In to el

TABLE OF CONTENTS

Chapter 1 Vector Analysis 1

Chapter 2 Electrostatics 22

Chapter 3 Special Techniques 42

Chapter 4 Electrostatic Fields in Matter 73

Chapter 5 Magnetostatics 89

Chapter 6 Magnetostatic Fields in Matter 113

Chapter 7 Electrodynamics 125

Chapter 8 Conservation Laws 146

Chapter 9 Electromagnetic Waves 157

Chapter 10 Potentials and Fields 179

Chapter 11 Radiation 195

Chapter 12 Electrodynamics and Relativity 219

Chapter 1

Vector Analysis

(a) Prom the diagram, |B + C| cos^s = |B| cos^i + |C| cos02. Multiply by |A|,
|A||B + C| cos 03 = |A||B| cos 01 + |A||C| cos 02-
So: A-(B + C) = A-B + A-C. (Dot product is distributive.)
Similarly: |B + C| sin 63 = |B| sin 61 + |C| sin62. Muhtply by |A| n.
|A||B + C| sin03 n = |A||B| sin0i n + |A||C| sin02 n.
If n is the unit vector pointing out of the page, it follows that
Ax(B + C) = (AxB) + (AxC). (Cross product is distributive.)
(b) For the general case, see G. E. Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
Section 8 (cross product).

The triple cross-product is not in general associative. For example,
suppose A = B and C is perpendicular to A, as in the diagram.
Then (BxC) points out-of-the-page, and Ax(BxC) points down,
and has magnitude ABC. But (AxB) = 0, so (AxB)xC = 0 ,1^
Ax(BxC).
Problem 1.3

A = +lx+ly-lz;A = /3;B = lx+ly + lz;5 = x/3. / , B
A-B = +1 + 1-1 = l = .45cos0 = /3/3cos0 ^cos0 = i.

|0 = cos-i(|) (w 70.5288"! Li X _J 1/
Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
we might pick the base (A) and the left side (B):
A = -lx + 2y + 0z;B = -lx + 0y + 3z.

CHAPTER 1. VECTOR ANALYSIS

X y z
AxB= -1 2 0 =6x + 3y + 2z.
I -1 0 3 I
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its

^ - AXB
|AxBi = VSe + 9 + 4 = 7. |x+|y+|z
Problem 1.5

I X y z I
Ax(BxC)= Ao; Ay ^
I iByC,-B,Cy) (B.C^-B^C,) (B^Cy-ByC^) I
= x[Ay{B.Cy - ByC.) " A,{B,C^ - B^C,)] + y() + z()
(I'll just check the x-component; the others go the same way.)
= ±{AyB^Cy - AyByC^ " A.B^C^ + A.B^C,) + y() + z().
B(A-C) - C(A-B) = [BMxC^ + AyCy + yl^C^) - C^{Aa,B^ + A3,5j, + A,B,)] x + () y + () z
= x(AygxCy + A.B^C, - AyByC^ - A.B^C^) + y() + z(). They agree.
Problem 1.6

Ax(BxC)+Bx(CxA)+Cx(AxB)=B(A.C)-C(A-B) + C(A-B)-A(C.B)+A(B-C)-B(C.A)=0.
So: Ax(BxC) - (AxB)xC = -Bx(CxA) = 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: Ax(BxC) = (AxB)xC <=4' either A is parallel to C, or B is perpendicular to A and C. I
Problem 1.7

>t=Dx + 6y + 8z)-Bx + 8y + 7z) = [2x-2y+ z|
-i = V4 + 4+l = [3]

Problem 1.8

(a) AyBy + AzBz = {cos(f>Ay +sm(j)Az){cos(f>By + sm(j)Bz) + {-smcpAy + cos(f>Ai){-sin(f>By +cos05j)
= cos^ (pAyBy + sin (f> cos (f>{AyBz + A^By) + sin^ (pA^B^ + sin^ 4>'^yBy - sin (f>cos (f>{AyB^ + A^By) +
cos^ (f)AzBz
= (cos^ (f) + sin^ (f>)AyBy + (sin^ (j) + cos^ (j))AzBz = AyBy + AzBz- /
(b) (Z.J + (Ay)^ + (Az)^ - ÛAA = Û {T.]îRijA,) (SLii?,fcAfc) = S,,A (S,iJ,,iJ,,) A,Ak.

This equals Al + AI + A provided '. T.UKR^'c-^ 0 •/ ]^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 = A,0,0). Then S^,*; (Sj RijRrk) AjAk — Sj RiiRii, and this must equal 1 (since we
wantZj,+Zj,+34j = 1). Likewise, SfîiZt2-Ri2 = S^=i-Rj3-Rt3 = 1- To check the case j 7^ k, choose A = A,1,0).
Then we want 2 = Sj,^ (Ej RtjRik) AjAk = Sj R,iRii + Sj Rz2Ri2 + S, RtiRt2 + Sj RtzRii- But we already
know that the first two sums are both 1; the third and fourth are equal, so Sj RtiRi2 = Sj R12R11 = 0, and so
on for other unequal combinations of j", k. / In matrix notation: RR = 1, where R is the transpose of R.

Looking down the axis: dk

A 120° rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax — Az,

0 0 1
i?= ( 1 0 0
0 1 0

Problem 1.10

(a) I No changeTj (A^ = Ax,^y — Ay, A^ = A^)
(b) IA —>■ -A7^ in the sense {Ax = —A^, Ay = -Aj,, ^^ : -A.)
(c) (AxB) —> (-A)x(-B) = (AxB). That is, if C = AxB, | C —^ C |. No minus sign, in contrast to
behavior of an "ordinary" vector, as given by (b). If A and B are pseudovectors, then (AxB) —> (A) x (B) =
(AxB). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector
and a pseudovector, one changes sign, the other doesn't, and therefore the cross-product is itself a vector.
Angular momentum (L = rxp) and torque (N = rxF) are pseudovectors.
(d) A-(BxC) —> (-A)-((-B)x(-C)) = -A.(BxC). So, if o = A.(BxC), then | o —? -^ a
changes sign under inversion of coordinates.
Problem 1.11

(a)V/ = = 2xx + 3y^y + 4z^ z

F)V/ = = 2a;y= z'^x + Sx^y'^z'^y-i Ax'^y^z^ z

(c)V/ = = e*si n 2/ In z X -f- e^ cos y Inzy-l-e^ smy{l/z)z

Problem 1.12

(a) V/i = W[{2y - 6x - 18) x + Ba; - 8y + 28) y]. V/i = 0 at summit, so

22y = 66=^y^3=^2x-2A + 2& = 0^^^x = ~2.
Top is 13 miles north, 2 miles west, of SouthHadleyJ
(b) Putting in a; = -2, y = 3:
h = 10(-12 - 12 - 36 -I- 36 -t- 84 -t-12) = | 720 ft. |
(c) Putting in a; = 1, y = 1: Vfe = 10[B - 6 - 18) x + B - 8 -f- 28) y] = 10(-22x 4- 22y) =: 220(- x -f- y).
|V/i| = 220v^ Pb 1311 ft/mile]; direction: | northwest. |

4 CHAPTER 1. VECTOR ANALYSIS

Problem 1.13

* - (a; - x') X + B/ - 2/') y + B - z') z; -> = y/{x - x')^ + {y - y'f + {z - z'f.
(a)V@-^[(a;-a:r + (y-yr + (-^-^')']x+|^()y + ^()i = 2(x-x')x + 2(y-2/')y + 2(z-z'U = 2*.
(b) V(i) = U{x- x'f + {y- y'? + {z - z'?]-h x + f^{)-h y + ^()-^ g
= _i()-|2(x-x')x-|()-i2(y-y')y-|()-^2(z-z'J
- -()-i[(x - x')i + B/ - y')y + (^ - ^')z] = -dA^)* = -(lA^)^
(c) ^(*") -n^"->f =n^"-nH2*.) =^^"-''i., so|v(^")^n^"^
Problem 1.14

y = +2/ cos 0 + z sin </>; multiply by sin </>: j/sin 0 = +j/ sin </> cos </> + z sin^ 0.
z — —y sin 0 + z cos 0; multiply by cos 0: ?cos 0 = — y sin 0 cos (f)-- z cos^ 0.
Add: 27sin0 + 2COS0 = z(sin^ 0 + cos^ 0) = z. Likewise, 27cos0 - ?sin0 = y.
So || = cos 0; ll = - sin 0; |= = sin 0; |= = cos 0. Therefore

M^ = iîi"'ii^"'"^'^f^"'^'"t^^{^ôV/transformsa.avector. qed
Problem 1.15

(o)V-v? = ^(a^') + |?Ca;^') + ^(-2a:2) - 2a; + 0 - 2a; -0.
F)V.V6 = ^(^^2/) + iBy^) + l -Cxz)=y + 2a; + 3a;.
(c)V-Vc = £B/') + li-^^y + z' + U^yz) = 0 + Ba;) + By) -2(x + y)-
Problem 1.16

V-V = f^{^) + l{^) + l,{^) = A [3:(a;2 + y2 + ^2)-|] +|. [j,(^2 + ^2 + ^2)-|] +^ [^(^2 + ^2 + ^2)-|J
= ()-§ + x(-3/2)()-t2x + ()-§ + y(-3/2)()-i22/ + ()-§ + z(-3/2)()-l22
- 3r-3 - 3r-5(x2 + y^ + ^2) ^ 3^-3 _ 3^-3 ^ q.
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the
origin. How, then, can V-v = 0? The answer is that V'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-v is infinite at
that one point, and zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17

Vy — COS (j)Vy + sm.(j)Vz;vz = — sin (f>Vy + cos 0 v^.
^ - ^ COS0 + % sin0 = (^f| + I^H) COS0+ (^1 + ^H) sin0. Use result in Prob. 1.14:
= (^ COS0+ ^ sin0) COS0+ (^ COS0+ ^ sin0) sin0.
^ = -^ sin0+ %- COS0 = - (î + î) sin0+ (î| + î) cos0
- - (-^ sin0+ ^ COS0) sin0+ (-^ sin0+ ^ cos0) cos0. So
^ + ^ ^ 1^ cos2 0 + ^ sin0cos0 + ^ sin0cos0 + %■ sin^ 0 + ^ sin^ 0 - |^ sin0cos0

-^ sm0cos0 + ^
^ (cos^0 + sin^0) + ^ (sm^0 + cos^0) = ^ + ^. /
X y z
0 a
(a) Vxv? = i
a? = x@ - 6x2) + y@ + 2z) + z{3z^ - 0) = -6xzyi + 2zy+ 32"^ z.
X^ 3xz' -2a;z
X z
y
0 a

1? 51
=
x@ - 2y) + y@ - 82) + z@ - a;) = |-2yx-3zy - a:z. |
a;2/ 2yz 3a;z

(c) Vxv, = = xB2 - 22) + y@ - 0) + zBy - 2y) = [o7]
{2xy + ^2) 2yz
Problem 1.19

V = y X + X y; on V = yz X + xz y + xy z; or V = {3x^z - 2^) x + 3 y + (a;^ - Zxz^) z;
or V = (sin a;) (cosh J/) x - (cosa;)(sinhj/) y; etc.
Problem 1.20

(i)V(/,) = Mx+M5. + M2=(/|.+,|£Mc+(/|| + ,g)y+(/if+,il)z
= f{^^+lly+^^)+9{U^+%y+U^)=fi'^9) + 9{'^f). qed
(iv) V.(AxB) = £ [AyB, - A,By) + §-^ {A,B, - A,B,) + ^ {A.By - AyB.)
=Ây'-t + B.'-t-A.'-t-By'-t + A.^ + Bj-^~A.'-t-B.'-^
Â^'-t + By'-t-^y'-t-B.'-t
-B.[^-'-t)-^By{^-^-t)^B.[^-^-^)-A.[^-^-§f)
-^.(l^-l^)-^^(^-^)=B-(VxA)-A.(VxB). qec
(V) Vx (/A) = (^ - 5IIM) 5t + (ÎM^ - ?i|M) y + BiM^ - ^^

= /(VxA)-Ax(V/). qed
Problem 1.21

(a)(A.V)B=(A.^ + ^,^ + ^.^)x+(^.^ + A,^ + A.^)y

[(f.V)fL = ^(x£+,4; + zf)-


Same goes for the other components. Hence: (f-V) f = 0 .

