Eletrodinamica griffiths resolução capitulo 2

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Chapter 2
Electrostatics
Problem 2.1
(a) I Zero.!
(b) IF =_4 1 q~,I wherer is the distancefromcenterto eachnumeral. F points towardthe missingq.11"1=0r
Explanation:by superposition,this is equivalentto (a), with an extra-q at 6 o'clock-sincetheforceof all
twelveis zero,thenetforceis thatof -q only.
(c) I Zero.I
(d)1-4 1 q~,I pointingtowardthemissingq. Samereasonas (b). Note,however,that if youexplained(b) as1r€0r
a cancellationin pairsof oppositecharges(1 o'clockagainst7 o'clock;2 against8,etc.),with oneunpairedq
doingthejob, thenyou'llneeda differentexplanationfor (d).
Problem 2.2
(a) "Horizontal"componentscancel.Net verticalfieldis: Ez = 4;<02~ cosO.
q I -q
Fromfar away,(z » d), thefieldgoeslikeE ~ 4;<0~z, which,asweshallsee,is thefieldof a.dipole.(If we
setd-+0,wegetE =0,asisappropriate;totheextentthatthisconfigurationlookslikeasinglepointcharge
fromfaraway,thenetchargeiszero,soE -+0.)
1 2qz A
Here 1-2=Z2 + (~)2; cosO=i, soI E =41r€0(Z2+ (~)2)3/2z.
Whenz» d you'resofar awayit just lookslikea singlecharge2q;thefield
shouldreduceto E =_4 1 ~ z. Andit aoes(justsetd-+0 intheformula).?r<oz
(b) This timethe "vertical"componentscancel,leaving
E =_4 1 2~sinOx,or?r<0'V
E 1 qd A
=41r€0(z2 + (~)2)3/2x.
E
x
E
x
22
23
Problem 2.3
Ez 1 rL), dx () ( 2 2 2411"fO JO ~ cos ; 1- =Z + x ; COS() =~)
/Ldq =)"dx Ex
---1-)..zrL 1 dx
411"fO JO (z2+x2)3/2
[ ]I
L
---1-)..Z 1 x - ---1-~~
411"fO Z2~ 0 - 411"fOz~'
1- rL )'dxsin() - 1-)..I xdx411"fOJo ~ - 411"fO (X2+z2)3/2
[ ]I
L
[ ]- ---1-).. - L- - - ---1-).. 1 - L-411"fO ~ 0 - 411"fO Z ~.x
--x
E 1)"
[( 1
Z
)
A
(
L
)
A
]
- -- - + x+ z
- 4m:0Z ..,jZ2+ L2 ..,jz2+ L2 . ,
Forz » L youexpectit to look likea pointcharge q =)"L:E -+_4 1 ¥-z. It checks,forwithz » L thex1I"fO Z
term-t 0,andthez term-+_4 1 ~kz.1I"fOz z
Problem 2.4
FromEx. 2.1,with L -+~andz -+vz2 + (~)2(distancefromcenterof edgeto F), fieldof oneedgeis:
El = ~ )..a .
41f1:0vz2 +a2vz2+a2 +a24 4 4
Thereare4 sides,andwewantverticalcomponentsonly,somultiplyby 4cos()= 4!i+?-:z2+ a4
E =~ 4)..az z.
47r£0(Z2+a42)VZ2 + ~2
Problem2.5
"Horizontal"componentscancel,leaving:E = 4;fO{I ~cos()}z.
Here,1-2= r2+ z2,cos()= ~(both constants),while Idl =27rr. So
E =~ )"(27rr)z A
47r£0(r2 +Z2)3/2z.
Problem2.6
Breakit intoringsof radiusr, andthicknessdr, anduseProb. 2.5to expressthefieldof eachring. Total
chargeof a ring is 0".27rr. dr =).. . 27rr,so ).. =O"dris the "line charge"of eachring.
1 (O"dr)27rrz 1 lR rEring=_4 3/2; Edisk=_4 27rO"z 3/2dr.7r£0(r2 +Z2) 7r£0 0 (r2 +Z2)
Edisk =_4 1 27rO"z[! - 1 ] z.7r£0 Z ..,jR2+ z2
24 CHAPTER 2. ELECTROSTATICS
ForR » z thesecondterm-+0,soEplane=_4
1 27Taz= I a z.11I"fO 21"0
1 1 ( R2 )-1/2 1 ( 1 R2 ) 1 1 1 R2 R2For z » R =- 1+~ ~ - 1- - ~ so[ ]~ - - - +- 73 =~'~z z z 2z' z z 2z 2z'
andE = ~ 21rR2".= ~g. whereQ =7TR2a. ,(411"fO~ 411"fOZ '
Problem 2.7 z
E is clearlyin thez direction.Fromthediagram,
dq = ada = a R2 sin 0 dOd<jJ,
1z-2= R2 + z2 - 2Rz cos 0,
cos'lj; = z-Rlz-cosO.
E - -2- / a R2 sin 0 dOd<jJ(z - R cos0)z - 47T€O (R2 +Z2 - 2Rzcos0)3/2 .
- 1 2
1
11" (z-RcosO)sinO _.X - . .
{
O=O=>u=+l
}- -(27TR a) (2 2 0)3/2dO. Let u - cosO,du - -smOdO, O
.
1
.
47T€O 0 R +z - 2Rzcos =7T=> u =-
= -2-(27TR2a)/1 (2 z2- R~ )3/2du. Integralcanbedoneby partialfractions-or look it up.47T€o -1 R + z - 2 zu
1 '
- -2- 27TR2a
[
.2.- zu - R
]
- -2- 27TR2a
{
(z - R) - (-z - R)
}- 47T€O( ) z2~R2+Z2- 2Rzu -1 - 47T€OZ2 Iz- RI Iz+RI .
