2
Chapter
Electrostatics
Problem (a)
2 .1
I Zero.!
(b) IF
=_ 411"1=0 r
q~,
where r is the the dist distan ance ce from from cent center er to each each nume numera ral. l. F poin points ts toward the missin missing g q. Explanation: by supe superp rpos osit itio ion, n, this this is equi equiva vale lent nt to (a), (a), with with an extra extra -q at 6 o'c o'clo lock ck-s -sin ince ce the the forc forcee of all all twel twelve ve is zero zero,, the the net net force force is that that of -q only. 1
(c) I Zero.
I
I
(d) 14 1r€0r 1 q~,
pointi pointing ng toward toward the missing missing q. Same Same reason reason as (b). (b). Note, Note, howe however ver,, that if you you expla explain ined ed (b) (b) as a cancel cancella lati tion on in pairs pairs of opp oppos osit itee char charges ges (1 o'cl o'cloc ock k agai agains nstt 7 o'c o'clo lock ck;; 2 agai agains nstt 8, etc.) etc.),, with with one one unpa unpair ired ed q doin doing g the the job, job, then then you' you'll ll need need a different explan explanati ation on for (d). (d). Prob roblem 2.2 E I
(a) "Horizo "Horizonta ntal" l" compon component entss cancel. cancel. Net vertic vertical al field field is: Ez = 4;<0 4;<02~ 2~ cos cos O. 1
Here Here 1-2 1-2
= Z2
+ (~)2 ; cosO = i, so E I
2qz
= 41r€0(Z2 +
A
(~)2 (~)2)3 )3/2 /2 z. x
When z» d you're you're so far far away it just just looks looks like like a single single charge charge 2q; the field field shou should ld redu reduce ce to E = 4 ?r
_
E
(b) This This time the "vertic "vertical" al" compon component entss cancel, cancel, leavi leaving ng
E= E
_ 2~ sinO sinOx, x, or 1 4 ?r<0 'V
qd (~)2)3/2 /2 x. 41r€0 (z2 + (~)2)3 1
=
A
q
I
x -q
which, h, as we shall shall see, see, is the the fiel field d of a .dip .dipol ole. e. (If we ~ z, whic
From From far away, (z » d), the the fiel field d goes goes like like E ~ 4;<0 4;<0
d -+ 0, we get E = 0, as is approp set d -+ appropria riate; te; to the extent extent that this this configur configurati ation onloo looks ks like like a singlepoint singlepoint charge charge from far away, the net charge is zero, zero, so E -+ 0.) 22
23 Prob Proble lem m
2.3 2.3 rL), rL), dx
1
Ez
~
411"fO 411"fO JO
cos
---1-)..z rL 411" 411"fO fO
/Ldq
= )"dx --
1-
Ex
x
[
1)"
- 4m:0 m:0 Z [
(
- 1+
..,j Z2 + L2 )
L
A
x+
( ..,j z2 + L2 )
For z » L you you expe expect ct it to look look like like a point point char char ge q
term-t term-t 0, andthe z Prob roblem lem
[
] 0 --
L-
2
+ x
; COS( COS())
= ~)
L
I
~
2
dx
~
- ---1-)..).. Z
x
= Z
rL )'dx sin() -
411"fO
E - --
1
1 Z2~
411"fOJo
x
2
(11-
;
(z2+x2)3/2
JO
---1-)..Z 411"fO
()
]
xdx I (X2+z2)3/2 411"fO
1-)..
L
I
---1 ---1-~ -~~ ~ O z~' 411"f z~'
[
- - ---1-).. 1 -
0
-
411"fO
Z
~.
L-
]
A
z
] .,
= )"L: E
-+ _ ¥-z. It check checks, s, for for with with z » L the x 4 1 ¥-z. 1I"fO
Z
1 ~kz. term -+ 4_1I"fOz 1I"fOz z
2.4 2.4
From Ex. 2.1, with L -+
~ and
z -+ vz2 + (~)2 (~)2 (dis (dista tanc ncee from from center center of edge edge to F), field of one edge edge is:
~
El =
.
)..a
41f1:0vz2
+
4 a2
vz2 +
4 a2
+
4 a2
There There are 4 sides, sides, and we want want vertic vertical al compon component entss only, only, so mult multipl iply y by 4 cos () = 4
E
=~
z2+ a4
z.
