4 j=1 sin(jπ/10)
Ω11 =
11 X 1 = 48.5757 4 j=1 sin(jπ/11)
Ω12 =
12 X 1 = 59.8074 4 j=1 sin(jπ/12)
10
11
If (n − 1) charges are on the circle (energy Ωn−1 q 2 /4π0 R), and the nth is at the center, the total energy is Wn = [Ωn−1 + (n − 1)]
q2 . 4π0 R
For n = 11 :
Ω10 + 10 = 38.6245 + 10 = 48.6245 > Ω11
n = 12 :
Ω11 + 11 = 48.5757 + 11 = 59.5757 < Ω12
Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at the center.
c
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CHAPTER 3. POTENTIAL
53
Chapter 3
Potential Problem 3.1 √ The argument is exactly the same as in Sect. 3.1.4, except that since z < R, z 2 + R2 − 2zR = (R − z), 1 q 1 q [(z + R) − (R − z)] = . If there is more than one charge instead of (z − R). Hence Vave = 4π0 2zR 4π0 R 1 Qenc , and the average due to exterior inside the sphere, the average potential due to interior charges is 4π0 R Qenc charges is Vcenter , so Vave = Vcenter + 4π . X 0R Problem 3.2 A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV . But we know that Laplace’s equation allows no local minima for V . What looks like a minimum, in the figure, must in fact be a saddle point, and the box “leaks” through the center of each face. Problem 3.3 Laplace’s equation in spherical coordinates, for V dependent only on r, reads: 1 d c dV c dV dV ∇2 V = 2 r2 = 0 ⇒ r2 = c (constant) ⇒ = 2 ⇒ V = − + k. r dr dr dr dr r r Example: potential of a uniformly charged sphere. 1 d dV dV dV c V = c ln s + k. In cylindrical coordinates: ∇2 V = s =0⇒s =c⇒ = ⇒ s ds ds ds ds s Example: potential of a long wire. Problem 3.4 z Refer to Fig. 3.3, letting α be the angle between r and the z axis. Obviously, Eave points in the −ˆ direction, so I Z 1 1 q 1 Eave = E da = −ˆ z 2 2 r 2 cos α da. 4πR 4πR 4π0 By the law of cosines, − 2 r z cos α
⇒
= R2 + z 2 − 2Rz cos θ
⇒
R2 = z 2 +
r
2
r
2
r 2 − R2 , 2r z cos α z 2 + r 2 − R2 z − R cos θ = = . 2 3 2 r 2z r (R + z 2 − 2Rz cos θ)3/2
cos α =
z2 +
c
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54
CHAPTER 3. POTENTIAL
Z z − R cos θ q R2 sin θ dθ dφ 2 2 2 16π R 0 (R + z 2 − 2Rz cos θ)3/2 Z π Z 1 qˆ z z − R cos θ z − Ru qˆ z sin θ dθ = − du =− 8π0 0 (R2 + z 2 − 2Rz cos θ)3/2 8π0 −1 (R2 + z 2 − 2Rzu)3/2
Eave = −ˆ z
(where u ≡ cos θ). The integral is 1 p 1 1 1 R2 + z 2 2 2 I = √ − R + z − 2Rzu + √ 2 2 2 2 2 R R + z − 2Rzu −1 2Rz R + z − 2Rzu −1 1 1 1 1 1 1 2 2 = z − R − (z + R) + (R + z ) − − − . R z − R z + R 2Rz 2 z − R z + R (a) If z > R, 1 1 1 1 1 1 2 2 (z − R) − (z + R) + (R − − + z ) − I = R z−R z+R 2Rz 2 z−R z+R 1 1 2R 2 2R = − −2R + (R2 + z 2 ) 2 = 2. R z 2 − R2 2Rz 2 z − R2 z So
1 q ˆ z, 4π0 z 2 the same as the field at the center. By superposition the same holds for any collection of charges outside the sphere. (b) If z < R, 1 1 1 1 1 1 2 2 I = − − (R − z) − (z + R) + (R + z ) − R R−z z+R 2Rz 2 R−z z+R 1 2z 2z 1 − = 0. = −2z + (R2 + z 2 ) 2 R R2 − z 2 2Rz 2 R − z2 Eave = −
So Eave = 0. By superposition the same holds for any collection of charges inside the sphere. Problem 3.5 H R Same as proof of second uniqueness theorem, up to the equation S V3 E3 · da = − V (E3 )2 dτ . But on either V3 = 0 (if V is specified on the surface), or else E3⊥ = 0 (if ∂V ∂n = −E⊥ is specified). So Reach surface, 2 (E3 ) = 0, and hence E2 = E1 . qed V Problem 3.6 identity: Z Putting U = T = V3 into Green’s I ρ ρ 2 V3 ∇ V3 + ∇V3 · ∇V3 dτ = V3 ∇V3 · da. But ∇2 V3 = ∇2 V1 − ∇2 V2 = − + = 0, and ∇V3 = −E3 . 0 0 V Z S I So E32 dτ = − V3 E3 · da, and the rest is the same as before. V
S
Problem 3.7 Place image charges +2q at z = −d and −q at z = −3d. Total force on +q is q −2q 1 1 29q 2 2q −q q2 1 1 F= + − ˆ z = − + + ˆ z = − ˆ z. 4π0 (2d)2 (4d)2 (6d)2 4π0 d2 2 8 36 4π0 72d2 c
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CHAPTER 3. POTENTIAL
Problem 3.8 (a) From Fig. 3.13: q0
r
0
r
=
55
√
r2 + a2 − 2ra cos θ;
r
0
=
√
r2 + b2 − 2rb cos θ.
Therefore:
R2 R q √ (Eq. 3.15), while b = (Eq. 3.16). a r2 + b2 − 2rb cos θ a q q . =− = −q q 4 2 2 R R a ar 2+ 2 − 2ra cos θ r − 2r cos θ + R R a2 a R =−
Therefore:
1 V (r, θ) = 4π0
q
r
+
q0
r
0
q = 4π0
(
1 1 √ −p 2 2 2 r + a − 2ra cos θ R + (ra/R)2 − 2ra cos θ
) .
Clearly, when r = R, V → 0. ∂V (b) σ = −0 ∂V (Eq. 2.49). In this case, ∂V ∂n ∂n = ∂r at the point r = R. Therefore, q 1 σ(θ) = −0 − (r2 + a2 − 2ra cos θ)−3/2 (2r − 2a cos θ) 4π0 2 −3/2 a2 1 2 2 R + (ra/R) − 2ra cos θ 2r − 2a cos θ + 2 2 R r=R −3/2 a2 q =− − a cos θ −(R2 + a2 − 2Ra cos θ)−3/2 (R − a cos θ) + R2 + a2 − 2Ra cos θ 4π R 2 q a = (R2 + a2 − 2Ra cos θ)−3/2 R − a cos θ − + a cos θ 4π R q = (R2 − a2 )(R2 + a2 − 2Ra cos θ)−3/2 . 4πR Z Z q (R2 − a2 ) (R2 + a2 − 2Ra cos θ)−3/2 R2 sin θ dθ dφ qinduced = σ da = 4πR π q 1 2 2 2 2 2 −1/2 = (R − a )2πR − (R + a − 2Ra cos θ) 4πR Ra 0 q 2 1 1 = (a − R2 ) √ −√ . 2 2 2 2a R + a + 2Ra R + a2 − 2Ra p But a > R (else q would be inside), so R2 + a2 − 2Ra = a − R. q 2 1 1 q q = (a − R2 ) − = [(a − R) − (a + R)] = (−2R) 2a (a + R) (a − R) 2a 2a = −
qR = q0 . a
(c) The force on q, due to the sphere, is the same as the force of the image charge q 0 , to wit: qq 0 R 2 1 q 2 Ra 1 1 1 − q F = = = − . 4π0 (a − b)2 4π0 a (a − R2 /a)2 4π0 (a2 − R2 )2 c
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4 4
CHAPTER 3. SPECIAL TECHNIQUES CHAPTER 3. SPECIAL TECHNIQUES
(c) The force on q, due to the sphere, is the same as the force of the image charge q ! , to wit: ! " ! (c) The force on q, due charge CHAPTER POTENTIAL qq ! sphere,1is the same R 2 as the 1force of the image q 2 Ra q , to3.wit: 1 to56the 1 − F = = q = − . ! " 2 2 2 2 2 2 ! 2 4π"01 (a −qq b) 4π"01 a R (a − R 1/a) 4π"01 (a −q RRa) − q2 F = = =− . 2 2 2 − b)we 4π" a (a − R /a) 4π"0 (a2 − R2 )2 0 (a 0 To bring bring qq in in from from infinity infinity 4π" to a, a, then, we do do work To to then, work To bring q in from infinity %&a $ #aa to a, then, we do work 2 2 2 q R 11 R 2RZ &&a%& = − 11 q R #a aa qq2 R qq2 R 11 W = da = − . $ & &a= − 2 2 2 2 2 2R2 ). W = 4π" da = − 2 2 2 4π" 2 4π" 2(a − (a − R ) (a − R ) 2 2 q R a q R 1 1 1 q R2 ) 2 0 0 0 2 2 2 ∞& 4π 2(a − R 4π 4π 2 (a − R ) (a − R ) 0 0 0 ∞ W = da = = − − . ∞ 4π"∞0 4π"0 2 (a2 − R2 ) &∞ 4π"0 2(a2 − R2 ) (a2 − R2 )2 ∞ Problem 3.8 Problem 3.9 2 CHAPTER 3. SPECIAL TECHNIQUES Place a 3.8 second image charge, q !! , at the center of the sphere; Problem 00 !!at the this will alter thecharge, fact that is an Place a second image q , qthe center of equipotential, thethe sphere; a−b Place a not second image charge, , atsphere the center of sphere; *' ( ) 1 q !! this will not alter the fact that the sphere is an equipotential, this will not alter the fact that the sphere is an equipotential, a−b but merely increase that potential from zero to V0 = 00 ;!! !! ! *' ) 1 q q q q( 4π"01 Rq but!!but merely increase thatthat potential from zerozero to to V0 V= ; ; ' !! () * merely increase potential from 0 = ! 4π4π" q qa q q = 4π"0 V0 R at center of sphere. 0 R0 R ' () * !! 4π0 V0 R at center of sphere. q 00 = q = 4π" V R at center of sphere. a 0 For a neutral0 sphere, q ! + q !! = 0. ! !! ! " 0 ! ! 00 !!" ! ForFor a neutral q + 0. 0. qq a neutral sphere, q qq+ q= = 1 sphere, 1 q 1 F = − q !2 !!+ = " ! 2+ " 2 ! ! 0 b) 0qq 4π" a00 q (a − 4π" a1 1 (a − b) q 1 2 01 0 1 q qq 1 q − + 2 q + + F F= =qq4π" = = 4π" ! q 2 − 2 a+ 2b)2 − −2b)2 −2 b)(a (a q(−Rq/a) /a)(2a2−(aR(a 0b(2a2 a − b) 4π(R −/a) b) 0 0 a = 4π0 ! 2 a = 2 2 2 2 2 2 2a (a 4π" (a − 4π"0 (R(R − R− /a) b(2a − b) q(−Rq/a) q(−Rq/a) /a)(2a −2R 0 a qq 0qq b(2a − b) b) /a)(2a R /a)/a) = = = 4π" 2 a2!(a − = 2 2 (a −2R2 /a) 2 " 2 2 2 b) 4π" a 3 4π0 qa02 (a −Rb) (2a2 − 4π R02 ) 0 a (a − R /a) . =− " ! 3 2 2 2 22 2 3 (a 2− R 4π" (2a −2)R q 2 q0 Ra R (2a −R ) ). = − =− 2 2. 2 a (a2(a− R −2R 4π4π" )2 ) 0 0 abecause (Drop the minus sign, the problem asks for the force of attraction.) (Drop the minus sign, because the problem asks for the force of attraction.) Problem 3.9 (Drop the minus sign, because the problem asks for the force of attraction.) Problem3.9 3.9 Problem (a) Image 3.10 problem: λ above, −λ below. Potential was found in Prob. 2.47: Problem z (a)(a) Image problem: −λ−λ below. was found in in Prob. 2.52: Image λ above, below. Potential was found Prob. 2.47: z 2λ λ above, λ Potential "problem: V (y, z) z= ln(s− /s+ ) = ln(s2− /s2+ ) " zy 4π" 4π" z s+ (y, z) 2λ λ 0 0 " 2 2 + 2λ λ V (y," z)# = y ln(s /s ) = ln(s /s ) " − + λ 2 2 + − +/s ) = y4π" V (z (y,+ z)d) =2 , ln(s ln(s− /s+ ) 4π" − + ss+− (y, z) 0 d λ 0! y 2 + # # 4π 4π !y + 0 = lnλλx+2 2 ,0 2 2 d 2 2 d 4π"0λ ! y + (z − d) y + (z + d) s − ! λ y + (z + d) !xx − y = ln d 4π"0 y 2 + (z − d)2 = 4π0 ln y 2 + (z − d)2 ∂V ∂V ∂V − (b) σ = −"0 . Here = , evaluated at z = 0. ∂n∂V ∂n∂V ∂z∂V + ,& (b) σ = −"∂V . Here = 0 ∂V , evaluated at z = 0.1 & λ ∂V∂n ∂n &,& (b) σ = −0= −" . 0Here = 1∂z , evaluated at − z = 0. σ(y) 2(z + d) 2(z − d) + & & 2 2 2 2 ∂n 4π"0λ ∂ny + ∂z (z + y + (z − 1 d) 1 d) z=0 & σ(y) = −"0+ 2(z + d) − 2(z − d) &z=0 + d)2, y 2 +1 (z − d)2 λ4π"0d y 2 +1 (z−d λd 2λ σ(y) = 2(z, + d) − − y22 + (z2 − = . 2 2(z − d) = − − 0 4π +2 y 2 2+−(z + 2 d)−d 4π2λ 0y + d y2 + d2 π(y +λdd ) d) z=0 − 2 2 = − . = − 2 2 2 2 4π y + d y + d π(y + d ) 2λ dinduced on −d λdl parallel to the y axis: Check: = Total − charge = of−width . − 2 a 2strip 2 + d2 2 + d2 ) 4π y y + d π(y ∞ $ of width %& # Check: Total charge induced on a strip l parallel to the y/ axis: . &∞ 1 lλd 1 lλd π π .0 lλd −1 y & ∞ dy = − = − y axis: tan − − q = − $ %& # induced ind Total charge ∞to the . / & 2 2 Check: on a strip of width l parallel & πlλd y + 1d πlλd d 1 πlλd 2 π 2 π .0 −1 d y −∞ & dy = − = − − tan − qind = − −∞ & πZ∞ y 2 + d2 π d 2 d ∞ −∞ lλdπh π 2 π i = −λl. λind = −λ, lλd Therefore lλd as1 it should −∞ 1 −1 y be. qind = − dy = − tan = − − − π y 2 + d2 λind = π−λ, das it should d be. π 2 2 = −λl. Therefore −∞
Chapter 1
Special Techniques
−∞
= −λl.