J d<jJ=27T.
y
So
Forz >R (outsidethesphere),Ez = 4:f ~ =_4
1 :\, so I E =_4
1 q
2
z.1.. 0 z 1I"fOz 7T€O Z
For z <R (inside),Ez=0,soI E =0.1
Problem 2.8
Accordingto Frob. 2.7,all shellsinteriorto thepoint (Le. at smallerr) contributeasthoughtheircharge
wereconcentratedat thecenter,whileall exteriorshellscontributenothing.Therefore:
E( ) - 1 Qint-r ---r,
47T€Or2
whereQintis thetotal chargeinteriorto thepoint. Outsidethesphere,all thechargeis interior,so
E=-2-Q-
47T€Or2 r.
Insidethesphere,only that fractionof thetotalwhichis interiorto thepointcounts:
!7Tr3 r3 1 r3 1
Qint =: R3Q =R3Q, so E =_4 R3Q2f =3"7T 7T€O r
Problem 2.9
(a) p =1"0V. E =€O~ tr(r2.kr3)=€o~k(5r4)=15€okr2.1
25
(b) By Gauss'slaw:Qenc=€ofE. da=€o(kR3)(41rR2)= I 41r€okR5. I
By direct integration: Qenc =J pdT =JoR(5€okr2)(41rr2dr) =201r€okJoR r4dr = 41r€okR5.,(
Problem2.10
Thinkof thiscubeasoneof 8 surroundingthecharge.Eachof the24squareswhichmakeup thesurface
ofthislargercubegetsthesamefluxaseveryotherone,so:
/E. da=2~ / E. da.
one whole
face large
cube
Thelatteris 1..q,by Gauss'slaw. Therefore
I /E. da=-.!L.<0 24€o
one
face
Problem 2.11
~GaUSsian. surface:Inside:f E. da=E(41rr2) =-!;Qenc=0 =>IE =O.I}G R2 (As in Frob. 2.7.)r - Gaussiansurface:Outside:E(41rr2)=1..(a41rR2)=> E =~ 2 r.<0 €or,-
Problem 2.12
Gaussiansurface .L'E. da =E. 41rr2= 1..Qenc= 1..!31rr3p. Soj <0 <0
JE=~prr.1
SinceQtot= ~1rR2p,E = 4';<0j&r (asin Frob. 2.8).
Problem2.13
Gaussia
~/ n ,unace" '
l
.L'E. da =E. 21rs.l = 1..Qenc= 1..>"l.Soj <0 <0
IE =~Sl (sameasEx. 2.1).21r€os
Problem 2.14
Gaussiansurface .L'E .da=E .41rr2=1..Qenc=1..J p dT =1..}(kf)(f2 sin Bdr dB d4»j <0 <0 <0
= 1..k 41rrr r3dr = 411'kr4 = 1I'kr4.
<0 Jo <0 4 <0
. 'E = ~1rkr2r.
41r€o
26 CHAPTER 2. ELECTROSTATICS
Problem 2.15
(i) Qenc= 0, soIE = 0.1
(H)f E. da = E(47IT2) = 1-Qenc= 1-f pdr=1-f Af2 sinOdfdOdphi<0 <0 <0 r
= M "df ~ "'(r - oj :.1E ~ ~(
r - a
)'.1<0 Ja <0 fO r2
(iii) E(411"r2)=~f:df = 4:rok(b- a), so
IE=H~)r.1
lEI
b ra
Problem 2.16
(i)
@
~
a fE.da-E 2- . 1I"S'[- 1Q
EE]
- psA - to enc= f-p7rs2[.- -8 0'
2fO .
)- Gaussiansurface
f G"""'an ,."face(H) f E. da= E. 211"s,[= 1-Qenc= 1-p7ra2[;<0 <02E = pa s.
2fOS
(Hi)
lEI
- Gaussiansurface
f E. da= E. 211"s,[= 1-Qenc= 0;<0
IE=O.I
\
~'-'
a 6 S
Problem 2.17
On the x z planeE =0 by symmetry.Setup a Gaussian"pillbox"with onefacein this planeandthe
otheraty.
Gaussianpillbox f E. da=E. A =1-Qenc= 1-Ayp;<0 <0
IE = PYY I (for Iyl < d).fO
27
Qenc= 1...Adp=}IE = P dy I (for y >d).<0 100
E
I!!!
'0 t""."'
-d
a y
Problem2.18
FromProb.2.12,thefieldinsidethepositivesphereis E+ = ~r+, wherer+isthevectorfromthepositive
centertothepointin question.Likewise,thefieldof thenegativesphereis - ...L3 r -. So the totalfieldis<0
P
E =-(r+ - r_)
3100
r-
~-
+But(seediagram)r+ - L =d. SoI E =~d'i
Problem2.19
VxE =4:100Vx J ~PdT =4:100 J [vx (~~)]pdT
=0 (sinceVx (~) =0,fromProb. 1.62).
(sincep dependson r', not r)
Problem2.20
x y Z
(1)VxE1 =k Itx ty tz1=k [x(O- 2y)+:9(0- 3z)+ z(O- x)] f:.0,
xy 2yz 3zx
soEl isan impossibleelectrostaticfield.
x y z
(2)VxE2=kItx ty tz1=k[x(2z- 2z)+y(O- 0)+z(2y- 2y)]=0,
y2 2xy+ z2 2yz
soE2is apossibleelectrostaticfield.
Let'sgobytheindicatedpath:
z
E. dl =(y2dx+ (2xy+ z2)dy+2yzdz)k
StepI: y =z =0;dy=dz=O.E .dl =ky2dx =O.