(c) (v?.V) V6 =(x^^+ 3xz^^ - 2xz^) [xy x + 2y2y + 3a;2 z)
= x^ B/ X + 0 y + 32 z) + 3x2^ (a; x + 22 y + 0 z) - 2x2 @ x + 2?/ y + 3a; z)
= (x^y + 3x22^) X + Fx2^ - 4xy2) y + Cx^2 - 6x^2) z
= I x2 B/ + 322) ^ ^ 2x2 C2^ - 22/) y - 3x^22 I
Problem 1.22 ~ ~~~~

(ii) [V(A.B)], = £(A.5. + AyBy + A,B.) = S^B. + A,^ + ^By + Ay^ + M.B, + a,^
[Ax{VxB)lÂyiVxB),-A.iVxB)y::.Ay{?t-^)-M^-^)
[Bx(VxA)L = B,(^-^)-B.(^-^)
[iA.y7)Bl = {A.£ + Ay^ + A.^^)B. = A.^ + Ay^ + A.^
[(B.V)A], = B,S^ + By^ + B,^
So[Ax(VxB)+Bx(VxA) + (A-V)B + (B-V)A]^
= Ay^ oxAy^ ay A,^ oz A,ôx By^ - y oy" ^ oz H" B,^
y - y - ^ + ^ + y ox By^ ^^^ * ox

= [V(A-B)]jj (same for y and 2)
(vi) [Vx(AxB)], = ^{AxB),--§-^{AxB)y = ^{A^By-AyB^)--l^{A,B^-A^B,)
- aj, ^y + ^X gy Sf^X Ay gy g^ H^ ^ Z -Q^ + Q^ ^J 4" ^X "gj^
[(B-V)A - (A.V)B + A(V.B) - B(V.A)]x
= B.^ + By^-^ + B.^-A.^-Ay^^-A.S^ + A4^-t + '-^ + '-t)-Bx{^ + '-^ + '-t)

+ M-^)+M-t)+B4^)
= [Vx(AxB)]j, (same for y and 2)
Problem 1.23

v(//ff) = -LUl9)^ + -§-y{fl9)y + Ufl9)i
^ ^fJT/ltx+îM^y + îi^z

V-(A/ff) = ^(A./5) + |;(A,/ff) + ^(A./5)

- 1 L ?M^ 4- Ma. 4. M^î ^( A ^4-/4 ^4-/4 ^"^1 - aV-A-A-Vg ,
- 5^ [5^^ ax + aj, + a^ ; V^â?+^2'aj,+^â^;J - ^ P ^- ^^^

IMvlg)

Problem 1.24

y z
(a) AxB = 2y Zz := xFxz) + y{9zy) + i{-2x^ - 6y^)
-2x 0

V.(AxB) = ^i6xz) + ^i9zy) + ^(-2x2 - 6y^) =6z + 9z + 0 = 15z
VxA = X [^(Zz) - ^By)) + y {^{x) - £Cz)) + z (£By) - ^(x)) = 0; B-(VxA) = 0
VXB = X A^@) - ^(-2x)) + y {^{Zy) ~ ^@)) + z (^(-2x) - ^{3y)) = -5z; A.(VxB) --
V-(AxB) ^ B.(Vx A) - A.(VxB) = 0 - (-15z) = 15^. /
(b) A-B = 3a;y - 4a;y = -xy ; V(A.B) = V{-xy) = x£(-a;y) + y|^(-xy) = -xy

Ax(VxB) = X 2y Zz = x(-10y)+yEa;); Bx(VxA) = 0
0 0-5

(A.V)B = (a;^ + 2y^ + Zz^) Cy x - 2a;y) = xFy) + y(-2a;)
(B.V)A = C2/^ - 2a;|;) (a; X + 2y y + Zz z) = xCy) + y(-4a;)
Ax(VxB) + Bx(VxA) + (A.V)B + (B-V)A
= -10yx + 5a;y+ 6yx-2a;y+ 3yx-4a;y = -yx-xy = V-(A-B). /

(c) Vx(AxB) - X (|^(-2x2 - 62/2) - ^(9^2/)) + y (^Fa;z) - ^(-2x2 - 6y2)) + z (£{9zy) - |^ Fxz))
= x(-12y - 9y) + yFx + 4x) + z@) = -21yx + lOxy

V-A = ^(x) + ^By) + ^Cz) = 1 + 2 + 3 - 6; V-B = ^Cy) + ^{-2x) - 0
(B-V)A - (A.V)B + A(V.B) - B(V-A) = 3y x - 4x y - 6y x + 2x y - 18y x + 12x y = -21y x
= Vx(AxB)./
Problem 1.25

(a) ^ = 2; ^ = ^ = 0 =!> I V2r? = 2.1
-3r6 =

(c) 0 = 25T, ; 0^ = -ler, ; ^ = -9T, ^ S/^T, = 0. |
id) ^^2;^ = ^=0 ^ W^v^=2
li^ = ^=0;|^-6x => V


Problem 1.26

■(V XV) -ai[-gf--gf) + si{^-^) + g^[-a^--S^)
- (& - fe) + (fe - fe) + (& - fe) = 0' by equality of cross-derivatives.
PromProb. 1.18: Vxvt =-2yx - 3zy - xz ^ V.(VXV6) = ^(-2y) + ^(-82) + ^(-a;) = 0. /
Problem 1.27 =========================^^ ====__

Vx(Vi)- _ ,->/ a't a't , ,%( a't a^i . a^ a''t e^t^

= 0, by equality of cross-derivatives.
In Prob. 1.11(b), V/ - 2xy'^z^ x + Zx'^y'^z^ y + 4x^^323 z, so
I X y z ■
Vx(V/)= ^ ^ ^
I 2xy^z'* 3a;^y3;j;4 4a;2j^3^
= xC • 4x^y^z3 _ 4 . 3a;2y2^3) ^ ^^D . 2xy^z^ - 2 • 4a;y3^3) + ^B . 3a:y2;^4 - 3 • 2xy^z'^) = 0. /
Problem 1.28

(a) @,0,0)—> A,0,0). a;:0-^l,y = z = 0;dl = da;x;v-dl = x2da;;/v-dl = /o'a;2da. = (a.3/3)|i^l/3
A,0,0) —> A,1,0). x= l,j/ :0-> l,z = 0;cfl = dj/y;v-dl = 2j/2d2/ = 0;/v-dl = 0.
A,1,0) —^ A, l,l).x = y = 1, z: 0 -> 1; dl = d^ z; v • dl = y^ ^z = dz; / v • dl = /(J dz = z|J = 1.
Total: / V • dl = A/3) + 0 + 1 = |4/3.|
(b) @,0,0) —> @,0,1). a; = 2/= 0,2:0->l;dl = d2z; vdl = 2/2 dz = 0;/v dl = 0.
@,0,1) —)• @,1,1). a; = 0,y : 0-> 1,2 = l;dl = dyy;v-dl = 2y2dy = 2j/dy;/ vdl =/o 2j/dy = j/2|i ^ 1
@,1,1) —)• A,1,1). X : 0 -> 1, y = z = 1; dl = da;x; v • dl = a;^ da;; / v • dl = /? a;^ da; = (a;3/3)|J = 1/3.
Total: / V • dl = 0 -)-1 + A/3) = 14/3. |
(c) X = y = z : 0 -> 1; da; = dy = dz; V • dl = a;2 da; + 2yz dy + y^dz = j? dx + 2a;^ da; -)- a;^ da; = 4a;^ da;;
/ V • dl = /o' 4a;2 da; = Da;3/3)|S - [l/s]
(d)/vdl=D/3)-D/3) = [0l| ____^__
Problem 1.29

x,y : 0 -> l,z = 0;da - dxdyz;v ■ da = y(z^ - Z)dxdy = -Zydxdy;Jv ■ da = -Zj^dx^^ydy =
-3(a;|o)Dlo) = -3B)B) = 112. | In Ex. 1.7 we got 20, for the same boundary line (the square in the a;y-
plane), 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 1.30

jTdr = J z^ dxdy dz. You can do the integrals in any order—here it is simplest to save z for last:

l''[l{h)'"h
The sloping surface isa; + y + z=l,so the x integral is /q ^ ^ da; = 1 - y - z. For a given z, y ranges from 0 to
1 - z, so the y integral is J^'-'^ {l-y-z)dy = [il-z)y- (y 72)]!^'^ = il-z)^-[{l- zf/l] = A - 2) V2 =

A/2) - z + (zV2). Finally, the z integral is ll zW - ^ + 4) dz = /o'D - z^ + 4) dz = (^ - ^ + '^) =
-e-
Problem 1.31
+ T6 = W^
T(b) = 1 + 4 + 2-7; T(a) = 0. => | T(b) - T(a) ^tTJ
VT = Ba; + 4y)x + Da; + 2z^)y + {6yz^)z; VT-dl = Ba; + 4y)da; + Da; + 2x^)dy + {6yz'^)dz

(a) Segment 1: x : 0-^1, y = z = dy = dz = 0. ^VT-dl = /^ Ba;) dx = x'^l = 1.
Segment 2: !/: 0 -> 1, a; = 1, 2 = 0, da; = dz = O./VT-dl = /q D)dy = 4y|J = 4. jl^T-di = 7. /
Segment 3: z:0-^l, a; = y = l, dar = dy = O./VT-dl = J^{Qz'^)dz = 2z3|J = 2.
(b) Segment 1; 0-^1, a; = y = da; = dy = 0. /VT-dl = /? @) dz = 0.
Segment 3; 00 -^ 1, ya;= = 0,1, dy1, dz = =0.dz/VT-dl = Jl{2x + 4) da; = 2y|J = 2. J^ VT-dl = 7. /
Segment
2: -> 1, 2 = 2 = = da; = O./VT-dl = Jq B)dy
(a;2+4a;)|; = l + 4 = 5.
(c) a;: 0 ->■ 1, y = x, z = x^, dy = dxdz = 2xdx.
VT-dl = {2x + 4x)dx + Da; + 2a;^)da; + Fa;a;'*Ja; da; = A0a; + 14a;?)dx.

f^ VT-dl = /o A0a; + 14a;?')da; = (Sa;^ + 2x'^)l = 5 + 2 = 7./
Problem 1.32

V-v = y + 2z + 3a;

Ji'V'v)dT = /(y + 2z + Zx) dxdydz = J J {/q (y + 2z + 3a;) dx}dydz
^[iy + 2z)x[y2+ +lx']l+ =6)y]l = +4 2z)+6+ =6)/{/oBj/16+ 4z + 6)dy}dz
M- {4z
2{y + 2Dz = 8z +
= f^iSz + 16)dz = D^2 + 16z) lo = 16 + 32 = [IsTI
Numbering the surfaces as in Fig. 1.29:

(i) da = dy dz X, a; = 2. v-da = 2y dy dz. /v-da = //2y dy dz = 2y^ |o — ^•
(ii) da = —dydz%x = 0. v-da = 0. Jv-d& = 0.
(iii) dsL — dxdzy,y = 2. v-da =:Azdx dz. JvdsL = JjAz dx dz = 16.
(iv) da = -da;dzy,y = 0. v-da = 0./v-da = 0.
(v) da = dx dy z, z = 2. v-da = 6a; da; dy. /v-da = 24.
(vi) da = —da; dy z, 2; = 0. v-da = 0. /v-da = 0.
^ /v-da = 8 + 16 + 24 = 48 /
Problem 1.33

Vxv = x@-2y) + y@ - 32) + z@ - a;) = -2yit-3zy - xz.
da = dy dz x, if we agree that the path integral shall run counterclockwise. So
(Vxv)-da = -2ydydz.


/(Vxv)-da = j{j^-'i-2y)dy}dz
^y^l-' = -{2-zr
= -J^i4-4z + z^)dz = -[4z-2z' + i)l
Meanwhile, vdl = (a;j/)da; + {2yz)dy + {Zzx)dz. There are three segments.

{l)x = z = Q; da; = dz = 0. y : 0 -^ 2. /vdl = 0.
B) a; = 0; z = 2-y dx = 0, dz = -dy, y : 2 -^ 0. vdl = 2yzdy.
/vdl = /° 2y{2 - y)dy = - J^i4y - 2y')dy = - {2y^ - y^) |J = - (8 - | • 8) = -|.
C) a; = 2/ = 0; da; = dy = 0; 2:2-^0. v-dl = 0. /v-dl = 0. So /v-dl = -|. /
Problem 1.34

By Corollary 1, /(Vxv)'da should equal |. Vxv = {4z^ - 2x)± + 2zz.
(i) da = dyd2x, a; = 1; y,2 : 0 -^ 1. (Vxv).da = {4z^ - 2)dydz; /(Vxv).da = /^ D^2 _ 2)dz

(ii) da = -dxdyz, z = 0; x,y:0^1. (Vxv).da = 0; /(Vxv)-da = 0.
(ill) da = da;d2y, y = 1; x,z:0^1. (Vxv)-da = 0; /(Vxv)-da = 0.
(iv) da = -dxdzy, y = 0; a;,z : 0 -^ 1. (Vxv).da = 0; /(Vxv)-da = 0.
(v) da = da;dyz, z = 1; a;,y : 0-^ 1. (Vxv)-da = 2dxdy; /(Vxv)-da = 2.
=i>/(Vxv)-da=-| + 2=|. /
Problem 1.35

(a) Use the product rule Vx(/A) = /(VxA) -Ax (V/) :

/"/(VxA).da=^5 ^5x (V/)] • da = / /A-dl+ /"[A x (V/)] • da. qed.
75
/" Vx(/A)-da+ /"[A ./P ^5

(I used Stokes' theorem in the last step.)
(b) Use the product rule V-(A x B) = B • (VxA) - A ■ (VxB) :

/"B-(VxA)dr= /" V.(AxB)dr+ /" A(VxB)dr= /(AxB)da+ / A-(VxB)dr. qed.
(I used the divergence theorem in the last step.)

Problem 1.36 r = i/a;2 +y^+ z"^; 9 = cos'^ P=tan-i(f),
Problem 1.37

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 = x^ + yy + zz = rsmOcos<^x + TsinOsm(f)y + rcosOz.