c $2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is No portion of this material may be $2005 Pearson Upper Saddle NJ. All reserved. This material is reproduced, in anyEducation, form or by Inc., any means, withoutRiver, permission inrights writing from the publisher. protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
under laws as they currently exist. Therefore λprotected as all it copyright should be. indc = −λ,
c
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c !2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be
58 In this case V0 (y) =
CHAPTER 3. POTENTIAL
+V0 , for 0 < y < a/2 −V0 , for a/2 < y < a
. Therefore,
( a/2 a ) Za Za/2 2V 2 cos(nπy/a) cos(nπy/a) 0 Cn = V0 sin(nπy/a) dy = − + sin(nπy/a) dy − a a (nπ/a) 0 (nπ/a) a/2 0
a/2
nπ nπ o 2V n nπ o 2V0 n 0 = − cos + cos(0) + cos(nπ) − cos = 1 + (−1)n − 2 cos . nπ 2 2 nπ 2 The term in curly n = 1 n=2 n=3 n=4
brackets is: 1 − 1 − 2 cos(π/2) = 0, 1 + 1 − 2 cos(π) = 4, etc. (Zero if n is odd or divisible by 4, otherwise 4.) 1 − 1 − 2 cos(3π/2) = 0, 1 + 1 − 2 cos(2π) = 0,
: : : :
Therefore
Cn =
8V0 /nπ, 0,
n = 2, 6, 10, 14, etc. (in general, 4j + 2, for j = 0, 1, 2, ...), otherwise.
So 8V0 π
V (x, y) =
∞ e−nπx/a sin(nπy/a) 8V0 X e−(4j+2)πx/a sin[(4j + 2)πy/a] . = n π j=0 (4j + 2) n=2,6,10,...
X
Problem 3.14 V (x, y) =
4V0 π
1 −nπx/a e sin(nπy/a) n n=1,3,5,... X
(Eq. 3.36);
σ = −0
∂V ∂n
(Eq. 2.49).
So σ(y) = −0 =
∂ ∂x
40 V0 a
4V0 X 1 −nπx/a 4V0 X 1 nπ −nπx/a = −0 e sin(nπy/a) (− )e sin(nπy/a) π n π n a x=0 x=0 X sin(nπy/a).
n=1,3,5,...
Or, using the closed form 3.37: 2V0 sin(πy/a) − sin(πy/a) π 2V0 1 −1 V (x, y) = tan ⇒ σ = −0 cosh(πx/a) 2 2 π sinh(πx/a) π 1 + sin 2(πy/a) sinh (πx/a) a x=0 sinh (πx/a) 20 V0 sin(πy/a) cosh(πx/a) 20 V0 1 = = . a sin2 (πy/a) + sinh2 (πx/a) x=0 a sin(πy/a) [Comment: Technically, the series solution for σ is defective, since termbyterm differentiation has produced a (naively) nonconvergent sum. More sophisticated definitions of convergence permit one to work with series of this form, but it is better to sum the series first and then differentiate (the second method.)]
c
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CHAPTER 3. POTENTIAL
59
Summation of series Eq. 3.36 V (x, y) =
X 1 4V0 I, where I ≡ e−nπx/a sin(nπy/a). π n n=1,3,5,...
Now sin w = Im eiw , so I = Im
X1 X1 e−nπx/a einπy/a = Im Z n, n n
where Z ≡ e−π(x−iy)/a . Now Z Z X ∞ ∞ X 1 X 1 Zn = Z (2j+1) = u2j du n (2j + 1) 0 j=0 1,3,5,... j=0 ZZ
1 1 1+Z 1 1 du = ln = ln Rei θ = (ln R + i θ)7 , = CHAPTER 3. SPECIAL 1 − TECHNIQUES u2 2 1−Z 2 2 0 1+Z i θi θ =1+Z whereReRe . Therefore where = 1−Z . Therefore 1−Z
soso
$ # −π(x+iy)/a $ # −π(x−iy)/a ! " e e−π(x−iy)/a 1 − e e−π(x+iy)/a 1+ 1− 11 11 1+ ZZ 1 + e−π(x−iy)/a 1+ 1+ e−π(x−iy)/a 1 + $# $ I I==Im RR ++ i θ) == == # −π(x−iy)/a i θ) == θ.θ. But But Im (ln(ln 22 22 1− ZZ 1 − 1− e−π(x−iy)/a 1− e−π(x−iy)/a 1 − e e−π(x−iy)/a 1 − e−π(x+iy)/a 1− 1− e−π(x+iy)/a $ −2πx/a # iπy/a −iπy/a −πx/a −πx/a 1+ e−πx/a −− e e−iπy/a−− e e−2πx/a 1 + 1+ e−πx/aeiπy/a e 2ie sin(πy/a) −− e−2πx/a 1+ 2ie sin(πy/a) e−2πx/a == == , , % 2 % 2 % 2%2 1%− % 1%− % e−π(x−iy)/a e−π(x−iy)/a 1− e−π(x−iy)/a 1− e−π(x−iy)/a
Therefore Therefore
−πx/a −πx/a 2e2e sin(πy/a) 2 sin(πy/a) sin(πy/a) 2 sin(πy/a) sin(πy/a) sin(πy/a) . . tan θ θ== == πx/a == tan −2πx/a −πx/a −2πx/a πx/a −πx/a sinh(πx/a) 1− e e − e sinh(πx/a) 1−e e −e
& ' & ' sin(πy/a) 1 0 0 −1 sin(πy/a) , and V (x, y) = 2V2V sin(πy/a) . −1−1 sin(πy/a) −1 I I== 1tan tan , and V (x, y) = π . tan sinh(πx/a) 2 2 tan sinh(πx/a) sinh(πx/a) π sinh(πx/a) Problem Problem3.15 3.14 2 ∂2V ∂ 2V2 ∂ V with boundary conditions (a)(a) 2 + + ∂2 V= 0, = 0, with boundary conditions ∂x∂x2 ∂y∂y 2 (i)(i) V V (x, 0)0)==0,0, (x, (ii) (x, a)a)==0,0, (ii)V V (x, (iii) (0,(0, y)y)==0,0, (iii)V V (iv) (b,(b, y)y)==V0V(y). (iv)V V 0 (y).
y
"
a
V0 (y)
V =0
b
z#
!x
AsAsininEx. Ex.3.4, 3.4,separation separationofofvariables variablesyields yields # kxkx −kx $ VV (x, y)y)==AeAe ++ Be kyky++ DD cos ky) . . (x, Be−kx(C(Csinsin cos ky)
Here Here(i)⇒ (i)⇒DD==0,0,(iii)⇒ (iii)⇒BB==−A, −A,(ii)⇒ (ii)⇒kakais isananinteger integermultiple multipleofofπ:π: / nπx/a −nπx/a 0 nπx/a −nπx/a sin(nπy/a) = (2AC) sinh(nπx/a) sin(nπy/a). VV (x, y) = AC e − e sin(nπy/a) = (2AC) sinh(nπx/a) sin(nπy/a). (x, y) = AC e −e
c But Pearson
2012 Inc., Upper River,general NJ. All rights This material is (2AC) Education, is a constant, and Saddle the most linearreserved. combination of separable protected under all copyright laws as they currently exist. No portion of this material may be (ii), (iii) inisany form or by any means, without permission in writing from the publisher. reproduced,
V (x, y) =
∞ 1
solutions consistent with (i),
Cn sinh(nπx/a) sin(nπy/a).
n=1
It remains to determine the coefficients Cn so as to fit boundary condition (iv): 1
2 Cn sinh(nπb/a) sin(nπy/a) = V0 (y). Fourier’s trick ⇒ Cn sinh(nπb/a) = a
2a
V0 (y) sin(nπy/a) dy.
60
CHAPTER 3. POTENTIAL
But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i), (ii), (iii) is ∞ X V (x, y) = Cn sinh(nπx/a) sin(nπy/a). n=1
It remains to determine the coefficients Cn so as to fit boundary condition (iv): X
Cn sinh(nπb/a) sin(nπy/a) = V0 (y). Fourier’s trick ⇒ Cn sinh(nπb/a) =
2 a
Za V0 (y) sin(nπy/a) dy. 0
Therefore 2 Cn = a sinh(nπb/a)
Za V0 (y) sin(nπy/a) dy. 0
2 (b) Cn = V0 a sinh(nπb/a)
Za
2V0 sin(nπy/a) dy = × a sinh(nπb/a)
0
V (x, y) =
4V0 π
0, if n is even, 2a nπ , if n is odd.
sinh(nπx/a) sin(nπy/a) . n sinh(nπb/a) n=1,3,5,... X
Problem 3.16 Same format as Ex. 3.5, only the boundary conditions are: (i) V = 0 when x = 0, (ii) V = 0 when x = a, (iii) V = 0 when y = 0, (iv) V = 0 when y = a, (v) V = 0 when z = 0, (vi) V = V0 when z = a. This time we want sinusoidal fuctions in x and y, exponential in z: √
X(x) = A sin(kx) + B cos(kx),
Y (y) = C sin(ly) + D cos(ly),
Z(z) = Ee
k2 +l2 z
√
+ Ge−
k2 +l2 z
.