Step II: x =Xo,Y : 0 -+Yo,z = O.dx = dz = O.
E .dl=k(2xy+z2)dy=2kxoYdy.
fII E. dl =2kxofrioy dy =kxoY5.
Step III: x =Xo,Y =Yo,z :0-+Zo;dx=dy=O.
(xo, Yo, ZO)
III
y
I
x
II
28 CHAPTER 2. ELECTROSTATICS
E. dl = 2kyzdz= 2kYozdz.
JIll E. dl= 2YokJozoz dz = kyoZ5.
(xo.Yo.zo) I IV(XO,Yo,ZO)= - J E .dl = -k(xoY5 +Yoz5), or V(x, y,z) = -k(xy2 +yz2).0
Check: -vv=k[lz(xy2+yz2) x+/v(xy2+yz2)y+/;(xy2+yz2) Z]=k[y2x+(2xY+Z2)y+2yzi]=E. ,(
Problem 2.21
{
Outsidethe sphere(r > R) : E = _4 1 :;\r.1I"fOr
V(r) = - Je: E .dl.
Insidethesphere(r < R) : E = 4;fO-J/:srr.
Soforr>R: V(r)=-Je: (4;f ~)df= 4;f q(to)lr =14q !,I0 0 00 11"100r
and for r <R: V(r) = - J:: (~~) df- J~ (4;fO-J/:sf)df = -dto[:k-:h (r22R2)]
q 1 (
r2
)=1411"1002R 3 - R2 .
Wh . R VV - -L- 8 (1) A - -L- 1 A E - VV - -L- 1 A ,(en r >, - 411"fO8r ;:: r - - 411"fOf=2"r,so - - - 411"fOf=2"r.
Whenr <R, VV =-dto2~tr (3- ~) r = -dto2k(-~) r = --dtobrj soE = -VV = 4;fO-J/:srr.,(
Problem 2.22
E =4;fO2;8 (Prob. 2.13). In this case we cannot set the referencepoint at 00, since the charge itself
extends to 00. Let's set it at s = a. Then
V(s) = - r (~2:)dB= --41 2,Xln(~).. a 1I"fO 11"100 a
(In this formit is clearwhya = 00wouldbenogood-likewisetheother"natural"point,a = 0.)
VV = --L2'xE.. (In (l. )) § = --L2,Xi§ = -E.,(411"fO 88 a 411"fO 8
Problem 2.23
V(O)= -J::OE.dl = -J:O(-!o(b;;a»)dr-Jba(-!o(r;;a»)dr-J~(O)dr= -!o(b~a)--!o(ln(%)+a(~-i))\
= li.{1-!!-ln (!!) -1+!! }=I~ln (~).fO b b b 100 a
Problem 2.24
UsingEq. 2.22andthefieldsfromProb.2.16:
V(b) - V(O) = - fb E. dl = - fa E. dl- J. bE. dl = _L faSds - e£..J. b idsJo Jo a 2fOJo 2fO a s
= - (~)8;I: +~Insl:=1-~:: (1+2ln(~) ) .1
Problem 2.25
(a)Iv = ~ 2q .
411"100Jz2 + (~)2
29
(b) V-I f L Adx A I ( Y 2 2) I
L
- ~ -L ~ =41UOn x + z +x -L
,\
[
£ +YZ2 +£2
]
I
=I-In = ~ln L+~
41r€0 -£ + YZ2 + £2 2"fO ( z ).
(c)V =4:fOJoR~;~:~~= 4:f021r<7(vr2 +z2)1:=I ~(JR2 +Z2- z) .1
x
In eachcase,by symmetry~~= ~~= O. :. E = _8VZ
A
8z .
(a)E =-~2q (-!) ( 2z 2t2 Z=1
-
4 1 2qz 2 3/2ZI(agreeswith Prob. 2.2a).
"fO z2+(~) 1r€0(z2+(~))
(b) E - A { 1 1 1 2 1 1 1 2 }
A
- - 41rfo (L+~) 2 v'z2+£2 z - (-L+~) 2~ z Z
- A z
{ -L+~-L-~ }A -12£'\ 1 A I(
. h E 2 1)- - ~ v' 2+L2 ( +L2 )-L Z- _4 Z agreesWIt x..."fO Z Z 1r€0ZY Z2 +£2
(c)E =--2(7 {
_
21 v'Ri 22z- I }z=12
a
[
1 - z
]
z I (agreeswithProb.2.6).
fO +z €o VR2 +Z2
Iftheright-handchargein(a)is-q, thenI V =0I,which,naively,suggestsE =- VV =0,incontradiction
withtheanswerto Prob. 2.2b. The point is that weonly knowV on thez axis,and fromthis wecannot
hopetocomputeEx = - ~~or Ey = - ~~.That wasOK in part (a), becauseweknewfromsymmetrythat
Ex =Ey =O. But now E points in the x direction, so knowing V on the z axis is insufficient to determineE.
Problem2.26
V(a) =~ rv'2h(
a21rr
)ch=21ra~(V2h) =ah.41r€0Jo 1- 41r€0V2 2€0
(where r =1-/V2)
1 l
v'2h
(
a21rr
) VV(b) =_4 -=.-- ch, whereii = h2+1-2- V2h1-.1r€0 0 1-
21ra 1 rv'2h 1-
=41r€0V2 Jo Jh2 +1-2- V2h1-ch
=~
[
Vh2+1-2- V2h1-+-
.
h In(2Vh2+1-2- V2h1-+21-- V2h)
]
v'2h
2V2€0 V2 0
= ~
[
h + ':nIn(2h+2V2h-V2h)- h- ':nIn(2h-V2h)
]
= ~ ':n[In(2h+V2h)-In(2h-V2h) ]2v2€0 v2 v2 2v2€0v2
=uk In
(
2+V2
)
=ah In
(
(2+V2)2
)
=ah In(1+ V2).