If I only vary r slightly, then dr = ^(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:

1^ = sin0cos0x + sin0sin0y + cos0z; |^| = sin^ 0cos^ 0 + sin^ 0sin^ ^ + cos^ 0 = 1.
If = r cos ^ cos 0X + r cos 0 sin 0y - >" sin 0 z; 11| | = r^ cos^ 6 cos^ <j!> + r^ cos^ 9 sin^ (p + r"^ sm^ 9 =
sin^z;

dit) flsin^x + rsin^cos^y; |0| = + r-' sin"" 6 cos'' (p - r^ sin"" (

f — sin 9 cos ^ X + sin 0 sin 0y + COS0Z.
e = cos 9 COS (f)jt + cos 9sm(f)y - sin 9 z.
0 = — sin^x + cos0y.
Check: f-f = sin^ ^(cos^ 0 + sin^ 0) + cos=^ 9 = sin^ 9 + cos^ 9 = 1, /
9-^=:-cos9sm(f)cos<l> + cos9sm(f)cos(f) = 0, / etc.
sin9r = sin^ 0cos0x + sin^ 9sm(j>y + sin9cos9z.
cos90 = cos^ 9cos(j)x. + cos^6sin0y — sin^cos9z.
Add these:
A) sinfif + cos^e ^+cos0x + sin^y;
B) 4> - -sm(f)Tt + cos<py.
Multiply A) by cos0, B) by sin0, and subtract:

h=- sinfl COS^f + COS0 COS 00 - sin 0 0.1
Multiply A) by sin0, B) by cos0,, and add:
and add:

|j == sinS sin 0 f + COS 9 sin 0 0 + cos (j) 0.1
cos^f = sin0cos0cos0x + sin0cos0sin0y+ cos2(
sin ^ 0 = sin 9 cos 0 cos 0 x + sin 9 cos ^ sin 0 y - sin^ (
Subtract these:


Problem 1.38

(a) V-v, = ^i:{r^r^) - ^4^^ = 4r
/(V-v,)dr = f{4r)ir^sin0drded,f>) ^ {4) J^r^dr J^ sin0d0 Jl^dcj, = D) (^) B)B7r) =^¥
/vi-da - /(r2f).(r2 sin^d^d^f) = r" /J" sin0d0f^' d<j) = 4nR'^ / (iVote; at surface of sphere r = /2.)
(b) V.V2-J.|:(r2 4,)=:0 =^ /(V.V2)dr = 0
/v2-da = /(^f) (r2sin6ld6l#f) = JsinOdOd^) = [i^
They rfon'i agree! The point is that this divergence is zero except at the origin, where it blows up, so our
calculation of /(V-V2) is incorrect. The right answer is 4ir.
Problem 1.39

V-v = 4,|,(r2rcos0) + ;:^^(sin0rsin0) + ^^(rsin0cos0)
— ^ 3r^ cos 9 + ^gl^g r 2 sin 0 cos 0 + j^^ r sin 0{— sin 0)
= 3cos0 + 2cos^ — sin0 = 5cos0 — sin0

J{V-v)dT = /E cos 0 - sin 0) r^ sin 0 dr d0 d(f> = f^ r^ dr j} U^""E cos 0 - sin 0) d^l dB sin 0
M'27rEcos0)
= (f )A07r)//sine cos 0,d0
-^-^1^ = 1

Two surfaces—one the hemisphere: da = i?^ sin 0 d0 d0 f; r =^ R; 0 : 0 ->■ 27r, 0 : 0 -> |.
Jv-da = J{rcos0)R^ sin0d0d(t) - R^ // sin0cos0d0 J^" d<j) = R^ (|) {2-k) = ttR^.
other the flat bottom: da = (dr)(r sin 0d0)(+0) = rdrd(j>e (here 0 = f). r : 0 -^ J?, 0 : 0 -> 27r.
/v.da = /(r sin 0)(r drd0) = J^ r"" dr J^"' d(f, = 2n^.
Total: Jv-da. = ttR^ + ^nR^ = ^nR^ /
Problem 1.40 Vt = (cos0+ sin0cos0)f + (-sin0 + cos0cos0H+ -:AT-(-si)if0sin0)^
W'H = V-{Vt)

= ^ (r2(cos0 + sin0cos0) + ^Jj^ (sin0(-sin0+ cos0cos0) + ^ri^ {-sin(j) — -^ 2r(cos0 + sin6cos0) + ^ l^ (-2sin0cos0 + cos^ 0cos0 - sin^ 0cos0) - ^:^t^ cos0
= 7^^ [2 sin 0COS 0 + 2 sin^ 0 cos 0 - 2 sin 0Cos 0 + cos^ 0 cos 0 - sin^ 0 cos 0 - cos 0]
= j^ [(sin^ 0 + cos^ 0) cos 0 - cos 0] = 0.
^ [Vh = 01
Check: rcos0 = z, rsin0cos(f> — x ^ in Cartesian coordinates t = x + z. Obviously, Laplacian is ze:

Gradient Theorem: f^ Vt-di = f (b) - f (a)
Segment i; 0 = f, 0 = 0, r : 0 -> 2. dl = drf; Vt-dl = (cos0 + sin0cos0)dr = @ + I)dr = dr.
fVt-d[ = f^dr = 2.
Segment g; 0 = f, r = 2, 0 : 0 -> f. dl = rsin0d00 = 2d0^.
Vf-dl = (-sin0)Bd0) = -2sin0d0. /Vf-dl = - f^ 2sin<pdcp = 2cos0|| = -2.

Segment 3: r = 2, 0 = f; 0 : f -^ 0.
dl = rdee = 2d0e; Vf-dl= (-sm0 + cos0cos0)Bd0) = -2smed9.
JVt-dl^-J°2smedd^ 2cos0|l =2.
Tota/; /^'' Vf-dl = 2-2 + 2 = [2]. Meanwhile, f (b) - f (a) = [2A + 0)] - [0( )] = 2. /
Problem 1.41 Prom Fig. 1.42, s = COS0X +sin</iy; 0 = — sin</ix + cos</iy; z = z
Multiply first by cos 0, second by sin <p, and subtract:
s cos 0 — ^ sin 0 = cos^ 0 x + cos 0 sin 0 y + sin^ 0 x — sin 0 cos 0 y = x(sin^ 0 + cos^ 0) = x.
So X = cos 0 s — sin 0 0.
Multiply first by sin 0, second by cos 0, and add:
s sin 0 + 0 cos 0 = sin 0 cos 0 x + sin^ 0 y - sin 0 cos 0 x + cos^ 0 y = y(sin^ 0 + cos^ 0) = y•
So y = sin0§ + cos00. |z = z. |
Problem 1.42

(a) V-v = i^(ssB + sin2 0))+i^(ssin0cos0) + ^Cz)
= i 2sB + sin^ 0) + 7 s(cos2 0 - sin^ 0) + 3
= 4 + 2 sin^ 0 + cos^ 0 - sin^ 0 + 3
= 4 + sin^ 0 + cos^ 0 + 3 = [sT]
(b) /(V-v)dr = /(8)sds d<pdz = 8 f^ s ds J^ d<j) f^ dz = 8B) (f) E) = 140n.
Meanwhile, the surface integral has five parts:

top: z = 5, da — sdsdcj) z; v-da = 3z s ds d0 = 15s ds d0. /v-da = 15 /^ s ds f^' d(f) = 157r. bottom: z = 0, da = —sdsdcpz; v-da = —Zzsds-dcj) = 0. /v-da = 0.
back: 0 = f, da = ds dz 0; V'da = s sin 0 cos (j)dsdz = 0. J v-da = 0.

left: 0 = 0, da = —ds dz 0; v-da = —s sin 0 cos (pdsdz = 0. fv'da = 0. front: s = 2, da = sd(j>dz s; v-da = sB + sin^ 0)s d0dz = 4B + sin^ 0)d0dz.
Jvda = 4/o? B + sin2 0)d0/o dz = D)Gr + f )E) = 257r.
So /vda = 157r + 257r = AOtt. /

(c) Vxv = (i^Cz)-^(ssin0cos0))s+(^(sB + sin2 0))-^Cz)H
+ J {mi^^sin0cos0) - ^ (sB + sin20))) z
= ^Bssin0cos0 - s2sin0cos0) = [qT]
Problem 1.43

(a) 3C2) _ 2C) _ 1 ^ 27 - 6 - 1 = [2071
(b)cos7r = m
(c) I zero. I
(d) ln(-2 + 3) = In 1 = [^eroT]
Problem 1.44

(a) f_^{2x + 3)|<J(a:) dx = |@ + 3) = [I]
(b) By Eq. 1.94, 5A - a;) = 5(x - 1), so 1 + 3 + 2 = [gT]


(d)|l(ifa>6), 0(ifa<6O|
Problem 1.45

(a) JZo /(^) [^^<^(^)] d^ = xnx)dix)C^ - jr^ £ ixfix)) 5{x) dx.
The first term is zero, since 5{x) = 0 at ±oo; ^ {x f{x)) -x^ +^f = x§^ + f.
So the integral is - /^ {x% + /) 5{x) dx ^ 0 - /(O) = -/(O) = - fZo fi^)î^) dx.
So, x£S{x) = -Six), qed
(b) !Zo m^dx = fix)eix)'^^ - fZ, fj{x)dx = /(CO) - /o°° £dx = /(oo) - (/(oo) - /(O))
- m - /.Too /(^)'^(a:) da:. So £ = <5(x). qed
Problem 1.46

(a) p{r)^q5^{r-r'). Check: /p(r)dr = q f5^{r - r') dr = q. /

(c) Evidently p{r) = AS{r — R). To determine the constant A, we require
Q^JpdT = jASir - iZL7rr2 dr = A47riZ2. So A = 4^. p{r) = ^5{r - R).
Problem 1.47

(a) a^ + a-a + a"^ - |3a^.|
(b)/(r-bJ^53(r)dr = ^62 = ^D2+32) = g]
(c) c^ = 25 + 9 + 4 = 38 > 36 = 6^, so c is outside V, so the integral is
(d) (e - Bx + 2y + 2z))^ = (Ix + Oy + (-1) z)^ = 1 + 1 = 2 < A.5J ^ 2.25, so e is ins
and hence the integral is e-(d - e) = C,2, l)-(-2,0,2) = -6 + 0 + 2 = [T]
Problem 1.48 ~~
First method: use Eq. 1.99 to write J = / e"'' D7r5^(r)) dr = 47re"° = |47r.|
Second method: integrating by parts (use Eq. 1.59).

J ^ - / 4 • "^(^"'■) dr+ ie-"-—- da. But V (e-"-) = (§ê~'') f = -
V 5

= / -ê~''A7rr^ dr + / e"""-^ • r^sinfld^d^f = 47r / e~''dr + e~^ / sin
47r (-e-'-) |~ + 4ne-^ = Att (-6"°° + e"") - 47r./ (Here R = -0.)

Problem 1.49 (a) V-Fi = £{0) + ^@) + ^ (x^) = [o]; V-F^ = ^ + ^ + ^ = 1+ 1 + 1 =[z]
X y z X y z

-y^(x^)=E23= ^x^^ =
^ .a. A a a a
Sx 8y dz dx di m = 0
0 0 a;2 x y z

IF2 is a gradient; Fi is a curl [ rt/2 = ^ (a;^ + 2/^ + 2^H would do (F2 = VU2).
For Ai, we want (^ - ^) = (^ - ^) - 0; ^-^=x^. Ay = ^, A^ = A, :^ 0 would do it.
Ai = |x^ y (Fi — Vx Ai). (But these are not unique.)

(b) V.F3 = ^(yz) + ^{xz) + f^ixy) = 0; VXF3 z x{x-x) + y {y-y) + z{z-z) = 0
yz xz xy
So F3 can be written as the gradient of a scalar (F3 = VC/3) and as the curl of a vector (F3 = VXA3). In
fact, Uz — xyz does the job. For the vector potential, we have

dy dz = yz, which suggests A^ - y'^z + f{x,z) Ay = -yz'^ + g{x,y)
= xz, suggesting Ax = z'^x + h{x,y); Az = -zx^ + Jiy,z)
= xy, so Ay = x^y+ k{y,z); Ax =-xy'^+l{x,y)

Putting this all together: A3 = j {x [z^ - J/^) x + y (a;^ — z^)y + z (y^ — a;^) z} (again, not unique).
Problem 1,50
(d) =i' (a): VxF = Vx(-VC/) = 0 (Eq. 1.44 - curl of gradient is always zero).
(a) ^ (c): fF-d[ = /(VxF) • da -: 0 (Eq. 1.57-Stokes' theorem).
(c)^(b):/^>.dl-/^V-'^l = /a>-^ + /bV.dl=i^F.dl = 0,so

/ F.dl= / Fdl.
■/a / Ja II

(b) ^ (c): same as (c) ^ (b), only in reverse; (c) =^ (a): same as (a)^ (c).
Problem 1.51
(d) ^ (a); V-F = V'(VxW) = 0 (Eq 1.46—divergence of curl is always zero).
(a) ^ (c); §F -da- /(V-F) dr = 0 (Eq. 1.56—divergence theorem).
(c) ^ (b): /^ F • da - /r^ F • da - f F • da = 0, so

/Fda= f I
Ji Jii

{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 1.52

In Prob. 1.15 we found that V-Vg = 0; in Prob. 1.18 we found that Vxvc = 0. So
Vc can be written as the gradient of a scalar; Vo can be written as the curl of a vector. E
(a) To find t:

A) ^=^y'^t^y^x + f{y,z)
B) IL={2xy + z^)
C) t - 2yz


Prom A) & C) we get % =2yz =^ f = yz^ + g{y) =^ t = y^x + yz^ + g{y), so ^ = 2xy + 2^ + || =
2xy + z^ (from B)) =*■ || = 0. We may as well pick 5 = 0; then | f = xy^ + yz^.
(b)TofindW: ^ - ^ = a^^l ^"^ : Zz'^x;
Pick ly^ = 0; then

dx = -3xz2 =>W, = -^x^z^ + f{y, z)
dWy = -2xz^Wy==-x'^z + g{y,z).
dx

= i^ + ^' - if = ^' ^ S - if = 0- May as well pick / = g = 0.