(i)⇒ B = 0; (ii)⇒ k = nπ/a; (iii)⇒ D = 0; (iv)⇒ l = mπ/a; (v)⇒ E + G = 0. Therefore p Z(z) = 2E sinh(π n2 + m2 z/a). Putting this all together, and combining the constants, we have: V (x, y, z) =
∞ X ∞ X
Cn,m sin(nπx/a) sin(mπy/a) sinh(π
p n2 + m2 z/a).
n=1 m=1
It remains to evaluate the constants Cn,m , by imposing boundary condition (vi): i p XXh V0 = Cn,m sinh(π n2 + m2 ) sin(nπx/a) sin(mπy/a). c
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CHAPTER 3. POTENTIAL
61
According to Eqs. 3.50 and 3.51: p 2 2 Za Za 2 2 V0 sin(nπx/a) sin(mπy/a) dx dy = Cn,m sinh π n + m = a 0
(
0
0, if n or m is even, 16V0 , if both are odd. π 2 nm
)
Therefore 16V0 V (x, y, z) = 2 π
√ sinh π n2 + m2 z/a 1 √ . sin(nπx/a) sin(mπy/a) nm sinh π n2 + m2 n=1,3,5,... m=1,3,5,... X
X
Consider the superposition of six such cubes, one with V0 on each of the six faces. The result is a cube with V0 on its entire surface, so the potential at the center is V0 . Evidently the potential at the center of the original cube (with V0 on just one face) is one sixth of this: V0 /6. To check it, put in x = y = z = a/2: 16V0 V (a/2, a/2, a/2) = 2 π
√ sinh π n2 + m2 /2 1 √ . sin(nπ/2) sin(mπ/2) nm sinh π n2 + m2 n=1,3,5,... m=1,3,5,... X
X
Let n ≡ 2i + 1, m ≡ 2j + 1, and note that sinh(2u) = 2 sinh(u) cosh(u). The double sum is then S=
∞ ∞ i h p (−1)i+j 1 XX sech π (2i + 1)2 + (2j + 1)2 /2 . 2 i=0 j=0 (2i + 1)(2j + 1)
Setting the upper limits at i = 3, j = 3 (or above) Mathematica returns S = 0.102808, which (to 6 digits) is equal to π 2 /96, confirming (at least, numerically) that V (a/2, a/2, a/2) = V0 /6. Problem 3.17 3 2 2 1 d3 1 d2 1 d2 2 2 x − 1 = 3 x − 1 2x = x x2 − 1 3 2 2 8 · 6 dx 48 dx 8 dx 2 i 1 d 2 1 d h 2 2 x − 1 + 2x x − 1 2x = x − 1 x2 − 1 + 4x2 = 8 dx 8 dx 1 1 d 2 2 = x − 1 5x − 1 = 2x 5x2 − 1 + x2 − 1 10x 8 dx 8 1 1 5 3 = 5x3 − x + 5x3 − 5x = 10x3 − 6x = x3 − x. 4 4 2 2
P3 (x) =
We need to show that P3 (cos θ) satisfies 1 d sin θ dθ where P3 (cos θ) =
1 2
sin θ
dP dθ
= −l(l + 1)P, with l = 3,
cos θ 5 cos2 θ − 3 .
1 1 dP3 = − sin θ 5 cos2 θ − 3 + cos θ(10 cos θ(− sin θ) = − sin θ 5 cos2 θ − 3 + 10 cos2 θ dθ 2 2 3 = − sin θ 5 cos2 θ − 1 . 2 c
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62 ∂ ∂θ
sin θ
1 d sin θ dθ
dP3 dθ
CHAPTER 3. POTENTIAL
3 3 d 2 sin θ 5 cos2 θ − 1 = − 2 sin θ cos θ 5 cos2 θ − 1 + sin2 θ (−10 cos θ sin θ) 2 dθ 2 2 2 = −3 sin θ cos θ 5 cos θ − 1 − 5 sin θ .
=−
dP sin θ dθ
= −3 cos θ 5 cos2 −1 − 5 1 − cos2 θ = −3 cos θ 10 cos2 θ − 6 = −3 · 4 ·
1 cos θ 5 cos2 θ − 3 = −l(l + 1)P3 . qed 2
Z1
Z1
(x)
P1 (x)P3 (x) dx =
1 1 1 1 5 5x3 − 3x dx = x − x3 −1 = (1 − 1 + 1 − 1) = 0. X 2 2 2
−1
−1
Problem 3.18 (a) Inside: V (r, θ) =
∞ X
Al rl Pl (cos θ)
(Eq. 3.66) where
l=0
(2l + 1) Al = 2Rl
Zπ V0 (θ)Pl (cos θ) sin θ dθ
(Eq. 3.69).
0
In this case V0 (θ) = V0 comes outside the integral, so (2l + 1)V0 Al = 2Rl
Zπ Pl (cos θ) sin θ dθ. 0
But P0 (cos θ) = 1, so the integral can be written Zπ
P0 (cos θ)Pl (cos θ) sin θ dθ =
0, if l 6= 0 2, if l = 0
(Eq. 3.68).
0
Therefore
Al =
0, if l 6= 0 V0 , if l = 0
.
Plugging this into the general form: V (r, θ) = A0 r0 P0 (cos θ) = V0 . The potential is constant throughout the sphere. ∞ X Bl Outside: V (r, θ) = Pl (cos θ) (Eq. 3.72), where rl+1 l=0
(2l + 1) l+1 Bl = R 2
Zπ V0 (θ)Pl (cos θ) sin θ dθ
(Eq. 3.73).
0
=
(2l + 1) l+1 R V0 2
Zπ
Pl (cos θ) sin θ dθ =
0, if l 6= 0 RV0 , if l = 0
.
0 c
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CHAPTER 3. POTENTIAL
Therefore V (r, θ) = V0
63
1 R (i.e. equals V0 at r = R, then falls off like ). r r
(b)
V (r, θ) =
∞ X l Al r Pl (cos θ), for r ≤ R (Eq. 3.78) l=0
∞ X Bl Pl (cos θ), for r ≥ R (Eq. 3.79) l+1 r
,
l=0
where Bl = R2l+1 Al
(Eq. 3.81)
and 1 Al = 20 Rl−1
Zπ σ0 (θ)Pl (cos θ) sin θ dθ
(Eq. 3.84)
0
1 = σ0 20 Rl−1
Zπ
Pl (cos θ) sin θ dθ =
0, if l 6= 0 Rσ0 /0 , if l = 0
.
0
Therefore
V (r, θ) =
Rσ0 , 0
for r ≤ R
.
2 R σ0 1 , for r ≥ R 0 r Note: in terms of the total charge Q = 4πR2 σ0 , 1 Q , for r ≤ R 4π0 R . V (r, θ) = 1 Q , for r ≥ R 4π0 r Problem 3.19 V0 (θ) = k cos(3θ) = k 4 cos3 θ − 3 cos θ = k [αP3 (cos θ) + βP1 (cos θ)] . (I know that any 3rd order polynomial can be expressed as a linear combination of polynomials; in this case, since the polynomial is odd, I only need P1 and P3 .) 5α 1 3 3 3 5 cos θ − 3 cos θ + β cos θ = cos θ + β − 4 cos θ − 3 cos θ = α 2 2 so 4=
5α 8 ⇒α= ; 2 5
the first four Legendre 3 α cos θ, 2
3 3 8 12 12 3 −3 = β − α = β − · = β − ⇒β= −3=− . 2 2 5 5 5 5
Therefore V0 (θ) =
k [8P3 (cos θ) − 3P1 (cos θ)] . 5
c
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64
CHAPTER 3. POTENTIAL
Now
V (r, θ) =
∞ X l Al r Pl (cos θ), for r ≤ R (Eq. 3.66) l=0
∞ X Bl Pl (cos θ), for r ≥ R (Eq. 3.72) rl+1
,
l=0
where (2l + 1) Al = 2Rl
Zπ V0 (θ)Pl (cos θ) sin θ dθ
(Eq. 3.69)
0
π Z Zπ (2l + 1) k = 8 P (cos θ)P (cos θ) sin θ dθ − 3 P (cos θ)P (cos θ) sin θ dθ 3 l 1 l 2Rl 5 0 0 2 2 k 1 k (2l + 1) 8 δ − 3 δ = [8 δl3 − 3 δl1 ] = l3 l1 5 2Rl (2l + 1) (2l + 1) 5 Rl 8k/5R3 , if l = 3 = (zero otherwise). −3k/5R, if l = 1 Therefore V (r, θ) = −
r 3k 8k 3 k r 3 rP1 (cos θ) + r P (cos θ) = P (cos θ) , 8 P (cos θ) − 3 3 1 3 5R 5R3 5 R R
or k 5
r r 31 k r r 2 3 2 8 5 cos θ − 3 cos θ − 3 cos θ ⇒ V (r, θ) = cos θ 4 5 cos θ − 3 − 3 R 2 R 5R R
(for r ≤ R). Meanwhile, Bl = Al R2l+1 (Eq. 3.81—this follows from the continuity of V at R). Therefore Bl =
8kR4 /5, if l = 3 −3kR2 /5, if l = 1
(zero otherwise).
So " # 2 4 −3kR2 1 8kR4 1 k R R V (r, θ) = P1 (cos θ) + P3 (cos θ) = 8 P3 (cos θ) − 3 P1 (cos θ) , 5 r2 5 r4 5 r r or k V (r, θ) = 5
R r
2
( ) 2 R 2 cos θ 4 5 cos θ − 3 − 3 r
c
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CHAPTER 3. POTENTIAL
65
(for r ≥ R). Finally, using Eq. 3.83: σ(θ) = 0
∞ X
(2l + 1)Al Rl−1 Pl (cos θ) = 0 3A1 P1 + 7A3 R2 P3
l=0
0 k 8k 3k 2 R P3 = [−9P1 (cos θ) + 56P3 (cos θ)] P1 + 7 = 0 3 − 5R 5R 5R3 0 k 56 0 k = −9 cos θ + 5 cos3 θ − 3 cos θ = cos θ[−9 + 28 · 5 cos2 θ − 28 · 3] 5R 2 5R =
0 k cos θ 140 cos2 θ − 93 . 5R
Problem 3.20 Use Eq. 3.83: σ(θ) = 0
∞ X
(2l+1)Al R
l−1
l=0
2l + 1 Pl (cos θ). But Eq. 3.69 says: Al = 2Rl
Zπ V0 (θ)Pl (cos θ) sin θ dθ. 0
Putting them together: Zπ
∞
0 X σ(θ) = (2l + 1)2 Cl Pl (cos θ), 2R l=0
with Cl =
V0 (θ)Pl (cos θ) sin θ dθ. qed 0
Problem 3.21 Set V = 0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the potential of a uniformly charged spherical shell: R3 1 Q V (r, θ) = −E0 r − 2 cos θ + . r 4π0 r Problem 3.22
∞ ∞ ∞ i X X X Bl Bl σ hp 2 Bl 2−r . r + R P (cos θ) (r > R), so V (r, 0) = P (1) = = (a) V (r, θ) = l l rl+1 rl+1 rl+1 20 l=0 l=0 l=0 p p 1 1 2 4 2 2 2 Since r > R in this region, r + R = r 1 + (R/r) = r 1 + (R/r) − (R/r) + . . . , so 2 8
2 ∞ X Bl σ 1 R2 1 R4 σ R R4 = r 1+ − + ... − 1 = − 3 + ... . rl+1 20 2 r2 8 r4 20 2r 8r l=0
Comparing like powers of r, I see that B0 =
σR2 σR4 , B1 = 0, B2 = − , . . . . Therefore 40 160
σR2 1 R2 V (r, θ) = − 3 P2 (cos θ) + . . . , 40 r 4r " # 2 σR2 1 R 2 = 1− 3 cos θ − 1 + . . . , 40 r 8 r c
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(for r > R).
66
(b) V (r, θ) =
∞ X
CHAPTER 3. POTENTIAL
Al rl Pl (cos θ) (r < R). In the northern hemispere, 0 ≤ θ ≤ π/2,
l=0
V (r, 0) =
∞ X l=0
Since r < R in this region,
Al rl =
i σ hp 2 r + R2 − r . 20
p p 1 1 r2 + R2 = R 1 + (r/R)2 = R 1 + (r/R)2 − (r/R)4 + . . . . Therefore 2 8 ∞ X l=0
Comparing like powers: A0 =
σ 1 r2 1 r4 + ... − r . Al r = R+ − 20 2R 8 R3 l
σ σ σ R, A1 = − , A2 = , . . . , so 20 20 40 R
σ 1 2 V (r, θ) = r P2 (cos θ) + . . . , R − rP1 (cos θ) + 20 2R (for r < R, northern hemisphere). r σR 1 r 2 2 3 cos θ − 1 + . . . , = cos θ + 1− 20 R 4 R In the southern hemisphere we’ll have to go for θ = π, using Pl (−1) = (−1)l .