4100 2- V2 4€0 2 2€0
a
:.1V(a)- V(b)=;€~[1-ln(1 +V2)].
30 CHAPTER 2. ELECTROSTATICS
Problem 2.27
Cut the cylinderinto slabs,as shownin the figure,and
useresultof Prob. 2.25c,with z -t x andC1-t Pdx:
z+L/2
V =!to J (VR2+X2- x)dx
z-L/2
=...£ 1. [XVR2 +x2 +R2ln(x + VR2 +X2) - X2]IZ+L/22<02 z-L/2
=I 4fo{(z+th/ R2+(z+t)2-(z- t h/R2+(z-t)2+R21n[::;::::~:::~2] -2ZL}.
( ( L )
2
(
L
)2 2 L2 2 L2Note: - z + 2" + z - 2" =-z - zL - "'4+z - zL + "'4=-2zL.)
1. 8 'r
x --do:
8V z
{/
H_H_
(
- L
)2 (z +L )
2
/
- n__
(
-- L
)
2
(z L )
2
E = -vv =-z- =--E... R2+ z +- + 2" R2+ z - - - - 2"
8z 4€0 2 VR2+(Z+t)2 2 VR2+(z-t)2
+L L1+ Z"2 1+ Z-"2
2
[
y'R2+(Z+t)2 y'R2+(Z-t)2
] }
+R - - 2L
z +t+VR2 +(z+t) 2 Z- t +VR2 +(z- t) 2
, ~ '
1 1
VR2 +(Z+t)2 - VR2 +(Z- t)2
E =--~ {2VR2+(Z+~)' -- 2VRZ +(Z--~)' -2£}
~,;,O [L-JR'+(Z+~)' +JR'+(Z-~)']Z
Problem 2.28
OrientaxessoP is on z axis. z
V =~ J B..dT.41T<0 1- {
Herep is constant,dT= r2sinedrded<jJ,
1- =vz2 + r2 - 2rzcose. y
V - -.1!.-J r2 sin IJ dr dIJdiP . f21T d'!" - 27r- 41T<0 vz2+r2-2rzcoslJ' JO 'f' - . x
f1T sin IJ de=..l.(vr2 + Z2- 2rzcose) 1 1T= ..l.(vr2+Z2+2rz- vr2+Z2- 2rz)Jo vz2+r2-2rzcoslJ rz 0 rz
{
2/z, ifr < z,
}= rlz(r+z -IT - zl) = 2/r, if r >z.
31
. V =--1!.-.211'. 2{fz lr2dr +fR lr2dr }
=.P...{l z3 +R2-z2 }=L (R2 - z2 ).. 411"<0 z r <0 z 3 2 2<0 3'0 z
ButP=Fw' soV(z) =~~ (R2 - ~)=-d!oR(3- ~); I V(r) =~ (3- ~)..I ,f
Problem 2.29
\72V=4;<0'V2f(*)dr = 4;<0f p(r')('V2k)dr (sincep is a functionair', not r)
= -L- fp(r')(-4mP(r - r')]dr =-1...p(r). ,f411"<0 <0 .
Problem 2.30.
(a)Ex.2.4:Eabove=-2u ft;Ebelow=--2u ft (ft alwayspointingup);Eabove- Ebelow= .£.ft.,f<0 <0 <0
Ex.2.5:At eachsurface,E = 0 onesideandE = .£.otherside,so6.E = .£..,f<0 <0
Prob. 2.11: Eout = ~f = .£.f;Ein=0; so6.E= .£.f.,f<or <0 <0
'IF" 5R Outside: §E. da=E(211's)l = 1...Qenc= 1...(211'R)l=}E = .£.B.§= .£.§ (at surface).: : <0 <0 <0s <0
'. ../ --' Inside: Qenc =0,soE =O.:. 6.E=.£.§. ,f.~ <0
l
(b) 0
(c) Vout=R2u= Ru (at surface); \In = Ru ; soVout=\In- ,f<or <0 <0
~ = _R2u = _.£. (at surface)- 8V;n=O' so ~ - 8V;n =_.£..,far ~ <0 ' 8r ' 8r 8r <0
Problem 2.31
( ) V - -L".!Ji.. - -L {=.!1. +-L + =.!1.} - -L- (-2 +...1-)a - 411"<0L" rij - 411"<0a V2a. a - 411"<oa V2 -
q2 (
1
).'.W4=qV =1-4 -2 + M -1I'foa V 2
(b)WI =0,W2=4;<0(=f); W3=4;<0(:;a - ~ ;W4=(see(a)).
1 2{ II }
1 2q2(
1
)Wtot = 411"<07 -1 +'Vi- 1- 2+ V2 = 411'fO-;;: - 2+V2 -
(1). (4).- +
+ -. .
(2) (3)
32 CHAPTER 2. ELECTROSTATICS
Problem 2.32
(a) W = !J pVdT. FromFrob.2.21(orFrob.2.28):V =f<o(R2 - r;)=4;€O m(3- ~)
1 1 q {R(
r2
) 2 qpW = 2P411"1002R Jo 3 - R2 411"rdr =4€oR
=qpR2= qR2~= ~ (~q2)5100 5100~11"R3 411"1005 R .
[
r3 1 r5
] I
R qp (
R3
)3"3 - R2"5 0 =4€oR R3 - 5
(b) W = !f J E2dT. Outside(r >R) E =4;€O ~rj Inside(r <R) E = 4;€OWrr.