4)..az
47r£0 (Z2 +
!i+?- :
V Z2 + ~2
a42)
Prob Proble lem m 2.5 2.5
~
"Horiz "Horizont ontal" al" compon component entss cancel cancel,, leavin leaving: g: E = 4;fO 4;fO {I cos ()} ()} z. (both h cons consta tant nts) s),, whil whilee I dl 27rr.. So dl = 27rr Here, 1-2= 1-2= r2 + z2, cos() = ~ (bot E
=~
)"(27rr)z
47r£0 (r2
A
+ Z2)3/2
z.
Prob Proble lem m 2.6 2.6 Break Break it it into into rings rings of radi radius us r, and thickn thickness ess dr, and and use use Prob Prob.. 2.5 2.5 to expre express ss the the fiel field d of each each ring ring.. Tota Totall char charge ge of a ring ring is 0" . 27rr . dr
=
)..
_1
. 27rr, 27rr, so
)..
the = O"dr is the
(O"dr)27rrz 27rrz
1 =_ 4
Edisk
l
=_ 47r£0 27rO"z 0
27rO"z
7r£0
of each each ring ring..
!-
[Z
r
R
1
Ering= 4 3/2; 7r£0 (r2 + Z2) Edisk
"lin "linee char charge ge""
1 ..,jR2 + z2 ]
(r2
z.
+ Z2)
3/2 dr.
CHAPTER 2. ELECTROSTATICS
24
For R
» z the secondterm -+ 0, so Eplane= _41I"fO
1 27Taz =
1
For z » R '~z
= -1
R2
-1/2
(1 + z ) ~
1
-z
~
1 R2
(1 -
-2z'~
)
I
21"0 a z.1
so [ ] ~
1 R2 R2 = ~2z' -z1 - -z1 + -2z 73
and E = ~ 21rR2". = ~g. Z ' where Q = 7TR2a. ,( 411"fO~ 411"fO Problem
z
2.7
E is clearly in the z direction. From the diagram, dq = ada
= a R2 sin 0 dO d
1z-2= R2 + z2 - 2Rz cos 0, cos'lj; =
z-Rlz-cosO.
So y
E - -2 z
-
/
a R2 sin 0 dO d
47T€O
1 - -(27TR 47T€O
2 a)
= -2-(27TR2a) 47T€o
- -2-
d
O=O=>u=+l _.X . . . O = 7T => u =- 1 } Let u - cosO, du - -smOdO, {
11" (z-RcosO)sinO 0 (2R + z 2 - 2Rzcos 0)3/2 dO.
1
/
-11 (2 2 zu )3/2du. R + zz2-- R~
27TR2a
- 47T€O (
J = 27T.
+ Z2 - 2Rz cos 0)3/2 .
(R2
1
.2.-
Integral can be done by partial fractions-or '
zu - R
- -2-
27TR2a
) [ z2 ~R2 + Z2- 2Rzu ] -1 - 47T€O Z2
:
For z > R (outside the sphere), Ez = 4 For z < R (inside), Ez = 0, so I E Problem 2.8
f
.. 0
= 4_
~
z
1I"fO z
1 :\,
(z - R) - (-z - R) { Iz - RI Iz+ RI } . 1
so
look it up.
I
E
= 0.1
= 4_
q
7T€O Z
2
z.1
According to Frob. 2.7, all shells interior to the point (Le. at smaller r) contribute as though their charge were concentrated at the center, while all exterior shells contribute nothing. Therefore: 1 QintE( r ) ---r, 47T€Or2 where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, so
E=-2-Q47T€O r2 r.
Inside the sphere, only that fraction of the total which is interior to the point counts:
Qint
Problem
(a) p
=
=
: R
!7Tr3
3"7T
3Q
r3
= R 3Q,
_ 1
V. E
1
so E = 4 R 3Q2f 7T€O r
2.9 1"0
r3
= €O~ tr (r2. kr3) = €o~k(5r4) = 15€okr2.1
=
.
25
(b) By
Gauss'slaw: Qenc= €ofE.
By direct integration:
Qenc
da = €o(kR3)(41rR2)=
I
41r€okR5.