Check: VxW = = X (x2) + y Cx22) + 2 (_2xz)./
0 -x^z -^xh
You can add any gradient (V*) to W without changing its curl, so this answer is far from unique. Some
other solutions:

W = xz^x — x^zy;
W = [2xyz + xz^) X + x'^y z;
W = xyz X - ^x^z y + ^x^ [y - 32^) g.
Probelm 1.53

W = l|.(r2r2cos0) + -l^^(sin0r2cos.^) = -^-|-(-r2cos0sm^)
= -x^r^ cos0 H r-^ cos0r^ cos 6 + —r-T f-r^ cosfl cos0)

= . [4 sin 0 + cos 0 - cos 0] = 4r cos 9.

l{V-)dT = I {4r cose)r^sinedr ded<f> = 4 I r^ dr I cose sine de I d^

Surface consists of four parts:
A) Curved: dsL = R^ sineded<pT; r = R. v ■ da = (J?^cose) (i?^;
7r/2 7r/2 ^
f vd& = R'' f cose sine de f d(i> = R''U (|)=r![|-.

B) Left: da = -rdrdd^; 0 = 0. v • da = (r^ cos^sin^) {rdrdB) = 0. /v • da = 0.
C) Back: da = rdrde^; cj) = 7r/2. v • da = (-r^ cos 9 sin 0) (r dr dB) = -r^ cos 0 dr d9

/"vda= Ir^dr f cos9de = -(^R^^{+l) = -^R*.
0 0

D) Bottom: da = r sin dr d(jN; 9 = i^j^. v • da = [r^ cos 0) (r dr d(f)).
R 7r/2

/ V • da = r^dr / cos^d^ = -i?''.
0 0

rota/; f V • da = 7riZV4 + 0 - ii?" + iij4 ^ 2l|l. /
Problem 1.54 =======================================^ =__
■ X y z I
£ ^ ^ =z(fe-a). So /(Vxv)-da=F-aOr/22.
ay bx 0 I
vdl= (ayx + 6a;y) • (tfccx+ dyy+ dzz) = aydx + bxdy; x^ + y"^ = R"^ =i' 2xdx + 2ydy = 0,
so dy = —{x/y) dx. So v • dl = aydx + bx{-x/y) dx = ^ [ay"^ — bx^) dx.
For the "upper" semicircle, y = y/R"^ — x^, so v • dl = ° vna- a " ^^-

|v.dl = |£^lz(i±M :|aiZ2sin-i(£)-(a + 6)[-|v^
a/^^
• a;-* + -r- sm
■(I)

= ^R-'ia - b) sn-xlR) ^ = Ra - b) (sin-i(-l) - sin-H+l)) = Ra - b) (-| - |)
= IwR-'ib-a).
And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so
§ydl = 7rR^{b-a). /
Problem 1.55 =========::==.=^...==.=..::===== = =
{I) x = z = 0; dx = dz = 0; y : 0 -^ 1. v ■ dl - {y + 3x) dy = y dy.
1 1

J vdl= I ydy = -.
0 0

B) X = 0; 2 = 2 - 2y; dz = -2dy; j/ : 1 -> 0. v • dl = (y + 3a;) dy + 6dz = ydy - 12dy = (y - 12) dy.
0

|v.dl = y(y-12)dy = -Q-12)=-i + 12.
C) X = y = 0; da; = dy = 0; z : 2 -^ 0. v • dl = 6dz;
0

fvdl= f6dz=-l2.


Total: ,fv-dl= ^-| + 12-12 = [or|
Meanwhile, Stokes' thereom says /v • dl = /(Vxv) • da. Here da = dydz it, so all we need is
(Vxv)^ = ^{6) - ^{y + 3x) = 0. Therefore /(Vxv) • da = 0. /
Problem 1.56
Start at the origin.
A) 0=1, 0 = 0; r:0->l. v • dl = (rcos^0) (dr) = 0. /v-dl = 0.

B) r=l, e = ^; (f>-.0-^71/2. v • dl = Cr)(rsin0d0) = 3d0. /v • dl = 3 / d0 = ^.
0

C) (f> — f; rsinO — y = 1, so r = ^^, dr = ^^^cosOdO, 0 : | -> j.
/ 9 ?N ,, ^ , ? ?^/ ,^x cos^0 / cosO , , cos0!
v-dl = (rcos^0)(dr -(rcos^sin^ (rd^ =^-Z- T^]dtheta -,
^ '^ ' ^ sm0 V sin^^y sin^
/cos^g cosg _ cosg /cos^g + sin'^gN cosg
Vsin^^ sin^y ~ sin0 I, sin^0 / ~ sin^0
Therefore

f rn- 7 ^°^^ r/ft- ^ r^"- ^ ^-1_i-l
/"' y sin^fl'"^" 2sin2 0L/2 .(l/2) 2-(l)-^ 2-2"
D) 61 = f, 0 = f; r : v^ -^ 0. v • dl = (r cos^ 9) (dr) = ^rdr.
r 1 r ir'^f 1 1

Stokes' theorem says this should equal /(Vxv) • da

+ - [|-(-rr cos0sin0) - ^ (rcos^^)] 4>

- Scoter-60.

A) Back face: da ^-r dr d9 4); (Vxv)-da-iO. /(Vxv)-da = 0.
B) Bottom: da - -rsinOdrdcp6; (Vxv) -da — 6rsm0drd<l). 0 = |, so (Vxv) • da = 6rd
n/2

j{Vxv)-da^ j Qrdr j d<j,A. Q---^ ^'^
'
= ?.!.___ 2 2 ■

Problem 1.57
V •dl = ydz.

A) Left side: z = a-x; dz = -dx; y = 0. Therefore / v • dl = 0.
B) Bottom: dz = 0. Therefore / v • dl = 0.

{3) Back: z = a-^y; dz = -l/2dy; y:2a^0. fvdl=fy{-^dy) = -^^° = ^ =^
Meanwhile, Vxv = x, so /(Vxv) • da is the projection of this surface on the xy plane = - a-2a — c?. /
Problem 1.58 ~~~~~ ~~~~ """

V.v = 41- ('•''•' si" ^) + ^ 4 (si" ^ 4r2 cos Q) + ^ ~ (r^ tan 0)
= 4r^ sin e + -^4r^ (cos" 9 - sin^ 9)^-^ (sin^ 0 + cos^ 9 - sin^ 0)
, cos2 0
smp

Av-v) dr ^ f Dr^) (r' sin ^ dr d0 d<j>) = / r3 dr f cos^ 9dd fd<P= {R') B7r) [^ + ^11
0 0 0

Surface consists of two parts:

A) The ice cream: r - R; (p :0 ^ 2n; 9 :0 ^ n/6; da = R'^ sin 9 d9 d(f> f; v-da = (J?^ sin 0) (J?^ sin 0 d0 d0) ==
/ZSin^ 61 d0d0.

/" v-da = /?"/' sin2 9 dJd j d<i> 0 [^9 - ^° = 27:/?" (^^ " | sin 60°^ " ^ ('^ ~ ^"T )
0
= {R"^) {2n) sin 2^] ^ ^ ^

B) Tfte cone: 0 = f; 0 : 0 -^ 27r; r : 0 -^ J?; da = rsin0d0dr0 = ^rdrd^S; v • da = x/Sr^ drd^

/"v • da = v/3 /r^dr /d0 = ^/3 ~ • 27r = ^Tri?".
0 0

Therefore / v • da = ^^ (f - ^^ + ^/3) = 2^ B7r + 3n/3) . /
Problem 1.59
(a) Corollary 2 says f (VT) -dl = 0. Stokes' theorem says f (VT) -dl = /[V x (VT)] -da. So /[V x (VT)] -da = 0,
and since this is true for any surface, the integrand must vanish: Vx(VT) = 0, confirming Eq. 1.44.


(b) Corollary 2 says/(Vxv)-da = 0. Divergence theorem says/(Vxv)-da = / V'(Vxv)dr. So/V'(Vxv)dr
= 0, and since this is true for any volume, the integrand must vanish: V(VXv) =0, confirming Eq. 1.46.
Problem 1.60
(a) Divergence theorem: /v • da = /(V-v) dr. Let v = cT, where c is a constant vector. Using product
rule #5 in front cover: V-v = V-(cT) = r(V-c) + c • (VT). But c is constant so V-c = 0. Therefore we have:
/c • (VT) dr — jTc- da. Since c is constant, take it outside the integrals: c • / VTdr = c • /Tda. But c
is any constant vector—in particular, it could be be x, or y, or z—so each component of the integral on left
equals corresponding component on the right, and hence

qed

(b) Let V ^ (v X c) in divergence theorem. Then / V'(v x c)dr = /(v x c) • da. Product rule #6 ^
V-(v X c) = c • (Vxv) - V • (Vxc) = c • (Vxv). (Note: Vxc = 0, since c is constant.) Meanwhile vector
identity A) says da • (v x c) = c • (da x v) = —c • (v x da). Thus /c • (Vxv) dr = — /c • (v x da). Take c
outside, and again let c be x, y, z then:

/ (Vxv) dr - - V xda.
(c) Let V = TVC/ in divergence theorem: / V-(TVf/) dr - J TVU-da. Product rule #E) => V-(TVC/) =
TV-(Vf/) + (Vf/) • (VT) = TV2[/ + (VC/) • (VT). Therefore

I {TV^U + (VC/) • (VT)) dr = [(TVU) ■ da. qed
(d) Rewrite (c) with T <^ U : J {UW^T + (VT) • (VC/)) dr = J{UVT)-dai. Subtract this from (c), noting
that the (VC/) ■ (VT) terms cancel:

I {TV^U - UV^T) dr = [{TVU - UVT) ■ da. qed
(e) Stoke's theorem: /(Vxv) • da = /v ■ dl. Let v = cT. By Product Rule #G): Vx(cT) = T(Vxc) -
c X (VT) = -c X (VT) (since c is constant). Therefore, - /(c x (VT)) ■ da = /To • dl. Use vector indentity
#1 to rewrite the first terra (c x (VT)) • da = c • (VT x da). So - / c • (VT x da) = /c • Tdl. Pull c outside,
and let c ->■ x, y, and z to prove:

/ VTxda = -ATdl. qed
Problem 1.61
(a) da = R^ sin 9dJ9dij)T. Let the surface be the northern hemisphere. The x and y components clearly integrate
to zero, and the z component of r is cos 6, so

a= f R^smecoseded(f>z = 2nR'^z r singcosgdg = 2nR^ z^^^"^^ = nR^ z.]
(b) Let T = 1 in Prob. 1.60(a). Then VT = 0, so /da = 0. qed.
(c) This follows from (b). For suppose aj 7^ a2; then if you put them together to make a closed surface,
/ da = ai - a2 7^ 0.
(d) For one such triangle, da = |(r x dl) (since r x dl is the area of the parallelogram, and the direction is
perpendicular to the surface), so for the entire conical surface, a = | /r x dl.

(e) Let T = c • r, and use product rule #4: VT = V(c • r) = c x (Vxr) + (c • V)r. But Vxr = 0, and (c-V)r= {cx-§i + ^?^ +Ci^)(a;x + ?/y = zz) - CxX + Cyy + c^z = c. So Prob. 1.60(e) says
/Tdl= i{c-T)d = - f{VT)xd&=- /cxda = -cx /da = -c X a = a X c.

Problem 1.62

For a sphere of radius R:

/vda = f {^t)-{R^ sine d9d<i>t)=Rf sine d0d<j) = A7rR. ]
(R I So divergence
fdrUf sineded<j>) = A-kR. f theorem checks.
Evidently there is no delta function at the origin.

^^ ('•"'^ = hi ('•''•") = hWr ('•"^'^ = ;^(" + 2)r"+> =|(n + 2)r"-^|
(except for n — —2, for which we already know (Eq. 1.99) that the divergence is 47r5^(r)).
B) 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
Prob. 1.60(b): If Vx(r"f) = 0, then /(Vxv)dr = 0 = -§v x da. But v = r"f and da =
I?sinedJ9dij)T sie both in the f directions, so v x da = 0. /

Chapter 2

Electrostatics

Problem 2.1

(a) I Zero. |

(b)
47reo T where r is the distance from center to each numeral. F points toward the missing q.
lanation: by superposition, this is equivalent to (a), with an extra -g at 6 o'clock—since the force of all
twelve is zero, the net folrce is that of —q only,

(c) I Zero. I

(d) 147reo pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) a
qQ r^
a cancellation in pairs of opposite charges A 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).
Problem 2.2

(a) "Horizontal" components cancel. Net vertical field is: Ez = '^r^'§^ cos9. y
2qz
-(r
4-0 (,2+AJ^
When z ? d you're so far away it just looks like a single charge 2q; the field
should reduce to E = j^^ff z. And it does (just set d —> 0 in the formula).
(b) This time the "vertical" components cancel, leaving
E=4i^2^sin0x,or

E = 47reo
Prom far away, {z ? d), the field goes like E w ^^Fj^fr z, which, as we shall see, is the field of a dipole. (If we
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.)