. - 100 1 2
{
roo 1 2 {R (
r
)2 2 }.. W - 2"(411"€O)2qJR r4(r 411"dr)+ Jo R3 (411"rdr)
1 q2
{(
1
)1
00 1
(
r5
)I
R
}
1 q2 (
1 1
) 13q2=411"100"2 -;: R + R6 "5 0 =411"100"2R + 5R =411"100:5R"/
(c) W =!f { is VE. da+Jv E2dT}, whereV is largeenoughto encloseall the charge,but otherwise
arbitrary.Let'susea sphereof radiusa >R. HereV =4;€O~'
W =100
{/(
-
4
1 ~)(-4 1 ~)r2sinOdOd4>+ {RE2dT+ {a(-41 ~)2 (411"r2dr) }2 11"100r 11"100r Jo J R 11"100r=a
100
{
q2 14 q2 411" 1 4 2 (
1
)l
a
}=2" (411"100)2~ 11"+ (411"100)25R + (411"100)21I"q . -;: R
1 q2
{
1 1 1 1
}
1 3q2
=411"100"2~+ 5R - ~+ R =411"100:5R ../
As a --t 00, the contributionfrom the surfaceintegral (4;€O~) goesto zero,while the volumeintegral
(~f.(~ - 1)) picksup theslack.
Problem 2.33
0~~
dW =dqV=dq(-4
1
)~, (q=chargeonsphereof radiusr).11"100r
4 r3
q=311"r3p = qR3 (q = totalchargeonsphere).
- 2 411"r2 3q 2
dq =411"r dr p = 43qdr =R3r dr.311"R
1
(qr3)
1
(3q 2 ) 1 3q2 4dW =411"100R3 ;: R3r dr =411"100R6 r dr
1 3q2 {R 1 3q2R5 1 (3q2)W =411"100R6 Jo r4dr=411"100R65 =411"100:5R ../
33
Problem2.34
(a)W =~f E2 dr. E =4;fO!-r (a <r <b), zero elsewhere.
W - ~ (~ )
2
J.
b
(
1
)24 2d - L J. b 1 -IL (~- ~)- 2 41TfO a;:2 1rr r - 81TfOa;:2 - 81rfO a b .
2 2
(b) WI =8;fO7' W2= 8;fOT, El = 4;fO!-rr (r >a), E2= 4;fO::;ir (r >b). So
E1.E2=(4;fO)2~, (r>b),andhencef E1.E2dr=- (4;fO)2q2go ;!r471T2dr=- 4::0b'
2
Wtot=WI +W2+100fEI .E2 dr =~q2 (~+!-~) =~ (~-!).,(
Problem2.35
I q -q q I(a) aR =4;R2;aa=~; ab=4;b2'
(b)V(O)=- f::O E. dl=- f::O(4;fO!-r)dr- fba(O)dr- faR(4;fO!-r)dr - f~(O)dr=I ~(t+~-~).1
(c)I ab ~ 0I (thecharge"drainsoff");V(O)= - f::O(O)dr- faR(~!-r )dr - f~(O)dr= I ~(~- ~).1
Problem2.36
( )
I qa II qb I qa + qba aa =-~; ab=-4;b2;aR = 41rR2.
(b)I 1 qa+qbA I fEout=-4 ~ r, wherer = vector romcenterof largesphere.1rfO r
1 qa A
-ra,
(c)lEa = 41rfOr~
Eb=_4 1 q~rb,Iwherera (rb) is thevectorfromcenterof cavitya (b).1rfOrb
(d) IZera.I
(e)aRchanges(butnotaa or ab);Eoutsidechanges(butnot Ea or Eb); forceonqaandqbstill zero.
Problem2.37
Betweentheplates,E =0;outsidetheplatesE =a/100=Q/foA. So
p =100E2 =100~ -1Q2l
2 2 f5A2 -~
Problem2.38
Inside,E = 0;outside,E = -41 ~r; so1TfOr
Eave= ~4;fO~r; fz =a(Eave)z;a = 41T~2'
Fz= J fzda = f( 41T~2) ~(4;fO~ ) cas() R2 sin () d() dq;
cME
- 1 (-9-)2 f1T/2. () ()
- 1
(9...)
2
(
1 . 2
()) 1
1T/2 - 1 (9... )2 _
I
Q2
- 2<641TR 21rJo sm()cas d - 1TfO4R 2"sm 0 - 211:fO4R- 321rR2fO.
34 CHAPTER 2. ELECTROSTATICS
Problem 2.39
Saythechargeon theinnercylinderis Q, for a lengthL. The fieldis givenby Gauss'slaw:
IE. da=E .21TS.L = .!..Qenc=.!..Q =>E =-9-2 L _81 s. PotentialdifferencebetweentheCylindersisEO EO "'EO
lb Q lb 1 Q (b)V(b) - V(a) =- E.d1=-- -dB=--In - .a 21T€oLa S 21T€oL a
As setup here,a is at thehigherpotential,soV = V(a) - V(b) =~ In(~).
C =~==~...t~)'so capacitanceper unit lengthis I 21T(~O) .1n " In Ii
Problem 2.40
(a) W =(force)x(distance)=(pressure)x(area)x(distance)=I ~E2A€.I
(b)W =(energyperunitvolume)x(decreasein volume)=(€O~2)(A€).
energylost is equalto theworkdone.
Problem 2.41
Sameas (a), confirmingthat the
FromProb. 2.4,thefieldat heightz abovethecenterof a squareloop(sidea) is
1 4Aaz Z.
E-- J 2- 41T€o(z2+ ~2) Z2 +a2 -~- da2 -~- da2
HereA -+ad2a(seefigure),andweintegrateovera from0 to ii:
E =~2az la alia
41T€O 0 (z2+a;)/z2 +a22
. -2
[ (
.