V (r, π) =
∞ X l=0
(−1)l Al rl =
i σ hp 2 r + R2 − r . 20
(I put an overbar on Al to distinguish it from the northern Al ). The only difference is the sign of A1 : A1 = +(σ/20 ), A0 = A0 , A2 = A2 . So:
V (r, θ) =
σ 1 2 r P2 (cos θ) + . . . , R + rP1 (cos θ) + 20 2R (for r < R, southern hemisphere).
r σR 1 r 2 = cos θ + 1+ 3 cos2 θ − 1 + . . . , 20 R 4 R
Problem 3.23
V (r, θ) =
∞ X Al rl Pl (cos θ), (r ≤ R) (Eq. 3.78), l=0
∞ X Bl Pl (cos θ), (r ≥ R) (Eq. 3.79), l+1 r l=0
c
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CHAPTER 3. POTENTIAL
67
where Bl = Al R2l+1 (Eq. 3.81) and 1 Al = 20 Rl−1
Zπ σ0 (θ)Pl (cos θ) sin θ dθ
(Eq. 3.84)
0
Zπ/2 Zπ 1 = σ P (cos θ) sin θ dθ − P (cos θ) sin θ dθ 0 l l 20 Rl−1 0 π/2 1 Z0 Z σ0 Pl (x) dx . P (x) dx − = l l−1 20 R
(let x = cos θ)
−1
0
Now Pl (−x) = (−1)l Pl (x), since Pl (x) is even, for even l, and odd, for odd l. Therefore Z0
Z0 Pl (x) dx =
−1
l
Z1
Pl (−x) d(−x) = (−1) 1
Pl (x) dx, 0
and hence Al =
σ0 1 − (−1)l l−1 20 R
Z1 Pl (x) dx = 0
0,
σ0 0 Rl−1
if l is even
Z1 0
Pl (x) dx, if l is odd
So A0 = A2 = A4 = A6 = 0, and all we need are A1 , A3 , and A5 . Z1
Z1 P1 (x) dx =
0
Z1
0
1 P3 (x) dx = 2
Z1
1 5x − 3x dx = 2 3
0
0
Z1
1 x2 1 x dx = = . 2 0 2 1 x2 1 5 3 x4 1 5 −3 = − =− . 4 2 0 2 4 2 8
1 1 x6 x4 x2 63x − 70x + 15x dx = 63 − 70 + 15 8 6 4 2 0 0 1 1 1 21 35 15 − + = (36 − 35) = . = 8 2 2 2 16 16
1 P5 (x) dx = 8
0
Z1
5
3
Therefore A1 =
σ0 0
1 σ0 1 σ0 1 ; A3 = − ; A = ; etc. 5 2 4 2 0 R 8 0 R 16
and B1 =
σ0 3 R 0
1 σ0 5 1 σ0 7 1 ; B3 = R − ; B5 = R ; etc. 2 0 8 0 16
c
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.
68
CHAPTER 3. POTENTIAL
Thus 1 r 4 1 r 2 σ0 r (r ≤ R), 20 P1 (cos θ) − 4 R P3 (cos θ) + 8 R P5 (cos θ) + ... , " # 2 4 V (r, θ) = 3 σ0 R 1 R 1 R P3 (cos θ) + P5 (cos θ) + ... , (r ≥ R). 20 r2 P1 (cos θ) − 4 r 8 r Problem 3.24 1 ∂ s ∂s
s
∂V ∂s
+
1 ∂2V = 0. s2 ∂φ2
Look for solutions of the form V (s, φ) = S(s)Φ(φ): 1 d dS 1 d2 Φ Φ s + 2 S 2 = 0. s ds ds s dφ Multiply by s2 and divide by V = SΦ: s d S ds
s
dS ds
+
1 d2 Φ = 0. Φ dφ2
Since the first term involves s only, and the second φ only, each is a constant: s d dS 1 d2 Φ s = C1 , = C2 , with C1 + C2 = 0. S ds ds Φ dφ2 Now C2 must be negative (else we get exponentials for Φ, which do not return to their original value—as geometrically they must— when φ is increased by 2π). C2 = −k 2 . Then
d2 Φ = −k 2 Φ ⇒ Φ = A cos kφ + B sin kφ. dφ2
Moreover, since Φ(φ + 2π) = Φ(φ), k must be an integer: k = 0, 1, 2, 3, . . . (negative integers are just repeats, but k =0 must be included, since Φ = A (a constant) is OK). d dS s s = k 2 S can be solved by S = sn , provided n is chosen right: ds ds s
d d snsn−1 = ns (sn ) = n2 ssn−1 = n2 sn = k 2 S ⇒ n = ±k. ds ds
Evidently the general solution is S(s) = Csk + Ds−k , unless k = 0, in which case we have only one solution to a secondorder equation—namely, S = constant. So we must treat k = 0 separately. One solution is a constant—but what’s the other? Go back to the diferential equation for S, and put in k = 0: dS dS dS C ds d s =0⇒s = constant = C ⇒ = ⇒ dS = C ⇒ S = C ln s + D (another constant). s ds ds ds ds s s So the second solution in this case is ln s. [How about Φ? That too reduces to a single solution, Φ = A, in the case k = 0. What’s the second solution here? Well, putting k = 0 into the Φ equation: d2 Φ dΦ =0⇒ = constant = B ⇒ Φ = Bφ + A. dφ2 dφ c
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1
CHAPTER 3. POTENTIAL
69
Contents
But a term of the form Bφ is unacceptable, since it does not return to its initial value when φ is augmented by 2π.] Conclusion: The general solution with cylindrical symmetry is ∞ X k V (s, φ) = a0 + b0 ln s + s (ak cos kφ + bk sin kφ) + s−k (ck cos kφ + dk sin kφ) . k=1 1 Special Techniques
Yes: the potential of a line charge goes like ln s, which is included. Problem Problem3.25 3.24 Picking V V= = 0 on thethe yz yz plane, with E0 E in0the direction, we have (Eq. 3.74): Picking 0 on plane, with in x the x direction, ! " (i)(i)V V== 0,0, when whens = s =R,R, (ii) −E ==−E φ,φ,forfors R.R. 0 x0 x 0 s0cos (ii)V V→→ −E −E s cos s#
y
− − − − −
2
" + + φ+ + +
s
!x Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1: $ # c1 cos φ. V (s, φ) = a1 s + s =⇒E0 Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1: # (i)⇒ c1 = −a1 R2 ; (ii)→ a1 = −E0 . Therefore c1 z V (s, φ) = a1 s + cos φ. s ( '% & & % 2 2 E R R 0 − 1 cos φ. (s, φ)a= (i)⇒ c1 = −a1 R2 ; V(ii)→ −E00s. + Therefore cos φ, or V (s, φ) = −E0 s 1 = −E s s " # 2 R )E0 R2 ) % & ) s V (s, φ) = −E0 s∂V +) cos φ, or 2 V (s, φ) = −E − 1 cos φ. ) s = −#0 E0 − R − 1 cos φ) 0 = 2# s 0 E0 cos φ. σ = −#0 ) ∂s )s=R s2 s=R ∂V R2 Problem 3.25 = 20 E0 cos φ. = −0 E0 − 2 − 1 cos φ σ = −* 0∞ ∂sk s=R s s=R Inside: V (s, φ) = a0 + s (ak cos kφ + bk sin kφ) . (In this region ln s and s−k are no good—they blow
k=1 Problem 3.26 ∞ up at s = 0.) X ∞k −k Inside: V (s, φ) = a0 + * s (a 1 k cos kφ + bk sin kφ) . (In this region ln s kand s are no good—they blow + (c cos kφ + d sin kφ). (Here ln s and s are no good at s → ∞). Outside: V (s, φ) = a0 k=1 k k sk k=1 up at s = 0.) ∞ X &) % 1 ) ln s and sk are no good at s → ∞). ∂Vin(Here Outside: V (s, φ) = a0 + (ck cos kφ +∂V dkout sin kφ). ) (Eq. 2.36). − sk σ = −#0 k=1 ∂s ∂s )s=R ∂Vout ∂Vin Thus σ = − − (Eq. 2.36). " 0 ∞ ! * ∂s ∂s s=R k a sin 5φ = −#0 − k+1 (ck cos kφ + dk sin kφ) − kRk−1 (ak cos kφ + bk sin kφ) . R k=1 Thus ∞ X k k−1 a sin 5φ = − (c cos kφ + d sin kφ) − kR (a cos kφ + b sin kφ) . c #2005 Pearson Education, Inc., Saddle River, NJ. All rights reserved. This material is 0 Upper− k k k k k+1 protected under all copyright laws as theyRcurrently exist. No portion of this material may be k=1 reproduced, in any form or by any means, without permission in writing from the publisher. 1 4 Evidently ak = ck = 0; bk = dk = 0 except k = 5; a = 50 d5 + R b5 . Also, V is continuous at s = R: R6 1 a0 +R5 b5 sin 5φ = a0 + 5 d5 sin 5φ. So a0 = a0 (might as well choose both zero); R5 b5 = R−5 d5 , or d5 = R10 b5 . R
c
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70
CHAPTER 3. POTENTIAL aR6 a ; d5 = . Therefore 4 100 R 100
Combining these results: a = 50 R4 b5 + R4 b5 = 100 R4 b5 ; b5 = a sin 5φ V (s, φ) = 100
s5 /R4 , for s < R, R6 /s5 , for s > R.
Problem 3.27 Since r is on the z axis, the angle α is just the polar angle θ (I’ll drop the primes, for simplicity). Monopole term: Z Z 1 ρ dτ = kR (R − 2r) sin θ r2 sin θ dr dθ dφ. r2 But the r integral is ZR R (R − 2r) dr = Rr − r2 0 = R2 − R2 = 0. 0
So the monopole term is zero. Dipole term: Z Z 1 r cos θρ dτ = kR (r cos θ) 2 (R − 2r) sin θ r2 sin θ dr dθ dφ. r But the θ integral is Zπ
π sin3 θ 1 = (0 − 0) = 0. 3 0 3
sin2 θ cos θ dθ =
0
So the dipole contribution is Quadrupole term: Z 3 cos2 θ − r2 2
likewise zero. 1 2
ρ dτ =
1 kR 2
Z
r2 3 cos2 θ − 1
1 (R − 2r) sin θ r2 sin θ dr dθ dφ. r2
r integral: Z
R
r2 (R − 2r) dr =
0
R r3 R4 r4 R4 R4 = R− − =− . 3 2 0 3 2 6
θ integral: Zπ
2

Zπ
2
3 cos θ − 1 {z } = 2
π 2
sin θ dθ − 3
sin θ dθ = 2
0 3(1−sin2 θ)−1=2−3 sin2 θ
0
−3
Zπ
2
3π 8
=π 1−
sin4 θ dθ
0
9 8
π =− . 8
φ integral: Z2π dφ = 2π. 0
The whole integral is: R4 π kπ 2 R5 1 kR − − (2π) = . 2 6 8 48 c
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CHAPTER 3. POTENTIAL
71
For point P on the z axis (r → z in Eq. 3.95) the approximate potential is V (z) ∼ =
Problem 3.28
1 kπ 2 R5 . 4π0 48z 3
(Quadrupole.)
z r
f'
R
r'
y
x For a line charge, ρ(r0 ) dτ 0 → λ(r0 ) dl0 , which in this case becomes λR dφ0 . r r0 r · r0 cos α
= = = =
r sin θ cos φ x ˆ + r sin θ sin φ y ˆ + r cos θ ˆ z, 0 0 R cos φ + R sin φ , so rR sin θ cos φ cos φ0 + rR sin θ sin φ sin φ0 = rR cos α, sin θ(cos φ cos φ0 + sin φ sin φ0 ).
n=0: Z
0
0
Z
ρ(r ) dτ → λR
2π
dφ0 = 2πRλ;
V0 =
0
n=1: Z Z Z r0 cos α ρ(r0 ) dτ 0 → R cos α λR dφ0 = λR2 sin θ
λ R 1 2πRλ = . 4π0 r 20 r
2π
(cos φ cos φ0 + sin φ sin φ0 )dφ0 = 0; V1 = 0.
0
n=2: Z Z Z h i 3 1 λR3 2 0 2 0 0 2 2 λRdφ0 = (r ) P2 (cos α) ρ(r )dτ → R cos α − 3 sin2 θ (cos φ cos φ0 + sin φ sin φ0 ) − 1 dφ0 2 2 2 Z 2π Z 2π Z 2π Z 2π λR3 3 sin2 θ cos2 φ cos2 φ0 dφ0 + sin2 φ sin2 φ0 dφ0 + 2 sin φ cos φ sin φ0 cos φ0 dφ0 − dφ0 = 2 0 0 0 0 3 3 λR πλR 3 1 = 3 sin2 θ π cos2 φ + π sin2 φ + 0 − 2π = 3 sin2 θ − 2 = −πλR3 cos2 θ − . 2 2 2 2 So V2 = −
λ R3 λ R3 2 3 cos θ − 1 = − P2 (cos θ). 80 r3 40 r3
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72
CHAPTER 3. POTENTIAL
Problem 3.29 p = (3qa − qa) ˆ z + (−2qa − 2q(−a)) y ˆ = 2qa ˆ z. Therefore V ∼ =
1 p·ˆ r , 2 4π0 r
and p · ˆ r = 2qa ˆ z·ˆ r = 2qa cos θ, so 1 2qa cos θ . 4π0 r2
V ∼ =
(Dipole.)
Problem 3.30 R R (a) By symmetry, p is clearly in the z direction: p = p ˆ z; p = zρ dτ ⇒ zσ da. Z p =
2
3
Zπ
(R cos θ)(k cos θ)R sin θ dθ dφ = 2πR k
2
3
cos θ sin θ dθ = 2πR k 0
=
2 3 4πR3 k πR k[1 − (−1)] = ; 3 3
p=
π cos3 θ − 3 0
4πR3 k ˆ z. 3
(b) V ∼ =
1 4πR3 k cos θ kR3 cos θ = . 2 4π0 3 r 30 r2
(Dipole.)