= J pdT = JoR(5€okr2)(41rr2dr)
I
= 201r€ok
JoR r4dr
= 41r€okR5.,(
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:
/ one face
The latter is 1..q, by Gauss's law. Therefore <0
Problem
2~
/
whole large cube
/
E. da.
E. da = -.!L.
I
24€o
one face
2.11
~
GaUSsian
G-
surface: Inside: f E. da
= E(41rr2) = -!;Qenc = 0
Gaussian surface: Outside: E(41rr2)
r
Problem
E. da =
.
= 1..(a41rR2) =>
E
=>
IE = O.I R2
=~
<0
2
r.
}
(As in Frob. 2.7.)
€or,-
2.12 j.L'E.
Gaussian surface
da
So = E. 41rr2= 1..Qenc = 1..! <0 <0 3 1rr3p.
JE= ~prr.1 Since Qtot = ~1rR2p, E = 4';<0j&r (as in Frob. 2.8). Problem 2.13 Gaussia
~/ "
.L'E. da
j
'
= 1..>"l. So 21rs.l = 1..Qenc <0 <0
n ,unace
IE
l Problem
= E.
21r€os =~Sl
(same as Ex. 2.1).
2.14
Gaussian surface
.L'E
j
.da = E .41rr2 = 1..Qenc= 1.. J p dT = 1.. } (kf)(f2 <0
<0
<0
= 1.. k 41r rr r3dr = 411'kr4 = 1I'kr4. <0
. ' E = ~1rkr2r. 41r€o
Jo
<0
4
<0
sin B dr dB d4»
CHAPTER 2. ELECTROSTATICS
26
Problem
2.15
(i) Qenc = 0, so IE = 0.1 sinOdfdOd phi (H) f E. da = E(47IT2) = 1-Qenc = 1<0 <0 f pdr = 1<0 f Af2 r r - a <0 Ja
= M
"df
E(411"r2)
(iii)
~
=~
"'(r - oj :.1E <0
f:
df
~
~ ( )'.1 fO
lEI
r2
= 4:rok(b - a), so
IE=H~)r.1 Problem
r
2.16
(i)
@ (H)
b
a
a
)-
Gaussiansurface
. 21I"S' [-
fE.da-E -
ps -- -8
- to enc= f-p7rs2[. 0'
A
EE]2fO .
~
f
= 1-p7ra2[; f E. da = E. 211"s,[= 1-Qenc <0 <0 2
G"""'an
-
1Q
,."face
E = pa s. 2fOS
Gaussiansurface = 0; f E. da = E. 211"s,[= 1-Qenc <0
(Hi)
IE=O.I
\ ~'-'
lEI
6
a Problem
S
2.17
On the x z plane E
=0
by symmetry.
Set up a Gaussian "pillbox" with one face in this plane and the
other at y.
Gaussian pillbox
= 1-Ayp; f E. da = E. A = 1-Qenc <0 <0 P Y Y (for Iyl < d). IE = fO I
27
<0 Qenc= 1...Adp P dy =} IE = 100
(for y
I
>
d).
E I!!! '0 t""."'
-d
a
y
Problem 2.18 From Prob. 2.12, the field inside the positive sphere is E+ = ~r+, where r+ is the vector from the positive r -. So the total field is centerto the point in question. Likewise, the field of the negative sphere is - ...L 3 <0 E
But (seediagram) r+ Problem 2.19 VxE
- L = d.
Problem 2.20
(since Vx
x
I
J ~PdT
= 4:100 Vx =0
So E
y
P = -(r+ 3100
r-
- r_)
~-
+
= ~d'i = 4:100
J [vx
(~) = 0, from
(~~)] pdT
(since p depends on r', not r)
Prob. 1.62).
Z
= k Itx
ty tz 1= k [x(O- 2y) +:9(0 - 3z)+ z(O- x)] f:. 0, xy 2yz 3zx soEl is an impossible electrostatic field. (1) VxE1
x (2)VxE2
y
= k Itx
y2
ty
2xy + z2
z
tz 1=k [x(2z -
2yz
2z) + y(O- 0)+ z(2y - 2y)]= 0,
so E2 is a possible electrostatic field.
z
Let's go by the indicated path: E. dl =(y2 dx + (2xy + z2)dy + 2yz dz)k
(xo, Yo, ZO)
= z = 0; dy = dz = O. E . dl = ky2 dx = O.