)]
0.2/4
1 a /4 du az 2 v2u +Z2=-4az =- -tan-l
41T€O 1 (u+z2)v2u+z2 1T€O Z Z 0
2
{ (
J a2 +z2
) }=1T: tan-l 2Z - tan-l(l) ;
2 -a+aa-a
. Let u ="4'so a da =2 duo
- a--..
2a
[
-1g 2 1T
]
AE =- tan 1+- - - Z.
1T€O 2z2 4
a -+00(infinitePlane): E =k [tan-l (oo) - 1!:.]=k (1!:.- 1!:.)=...fL.,f"'EO 4 "'EO 2 4 2EO
z » a (pointcharge):Let f(x) =tan-l vI +x- 7'andexpandasa Taylorseries:
1
f(x) =f(O) +x!,(O)+"2x21"(0)+...
35
f( ) -1 (1)
1< 1< 1<- O. f '( ) - 1 1 1- - 1 SO 1'(0) =1. SOHere 0 =tan - 4"=4"- 4"-, x - 1+(1+x)2v'I+X- 2(2+x).,!1+x' 4'
1
f(x) = -x +( )x2+( )x3+...4
Th (
. a2
I)E
20" (1 a2 ) 1 O"a2 1 q /us SinceW =x« , ~ 1rlo4W = 41<EOzr = 41<EO~'v
Problem2.42
- V E -
{
I 8
(
2A
)
1 8
(
BSinocos</»
}p - fO . - fO - - r - +--r28r r r sin08</> r
[
1 1 B sin0 .
] I fO I
=fO 2A + :- 0-( - sm</»=2(A - B sin</».r r sm r r
Problem2.43
FromProb.2.12,thefieldinsidea uniformlychargedsphereis: E =4;EO~r. Sotheforceperunit volume
isf=pE = (~~3)(41<~R3)r= ~(41<~3)2r,andtheforcein thez directionon dr is:
3
( Q )
2
dFz = fz dr = fO 471-R3 r casO(r2sin0drdOd</».
Thetotalforceonthe "northern"hemisphereis:
f
3
( Q )
2
1
R
1
1f/2
1
21f
Fz = fzdr = fO 471'R30 r3dr 0 cas0sin0dO 0 d</>
3
( Q )
2
(
R4
)(
Sin2°
1
1</2
)
13Q2
= fO 471'R3 "4 ~ 0 (271')=1~OR2'
Problem2.44
1
f
a i a
f 1 a 2 aRVcenter= 4- -da=-4 R da=4- R (271'R)=-27I'fO 1- 7I'fO 7I'fO fO
v. - 1 f
a
d
' h
{
da=271'R2sinOdO,
pole - - - a, Wit471'fO1- 1-2= R2+R2 - 2R2cosO= 2R2(1- cosO).
1 a(271'R2)
1
1f/2 sinOdO aR .
1
1f/2
= - =- (2Vl - cas0)
471'fORV2 0 VI - cas0 2V2fO 0
aR aR aR In
:::: M (1- 0)= In' :. Vpole-Vcenter= -2 (v2 - 1).v 2fO V 2fO fO
Problem 2.45
First let's determinethe electric field inside and outside the sphere,using Gauss's law:
fO f E .da =f0471'r2E =Q enc= f P dr =pk1')1'2sin0d1'dOd</>= 471'k lr 1'3d1'= { : ~~4
(r <R),
(r >R).
36 CHAPTER2. ELECTROSTATICS
So E =4~0r2 f (r <R); E = 4~~:2f (r >R).
MethodI:
€Qj €Q {R (
kr2
)2 2 €Q {'X>(
kR4
)
2
W ="2 E2dr (Eq. 2A5)="2 lQ 4€Q 47rrdr +"2 1R 4€Qr2 47rr2dr
€
(
k
)
2
{l
R
1
00 1
}
7rk2
{
R7
(
1
)1
00
}
7rk2
(
R7
)=47r"'£- r6dr+R8 -dr =- -+R8 -- =- -+R72 4€Q Q R r2 8€Q 7 r R 8€Q 7
=l7rk2 R7 .17€Q
Method II:
Forr <R,
w =~j pV dr (Eq. 2A3).
l
r
l
R
(
kR4
) 1
r
(
kr2
)
k
{ (
1
)I
R r3
1
r
}
V(r)=- E.d1=- --"2 dr- - dr=-- R4 -- +-
00 00 4€Qr R 4€Q 4€Q r 00 3 R
=- ~ (- R3 +r3 - R3)=~ (R3 - r3).4€Q 3 3 3€0 4
l
1
R
[
k
(
r3
)]
27rk2
1
R
(
1
):. W =- (kr) - R3- - 47rr2dr=- R3r3 - -r6 dr2 0 3€0 4 3€0 0 4
=27rk2
{
R3 R4 _!R7
}
=7rk2R7 (~)=7rk2R7 . .(3€0 4 4 7 2. 3€0 7 7€0
Problem 2.46
~
(
-Ar
) {(
' ) -Ar -Ar
} I
A
E VV - A v e A - A r -A e - e A - A -Ar(1 ' )
r=- - - - - r - - r - e +Ar -.
~ r ~ ~
p =€o V. E =€oA{e-Ar(1+Ar)V. (~) +~. V (e-Ar(1+Ar))}. But V. (~) = 4m53(r)(Eq.1.99),and
e-Ar(1+Ar)(P(r)=83(r)(Eq.1.88).Meanwhile,
V (e-Ar(1+Ar)) =ftr (e-Ar(I+Ar)) =f{-Ae-Ar(I+Ar)+e-ArA} =f(-A2re-Ar).
So~. V (e-Ar(1+Ar)) = -Ar2e-Ar,andIp =€oA[47r83(r)- ~2e-Ar}.