This is also the exact potential. Conclusion: all multiple moments of this distribution (except the dipole) are exactly zero. Problem 3.31 Using Eq. 3.94 with r0 = d/2 and α = θ (Fig. 3.26): 1
r for
r
−,
+
n ∞ 1X d = Pn (cos θ); r n=0 2r
we let θ → 180◦ + θ, so cos θ → − cos θ: 1
r
−
=
n ∞ 1X d Pn (− cos θ). r n=0 2r
But Pn (−x) = (−1)n Pn (x), so V =
1 q 4π0
1
r
−
+
1
r
=
−
n ∞ 1 1X d 2q q [Pn (cos θ) − Pn (− cos θ)] = 4π0 r n=0 2r 4π0 r
Therefore Vdip = Voct =
2q 4π0 r
d 2r
2q 1 d qd cos θ P1 (cos θ) = , 4π0 r 2r 4π0 r2
3 P3 (cos θ) =
while
X n=1,3,5,...
d 2r
n Pn (cos θ).
Vquad = 0.
2q d3 1 qd3 1 5 cos3 θ − 3 cos θ = 5 cos3 θ − 3 cos θ . 4 4 4π0 8r 2 4π0 8r
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CHAPTER 3. POTENTIAL
73
Problem 3.32 1 2q 3qa cos θ . + = + z, (ii) p = 3qa ˆ 4π0 r r2 1 2q qa cos θ (ii) p = qa ˆ z, (iii) V ∼ . + = 4π0 r r2 2q 3qa sin θ sin φ 1 (ii) p = 3qa y ˆ, (iii) V ∼ + (from Eq. 1.64, y ˆ ·ˆ r = sin θ sin φ). = 4π0 r r2 (iii) V ∼ =
(a) (i) Q = 2q, (b) (i) Q = 2q, (c) (i) Q = 2q,
1 4π0
h
Q r
p·ˆ r r2
i
Problem 3.33 (a) This point is at r = a, θ =
π 2,
φ = 0, so E =
pq p p θˆ = (−ˆ z); F = qE = − ˆ z. 3 3 4π0 a 4π0 a 4π0 a3
2p p 2pq (2ˆ r) = ˆ z. F = ˆ z. 4π0 a3 4π0 a3 4π0 a3 π i qp h pq . (c) W = q [V (0, 0, a) − V (a, 0, 0)] = cos(0) − cos = 4π0 a2 2 4π0 a2 (b) Here r = a, θ = 0, so E =
Problem 3.34 Q = −q, so Vmono =
1 −q ; 4π0 r
q 4π0
V (r, θ) ∼ =
p = qa ˆ z,
1 a cos θ − + r r2
.
so Vdip =
E(r, θ) ∼ =
1 qa cos θ . Therefore 4π0 r2 q a 1 ˆ r + 3 2 cos θ ˆ − 2ˆ r + sin θ θ . 4π0 r r
Problem 3.35 The total charge is zero, so the dominant term is the dipole. We need the dipole moment of this configuration. It obviously points in the z direction, and for the southern hemisphere (θ : π2 → π) ρ switches sign but so does z, so Z p =
Z
π/2 2
zρ dτ = 2ρ0
r cos θ r sin θ dr dθ dφ = 2ρ0 (2π)
θ=0 π/2
R4 sin2 θ 4 2 0
= 4πρ0
Z
=
R 3
Z
π/2
r dr 0
cos θ sin θ dθ 0
πρ0 R4 . 2
Therefore (Eq. 3.103) E≈
πρ0 R4 ˆ . 2 cos θ ˆ r + sin θ θ 8π0 r3
Problem 3.36 ˆ θˆ = p cos θ ˆ p = (p · ˆ r) ˆ r + (p · θ) r − p sin θ θˆ (Fig. 3.36). So 3(p · ˆ r) ˆ r − p = 3p cos θ ˆ r − p cos θ ˆ r + p sin θ θˆ = ˆ 2p cos θ ˆ r + p sin θ θ. So Eq. 3.104 ≡ Eq. 3.103. X Problem 3.37 Vave (R) =
1 4πR2
Z V (r) da, where the integral is over the surface of a sphere of radius R. Now da =
c
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74 R2 sin θ dθ dφ, so Vave (R) =
1 4π
CHAPTER 3. POTENTIAL
Z V (R, θ, φ) sin θ dθ dφ.
Z Z Z 1 dVave 1 ∂V 1 (∇V · ˆ r) sin θ dθ dφ = (∇V ) · (R2 sin θ dθ dφ ˆ = sin θ dθ dφ = r) dR 4π ∂R 4π 4πR2 Z Z 1 1 (∇V ) · da = (∇2 V ) dτ = 0. = 4πR2 4πR2 5 (The final integral, from the divergence theorem, is over the volume of the sphere, where by assumption the Laplacian of V is zero.) So Vave is independent of R—the same for all spheres, regardless of their radius—and 2 hence (takingcan thealso limit R → 0), Vdirectly. V (0). ave (R) =Let This integral beas integrated x = uqed ; dx = 2u du. Problem 3.38 At a point (x, y) on the plane the field of q is " $ %&'0 !0 √ !0 ' u2 u# u x d −1 2 1 q ' = −d sin−1 (1) = −d π . √ √ E =du = 2 − rˆ , d and √ dx = 2 − u +r = sinx x '√ ˆ + y y ˆ − d ˆ z, 2 q 2 2 2 d−x d − u 4π0 r 3 d d √ d
d
q d σ . Meanwhile, the field of σ (just below the surface) is − , 2 2 2 3/2 4π0 (x(+ y + d )) 20 ) (Eq. 2.17). (Of course, this is for a duniform surface but as long πd d π 2 d2 charge, 2π 3as d3 "we 0 mare infinitesimally far away σ is 16π" . t =field inside=the conductor 0 m =so effectively uniform.) The total is zero, 2 2 2A 2 4 2q q
so its z component is − Therefore
Problem 3.35
−
Problem 3.39 x!
q σ d − =0 4π0 (x2 + y 2 + d2 )3/2 20
+
−
+
−
+
⇒
σ(x, y) = −
+
−
+
−
q
qd . X 2π(x2 + y 2 + d2 )3/2
+
−
−
The image configuration is shown in the figure; the positive image charge forces cancel in pairs. The net force of the negative image charges is: *( 1 2 11 1 11 11 + + 2+ 2+ 2 + ... FF == 4π0qq2 2 2 2 + ... [2(a − x)] [2a + 2(a − x)] [4a + 2(a − x)] 4π"0 [2(a − x)] [2a + 2(a − x)] [4a + 2(a − x)] & 1 1 1 1 1 1 −− (2x)22−− (2a + 2x)22−− (4a + 2x)22−−. .. .. . (2x) (2a + 2x) (4a + 2x) 2"+ , + 1 ,& 11 11 11 11 1 + 11 11 qq2 = + + + . . . − + + . . . . = 4π0 4 (a−−x) x)22+ (2a (2a−−x) x)22+ (3a (3a−−x) x)22+ . . . − xx22+ (a (a++x) x)22+ (2a (2a++x) x)22+ . . . . 4π"0 4 (a 1 1 q2 2 When a → ∞ (i.e. a x) only the 1 2 term survives: F = − 1 X (same as for only one plane— q When a → ∞ (i.e. a % x) only the x2 term survives: F = − 4π0 (2x)22 ! (same as for only one plane— x 4π"0 (2x) Eq. 3.12). When x = a/2, Eq. 3.12). When x = a/2, 1 q 2"+ 1 1 1 1 1 1 , + ,& F = 1 q2 = 0. X 1 2 + (3a/2) 1 2 + (3a/2) 1 2 + (5a/2) 1 2 + . . . − (a/2) 1 2 + (5a/2) 1 2 + ... (a/2) F = 4π0 4 + + + ... − + + + ... = 0. ! 2 2 2 2 2 2 4π"0 4 (a/2) (3a/2) (5a/2) (a/2) (3a/2) (5a/2) Problem 3.40 Problem 3.36Prob. 2.52, we place image line charges −λ at y = b and +λ at y = −b (here y is the horizontal Following axis, z vertical). Following Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the horizontal axis, z vertical). z Upper Saddle River, NJ. All rights reserved. This material is c
2012 Pearson Education, Inc., protected under all copyright# laws as they currently exist. No portion of this material may be P reproduced, in any form or by any means, s1 without permission in writing from the publisher. s2 −λ
−b
s3
b
+λ R
$ 
 ./ 0  ./ 0 λ a−b a−b 2 2 ./ 0 y0 ./ 0 a
−λ
c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
s4 "y
When a → ∞ (i.e. a % x) only the
1 1 q2 term survives: F = − ! (same as for only o 2 x 4π"0 (2x)2
Eq. 3.12). When x = a/2, F =
1 q2 4π"0 4
"+
, + ,& 1 1 1 1 1 1 + + + . . . − + + + . . . = (a/2)2 (3a/2)2 (5a/2)2 (a/2)2 (3a/2)2 (5a/2)2
Problem 3.36 CHAPTER 3. POTENTIAL 75 Following Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the axis, z vertical). z # P s1 s2 −b
−λ
b
+λ R
$ 
In the solution to Prob. 2.52 substitute:
s3
 ./ 0  ./ 0 λ a−b a−b 2 2 ./ 0 y0 ./ 0 a
−λ
s4 "y
c #2005 Pearson Education, Inc., Upper Saddle material is NJ. All 2 River, 2 rights reserved. This protected all copyright may be + b No portion R2 a − under b a + b laws asa they − b currentlyaexist. 2 of this material = permission − R ⇒ from b = the publisher. areproduced, → ,inyany . form or byso any means, without in writing 0 →
2
2
2
2
a
2 2 2 2 λ s s λ s s ln 32 + ln 12 = ln 12 32 4π0 s4 s2 4π0 s s 4 2 2 2 2 2 λ [(y + a) + z ][(y − b) + z ] , or, using y = s cos φ, z = s sin φ, = ln 4π0 [(y − a)2 + z 2 ][(y + b)2 + z 2 ] 2 λ (s + a2 + 2as cos φ)[(as/R)2 + R2 − 2as cos φ] ln . = 4π0 (a2 + a2 − 2as cos φ)[(as/R)2 + R2 + 2as cos φ]
V =
Problem 3.41 Same as Problem 3.9, only this time we want q 0 + q 00 = q, so q 00 = q − q 0 : 00 q q q0 q2 qq 0 1 1 F = + = + − 2+ . 4π0 a2 (a − b)2 4π0 a2 4π0 a (a − b)2 The second term is identical to Problem 3.9, and I’ll just quote the answer from there: 2 2 q2 3 (2a − R ) F = a−R . 4π0 a3 (a2 − R2 )2 (a) F = 0 ⇒ a(a2 − R2 )2 = R3 (2a2 − R2 ), or (letting x ≡ a/R), x(x2 − 1)2 − 2x2 + 1 = 0. We want a real √ root greater than 1; Mathematica delivers x = (1 + 5)/2 = 1.61803, so a = 1.61803R = 5.66311 ˚ A. (b) Let a0 = x0 R be the minimum value of a. The work necessary is Z a0 Z ∞ Z ∞ 2 2 q2 (2x2 − 1) 1 q2 1 3 (2a − R ) W =− F da = − a − R da = dx 4π0 a0 a3 (a2 − R2 )2 4π0 R x0 x2 x3 (x2 − 1)2 ∞ q2 1 + 2x0 − 2x30 = . 4π0 R 2x20 (1 − x20 ) √ Putting in x0 = (1 + 5)/2, Mathematica says the term in square brackets is 1/2 (this is not an accident; see q2 . Numerically, footnote 6 on page 127), so W = 8π0 R W =
(1.60 × 10−19 )2 J = 2.03 × 10−19 J = 1.27 eV. 8π(8.85 × 10−12 )(5.66 × 10−10 )
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76
CHAPTER 3. POTENTIAL
Problem 3.42 y V0 b
a
y
V1
=
V0 0
b
a
V0
x
y
0
0
b
+
V0 x
b
0 b
a
V1 0 0
b
x
The first configuration on the right is precisely Example 3.4, but unfortunately the second configuration is not the same as Problem 3.15: y 0
a 0
V0 0
x
b
We could reconstruct Problem 3.15 with the modified boundaries, but let’s see if we can’t twist it around by an astute change of variables. Suppose we let x → y, y → u, a → c, b → a, and V0 → V1 : y
V1
a 0
0 0
u
c
This is closer; making the changes in the solution to Problem 3.15 we have (for this configuration) V (u, y) =
4V1 π
sinh(nπy/c) sin(nπu/c) . n sinh(nπa/c) n=1,3,5... X
Now let c → 2b and u → x + b, and the configuration is just what we want: y a
0
V1 0 0
b
b
x
The potential for this configuration is V (x, y) =
4V1 π
sinh(nπy/2b) sin(nπ(x + b)/2b) . n sinh(nπa/2b) n=1,3,5... X
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CHAPTER 3. POTENTIAL
77
(If you like, write sin(nπ(x + b)/2b) as (−1)(n−1)/2 cos(nπx/2b).) Combining this with Eq. 3.42, V (x, y) =
4 π
1 cosh(nπx/a) sin(nπy/a) sinh(nπy/2b) sin(nπ(x + b)/2b) V0 + V1 . n cosh(nπb/a) sinh(nπa/2b) n=1,3,5... X
Here’s a plot of this function, for the case a = b = 1, V0 = 1/2, V1 = 1:
1.0 1.0 0.5
0.0 1.0
0.5 0.5 0.0 0.5 1.0
0.0
Problem 3.43
Bl Since the configuration is azimuthally symmetric, V (r, θ) = Al r + l+1 Pl (cos θ). r X Bl (a) r > b: Al = 0 for all l, since V → 0 at ∞. Therefore V (r, θ) = Pl (cos θ). rl+1 X Dl a < r < b : V (r, θ) = Cl rl + l+1 Pl (cos θ). r < a : V (r, θ) = V0 . r We need to determine Bl , Cl , Dl , and V0 . To do this, invoke boundary conditions as follows: (i) V is 1 ∂V = − σ(θ) at b. continuous at a, (ii) V is continuous at b, (iii) 4 ∂r 0 X X Bl Dl Bl Dl l Pl (cos θ) = Cl b + l+1 Pl (cos θ); = Cl bl + l+1 ⇒ Bl = b2l+1 Cl + Dl . (1) (ii) ⇒ bl+1 b bl+1 b Dl l X Cl a + l+1 = 0, if l 6= 0, D = −a2l+1 C , l 6= 0, Dl l l a (i) ⇒ Cl al + l+1 Pl (cos θ) = V0 ; (2) D0 D = aV − aC a 0 0 0 0. C0 a + = V , if l = 0; 0 a1 Putting (2) into (1) gives Bl = b2l+1 Cl − a2l+1 Cl , l 6= 0, B0 = bC0 + aV0 − aC0 . Therefore X
l
Bl = b2l+1 − a2l+1 Cl , l 6= 0, (10 ) B0 = (b − a)C0 + aV0 . (iii) ⇒
X
Bl [−(l + 1)]
1
b
P (cos θ) − l+2 l
X
Cl lbl−1 + Dl
−(l + 1) bl+2
Pl (cos θ) =
−k P1 (cos θ). So 0
−(l + 1) (l + 1) l−1 + Dl = 0, if l 6= 1; − l+2 Bl − Cl lb b bl+2 or −(l + 1)Bl − lCl b2l+1 + (l + 1)Dl = 0;
(l + 1)(Bl − Dl ) = −lb2l+1 Cl .