^ +x^; cos9 = |) )se; (-^2

dq = Xdx

[{-'*7A^)^^{7^)i
For z^ L you expect it to look like a point charge q — XL: E - -^z. It checks, for with 2 ? L the x
term -^ 0, and the z term -¥ 4^^772-
Problem 2.4

From Ex. 2.1, with L -^ | and z -> yz^ + (|) (distance from center of edge to P), field of one edge h
1 Ao El-
47reo ^22 + ^^^2 + 0^ + 3^'
There are 4 sides, and we want vertical components only, so multiply by 4cos0 = 4 , '

4Aaz

E= z 1 (-^ + f)
47reo

"Horizontal" components cancel, leaving: E = ^^ {/^ cos^} z.
Here, -i^ = r^ + z^, cos0 = | (both constants), while Jdl = 2irr. So

1 AB7rr)z
' 4^ (^2 + ^2K/2

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 a ■ 27rr • dr = X- 2nr, so A = adr is the "line charge" of each ring.

1 {adrJnrz _ ^ ^ [^ r ,

CHAPTER 2. ELECTROSTATICS

For i? > 2 the second term -^ 0, so Epiane = 4^'^''^^''

For z ?/2, ^^gl^ - i A + f )~'^'? i A - If^), so [
and £; = ji^^^ =i^§s, where Q - TriZV. /
Problem 2.7

E is clearly in the z direction. Prom the diagram,
dq — ada — aR^ sin $
-i2^i?2+z2-2iZzc(

= J_ f crR^ sin 2/2z cos 61K/2
^ ~ 47r£o y (i?2 + Jd(f> = 27r.
B-i?cosg)sing 0^u = +l 1
= COS 9 du = - sin
22 -2iJz cos 61K/2

47reo - 2Rzu)y- r^du. Integral can be done by partial fractions—or look it up.
1 zu-R
47r£o Vi?2 + z2 _ 2Rzu. 47reo z^ (jz-R) i-z +- R J'
1 27rJZV {z-R z R) ]
For z > iZ (outside the sphere), -E^ = 5^
For z < /2 (inside), E^ = 0, so |E = 0. |
Problem 2.8
According to Prob. 2.7, all shells interior to the point (i.e. at smaller r) contribute as though their charge
were concentrated at the center, while all exterior shells contribute nothing. Therefore:

E(r) = 47reo -f,
where Qmt is the total charge interior to the point. Outside the sphere, all the charge is interior, so

E = - 47reor
Inside the sphere, only that fraction of the total which is interior to the point counts:

" |7riZ3^ " R^ Q, so E = 47reoiZ3^r2 47reo R^
Problem 2.9

(a) p ^ eo V- E =: eo;nr^ (r^ • kr^) = eo^k{5r^) = | Sepfcr^ [

(b) By Gauss's law: Qenc = co/E • da = eo{kR^)i4nR'^) = |47reofci?^|
By direct integration: Qenc = fpdr = J^{5 okr^)i4nr'^dr) = 20n ok J^r'^dr = AneokR^V
Problem 2.10
Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface
of this larger cube gets the same flux as every other one, so:

/ E • da = — / E • da.

The latter is ^-q, by Gauss's law. Therefore / ■ 24eo'

/;;;;ZIZ:;X^ Gaussian surface: Inside: / E • da = Eiinr"^) = ^Qenc = 0 =i? | E ^ 0. L
( ( "F7 ) " Gaussian surface: Outside: E{A7rr^) = ^(aiTtR^) ^ E = ^ f. J >- (As in Prob. 2.7.)

Problem 2.12

- Gaussian surface /E-da = E ■ 47tr^ ^ ^Qenc = = i|7rrV So

u ^=i-o^'-
^ ^
Problem 2.13
Since (Jtot =3"' ^'p^^=ik ;^f (as in Prob. 2.8).

Gaussian surface
t /
^

(] [) Y J E = -^s (same as Ex. 2.1).
2neos
7"
Problem 2.14

Gaussian surface /E • da = S • Anr"^ = j^Qenc = 7^ Jp^t = 7^ K^^)(^^ ^^"^'^^'^

E = -±-nkrH
47reo


Problem 2.15

(i) Qenc = 0, SO I E = 0. I
(ii) §E-da = E{A7tr^) = ^Qenc = j^fpdT = j^J^f^sin6dfd9dphi
= ^Jldf=^{r-a): '-ii:-^y
(iii) £D7rr2) = ^J^df=îb- a), so

Problem 2.16

•(^m - Gaussian surface §E-da = E-2ns-l = j^Qeuc = j^P^s'^h

- Gaussian surface

,., Q Fi §E.dsi = E-2ns-l= Ôenc = îOTra^/;

Q Yyy
Gaussian surface

(iii)

r—^

Problem 2.17
On the xz plane E — Qhy symmetry. Set up a Gaussian "pillbox" with one face in this plane and the
other at y.

Gaussian pillbox / E • da = g • ^ = ^Qenc = TÂyp;
(for y < d).

<?enc = iiAdp =^ E = —dy (for y > d).

Problem 2.18

Rrom Prob. 2.12, the field inside the positive sphere is E+ = 3f-r+, where r+ is the vector from the positive
center to the point in question. Likewise, the field of the negative sphere is — g^-r-. So the total field is

VxE = Vx / -^pdr = / Vx [ -T I pdr (since p depends on r', not r)
47reo J i^ 47reo 7 [ * / J

= 0 (since Vx ( ^ j =0, from Prob. 1.62).
Problem 2.20

A) VxEi = = k [x@ - 2y) + y@ - Zz) + z@ - a;)] ^ 0,
xy 2yz Zzx
0 El is an impossible electrostatic field.

B) VxEz = fc = k [x{2z - 2z) + y@ - 0) + zBy - 2y)] =
2xy + z^ 2yz
so E2 is a possible electrostatic field.
Let's go by the indicated path:

E • dl = (y2 dx + {2xy + z^)dy + 2yzdz)k •(a;o.yo,2o)
Step I: y = z = 0;dy = dz = O.E-dl. = ky^^ = 0.
Step II: X = xo, y : 0 -^ yo, z = 0. dx = dz — 0.
E-dl = k{2xy + z'^)dy = 2kxoy dy.
JjjE'dl = 2kxoJ^°ydy^kxoyl X"
Step III: X = xo, y = yo, z : 0 -> zq; dx = dy = 0.


E . dl = 2kyzdz = 2kyQzdz.
Jjjj E . dl = 2yok J^° zdz = kyoz^.

V{xo,yo,zo) = - 7 E . dl = -k{xoyl + yoz^), or v{x,y,z) ^ -kixy"^ + yz'^).
Check: -■VV=k[-^{xy''+yz')it+-^{xy''+yz^)$+-^{xy'+yz'')i]=kly'S.+Bxy+z'')y+2yzZ]=B /
Problem 2.21

nr) = -X;E.dl. {Outside thethe sphere {r i?)R)::'E^-^î^r.
Inside
sphere (r > < E = ^—-^rr.
Q 1
So for r > iZ: F(r) = -/; (jJ^^) df = j5^9 (i) [ 47r£o r'

and for r < R: F(r) = - J^ (^^^) dr - f^, {^-^r) df=^^[^--^ {^

When r>R,VV=-^^{i:)T= —^^r, so E = -VF = j^^rf. /
When r<R,VV = j^^j: {s-^)f= j^ A J-w) ? - -J^S^?; ô ^ = "W ^ ^^^rf./
Problem 2.22

E = 45fj-^s (Prob. 2.13). In this case we cannot set the reference point at oo, since the charge i
extends to oo. Let's set it at s = a. Then

V I Ja V47r,Eo s J 4^^^ Va/
(In this form it is clear why a = oo would be no good—likewise the other "natural" point, o = 0.)
VV = - j^2A4 (In (f)) s = -7;^2Ais = -E. /
Problem 2.23

V{0) = -/Ê . dl = -XL(^^)dr-/;(^^)dr - f^{0)dr = { -^(ln(f)+a(i-i))
Hl-f-ln(f)-l + f}=|^<^)-
Problem 2.24
Using Eq. 2.22 and the fields from Prob. 2.16:

FF)-F@) = -/o'E.dl = -/;E.dl-/?*E.dl=-2f^/;sds-|f^/?*ids

=-F)C-tS'-i'=-^(^^""c)
Problem 2.25

(a)
V-^+df

i^)y=^j-LM^ =

W ^ = 4^/o''5? = 4i^2^^ (V^^^"^)ir - ^ (n/^^
In each case, by symmetry |^ = ^ = 0.

(agrees with Prob. 2.2a).

47rfo y/z^+L^ (agrees with Ex. 2.1).

WE = -4{l7H^2.-l}z with Prob. 2.6).

If the right-hand charge in (a) is -q, then v = o], which, naively, suggests E = - VV = 0, in contradiction
with the answer to Prob. 2.2b. The point is that we only know V on the z axis, and from this we cannot
hope to compute -Sx = -f^ or J5j, = — ^. 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
2iTa 1 , /T-,, ah

- I V ,.- -r ■<■' - V ^■<^'- -r -r= iiiv^ V/l2 + -^2 - V^/W + 2? - V2/1)

= _^ [ft+-^lnB/H-272/1-^2/1)-ft-y=lnB/i-V2/i)] - —^= ~ [lnB/H-V2/i) - lnB/i-V2/1)]
0 V2-V2; 4eo ^ 2 J - ln(l + V2).
.'. F(a) - F(b) = - In(l + V2)] .


Problem 2.27

Cut the cylinder into slabs, as shown in the figure, and
use result of Prob. 2.25c, with z -¥ x and a -> pdx:

J (ViZ2 + x2 - x) dx

2f^i [xVE"" +x^ + K' ln(x + ^R^ + x2) --^Ji; -L/2

i%|(-+*)/?='+(-+*)='-(-t)/?=+(-t [•+^+/?=+(-+^ri -4
='+fl=ln L^-*+/?=+(^-^rJ
{Note: -{z+ f)' + (z - f)' = -z^ - zL - ^ + ^2 _ ;jL + ^ ^ _2zL.)

U+| + ^2+(;,+ |J ,_L+^^2+(,_|Jj J
1 1

L-^R^.[z.l)^R^.(z-iy
Problem 2.28

Orient axes so P is on 2; axis.

_ ^ pg f Here p is constant, dr = r^ sin 9 drd
^^i^J* { ^ = Vz'^+T'^-2rzcos9.
^^■dro I Vz^+J-lrzcole 'Jo '^'f> = 27r.

lo v.^+r^'-Lcose'^^^^z (Vr2 + z2 _ 2rzcos0) 1^ = 5^ (Vr^ + z^ + 2rz - Vr^ + 22 _ 2rz)

'•-^'^^^ |r z) I 2/r ,ifr>z. J

Butp= ^, so V{z) = ^jly [r^ _ 4) = 5-2.^ C- ^) V{r] SttcoR V ^V
Problem 2.29

V^y = 4;^V2/(f )dr = ji^ //9(r')(V2|)dr (since p is a function of r', not r)
= ^fpin-^^SHr - r')] dr = -^p(r). / ___=^___
Problem 2.30.

(a) Ex. 2.4: Eabove = af^n; Ebeiow = "if^" (n always pointing up); Eabove - Ebeiow = ^n- /
Ex. 2.5: At each surface, J5 = 0 one side and E = f- other side, so AE — f^. /

Prob. 2.11: Eout = ^f = ^f; Esn = 0; so AE = ^f. /

(b) ■ L"'- j^ Outside: /E • da = E{27rs)l - j^Qenc = ^B7ri?)/ ^ E = ^f i = ^i (at surface).
= 0, so E = 0. .-. AE = -2^8. /

(c) Fout = S = ^ (at surface); T^? = ^^ ; so Kut = Kn- /
^ = -& = -^ (at surface); ^ = 0 ; so ^ - ^ = -^. /
Problem 2.31

.-. ^4=9^- 47reoa

(b) W^i = 0, W^2 = 5^^ {^); W,^^^ (;fe-^);^4 = (see(a)).


Problem 2,32

(a) ly = I JpVdT. Prom Prob. 2.21 (or Prob. 2.28): V = ^ (^R"^ - ^^ = ^^^ (z- ^^
-IpJ-J-T '^-^-=^4-mr&i^<
2^47reo2J?yo

5eo 5eo fTrii^ 47reo VS-R;

(b) ly = f jE^dr. Outside {r > R) E = ^^^r ; Inside {r < R) E = ^^-^rr.