Q= jPdr = €OA{47r/83(r)dr-A2 je~Ar 47rr2dr}=€OA(47r-A247r100re-Ardr).
But It re-Ardr =~, so Q =47r€oA(1- ~) =Izero. I
Problem 2.47
(a) Potentialof +A is V+= - 2:'0In(~), wheres+isdistancefromA+(Prob.2.22).
Potentialof -A is V- =+-2 A In (!=.), wheres- isdistancefromA-.?TfO a
37
z
(X,Y,Z).'.TotallV= ~ln (
s_
).21T1:0 s+
Nows+=yI(y- a)2+Z2,ands- = yI(y+a)2+z2,so
Vexy z)=~ In(y'(y+a)2+z2)= ~ In [ (y +a)2+z2]" 21T<0 y'(y-a)2+z2 471"Eo (y - a)2+ z2 . -). ).
y
(b) Equipotentialsaregivenby (y+a)2+z2= e(41T<0Va/A)= k = constantThat is'(y-a)2+z2 .,
y2+2ay+a2+ Z2=k(y2- 2ay+a2+ Z2)=>y2(k- 1)+z2(k- 1)+a2(k- 1)- 2ay(k+1)=0,or
y2+Z2+a2- 2ay(fI:t) =O.The equationfor a circle,withcenterat (Yo,0) andradiusR, is
(y- yo?+z2=R2,ory2+z2+ (Y5- R2)- 2yyo=O.
Evidentlytheequipotentialsarecircles,withYo=a (fI:t) and
a2- Y2- R2=>R2 - Y2- a2- a2(!:tl )2 - a2- a2 (k2+2kH-k2+2k-l) - a2 4k Or- 0 - 0 - k-l - (k-l)2 - (k-l)2,
R = 2av'k. or in termsof Vr.Ik-ll' , o.
- e41T<OVO/A+ 1 - e21T<OVO/A+ e-21T<OVO/A-
I (
271"EOVO )Yo- a - a / / - acoth - .e41T<0VolA-I e21T<0VoA - e-211"<0VoA ).
e211"<OVO/A 2 a
I (271"EOVO )R =2a =a = =acsch - .e41T<OVO/A-1 (e211"<OVO/A- e-21T<OVO/A)sinhe1l"',Xvo) ).
z
y
Problem2.48
d2V 1
(a) V'2V=_L (Eq.2.24),so
I
-
d 2 =-'--po<0 X EO
()
-, , I-tqv Ib qV - 2mv -+ v - --:;;;.
(c)dq=Apdx j ¥i =ap~~=I Apv =I I (constant).(Note:p,hencealsoI, is negative.)
38 CHAPTER 2. ELECTROSTATICS
(d) d2V =-.1..p =-.1..1- = LV m :::} I tPV ={3V-1/21where{3= L Im..-;rxr <0 <0Av <oA 2qV dx2 ' <oAV2q
(Note: [ is negative,so{3is positive;q is positive.)
(e) Multiply by V' =~~:
v' dV'={3V-1/2dV :::}IV'dV'={3IV-1/2 dV :::} ~V'2=2{3V1/2+constant.~ ~ 2
But V(O)=V'(O)=0 (cathodeis at potentialzero,andfieldat cathodeis zero),sotheconstantis zero,and
V'2=4{3V1/2:::}dV =2VP V1/4:::}V-1/4dV=2vp~'dx '
IV-1/4 dV =2VP Idx :::} ~ V3/4 =2VP x + constant.
But V(oy= 0,sothis constantis alsozero.
3
(
3
)
4/3
(
9
)
2/3
(
81I2
)
1/3
V3/4 = 2VPx, soVex)= 2VP X4/3,orVex)="4{3 x4/3= 32€~A~q x4/3.
(
X
)
4/3
Intermsof Vo (insteadof I): I Vex)=Vo d I (seegraph).
Withoutspace-charge,V wouldincreaselinearly:Vex)=Vo(~). v.
d2V 1 4 1 -2/3_
I
4€oVo
p =-€o dX2 =-€o Vod4/33. 3x - - 9(d2x)2/3.
f2q
I (
X
)
2/3
V =V m:vv= v2qVo/md .
(
2
)
1/3 4 2 2
(f) V(d) =Vr = 811m d4/3 :::}11,3= 81md [2. [2 = 32<oA4qV,3.0 32<5A2q 0 ~, 81md 0 ,
[ - 4V2<oAVQTT3/2 - KTT3/2 h
I
K - 4€oAf2q
- 9v'1nd2vo - vo ,were - 9d2Vm:'
d x
Problem 2.49
1 I
A
(a) IE=~ fJ'I-.(1+~)e-1J-/Adr.47f€o 1-2 A
(b) I Yes.!The fieldofa pointchargeat theoriginis radialandsymmetric,soV XE = 0,andhencethisis also
true (bysuperposition)for any collectionof charges.
ir 1 ir 1 rV =- E .dl =--q "2(1+- )e-r/Adr00 47f€o 00r A
1
{OO 1 r q {{
OO 1 1
1
00 1
}
=-q - (1+- )e-r/Adr = - -e-r/Adr +- .' -e-r/)..dr .'47f€o r r2 A 47f€o r r2 A r r .
(c)
39
I 1 /A -r/). 1 I -r/).Now ~e-r dr=-~ - X ~dr t-- exactlyrighttokill thelastterm.Therefore
q
{
e-r/A
l
oo
} I
q e-r/A
V(r)=- -- ---
41rfO r r - 41rfO r .
(d) 1E .da =~q~ (1+~)e-R/A# I)? =!J...(1+ R)e-R/\Js MfO l)!l-,X fO'x
! q lR e-r/A 2 q lR - q [ e-r/\ ( r )]
R
Vdr=- -r #dr=- re r/Adr=- - ---1
V MfO 0 r fO 0 fO (1/ ,X)2,X 0
=,X2f~{-e-R/A (1 +~)+I}.