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78
B1 (+2)
CHAPTER 3. POTENTIAL
1 −2 k + C + D = , for l = 1; 1 1 2 b2 b 0
C1 +
2 (B1 − D1 ) = k. b3
Therefore (l + 1)(Bl − Dl ) + lb2l+1 Cl = 0, for l 6= 1, 2 k (3) C1 + 3 (B1 − D1 ) = . b 0 Plug (2) and (10 ) into (3): For l 6= 0 or 1: (l+1) b2l+1 − a2l+1 Cl + a2l+1 Cl +lb2l+1 Cl = 0; (l+1)b2l+1 Cl +lb2l+1 Cl = 0; (2l+1)Cl = 0 ⇒ Cl = 0. Therefore (10 ) and (2) ⇒ Bl = Cl = Dl = 0 for l > 1. 2 For l = 1: C1 + 3 b3 − a3 C1 + a3 C1 = k; C1 + 2C1 = k ⇒ C1 = k/30 ; D1 = −a3 C1 ⇒ b D1 = −a3 k/30 ; B1 = b3 − a3 C1 ⇒ B1 = b3 − a3 k/30 . For l = 0: B0 −D0 = 0 ⇒ B0 = D0 ⇒ (b−a)C0 +aV0 = aV0 −aC0 , so bC0 = 0 ⇒ C0 = 0; D0 = aV0 = B0 . b3 − a3 k aV0 aV0 k a3 Conclusion: V (r, θ) = + r ≥ b. V (r, θ) = + cos θ, a ≤ r ≤ b. cos θ, r − r 3r2 0 r 30 r2 ∂V k k 0 aV0 a3 V0 (b)σi (θ) = −0 = − + cos θ = − + − 1 + 2 − cos θ = −k cos θ + V0 . 0 0 2 3 ∂r a 3 a a a 0 0 a Z V0 0 aV 1 1 aV Q 4πa V ? 0 0 0 0 (c)qi = σi da = 4πa2 = 4πa0 V0 = Qtot . At large r: V ≈ = = = .X a r 4π0 r 4π0 r r Problem 3.44 dr' r' q'=q
Use multipole expansion (Eq. 3.95): ρ dτ 0 → λ dr0 ; Q λ= ; the r0 integral breaks into two pieces: 2a
r
q'=pq
a Z Za ∞ 1 X 1 V (r) = (r0 )n Pn (cos θ0 )λ dr0 + (r0 )n Pn (cos θ0 )λ dr0 . 4π0 n=0 rn+1 0
0
0
0
In the first integral θ = θ (see diagram); in the second integral θ = π − θ, so cos θ0 = − cos θ. But Pn (−z) = (−1)n Pn (z), so the integrals cancel when n is odd, and add when n is even. a
Z ∞ X 1 Q 1 V (r) = 2 Pn (cos θ) xn dx. 4π0 2a n=0,2,4,... rn+1 0
The integral is
an+1 , so n+1 Q 1 X 1 a n V = Pn (cos θ) . 4π0 r n=0,2,4,... n + 1 r
c
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CHAPTER 3. POTENTIAL
79
Problem 3.45 Use separation of variables in cylindrical coordinates (Prob. 3.24): V (s, φ) = a0 + b0 ln s +
∞ X k s (ak cos kφ + bk sin kφ) + s−k (ck cos kφ + dk sin kφ) . k=1
P∞ s < R : V (s, φ) = k=1 sk (ak cos kφ + bk sin kφ) (ln s and s−k blow up at s = 0); P∞ s > R : V (s, φ) = k=1 s−k (ck cos kφ + dk sin kφ) (ln s and sk blow up as s → ∞). (We may as well pick constantsXso V → 0 as s → ∞, and hence a0 = 0.) Continuity at s = R ⇒ X Rk (ak cos kφ + bk sin kφ) = R−k (ck cos kφ + dk sin kφ), so ck = R2k ak , dk = R2k bk . Eq. 2.36 says: ∂V 1 ∂V − = − σ. Therefore ∂s R+ ∂s R− 0 X X −k 1 (c cos kφ + d sin kφ) − kRk−1 (ak cos kφ + bk sin kφ) = − σ, k k Rk+1 0 or: X
2kRk−1 (ak cos kφ + bk sin kφ) =
σ0 /0 (0 < φ < π) −σ0 /0 (π < φ < 2π)
.
Fourier’s trick: multiply by (cos lφ) dφ and integrate from 0 to 2π, using Z2π
Z2π sin kφ cos lφ dφ = 0;
0
Then
cos kφ cos lφ dφ =
0, k 6= l π, k = l
.
0
π ( π 2π ) Z Z2π sin lφ sin lφ σ σ 0 0 − cos lφ dφ − cos lφ dφ = = 0; 2lRl−1 πal = 0 0 l 0 l π 0
al = 0.
π
Multiply by (sin lφ) dφ and integrate, using
2π R
sin kφ sin lφ dφ =
0
0, k 6= l : π, k = l
π ( π 2π ) Z Z2π σ cos lφ cos lφ σ0 σ0 0 l−1 + sin lφ dφ − sin lφ dφ = − = (2 − 2 cos lπ) 2lR πbl = 0 0 l l l 0 0 π π 0 0, if l is even 0, if l is even = ⇒ bl = . 4σ0 /l0 , if l is odd 2σ0 /π0 l2 Rl−1 , if l is odd Conclusion: V (s, φ) =
2σ0 R π0
X k=1,3,5,...
1 sin kφ k2
(s/R)k (s < R) (R/s)k (s > R)
Problem 3.46 ∞ 1 X Pn (cos θ) Use Eq. 3.95, in the form V (r) = In ; 4π0 n=0 rn+1
Za In =
z n λ(z) dz.
−a
c
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.
Contents 1
Contents Problem 3.40
80
Contents Z a
CHAPTER 3. POTENTIAL
h i 4ak πz a 2a = 2ak sin π − sin − π dz = k = sin . Therefore: π 2a π 2 2 π −a
πz Problem (a) I0 = k 3.40 cos 2a −a
λ(z) "
V (r, θ) ∼ =
1 4π0
4ak π
1 . r
(Monopole.)
!z Problem 3.40 −a a (b) I0 = 0. a πz az πz a λ(z) Z a 2 " I1 = kλ(z)z sin(πz/a) dz = k sin − cos π a π a " −a −a ! z 2 z −a a a! a2 a2 2a2 −a = k a [sin(π) − sin(−π)] − cos(π) − cos(−π) = k ; π π π π λ(z) λ(z) "" 2 1 2a k 1 ∼ λ(z) V (r, θ) cos θ. (Dipole.) = !! " z 4π0 π r2 −a−a aa z −a a! z (c) I0 = I1 = 0. λ(z) a Z" πz a πz 2z cos(πz/a) (πz/a)2 − 2 2 I2 =λ(z) k z cos + sin ! z a dz = k 2 3 (π/a) (π/a) a " −a −a a −a ! −a a a z2 4a3 k = 2k [a cos(π) + a cos(−π)] = − 2 . π π λ(z) " a!
−a
z
V (r, θ) ∼ =
1 4π0
−
4a3 k π2
1 3 cos2 θ − 1 . 3 2r
(Quadrupole.)
2
Problem 3.47
c !2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright currently exist. qNo (a) The average field laws dueastothey a point charge atportion r is of this material may be Problem 3.41 reproduced, in any form or by any means, without permission in writing from the publisher.
r
!
r q
Eave =
1 4 3 3 πR
Z E dτ, Z
where E =
rˆ r
1 q rˆ , 4π0 r 2
1 1 dτ c !2005 Pearson Education, Inc., Upper Saddle NJ.All rights reserved. This material is q 2 dτ. so E ave =River, 4 3 No 4π protected under all copyright laws as they currentlyπR exist. portion of this material may be 0 3 reproduced, in any form or by any means, without permission in writing from the publisher.
(Here r is the source point, dτ is the field point, so r goes from r to dτ .) The field at r due to uniform Z rˆ 1 charge ρ over the sphere is Eρ = ρ 2 dτ. This time dτ is the source point and r is the field 4π 0 c !2005 Pearson Education, Inc., Upper Saddle River, NJ.rAll rights reserved. This material is protected under copyright as they currently exist. No portion of this material r allgoes point, so from laws dτ to r, and hence carries the opposite sign.may So be with ρ = −q/ 43 πR3 , the two reproduced, in any form or by any means, without permission in writing from the publisher. expressions agree: Eave = Eρ .
(b) From Prob. 2.12: Eρ =
1 q r p ρr = − =− . 3 30 4π0 R 4π0 R3
c
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CHAPTER 3. POTENTIAL
81
(c) If there are many charges inside the sphere, Eave is the sum of the individual averages, and ptot is the p sum of the individual dipole moments. So Eave = − . qed 4π0 R3 (d) The same argument, only with q placed at r outside the sphere, gives 4 3 1 −q 1 3 πR ρ ˆ r (field at r due to uniformly charged sphere) = ˆ r. Eave = Eρ = 2 4π0 r 4π0 r2 But this is precisely the field produced by q (at r) at the center of the sphere. So the average field (over the sphere) due to a point charge outside the sphere is the same as the field that same charge produces at the center. And by superposition, this holds for any collection of exterior charges. Problem 3.48 (a) p ˆ (2 cos θ ˆ r + sin θ θ) 4π0 r3 p = [2 cos θ(sin θ cos φ x ˆ + sin θ sin φ y ˆ + cos θ ˆ z) 4π0 r3 + sin θ(cos θ cos φ x ˆ + cos θ sin φ y ˆ − sin θ ˆ z)]
Edip =
=
p z . ˆ + 3 sin θ cos θ sin φ y ˆ + 2 cos2 θ − sin2 θ ˆ 3 sin θ cos θ cos φ x 3 4π0 r  {z } =3 cos2 θ−1
Eave = =
1 4 3 3 πR
p 4π0
Z
2 1 3 sin θ cos θ(cos φ x ˆ + sin φ y ˆ) + 3 cos2 θ − 1 ˆ z r sin θ dr dθ dφ. r3
Z2π cos φ dφ =
0
Edip dτ
1 4 3 πR 3
Z2π But
Z
sin φ dφ = 0, so the x ˆ and y ˆ terms drop out, and
2π R
dφ = 2π, so
0
0
Eave =
1 4 3 πR 3
p 4π0
ZR
2π
0
1 dr r
Zπ
3 cos2 θ − 1 sin θ dθ .