1 Q

47reo 2 I V t-/ U ^^ 5 / |o J 47reo 2 R 5R) 47reo 5 iZ''
47reo 2

(c) W — ^ { jij VE • da + JyE'^dr}, where V is large enough to enclose all the charge, but otherwise
arbitrary. Let's use a sphere of radius a> R. Here V = 4^^-

4n

1 eof q' K , 4nq^ -m
' 2 D7reoJ o '^ D7reoJ 5iZ "^ (Anco)^

- J_^/I 4. J_ _ 1 + i = J_^g! ' 47reo 2la5J? a R) 47reo 5 i?
As o -> 00, the contribution from the surface integral f^^^fs) goes to zero, while the volume integral
(i^^fsdl - 1)) picks up the slack.
Problem 2.33

dW = dqV = dq{ J -, {q = charge on sphere of radius r).
47reo/ r

{q = total charge on sphere).

r^dr. dq = Anrdrp^j;^qdr^-
1 (qr'l(Zq , _ 1 Zq,

W=J-^-^rr^dr=-^^^^^(^'-]/
47reo R^ Jo 47reo -R^ 5 4neo5 RJ'

Problem 2.34

(a.)W = f J E'^ dr. E = ^^ (o < r < 6), zero elsewhere.
rb 1
Sttco :i-i)
(b) W, ■■ ^4.^^2 = 8^^, ^^ = j^^rir>a),E2 = j^^r{r>b). So
^1 • ^2 = D;?^) 1^. ('^ > *). and hence /Ej • E2 dr = - C;^) q^ /~ ^47rr2dr =
Problem 2.35

(b) y(o) =
-C^''^--lLiô^)dr-i:iO)dr-J^^{^^)dr-fîO)dr = ^^{l + ^-l)
(c) r^TTo] (the charge "drains off"); V{0) = - f^{0)dr - !^{^^^)dr - f^{0)dr = U- (| - ^)
obk m2.36

(a)
1 9a + ?6
4x0^'

(b) *'out - ~. 0 r, where r = vector fron

(c) ô ~ 1 'Tâ> ■'^6 ~ 7 2^6' where Fo (rt) is the vector from center of cavity a [b).
where Fo
1 cu / a M/i CO / 6 I

(d)[z^
(e) ffji changes (but not Ga or o-ft); Eoutside changes (but not Ea or Eft); force on qa and qb still zero.
Problem 2.37
Between the plates, E = 0; outside the plates E = a/ta = Q/eoA. So

p_£^p2_£o_Q^.
2 e2^2 - 2eoA2 ■

Problem 2.38

Inside, E = 0; outside, E = j^J^^f; so

F, = lf,da = j{^m^^§,)coseRHmeded<f>
i-A^f^-i:''^^^ôsede = ^^{^f {'jsinH);/' = ^^{^f 32nR-êo'

34 CHAPTER 2. ELECTROSTATICS

Problem 2.39
Say the charge on the inner cylinder is Q, for a length L. The field is given by Gauss's law:
/E • da = J5 • 27rs • L — -j^Qenc = j-Q ^ E = ^^^ ^ j s. Potential difference between the cylinders is

m-V(a) = -/'E.,iI = -5^^'l* = - 2ntQL

As set up here, a is at the higher potential, so V = V'(a) - V{h) = ^^ ^ In (^).

C = ^ = YJhJ' ^° capacitance per unit length is 27reo l"(i)'
Problem 2.40

(a) W — (force) x (distance) = (pressure) x (area) x (distance)

(b) W — (energy per unit volume) x (decrease in volume) = (eo'x) (-^^)- Same as (a), confirming that the
energy lost is equal to the work done. ^^
Problem 2.41

From Prob. 2.4, the field at height z above the center of a square loop (side a) is
i
'Ai///y//y///////////////.
E=J: B ^^°^7,2 + si
47reo J. si) ,
I ^"T _^
Here X -¥ a^ (see figure), and we integrate over a from 0 to a:
'.iiiiii:""i"^ii^
1 r ada a^
E = 2az I . . Let u = —, so ada =
47reo Jo B + ^ ^ L2 , ^ 4
V' ^ A ) -^ ^2

= 47reo Jo (u + %= = ^ '- tan- z ^ z j
-1-4.. r' z'^)/2u + z^ ttco (:/^±Z]
= ^tan-M -^ -tan ^1) ?;

tan-^Vl + ^,-J
a -> 00 (infinite plane): E — ^ [tan ^@0) — f ] = ^ (f - f) = ^- /
z > a (point charge): Let f{x) = tan~^ VH-a; — f, and expand as a Taylor series:

Here /(O) = tan-^l) - f = f - f = 0; /'(x) = rR^|;^ = 2B^jy7TT^. ?0 /'(O) = i. so

Thus (since ^ = ^ ? 1), g <:. |^ (j^) = 3^^ = ^4r- /
Problem 2.42

? ^ ( I d f 2^^ 1 ^ /5sin0cos01

^{A-Bsm(j)).

Problem 2.43

Rrom Prob. 2.12, the field inside a uniformly charged sphere is: E = 4;f7"^r. So the force per unit volume
isf = pE = A^3) Da-t^ita)r = ^C;^) ""i ^"d the force in the z direction on dr is:

dF, = f,dT=j- D^) rcos^Cr^sin^drd^d^).
The total force on the "northern" hemisphere is:

3 / Q y fR' l^sin^gr B7r) = 3Q^
64neoRP''

Problem 2.44

Kenter47reo -da = / da = R-5{27tR^)R =2eo
= / J '!■ 47reo J 47reo
—-

sine dO,
Vpole= 1^J-J^'^^''^[,2 == n^ + R:' 2J?2cos6l = 2J?2(l-cos0).
y^^j ■ L. i da 27rR^ si
_ 47reo J?v^ 70 ^/^ -cos0 ~ 2V2eo BvT^
1 ai27rR'^) VI gingdg _ (xR

■^<'-°' "' .-. Fpole - Vcenter = g(^-'-
/2eo'
Problem 2.45
First let's determine the electric field inside and outside the sphere, using Gauss's law:

eo$E = eo^Ttr'^E = Q
= /.* = /(^)r= sin.*-d.<i* = 4.*|f3* = H'^, J'^ < ?|;


SoE^J^r^f{r<R); E=^f(r>iZ).
Method I:

'Hi^rii>^î:>}-m^'^{-i):hm^-:

W = ^fpVdT (Eq. 2.43).

27rfc2 ( Ri ilf^ nk^R-r .
3eo 1 4 4 7 J 2 - 3eo V^V 7eo
Problem 2.46

E = -W =
,(^),...(:hil.^^^}. Ae-^'-(H-Ar)-

p = £o V- E = eoA {e-^^(l + Ar) V- (^) + ^ • V (e-^'-(l + Ar))}. But V- {^) = 47r(j3(r) (Eq. 1.99), and
e-^'-(H-ArM3(r) =<J3(r) (Eq. 1.88). Meanwhile,
V (e-^'-(l + Ar)) = r|r (e-^'-(l + Ar)) = f {-Ae-^'-(l + Ar) + e-^^} = f(-AVe--
So ^ ■ V (e-^'-(l + Ar)) - -ê"- p = eoA 47r<J3(r) A2

Q= IpdT = toA Ia-k fdîr) dT - A^ f^—47rr^dr = eoA Utt - A247r / re-^''dr j .
But /o°° re-^''dr = ^,soQ = AttôA (l- ^^ = | zero. |
Problem 2.47

(a) Potential of +A is V^ = ~2^^'^ (^)' where s+ is distance from A+ (Prob. 2.22).
Potential of —A is VL = +5^^" (^), where s_ is distance from A_.

(x,2/,^)
27reo V?+/J
Now s+ = y/iy -a)^ + z"^, and s_ - /(y + aJ +z^, so
jy + a)'
{y - aJ + z2

(b) Equipotentials are given by [^+°]5:]:^2 - e('*"^°^°/''^) =k^ constant. That is:
y2 + 2ay + a^ + z^ = A;(y2 - 2ay + o?+ z^) =;> y2(^ _ j) ^ ^2(^ _ 1) ^ ^2(^ _ 1) _ 2ay{k + 1) = 0, or
j/^ + 2^ 4- a2 _ 2^2/ [l^ j = 0. The equation for a circle, with center at (j/o, 0) and radius R, is
(j/ - 2/0)' + Z2 = /?2, or y2 + Z2 + (^2 _ JJ2) _ 2yy^ = Q.
Evidently the equipotentials are circles, with yo = a( |^ j and
.2 _ ,,2 _ d2 _. d2 _ ,,2 _ ?2 _ ?2 /'Mi'^ - o^ — ?2 (fc^+2fc+l-*;^+2fc-l) _ 2 4*
i? = ^^ ; or, in terms of Vq:

Problem 2.48

(a)V2y =:-f (Eq. 2.24), so

(b) qV = mv^

{c) dq - Ap dx ; ^ = ap^ = I Apv = 11 (constant). {Note: p, hence also /, is negative.)

38 CHAPTER 2. ELECTROSTATICS

cPV
eoAy 2qV " dx^ pv~'/^ where /3
{Note: I is negative, so /3 is positive; q is positive.)
(e) Multiply by F' - ^ :

^'^ " ^^"'^'^ ^ Z"^'*^^' = /^/^"'^'^ ^ ^^'' = 2/3y'/2 + constant.
But F@) == V'{0) — 0 (cathode is at potential zero, and field at cathode is zero), so the constant is zero, and

V-2 ^ 4^^1/2 ^^ = 2y/^V'/* => V-'/UV = 2,/^dx;
/"y-i/4 ^ ^ 2v^ fdx ^ ^F^/" = 2y^x + constant.
But V{0) = 0, so this constant is also zero.

^/^. ^Vix)=(IM xV r Vix) = '4/3 ^ / 81J^m N
S2elA-^qJ

F(.)=FoQ) (see graph). Interms of Vq (instead of /):
Without space-charge, V would increase linearly: V{x) — Vq (^).
1 4 1 4eoFo
9(d2xJ/3 •

/2^VV^Q) '
(f) V{d) ^ Fo - i^^y^'d^^' => Vo' = ^^,I';l' = '-I^Vo^:
I = 1^^V'V2 ^ ;^y^3/2^ ^^^^^ _ 4eoA /2g
Problem 2.49

(a) E = 1 47reo I${^-iy -^/^dr.
(b) I Yes. I The field of a point charge at the origin is radial and symmetric, so V X E = 0, and hence this is also
true (by superposition) for any collection of charges.

Now /;ê ^/^dr = —^-^ /^~?—'^'" ^— exactly right to kill the last term. Therefore

47reo I >" Ir J I 47reo r

■^•^=4^'?0n)-^'***# =

(e) Does the result in (d) hold for a nonspherical surface? Suppose we
make a "dent" in the sphere—pushing a patch (area B? sin 9
from radius R out to radius S (area S^ sin 9 d9 d0).

{R^ sin 9 d9d(f>)

4A /^ •r=^sin0 ,drd9d(f)^ A^ 47reo
A^ 47reo Jr
■'/^dr

47reo

So the change in p- JV dr exactly compensates for the change in Ê • da, and we get j-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 j-Qenc- Charges outside do not contribute (in the argument above we found that C^" for this
volume Ê ■ da + p- JV dr = 0—and, again, the sum is not changed by distortions of the surface, as long as q
remains outside). So the new "Gauss's Law" holds for any charge configuration.

(f) In differential form, "Gauss's law" reads: V-E + - or, putting it all in terms of E:

V'E - Y2 / E • dl = —p. Since E = -W, this also yields "Poisson's equation": -V^F + --^V


Problem 2.50

/9 = eo V- E = o^{ax) = |eoo| (constant everywhere).
The same charge density would be compatible (as far as Gauss's law is concerned) with E = ayy, for
instance, or E =; (|)r, etc. The point is that Gauss's law (and VxE = 0) by themselves do not determine
the field—like any differential equations, they must be supplemented by appropriate boundary conditions.
Ordinarily, these are so "obvious" that we impose them almost subconsciously ("J5 must go to zero far from
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 plane of surface charge"). But in this case there are no natural
boundary conditions, and no persuasive symmetry conditions, to fix the answer. The question "What is the
electric field produced by a uniform charge density filling all of space?" is simply ill-posed: it does not give
us sufficient information to determine the answer. (Incidentally, it won't help to appeal to Coulomb's law
(E = jJ^ fp^dr)—the integral is hopelessly indefinite, in this case.)
Problem 2.51

Compare Newton's law of universal gravitation to Coulomb's law:

miTn2 __ 1 9192 J.
t = —G-—5—^r t = 5-r.

> G and q -¥ m. The gravitational energy of a sphere (translating Prob. 2.32) is therefore Evidently j^

Now, G - 6.67 X 10-1^ N mVkg^, and for the sun M = 1.99 x 10^° kg, R = 6.96 x 10^ m, so the sun's
gravitational energy sW = 2.28 x 10*^ J. At the current rate, this energy would be dissipated in a time

'' P '' < lO^'* s = 11.87 X 10'^ years. I

Problem 2.52
First ehrainate z, using the formula for the ellipsoid:

^(a;,y) = ^ 47ra6 yc2(a;2/a4) + c'^{y^/b^) + 1 - (xVa2) - {y^/b"^)'
Now (for parts (a) and (b)) set c ->■ 0, "squashing" the ellipsoid down to an ellipse in the xy plane:

''^''' ^^ li^ab ^-{xlaY-{ylbY
(I multiplied by 2 to count both surfaces.)