:. 1E. da+~rV dr =!J...
{(I + R)e-R/A- (1+ R)e-R/A +I } =!J....Js ,X2Jv fO'x ,X fO qed
(e)Doestheresultin (d)holdfora nonsphericalsurface?Supposewe
makea "dent"in the sphere-pushinga patch(areaR2sinBdBd</»
fromradiusR out to radiusS (areaS2sinBdBd</».
D.fE. da=4:fO {;2 (1 +~) e-S/A(S2sinBdBd</»- ~2(1 + ~) e-R/A(R2sinedBd</»}
=4:fO [(1 +~)e-s/A - (1 + ~) e-R/A] sinBdBd</>.
1 I 1 I -rIA 1 I
s
- - -~ ~ 2 . - -~ . -rIA
D.,X2 V dr - ,X241rfO r r smB ,drdBd</>- ,X241rfOsmBdBd</>R re dr
=- 4:fO sinBdBd</>(e-r/A (1 +i))I:
= - 4:fO[(1 +~) e-S/A- (1 + ~) e-R/A]sinBdBd</>.
Sothechangein -b IV dr exactlycompensatesforthechangein §E .da, andweget t;q for thetotal using
thedentedsphere,just as wedid with theperfectsphere.Any closedsurfacecanbe built up by successive
distortionsof thesphere,sotheresultholdsfor all shapes.By superposition,if therearemanychargesinside,
thetotalis .!..Qenc.Chargesoutsidedo not contribute(in the argumentabovewefoundthat0 for thisEO
volumefE .da+-b IV dr =a-and, again,thesumisnotchangedbydistortionsofthesurface,aslongasq
remainsoutside).Sothenew"Gauss'sLaw" holdsfor anychargeconfiguration.
(f) In differentialform, "Gauss's law" reads: IV.E + ~V = ~p, lor, puttingit all in termsof E:
V.E - ,12IE. d1=~p.SinceE =-VV, thisalsoyields"Poisson'sequation":_\72V+ ,12V =~p.A ~ A ~
40 CHAPTER 2. ELECTROSTATICS
~
,,~..
.~"
-;$i"
~
-'!;
""co
-;,
"'A
E=-VV
V=- JE.dl
Problem 2.50
p =100V. E =100tx(ax)=I foa I (constanteverywhere).
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'slaw (andVxE = 0) by themselvesdo not determine
thefield-like any differentialequations,they mustbe supplementedby appropriateboundaryconditions.
Ordinarily,theseareso "obvious"that weimposethemalmostsubconsciously("E mustgoto zerofar from
the sourcecharges")-or weappealto symmetryto resolvethe ambiguity("thefieldmustbe the same-in
magnitude-on both sidesof an infiniteplaneof surfacecharge"). But in this casethereare no natural
boundaryconditions,andno persuasivesymmetryconditions,to fix theanswer.The question"What is the
electricfieldproducedby a uniformchargedensityfillingall of space?"is simplyill-posed:it doesnot give
us sufficientinformationto determinethe answer. (Incidentally,it won'thelp to appealto Coulomb'slaw
(E =~ Jp~dT)-the integralis hopelesslyindefinite,in thiscase.)
Problem 2.51
CompareNewton'slawof universalgravitationto Coulomb'slaw:
F =_Gmlm2 A
~r;
1
F =- qlq2 A
471"100-;:2 r.
Evidently4;<0-t G andq -t m. Thegravitationalenergyofa sphere(translatingProb. 2.32) is therefore
IWgrav=~G~.1
Now,G = 6.67X 10-11N m2/kg2,andforthesunM = 1.99X 1030kg,R = 6.96X 108m,sothesun's
gravitationalenergyis W =2.28X1041J. At thecurrentrate,thisenergywouldbedissipatedin atime
- W - 2.28X 1041=5.90X 1014s =11.87X 107years.!t - P - 3.86X 1026
41
Problem 2.52
First eliminatez,using the formula for the ellipsoid:
Q 1
a(x,y)=_m .
47rabyI C2(X2f a4) +C2(y2fb4) + 1- (x2fa2)- (y2fb2)
Now(forparts(a) and(b)) setc -+0, "squashing"theellipsoiddownto anellipsein thexy plane:
Q 1
a(x,y)=- .
27rabyl1- (xfa)2- (yfb)2
(1multipliedby 2 to countbothsurfaces.)
(a)For the circular disk, set a = b = R and let r ==ylx2 + y2.1a(r) =2
Q
R 1 .17r ..;R2- r2
(b)Fortheribbon,let Qfb ==A, andthentakethe limit b-400:1a(x) =2A 1 .\7r..;a2- x2
(c)Letb=c,r ==yly2+ z2,makinganellipsoidof revolution:
X2 r2 . Q 1- +- =1, wItha=--,-- .
a2 C2 47rac2yIx2fa4+r2fc4
Thechargeona ringof widthdx is
dq=a27rrdB, whereds= yldx2+dr2=dxJ1 +(drfdx)2.
2xdx 2rdr dr c2x
V
C4X2 c2
Now~ +~ =0 ~ -d =-~, sods=dx 1+42 =dx-Jx2fa4 + r2fc4. Thusa c x ar ar r
~ Q 1 2
I
Q
A(X) =-d =27rr-4 2 -ylx2fa4 +r2fc4 = -2 ' (Constant!)x 7racylx2fa4+r2fc4 r a
0'(r)
0'(r)
r ' I .. rR -a a
(a) (b)
1A(X)JI X LxI -a aII
(c) (d)