0

{z
}
(− cos3 θ+cos θ)π 0 =1−1+1−1=0
Z
R
1 dr, r blows up, since ln r → −∞ as r → 0. If, as suggested, we truncate the r integral at r = , then it is finite, and the θ integral gives Eave = 0.] (b) We want E within the sphere to be a delta function: E = Aδ 3 (r), with A selected so that the average field is consistent with the general theorem in Prob. 3.47: Z p 3 1 A p p δ (r). Eave = 4 3 Aδ 3 (r) dτ = 4 3 = − ⇒A=− , and hence E = − 3 30 4π0 R 30 3 πR 3 πR Evidently Eave = 0, which contradicts the result of Prob. 3.47. [Note, however, that the r integral,
0
c
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82
CHAPTER 3. POTENTIAL
Problem 3.49 We need to show that the field inside the sphere approaches a deltafunction with the right coefficient (Eq. 3.106) in the limit as R → 0. From Eq. 3.86, the potential inside is V =
k k r cos θ = z, 30 30
so
E = −∇V = −
k ˆ z. 30
From Prob. 3.30, the dipole moment of this configuration is p = (4πR3 k/3) ˆ z, so k ˆ z = 3p/(4πR3 ), and hence the field inside is 1 p. E=− 4π0 R3 Clearly E → ∞ as R → 0 (if p is held constant); its volume integral is Z 1 4 1 E dτ = − p πR3 = − p, 4π0 R3 3 30 which matches the deltafunction term in Eq. 3.106. X Problem 3.50 Z (a) I = (∇V1 ) · (∇V2 ) dτ . But ∇·(V1 ∇V2 ) = (∇V1 ) · (∇V2 ) + V1 (∇2 V2 ), so Z I= Problem 3.41
2 Z Z I 1 2 V1 ρ2 dτ. ∇·(V1 ∇V2 ) dτ − V1 (∇ V2 ) = V1 (∇V2 ) · da + 0 S
Z 1 But the surface integral is over a huge sphere “at infinity”, where V1 and V2 → 0. So I = V1 ρ2 dτ . By 0 Z Z Z r 1 the same with 1 and 2 reversed, I = V2 ρ1 dτ . So V1 ρ2 dτ = V2 ρ1 dτ . qed q ! argument, 0 R R dτ Situation (1 ) : Qa = a ρ1 dτ = Q; Qb = b ρ1 dτ = 0; V1b ≡ Vab . (b) R R Situation (2 ) : Qa = a ρ2 dτ = 0; Qb = b ρ2 dτ = Q; V2a ≡ Vba . R R R V1 ρ2 dτ = V1a a ρ2 dτ + V1b b ρ2 dτ = Vab Q. R
V2 ρ1 dτ = V2a
R a
ρ1 dτ + V2b
R b
ρ1 dτ = Vba Q.
Green’s reciprocity theorem says QVab = QVba , so Vab = Vba . qed Problem 3.51 (a) Situation (1): actual. Situation (2): right plate at V0 , left plate at V = 0, no charge at x. Z V =0 V =0 x V1 ρ2 dτ = Vl1 Ql2 + Vx1 Qx2 + Vr1 Qr2 . "x d 0 q R But Vl1 = Vr1 = 0 and Qx2 = 0, so V1 ρ2 dτ = 0. Z V2 ρ1 dτ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 . But Vl2 = 0 Qx1 = q, Vr2 = V0 , Qr1 = Q2 , and Vx2 = V0 (x/d). So 0 = V0 (x/d)q + V0 Q2 , and hence Q2 = −qx/d. c
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CHAPTER 3. POTENTIAL
83
Situation (1): actual. Situation (2): left plate at V0 , right plate at V = 0, no charge at x. Z Z V1 ρ2 dτ = 0 = V2 ρ1 dτ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 = V0 Q1 + qVx2 + 0. x But Vx2 = V0 1 − , so d Q1 = −q(1 − x/d). (b) Situation (1): actual. Situation (2): inner sphere at V0 , outer sphere at zero, no charge at r. Z V1 ρ2 dτ = Va1 Qa2 + Vr1 Qr2 + Vb1 Qb2 . R But Va1 = Vb1 = 0, Qr2 = 0. So V1 ρ2 dτ = 0. Z V2 ρ1 dτ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = Qa V0 + qVr2 + 0. But Vr2 is the potential at r in configuration 2: V (r) = A + B/r, with V (a) = V0 ⇒ A + B/a = V0 , or aA + B = aV0 , and V (b) = 0 ⇒ A + B/b = 0, or bA + B = 0. Subtract: (b − a)A = −aV0 ⇒ A = aV0 b −aV0 /(b − a); B a1 − 1b = V0 = B (b−a) ab ⇒ B = abV0 /(b − a). So V (r) = (b−a) r − 1 . Therefore aV0 Qa V0 + q (b − a)
b − 1 = 0; r
qa Qa = − (b − a)
b −1 . r
Now let Situation (2) be: inner sphere at zero, outer at V0 , no charge at r. Z Z V1 ρ2 dτ = 0 = V2 ρ1 dτ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = 0 + qVr2 + Qb V0 . B with V (a) = 0 ⇒ A + B/a = 0; V (b) = V0 ⇒ A + B/b = V0 , so This time V (r) = A + r bV0 qb a bV0 a a V (r) = Qb = − 1− . Therefore q 1− + Qb V0 = 0; 1− . (b − a) r (b − a) r (b − a) r Problem 3.52 Z 3 3 3 X X 1 X 0X 3 ˆ ri ri ˆ rj rj0 − (r0 )2 ˆ riˆ rj δij ρ dτ 0 (a) ˆ riˆ rj Qij = 2 i=1 i,j=1 j=1 i,j But
3 X
ˆ ri ri0 = ˆ r · r0 = r0 cos α =
i=1
3 X j=1
ˆ rj rj0 ;
X
ˆ riˆ rj δij =
X
ˆ rj ˆ rj = ˆ r·ˆ r = 1.
So
i,j
Z 1 1 1 02 2 02 Vquad = 3r cos α − r ρ dτ 0 = the third term in Eq. 3.96. X 4π0 r3 2 √ (b) Because x2 = y 2 = (a/2)2 for all four charges, Qxx = Qyy = 12 3(a/2)2 − ( 2a/2)2 (q − q − q + q) = 0. √ Because z = 0 for all four charges, Qzz = 12 [−( 2a/2)2 ](q − q − q + q) = 0 and Qxz = Qyz = Qzx = Qzy = 0. This leaves only Qxy = Qyx =
a a a a a a i 3 h a a 3 q+ − (−q) + − (−q) + − − q = a2 q. 2 2 2 2 2 2 2 2 2 2
c
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84
CHAPTER 3. POTENTIAL
(c) Z
3(ri − di )(rj − dj ) − (r − d)2 δij ρ dτ (I0 ll drop the primes, for simplicity.) Z Z Z Z Z = 3ri rj − r2 δij ρ dτ − 3di rj ρ dτ − 3dj ri ρ dτ + 3di dj ρ dτ + 2d · rρ dτ δij Z 2 − d δij ρ dτ = Qij − 3(di pj + dj pi ) + 3di dj Q + 2δij d · p − d2 δij Q.
2Qij =
So if p = 0 and Q = 0 then Qij = Qij . qed (d) Eq. 3.95 with n = 3: Z 1 1 Voct = (r0 )3 P3 (cos α)ρ dτ 0 ; 4π0 r4 Voct =
P3 (cos θ) =
1 5 cos3 θ − 3 cos θ . 2
1 1 X ˆ riˆ rj ˆ rk Qijk . 4π0 r4 i,j,k
Define the “octopole moment” as Qijk ≡
1 2
Z
0 0 0 5ri rj rk − (r0 )2 (ri0 δjk + rj0 δik + rk0 δij ) ρ(r0 ) dτ 0 .
Problem 3.46 Problem 3.53 1 1 1 1 1 0 V = q r 1 − r 2 +q r 3 − r 4 4π0
r r r r
1
=
2
=
3
=
1
=
r −q!
Expanding as in Ex. 3.10:
1
r
− 3
But q 0 = −
1
r
4
r
− 1
1
r
2
r3
b
"# a
r1
θ
+q
!"#$ !"#$+q!
−q !
1
r4
r2
p r2 + a2 − 2ra cos θ, p r2 + a2 + 2ra cos θ, p r2 + b2 − 2rb cos θ, p r2 + b2 + 2rb cos θ.
r ! " #
%$b
$!
"# a
$
2r ∼ = 2 cos θ (we want a r, not r a, this time). a
2b ∼ = 2 cos θ (here we want b r, because b = R2 /a, Eq. 3.16) r 2 R2 = cos θ. a r2
R q (Eq. 3.15), so a V (r, θ) ∼ =
2r 2q R3 1 R 2 R2 1 q 2 cos θ − q r − cos θ. cos θ = 4π0 a a a r2 4π0 a2 r2 c
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CHAPTER 3. POTENTIAL
Set E0 = −
85
1 2q (field in the vicinity of the sphere produced by ±q): 4π0 a2 R3 V (r, θ) = −E0 r − 2 cos θ r
(agrees with Eq. 3.76).
Problem 3.54 The boundary conditions are (i) V = 0 when y = 0, (ii) V = V0 when y = a, (iii) V = 0 when x = b, (iv) V = 0 when x = −b. Go back to Eq. 3.26 and examine the case k = 0: d2 X/dx2 = d2 Y /dy 2 = 0, so X(x) = Ax + B, Y (y) = Cy + D. But this configuration is symmetric in x, so A = 0, and hence the k = 0 solution is V (x, y) = Cy + D. Pick D = 0, C = V0 /a, and subtract off this part: V (x, y) = V0
y + V¯ (x, y). a
The remainder (V¯ (x, y)) satisfies boundary conditions similar to Ex. 3.4: (i) V¯ = 0 when y = 0, (ii) V¯ = 0 when y = a, ¯ (iii) V = −V0 (y/a) when x = b, (iv) V¯ = −V0 (y/a) when x = −b. (The point of peeling off V0 (y/a) was to recover (ii), on which the constraint k = nπ/a depends.) The solution (following Ex. 3.4) is V¯ (x, y) =
∞ X
Cn cosh(nπx/a) sin(nπy/a),
n=1
and it remains to fit condition (iii): V¯ (b, y) =
X
Cn cosh(nπb/a) sin(nπy/a) = −V0 (y/a).