(a) For the circular disk, set a = 6 = iZ and let r = yjx^2nR^R2-:,^-
^
+ y^, . Q 1
(b) For the ribbon, let Qjb = A, and then take the limit b -> c ( )- ^ 1
(c) Let b = c, r = ^/y"^ + z'^, making an ellipsoid of revolution:

a2+c2-l' ^'^^''- 4na(P^xya^ + ryc*'
The charge on a ring of width dx is

J = a2Trrds, where ds = ydx"^ + dr^ — dxy/l + (dr/dx)^.
dr <?x I -4-2 -2
= dxJl + ^ = dx-y/xy^^Try^. Thus
(Constant!)
A(x) = dx 47rac2 ^x^ja^ + r2/c'* r s/^J^^H^ =

Chapter 3

Special Techniques

Problem 3.1
The argument is exactly the same as in Sect. 3.1.4, except that since z < R, y/z^ + R^ — 2zR = {R — z),
instead of {z - R). Hence Vkve = -r~7ryi [i^ + R)-{R- z)]
47reo 2zR If there is more than one charge 1 g 47reo R
inside the sphere, the average potential due to interior charges is47reo R and the average due to exterior
—^— ^^|j^,
charges is Fcenter, so Fave = Vcenter + ^f^- ^
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:
dV c
= c (constant) => —r- = -^ =>
' dr r^ r

Example: potential of a uniformly charged sphere,

In cylindrical coordinates: sdsVdV J ? dy1 dV0 c ds ds F ^ cln s + k.
o Id/ = ds s—r- = J ^ 1 s
V --7-
Example: potential of a long wire.
Problem 3.4

Same as proof of second uniqueness theorem, up to the equation /^ V3E3 • da = -f^{E3)^dT. But on
each surface, either V3 = 0 (if V is specified on the surface), or else Esj^ = 0 (if ^ = -E± is specified). So
JyiEs)^ = 0, and hence E2 = Ej. qed
Problem 3.5
Putting U = T = V3 into Green's identity:
f [V3VVs + Ws • Wa] dT= i Vg VVs • da. But V^Vs = VV, - VVz =
So / E^dr = — f V2E3 • da, and the rest is the same as before.

Problem 3.6
Place image charges +2q a.t z —d and —q&tz — -3d. Total force on --q is

g [ -2g 2g _-£_■ ' 47reocP
' 47reo [BdJ ^ DdJ FdJ 47reo V72(i2y ■

(a) FVom Fig. 3.13: i = -y/r^ + a2 - 2ra cos 0; V = Vr^ + &2 _ 2r& cos 0. Therefore:
g'
(Eq. 3.15), while b = — (Eq. 3.16).

(f)/'-' + ^-2^f cosfl y(f)' + iZ2-2racos0

- 2ra cos 0 y/R^ + (ra/R)^-2rao
Clearly, when r = iJ, V -^ 0.
(b) a = -eof^ (Eq. 2.49). In this case, |^ = I7 at the point r = R. Therefore,

(T(e) = -eo D^) {-^('•^ + "^ - 2mcos0)-3/2B^ _ 2acos0)
+ i (ij2 + (ro/iZJ - 2racos0)"^^' (^2r - 2a cos0^ 11
= —^S.-{R''+a''-2Racose)-^^R-acose) + {R^ + a''-2Racose)~^^'' ( ^)}
= ^(i?2 + a^ - 2Racose)-^l'^ r - acosd - ^ + acos^l
-(ij2_a2)(^2_j.^2_2ijacos0;

induced = fada= -^{R^ - a') [(R^ + a' - 2Racos9)-^l''R'' smOd,

2a (a2 - ij2) ViZ2 + a2 + 2iZa V-R^ + a2 - 2iZaJ
But a > J? (else g would be inside), so /j?2 + a2 - 2jRa = a-R.

^y -?'' [(^ - (?^] - ^ '<° - ^' -'°+?"=s'-^^'
I 9fi _ ?, I

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:
1 qq' _ 1 f R 2 1 _ 1 q'^Ra
47reo (o - b)^ 47reo (a - i?V?)' 47reo (a^ - R-^)-
To bring q in from infinity to o, then, we do work

q'R
4neoJ (a2-iZ2J" 4^eo [ 2 E^ - i?2) 47reo 2(a2 - 7^2)"

Problem 3.8
Place a second image charge, q", at the center of the sphere;
this will not alter the fact that the sphere is an eqmpotential,
1 g".
Aireo R ' but merely increase that potential from zero to Vo
q" — 47reoVoi? at center of sphere.
For a neutral sphere, q' + q" = 0.
1

AntQ^ o? ' {a-by ^J_9g^ (_47reo V^ "^ " (a-6J J
{_ "^ ~ -by) 47r6o
F =

qq' b{2a - b) _ qj-Rq/a) {R^/a){2a - R^/a)
47reoa2(a-6J ~ 47reo a^ia-R'^laf

q^ f rV {2a''- R^)
4neo a) (a2 - R-^y '
(Drop the minus sign, because the problem asks for the force of attraction.)
Problem 3.9

(a) Image problem: A above, -A below. Potential was found in Prob. 2.47:
2A , , , , A
n?..) = 5^1n(.-K) = 3^1n(.iAl) {y,z)

-In 47reo[y^ ++ {z-d)^j
ly^ jz + dy] ^ /s-
{h)a-- dn' dn dz

^<i-A I y^ + {z + d)^<^)~ +2 {z-d)^ '] 1^^,
47reo
2 l/^^X2^(^ + y^ l/ ^2(z-d)|
Ad
2A f d -d
47rU2 + d2 y2 + ^J = 7r(j/2 + d2)-
Check: Total charge induced on a strip of width / parallel to the y axis:

qnd
_ IXd FTT / Try
7t J y^ + cP TT [d d. ~~~V [2 ~ V 2I
-XI. Therefore Ajnd = -A, as it should t

Problem 3.10
The image configuration is as shown.
9. ?■ .9

V{x,y) -- 1 i 1 , 1
47reo ( v^(xaJ- +aJiy-B/- 6J+ +z2^2^{x - aJ+ a)^(y+ {y 6Jb)^ ^2 z^J ]
^/ix +
+ 6J ^(^j; + + + + + 9- - •-?

For this to work, I Q must be and integer divisor of 180°. Thus 180°, ' , 45°, etc., are OK, but no
others. It works for 45°, say, with the charges as shown.
(Note the strategy: to make the x axis an equipotential (F — 0), f B) y 4?°
you place the image charge A) in the reflection point. To make the
45° line an equipotential, you place charge B) at the image point.
-.
/^ '•+ ^
But that screws up the x axis, so you must now insert image C) to +*. V •-(!)
balance B). Moreover, to make the 45° line V — ^ you also need D),
to balance A). But now, to restore the x axis to F = 0 you need E) E C) ^
to balance D), and so on.

The reason this doesn't work for arbitrary angles is that you are
eventually forced to pla.ce an image charge within the original region of
interest, and that's not allowed—all images must go outside the
region, or you're no longer dealing with the same problem at all.)

Problem 3.11

Prom Prob. 2.47 (with j/o - jx + ay+y^' , where a^ - yg"^ - Er ^
47reo "' L(^~ ^y +2/^.

■ ocschB7reoVo/A) =(dividing) ^^= R A /,
"'''''!:}''''fiS =i]^ J? J ^ 4 cosh f^) -'id/RY

m Problem 3.12

oo 2 ?
Vix,y) = Y^C?e-"'^='/''sininny/a) (Eq. 3.30), where <^n - - / Vb(y)sm{mry/a)dy (Eq. 3.34).

I this case Voiy) = _y^^ forf o/2< <^Z2/< <a/2 J1 Therefore,
uvj/; -j^
( +^'' ? a

n 2, (T. / / ^J ? • / / N^ 1 2^0/ cos{mry/a)r^^ , cos(n7r2//a) I" 1 C? = -Vo< sm{niTy/a)dy - / sm{mry a)dy } =--< / .[ + ^,[ >
^{-cos(^).cos@, + c?,??-cos(f)} = ^{l + (-ir-2co?(f)}.

CHAPTER 3. SPECIAL TECHNIQUES

The term in curly brackets is:
n = l l-l-2cosGr/2) = 0,
n = 2
n = Z -, _■, _<y (f> /n'_ 0 } etc. (Zero if n is odd or divisible by 4, otherwise 4.)
1 + 1-2 cosB7r) = 0,

r - f SFo/nTT, n = 2,6,10,14,etc. (in general, 4j + 2, for j = 0,1,2,...),
" ~" 1 0, otherwise.

v(x,y) = ^ y ^~"'^^^°^i"("^y/°) 8Fo ^ e-(''^+^)'^'''/°sin[Dj2)+ 2)ny/a]
Dj +

Problem 3.13

V{x,y) = ^ ^ ^e-""^/''sin(n7ry/a) (Eq. 3.36); ^ =-eo|^ (Eq. 2.4!

^B/) ~^° a^ l~^n^ ""''/"sin(n7r2//a)I = -cq "^Z^-l-—)e ""''/"sin(n7ry/a)
y^ sin(n7rj//a).

Or, using the closed form 3.37:

,w , 2Fo^ _i / sinGry/a) 2Fo 1 / -sinGrj//a) tt ^, , ,| V{x,y) = tan M . ' ' ' ] ^ a =- q .in2f^?/?^ . ^2/ , [ -coshGra:/o)
_ 2eoFo sinGrj//o) coshGra;/o) I
a sin2Gry/a) + sinh2Gra;/a) l^^^ a2eoK) 1
sm{7ry/a)'

Summation of series Eq. 3.36

Vix,y) = —I, where / = ^ -e-?"^/''sin(n7ry/a).
Now sinio = Im (e**"), s

where Z = g-'^C^-^^')/". Now

where Re*^ - 7^. Therefore

/ = Im-inR + i9) = ^e. But f^^ =^ , ^= , ■ ^ , ^
1 + e-'^^/" (e^'^^^/" - e-'-'^y/") - e-^^^/" _ 1 + 2ze-'^'/''sinGr^/Q) - e-^^^/°
|l_e-'r(:t-tJ/)/a|2 ~ 11 _ e-,r(x-tj,)/a|2
2e~'^^''"sinGr2//o) _ 2sinGr2//a) _ sm{ny/a)
rx/a _ g-TTx/o sinhGrx/a)'

/=i.an-r-'°<r?/f'; , and F(a;, y) _i
_i / smGry/o , Lr, X 2K), =
tan . ■ , :
f sin{iTy a)
tan . -V , .

Problem 3.14

(a) -7-^ + ——- = 0, with boundary conditions
OX-' oy^

(i) F(a:,0) = 0,
(ii) V{x,a)=Q,
(iii) F@,y) = 0,
(iv) FF,y) = Fo(y). J ;j-
As in Ex. 3.4, separation of variables yields

V{x, y) = (Ae*^ + 5e-*^) (C sin ky + D cos A;y).
Here (i)::^ D = 0, (iii)=i' B = -A, (ii)=^ fca is an integer multiple of tt:

V{x,y) = AC (e"'^^/" - 6-""="/") sin(n7ry/a) = B^C) sinh(n7ra;/a) sin(n7ry/a).
But BAC) is a constant, and the most general linear combination of separable solutions consistent with (i),
(ii). (iii) is
V{x,y) — ^C?sinh(n7ra;/a)sin(n7rj//a).

It remains to determine the coefficients C? so as to fit boundary condition (iv):

^C?sinh(n7r6/o)sin(n7r2//a) — Va{y). Fourier's tricky C?sinh(n7r6/a) = - Vo{y)sm{nny/a)dy.
0

Therefore

Cn = —r-TT—TT-T / Vo{y)sm{mry/a)dy.

5 CHAPTERS. SPECIAL TECHNIQUES

,. , ^ 2 rr f ■ , I j 2Fo f 0, if n is even, 1
asmh{nnb/a) J asmh{nnb/a) { j^, if n is odd. J

sinh(n7ra;/a) sm{nny/a)
n?,?) = ^^ E = 1,3,5,.
nsinh(n7r6/a) '

Problem 3.15
Same format as Ex. 3.5, only the boundary conditions are:

[) V = 0 when a; = 0, "j li) V = 0 when x = a,
iii) V = 0 when y = 0,
) F — 0 when y = a,
v) F = 0 when z = 0,
vi) V = Vo when 2 = a.

This time we want sinusoidal functions in x and y, exponential in z:

Xix) = ^sin(fca;) + 5cos(A;a;), Y{y) = Csin{ly) + Dcos{ly), Z{z) = Ee"^^"^^ + Ge-^/*'+^^
(i)^ B = 0; (ii)=> k = nn/a; (iii)^ D = 0; (iv)^ / = mn/a] (v)=^ E + G = 0. Therefore

Z{z) = 2Esmhi7ry/n^+m^z/a).

Putting this all together, and combining the constants, we have:

V{x,y,z) = V^ "S"^ C?,m sin(n7ra;/a) sin(m7rj//a) sinhGrvn^ +m'^z/a).
n=l m=l

It remains to evaluate the constants C?,m. by imposing boundary condition (vi):

Vo = ^2^2 r^?."'sinhGrvV+m^) sin(n7ra;/a)sm{mny/a).
According to Eqs. 3.50 and 3.51:

, V /'2^ } } f ^' if n or m is even, "j
Cn,m sinh (ny/ri^ + m^ 1 = I - ) Vq / sin{n7rx/a) sm{m7ry/a) dx dy = < 16Vb -r u i, jj f
^ '^ 0 0 ^ TT^nm' ^ ■ J

F(x,j/,2) = —TT- V-> 1sm(n7ra;/a)sm(m7rj//a) i——. ?
, 16Fo V- > w / ^ w / ,sinhGrVn2+m22/ ■■M.
^2 ?=i^5 . ,?=f^5_... ""^ / y V ?/ y sinh (ttVu^ + m^]

^ , ^ lrf^/9 ,3 Id^ 2? Id^ ., .2

= li [(^' - 1) (^^' -!)]=! f2x Ex2 _ 1) + (^2 _ 1) 10^]
5 3 3
= iEx3-x + 5x3-5x)-J(l0x3-6x)- -X-* - -X. 2 2

We need to show that P3(cos0) satisfies

Id/. ?dP -1A + 1)P, with / = 3,
where P3(cos0) = ^cosO E cos^ 0 -

— - - ^ 1 0 E cos2 0 - 3) + cos 0A0 cos ${- sin 0)] = -1 sin $ E cos^ 0-3+10 cos^ 6)
d^
de r- sin
2

= --sin0Ecos20-l).

fsin0-^J = -~[sin2 0Ecos2 0-l)] =-^ [2sin0cos0 Ecos2 0-l) + sin2 0(-lOcos0sin0)]
= -3sin0cos0[5cos20-l-5sin20] .