Invoke Fourier’s trick: X
Z Cn cosh(nπb/a)
a
sin(nπy/a) sin(n0 πy/a) dy = −
0
a V0 Cn cosh(nπb/a) = − 2 a
Z
V0 a
Z
a
y sin(n0 πy/a) dy,
0
a
y sin(nπy/a) dy. 0
a ay 2V0 a 2 Cn = − 2 sin(nπy/a) − cos(nπy/a) a cosh(nπb/a) nπ nπ 0 2 a 2V0 (−1)n 2V0 cos(nπ) = . = 2 a cosh(nπb/a) nπ nπ cosh(nπb/a) c
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86
CHAPTER 3. POTENTIAL "
V (x, y) = V0
# ∞ y 2 X (−1)n cosh(nπx/a) + sin(nπy/a) . a π n=1 n cosh(nπb/a)
Alternatively, start with the separable solution V (x, y) = (C sin kx + D cos kx) Aeky + Be−ky . Note that the configuration is symmetric in x, so C = 0, and V (x, 0) = 0 ⇒ B = −A, so (combining the constants) V (x, y) = A cos kx sinh ky. But V (b, y) = 0, so cos kb = 0, which means that kb = ±π/2, ±3π/2, . . . , or k = (2n − 1)π/2b ≡ αn , with n = 1, 2, 3, . . . (negative k does not yield a different solution—the sign can be absorbed into A). The general linear combination is ∞ X V (x, y) = An cos αn x sinh αn y, n=1
and it remains to fit the final boundary condition: V (x, a) = V0 =
∞ X
An cos αn x sinh αn a.
n=1
Use Fourier’s trick, multiplying by cos αn0 x and integrating: Z
b
V0
cos αn0 x dx = −b
V0
∞ X
Z
b
An sinh αn a
cos αn0 x cos αn x dx, −b
n=1
∞ 2 sin αn0 b X An sinh αn a(bδn0 n ) = bAn0 sinh αn0 a. = αn0 n=1
2V0 sin αn b So An = . But sin αn b = sin b αn sinh αn a
V (x, y) = −
2n − 1 π 2
= −(−1)n , so
∞ 2V0 X sinh αn y (−1)n cos αn x. b n=1 αn sinh αn a
Problem 3.55 (a) Using Prob. 3.15b (with b = a): V (x, y) =
4V0 X sinh(nπx/a) sin(nπy/a) . π n sinh(nπ) n odd
c
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CHAPTER 3. POTENTIAL
87
4V0 X nπ cosh(nπx/a) sin(nπy/a) ∂V = −0 ∂x x=0 π a n sinh(nπ) x=0 n odd 40 V0 X sin(nπy/a) =− . a sinh(nπ) n odd Z a Z a 40 V0 X 1 sin(nπy/a) dy. σ(y) dy = − λ = a sinh(nπ) 0 0 n odd Z a a a a 2a sin(nπy/a) dy = − But cos(nπy/a) 0 = [1 − cos(nπ)] = (since n is odd). nπ nπ nπ 0 0 V0 80 V0 X 1 ln 2. =− = − π π n sinh(nπ)
σ(y) = −0
n odd
[Summing the series numerically (using Mathematica) gives 0.0866434, which agrees precisely with ln 2/8. The series can be summed analytically, by manipulation of elliptic integrals—see “Integrals and Series, Vol. I: Elementary Functions,” by A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (Gordon and Breach, New York, 1986), p. 721. I thank Ram Valluri for calling this to my attention.] Using Prob. 3.54 (with b = a/2): # " 2 X (−1)n cosh(nπx/a) sin(nπy/a) y + . V (x, y) = V0 a π n n cosh(nπ/2) σ(x) =
=
λ =
=
# " 2 X nπ (−1)n cosh(nπx/a) cos(nπy/a) ∂V 1 −0 = −0 V0 + ∂y y=0 a π n a n cosh(nπ/2) y=0 # " # " X (−1)n cosh(nπx/a) 0 V0 1 2 X (−1)n cosh(nπx/a) + =− 1+2 . −0 V0 a a n cosh(nπ/2) a cosh(nπ/2) n " # Z a/2 X (−1)n Z a/2 0 V0 a+2 σ(x) dx = − cosh(nπx/a) dx . a cosh(nπ/2) −a/2 −a/2 n a/2 Z a/2 2a a = sinh(nπx/a) sinh(nπ/2). cosh(nπx/a) dx = But nπ nπ −a/2 −a/2 " # " # 0 V0 4a X (−1)n tanh(nπ/2) 4 X (−1)n tanh(nπ/2) − a+ = −0 V0 1 + a π n n π n n
= −
0 V0 ln 2. π
[The numerical value is 0.612111, which agrees with the expected value (ln 2 − π)/4.] (b) From Prob. 3.24: ∞ X 1 k V (s, φ) = a0 + b0 ln s + ak s + bk k [ck cos(kφ) + dk sin(kφ)]. s k=1
In the interior (s < R) b0 and bk must be zero (ln s and 1/s blow up at the origin). Symmetry ⇒ dk = 0. So V (s, φ) = a0 +
∞ X
ak sk cos(kφ).
k=1 c
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4 Problem 3.48
88 y
CHAPTER 3. POTENTIAL
" V0 !x
At the surface: V (R, φ) =
X
k
ak R cos(kφ) =
k=0
V0 , if − π/4 < φ < π/4, 0, otherwise.
Fourier’s trick: multiply by cos(k 0 φ) and integrate from −π to π: Problem 3.49 π/4 Z π Z π/4 ∞ X = (V0 /k 0 ) sin(k 0 π/4), if k 0 6= 0, V0 sin(k 0 φ)/k 0 0 0 k cos(kφ) cos(k φ) dφ = V0 cos(k φ) dφ = ak R −π/4 V π/2, if k 0 = 0. −π −π/4 k=0 0 z
But
" θ φ
0, if k 6= k 0 0 cos(kφ) cos(k φ) dφ = 2π, if k = k 0 = 0, −π π, if k = k 0 6= 0.
Z
l
T$
π
mg # So 2πa0 = V0 π/2 ⇒ a0 = V0 /4; πak Rk = (2V0 /k) sin(kπ/4) ⇒ ak = (2V0 /πkRk ) sin(kπ/4) (k 6= 0); hence
" V (s, φ) = V0
# ∞ 1 2 X sin(kπ/4) s k + cos(kφ) . 4 π k R k=1
Using Eq. 2.49, and noting that in this case n ˆ = −ˆ s: σ(φ) = 0
∞ ∞ 20 V0 X 2 X sin(kπ/4) k−1 ∂V = 0 V0 ks cos(kφ) = sin(kπ/4) cos(kφ). ∂s s=R π kRk πR s=R k=1
k=1
We want the net (line) charge on the segment opposite to V0 (−π < φ < −3π/4 and 3π/4 < φ < π): Z π ∞ 40 V0 X sin(kπ/4) cos(kφ) dφ π 3π/4 3π/4 k=1 ∞ ∞ 40 V0 X sin(kφ) π 40 V0 X sin(kπ/4) sin(3kπ/4) = sin(kπ/4) = − . π k π k 3π/4 Z
λ =
Z
σ(φ)R dφ = 2R
k=1
π
σ(φ) dφ =
k=1
k sin(kπ/4) sin(3kπ/4) product √ √ 1 1/ 2 1/ 2 1/2 2 1√ 1 1 √ 3 1/ 2 1/ 2 1/2 4 0√ 0√ 0 5 1/ 2 1/ 2 1/2 c 6Pearson Education, 1√ 1√ Saddle River, 1 NJ. All rights reserved. This material is !2005 Inc., Upper protected under all copyright laws as they currently exist. No portion of this material may be 7 1/ 2 1/ 2 1/2 reproduced, in any form or by any means, without permission in writing from the publisher. 8 0 0 0 c
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CHAPTER 3. POTENTIAL
" # 40 V0 1 X 1 1 X 1 =− = 0. − π 2 1,3,5... k 2 1,3,5,... k P Ouch! What went wrong? The problem is that the series (1/k) is divergent, so the “subtraction” ∞ − ∞ is suspect. One way to avoid this is to go back to V (s, φ), calculate 0 (∂V /∂s) at s 6= R, and save the limit s → R until the end: ∞ ∂V 20 V0 X sin(kπ/4) ksk−1 σ(φ, s) ≡ 0 cos(kφ) = 4 ∂s π k Rk 40 V0 λ=− π
"
89
X 1 1 X 1 − 2 1,3,5... k 2,6,10,... k
#
k=1
∞ 20 V0 X k−1 x sin(kπ/4) cos(kφ) Problem =3.48 πR
(where x ≡ s/R → 1 at the end).
k=1
∞ 40 V0 X 1 k−1 x sin(kπ/4) sin(3kπ/4) π k k=1 y " 40 V0 1 x3 x5 1 x2 x6 x10 x+ + + ··· − + + + ··· =− π V 2x 3 5 x 2 6 10 0 20 V! x3 x5 x6 x10 0 x x+ =− + + · · · − x2 + + + ··· . πx 3 5 3 5 1+x x3 x5 But (see math tables) : ln =2 x+ + + ··· . 1−x 3 5 2 1+x 1 1+x 1+x 1 + x2 20 V0 1 0 V0 ln − ln ln =− = − πx 2 1−x 2 1 − x2 πx 1−x 1 − x2 Problem 3.49 0 V0 (1 + x)2 −0 V0 =− ln ln 2. ; λ = lim λ(x) = 2 x→1 πx 1+x π
λ(x) ≡ σ(φ, s)R dφ = −
Problem 3.56 z
" θ φ
F = qE = l
T$
qp ˆ (2 cos θ ˆ r + sin θ θ). 4π0 r3
mg #
Now consider the pendulum: F = −mg ˆ z−Tˆ r, where T − mg cos φ = mv 2 /l and (by conservation of 2 2 energy) mgl cos φ = (1/2)mv ⇒ v = 2gl cos φ (assuming it started from rest at φ = 90◦ , as stipulated). But cos φ = − cos θ, so T = mg(− cos θ) + (m/l)(−2gl cos θ) = −3mg cos θ, and hence ˆ + 3mg cos θ ˆ ˆ F = −mg(cos θ ˆ r − sin θ θ) r = mg(2 cos θ ˆ r + sin θ θ). This total force is such as to keep the pendulum on a circular arc, and it is identical to the force on q in the field of a dipole, with mg ↔ qp/4π0 l3 . Evidently q also executes semicircular motion, as though it were on a tether of fixed length l. Problem 3.57 Symmetry suggests that the plane of the orbit is perpendicular to the z axis, and since we need a centripetal force, pointing in toward the axis, the orbit must lie at the bottom of the field loops (Fig. 3.37a), where the z component of the field is zero. Referring to Eq. 3.104, c
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90
CHAPTER 3. POTENTIAL
p √ E·ˆ z = 0 ⇒ 3(p·ˆ r)(ˆ r ·ˆ z)−p·ˆ z = 0, or 3 cos2 θ −1 = 0. So cos2 θ = 1/3, cos θ = −1/ 3, sin θ = 2/3, z/s = √ The field at the orbit is (Eq. 3.103) tan θ ⇒ z = − 2 s. ! r 1 p 2ˆ E= −2 √ ˆ r+ θ 4π0 r3 3 3 r h i √ p 2 − 2(sin θ cos φ x ˆ + sin θ sin φ y ˆ + cos θ ˆ z ) + (cos θ cos φ x ˆ + cos θ sin φ y ˆ − sin θ ˆ z ) = 4π0 r3 3 r h √ √ i √ p 2 = − 2 sin θ + cos θ cos φ x ˆ + − 2 sin θ + cos θ sin φ y ˆ + − 2 cos θ − sin θ ˆ z 3 4π0 r 3 " ! ! # r r r √ 1 √ p 2 2 1 2 2√ − = − 2 −√ (cos φ x ˆ + sin φ y ˆ) + ˆ z 4π0 r3 3 3 3 3 3 r h i √ p √ p p 2 3 (cos φ x ˆ + sin φ y ˆ ) =− 2ˆ s=− √ ˆ s. = − 3 3 4π0 r 3 4π0 r 3 3π0 s3 p (I used s = r sin θ = r 2/3, in the last step.) The centripetal force is qp mv 2 F = qE = − √ =− s 3 3π0 s3
qp v2 = √ 3 3π0 ms2
⇒
⇒
v=
1 s
r
qp √ . 3 3π0 m
The angular momentum is r L = smv =
qpm √ , 3 3π0
the same for all orbits, regardless of their radius (!), and the energy is W = Problem 3.58 Potential of q:
q p cos θ qp 1 1 qp qp √ √ = √ + − = 0. mv 2 + qV = 2 2 2 2 2 3 3π0 s 4π0 r 6 3π0 s 4π0 3(3/2)s2
Vq (r) =
1 q , 4π0 r
Equation 3.94, with r0 → a and α → θ:
Vq (r, θ) =
Meanwhile, the potential of σ is (Eq. 3.79)
r
= a2 + r2 − 2ar cos θ. ∞ 1 1 X a n = r r n=0 r Pn (cos θ). So
where
2
∞ q 1 X a n Pn (cos θ). 4π0 r n=0 r
Vσ (r, θ) =
∞ X Bl Pl (cos θ). rl+1 l=0
Comparing the two (Vq = Vσ ) we see that Bl = (q/4π0 )al , and hence (Eq. 3.81) Al = (q/4π0 )al /R2l+1 . Then (Eq. 3.83) " ∞ # ∞ ∞ l a l X a l X q a q X Pl (cos θ) = Pl (cos θ) + Pl (cos θ) . σ(θ) = (2l + 1) 2 l 4πR2 R 4πR2 R R l=0
l=0
l=0
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 3. POTENTIAL
91
Now (second line above, with r → R) ∞
√
1 X a l 1 = Pl (cos θ). R R a2 + R2 − 2aR cos θ l=0
Differentiating with respect to a: d da
1 √ 2 2 a + R − 2aR cos θ
∞
=−
(a − R cos θ) 1 X a l = l Pl (cos θ). aR R (a2 + R2 − 2aR cos θ)3/2 l=0
Thus R (a − R cos θ) q + −2aR 4πR2 (a2 + R2 − 2aR cos θ)3/2 (a2 + R2 − 2aR cos θ)1/2 q −2a(a − R cos θ) + (a2 + R2 − 2aR cos θ) q (R2 − a2 ) = . = 2 2 2 3/2 4πR (a + R2 − 2aR cos θ)3/2 4πR (a + R − 2aR cos θ)
σ(θ) =
c
2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.