1 d ( . ?dP
sine d9 e^j = -3cos0[5cos2-l-5(l-cos2 0)] = -3cos0(lOcos2 0-6)
= -3-4-icos0Ecos20-3) =-/(/ + l)P3. qed

|Pi(x)P3(x)dx = |(x)^ Ex3 -3x) ^^ ^ 1 (x'-x')'_^ = i(l- 1 + 1 - 1) = 0. V
Problem 3.17 '^
(a) Inside: F(r,0) = JZ ^jr'Pj(cos0) (Eq. 3.66) where

Ai = ^^^|j^ /"v'o@)P,(cos0) sin0d0 (Eq. 3.69).
In this case Vb@) = Vb comes outside the integral, so

. _ B/ + l)Fo /Pj(cos0)sin


But Po(cos0) = 1, so the integral can be written

0
/Poicos9)Piicose)sin^d^ = I 2' Jf J f q } ^^^- ^•^^)-
Therefore
r 0, if / ^ 01
^~ Vo, if / = 0 / •
Plugging this into the general form:

y (r, e) = Ao r°Po(cos 6) = [^
The potential is constant throughout the sphere.

Outside: Vir,0) = ^ ~Pi{cose) (Eq. 3.72), where

Bi = ^^^-^R'+^ fvo{e)Piicose)smede (Eq. 3.73).
0

B1 + 1)
2

Therefore Vir,e) = Vo- (i.e. equals Vb at r = R, then falls off like -).
(i.e. equals
(b)

Yl Ajr'P, (cos e), for r < P (Eq. 3.78)
j=o

F(r,0) = ^-^P,(cos0), forr>P (Eq. 3.79)
1=0 '"
where
Bi = p2i+i^j (Eq. 3.81)
and

^' = -^r—^izi Icro{S)Pi{cose)smedd (Eq. 3.84)
0

' R<Jo
for r

V{t e) = PVqI >.| forr

Note: in terms of the total charge Q = AnR'âo,

v{r,e) =

Fo@) = A;cosC0) = k [4cosê-Zcose]=k[aPs(cos0) + /3Pi(cos0)].
(I know that any 3'''^ order polynomial can be expressed as a linear combination of the first four Legendre
polynomials; in this case, since the polynomial is odd, I only need Pj and P3.)

4cosê-Zcose = a - Ecos^0 - 3cos0) +/!3cos0 = -^cos^0+ f/3 - -a) cos0,
3 8
-3 = /3-â = '2'5 ' 125 _
"
Therefore

Now

Y^ Air'Pi (cos e), iorr<R (Eq. 3.66)
1=0
vir,e) =
^^P,(cos0), forr>P (Eq. 3.71)

Ai = ^^^^^ fvo{e)Pi{cose)smede (Eq. 3.69)
0

= ^^^J^ I 1 ^ / ^^ ^^°^ ^^^' ^^°^ 0)smedS-Z I Pi (cos 0)P (cos 0) sin 0 d0 I
k {21 +1) f 2 o_A_x 1 ^ 1 fR^ -^^ 1

' -3fc/5-R, if/==3l1 J, ^^ otherwise).
8A;/5P^ if / zero

V{r,e) = -||rPi(cos0) + ^r'p,{coê) -- 8(^)'p3(cos0)-3(^)Pi(cos0)

8 (■^)^ ^ [5cos3 0 - 3COS0] _ 3 f^Vj cosflj ^ F(r,0) = ^^ cosfl |4 (^)^ [Scos^ 0 - 3] - 3!


(for r < R). Meanwhile, Bi = AiR^^+^ (Eq. 3.81—this follows from the continuity of V at R). Therefore

^' = -3fciiv5, if; = l/ (==ero otherwise).

V(r,e) = -^lpi(cos0) + §^lp3(cos0) = ■ij)' 8 - P3(cos0)-3 - Pi(cos0)

Vir,e) = I (^)'cos0 L (^y [5cos2 0-3] -3!
(for r>R). Finally, using Eq. 3.83:

a{e) = eo X^BZ + l)A,iZ'-iPj(cos0) = eo [3AiPi + TylsiJ^Ps]

[-9Pi(cos0) + 56P3(cos0)]

^ [-9COS0 + y Ecos^ 0 - 3COS0)] = ^ cos0[-9 + 28 • Scos^ 0-28-3]
- cos 61 [140 cos^ 61-93].

Problem 3.19

Use Eq. 3.83: a{e) = eo ^BZ+l)^jiJ'~^P(cos 61). But Eq. 3.69 says: Ai = ?^^ f Voie)Pi{cos0) sin0A9.
1=0 0
Putting them together:

aie) = -^Yli2l + lfCiPiicose), with Ci = fvoie)Pi{cose)sinede. qed
1=0 0

Problem 3.20
Set y = 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:

Vii ,e) = -Eo(r-^)cose+-^^.
r^ J 47reo r

(a) V{r,e) = f;-|Lp,(cos0) (r > R), so F(r,0) = Y^-^m) = £ 4t = |- [V^^TW - r] .
1=0 1=0 1=0 °

Since r > iZ in this region, s/r'^ + i?2 = rx/l + (i?/rJ = r [iH- ^{R/rf - ^{R/rf + .. .1, so
'r'+i 2eo''[ 2r2 8 r" ^

Comparing like powers of r, I see that Bo = -;—, Bi =0, B2 = -7^—,.... Therefore
4eo loeo

[ !
aR""
•
V{r e) =

460
I -S-*X)S0) + ...
[ (for r>R).
=

4eor
-
-Kf)^ -,....],
Ccos2 0-

(b) Vir,e) -"^AyPiicose) (r < R). In the northern hemispere, 0 < 0 < 7r/2,
j=o

F(r,0) = f;^,r' = ^ [A/^^^TR^-rl .

Since r < i? in this region, /r^ + R^ = Ry/T+JrJW = J? [1 + {rlRf - ^{t/RY + ...]. Therefore

Comparing like powers: Aq = -—R, Ai = --—, A2 = -—=;,... , so
2eo 2eo 2eoii

V{r,e) = ^[iZ-rPi(cos0) + ^P2(cos0)+ ...], (for r < R, northern hemisphere).

oP [ 260 I
In the southern hemisphere we'll have to go for 0 = tt, using Pj(—1) = (-!)'•

y(r,^) =f:(-l)%r' = ^ [./^^TW-r] .


(I put an overbar on Ai to distinguish it from the northern Ai). The only difference is the sign of 'Ai:
'Ai = +(o-/2eo), lo = Ao, A^ = A-i. So:

vir,e)
^ 2^[^ + '--Pi(cos0) + ^r2P2(cos0) + ...], (for r < R, southern hemisphere).

= gh(^)-^-i(sr(^-^^-)-4
Problem 3.22

Y^Air'Piicosd), (r < R) (Eq. 3.78),
Vir,0) =

Yl -^-Pi(cos0), (r > R) (Eq. 3.7{ I. 1=0
where Bi = AjiZ^'+i (Eq. 3.81) and

Ai = -^^^^zT f(^oiO)Pi{cose)sinede (Eq. 3.84)
0

= p|_i o"o < Piicose)smede- Pi{cos0) sined9 (let a; = cos0)

— nPiix)dx-JPiix)dx.
2eoR^-

Now Pi{—x) = (-l)'Pj(a;), since Pi{x) is even, for even /, and odd, for odd /. Therefore

0 0 1

j Pi{x)dx = jPi{-x)d{-x) = {-!)' j Pi{x)dx,

if / is even

^' = 2iS^t^-(-i)']/'''^^^'^H T^l^d^)^^^ if f I is odd

So Ao = A2 = Ai = Ae = 0, and all we need are Ai, A3, and ^5.

r r x2p 1

0 0 "

/P3(.,.. = i/p.-3.,.= iD-3f^)[.lg-2)=-l.
0 0

/ P5 (x) dx = ^f F3x5 - 70x3 ^ 15^) ^^ ^ 1 /g3^ _ 70^ + l5^^ I
,2 2 27 16'
I- Therefore

-'^=SG)=-^-=5^D)'-^--3?(^)-
^-^^G)=--^^B)^--^1i^)--

g [f.(cos.) - {Ly P3(cos.) + 1 {Ly p,^(eos.) + . .] , (. < i ), ig Ufcos^) - (f) VsCcos^) + 1 (f) VsCcos^) + . .1 , (r > P).
Look for solutions of the form V{s,(p) = 5(s)$(</i):

1$A /^s^^ Ic^ -
s ds ds J s^ (i02
Multiply by s^ and divide by V = 5$:

5 ds V''dsJ'*'$d
Since the first term involves s only, and the second cp only, ea is a constant:
the second (p only, each

d f dS ^ 1 d^$ ^
d;(^d7J=^- ¥d0^ = ^- -^*^^


Now 6*2 must be negative (else we get exponentials for $, which do not return to their original value—as
geometrically they must— when 0 is increased by 27r).

C2 - -k^. Then —-^ - -k'^i =^ $ = Acosk(j) +Bsinkcp.

Moreover, since #(</> + 2n) = $(</>), k must be an integer: fc = 0,1,2,3,... (negative integers are just repeats,
but A: = 0 must be included, since $ = A (a constant) is OK).
s-r I s-T- 1 = k^S can be solved by 5 = s**, provided n is chosen right:
as as J

s-r- (sns"~^)as (s") = n^ss"'~^ = ri^s" — k^S ^n — ±k.
as
= ns-r-

Evidently the general solution is S{s) — Cs^ + Ds~^, unless A: = 0, in which case we have only one solution
to a second-order 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 differential equation for 5, and put in fc = 0:

So the second solution in this case is In s. [How about $? That too reduces to a single solution, ^ = A,'m the
case k = 0. What's the second solution here? Well, putting k = 0 into the $ equation:

-—T -0^ -rr - constant - B ^ ^ - BA + A.
d(j)^ acp

But a term of the form B(f> is unacceptable, since it does not return to its initial value when 0 is augmented
by 27r.] Conclusion: The general solution with cylindrical symmetry is

V{s, (j)) = oo + 60 In s + ^ [s* (oft cos kcj) + bk sin k(j)) + s * (cfc cos k(j) + dk sin A;^)].

Yes: the potential of a line charge goes like In s, which is included.
Problem 3.24
Picking F = 0 on the yz plane, with Eq in the x direction, we have (Eq. 3.74):

(i) ^^ = 0, when s = R,
(ii) V -> -Eqx = s cos (j>, for s'3> R.
Evidently a^ = bo — bk — dk — Q, and ak—Ck—G except for A; — 1:

y{s,4>) = (oi? + —)COS(f).
(i)=> ci = -oii?^; (ii)->- ai — -Eq. Therefore

V{s,<j>) = V{s,,f>)^-Eos
0''] COS(j).

R-"
ar = - o-^ =-eo-Eo f-^ - 1 ] cos0| = [ 2eoE'o cos0
Problem 3.25

Inside: F(s, 0) = Qq + y^s* (at cosfc0 + 6fc sin A;0). (In this region Ins and s~* are no good—they blow
k=l

up at s = 0.)

Outside: V{s, 0) = Oq + ^ -j^ (cft cos fc0 + dk sin fc0). (Here In s and s'' are no good at s -> oo).

(Eq. 2.3
a = -eo ' ' ds ^
Thus

a sin 50 = —eo Yj i — ^, (cfc cos fc0 + d^ sin fc0) — kR^ ' (ofc cos fc0 + bk sin fc0) >

Evidently Ofc = Cft = 0; 6fc = dfc = 0 except A; = 5; a = 5eo ( 'Ee^s + ^*h 1 Also, V is continuous at s = iJ:
1

V(s,0) _ gsin50 f 1 i^Vs^ for s < R, _
lOeo
s^/J?'*, ioTS>R. '
Problem 3.26
Monopole term:

Q= I pdT=zkR I -2(J?-2r)sin0 r"^sin6drded(f>.
K But the r integral is
R

[{R - 2r) dr = {Rr - r^) ^ = R^-R'^=0. So Q = 0.
0

IrcosOpdr-kR Mrcos^) —{R-2r)smer'^sm6drded>
i' But the 6 integral is

Lm^ecosede='^ =i(o-o) = o.
0

So the dipole contribution is likewise zero,
wise zero.
Quadrupole term:

/r^ (^cos'^e-^^pdT=^kR f fr''[Z CDs'" e-1) f ^ (i? - 2r) sin ^l r^sin^drd

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