Step I: y
III
= Xo, Y : 0 -+ Yo, z =
O. dx = dz = O. E dl = k(2xy + z2)dy = 2kxoY dy.
Step II: x
fII
Step III: x
.
E. dl
y dy = kxoY5. = 2kxo frio
= Xo,Y = Yo,z : 0 -+ Zo;dx = dy = O.
I x
II
y
28
CHAPTER 2. ELECTROSTATICS E. dl = 2kyzdz = 2kYozdz.
E. dl = 2Yok Jozoz dz = kyoZ5.
JIll
.
V(XO, Yo,ZO)= (xo.Yo.zo) E dl = -k(xoY5 - J 0 Check:
-vv=k[lz(xy2+yz2)
2.21
Problem
V(r) =
-
Soforr>R:
Yoz5), or V(x, y, z) = -k(xy2
.dl.
{ Inside the sphere(r < R) :
(
V(r)=-Je:
4;f
0
~ ) df=
q
1
dfr2
(
4;f q(to) 0
I
Z]=k[y2 x+(2xY+Z2) y+2yz i]=E. ,(
Outside the sphere (r > R ) : E =
Je:E
+ yz2).
I
x+/v(xy2+yz2)y+/;(xy2+yz2)
< R: V(r) = - J:: (~~)
and for r
+
41I"fOr
E = 4;fO-J/:srr.
l =1 4 q 00 r
_ 1 :;\r.
!,I
11"100r
J~ (4;fO -J/:sf)df = -dto [:k-:h
(r22R2)]
)
2R 3 - R2 . =1411"100 . - -L- 1 1 8 1 Wh en r >, R VV -- -LE -- - VV -- -L,( 411"fO 8r ( ;::) r - - 411"fO f=2"r,so 411"fO f=2"r. A
A
A
Whenr < R, VV = -dto2~tr (3 - ~) r = -dto2k (-~) Problem 2.22
r = --dtobrj
soE = -VV = 4;fO-J/:srr.,(
=
E 4;fO2;8 (Prob. 2.13). In this case we cannot set the reference point at 00, since the charge itself extends to 00. Let's set it at s = a. Then
= V(s) .
r (~2: ) dB = a
1I"fO
( ).
-- 1 2,Xln ~ 411"100 a
(In this form it is clear why a = 00 would be no good-likewise the other "natural" point, a = 0.) In ( l. )) VV = --L2'xE.. 411"fO 88 ( a
§
= --L2,Xi§ = -E.,( 411"fO 8
2.23
Problem
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))\ !! ( ) -1+!!
= li. { 1-!!-ln fO
b
b
b
I
} = ~ln 100
(~) . a
2.24
Problem
Using Eq. 2.22 and the fields from Prob. 2.16:
-
V(O) =
=
-
V(b)
Problem
(~)
-
fb Jo
E. dl = -
8;I: +
411"100 Jz2
2q + (~)2
E. dl-
b E. J.a
b ids dl = _L2fO Jo fa S ds - e£.. 2fO J.a s
~ Insl: = 1- ~:: (1 + 2ln (~) ) .1
2.25
(a)Iv = ~
fa Jo
.
29 (b) V-I-
L Adx ~ f -L ~
=
A I 41UOn (x
£ + YZ2 + £2
,\ =I-In 41r€0
[
-£
L
+ Yz2 + x2 ) -L I
= ~ln
+ YZ2 + £2 ]
(
2"fO
I
L+~
).
z
x (c)V
= 4:fO JoR~;~:~~
+ z2)1: =
= 4:f021r<7(vr2
~
I
(JR2 + Z2- z)
.1
Z :. E = _8V 8z . A
In each case, by symmetry ~~ = ~~ = O.
(a)E = -~2q "fO (b) E
A 1 - - 41rfo { (L+~) -
- (c)E
2z (-!) (z2+(~)
= --2
A
z
1
2 v'z2+£2
~"fO v' Z2+L 2
(7
fO {
_ 21 v'Ri
+z
1
2
I
2qz 2 3/2 Z (agrees with Prob. 2.2a). 1r€0(z2+(~))
41
Z=
2t2
1
z
-
1
- (-L+~)
2~
1r€0 ZY Z21
A
a
z= I} 12 €o
2 z Z
1
}Z--12£'\4_
-L+~-L-~ ( + L2 ) -L { Z
22z
}
A
1
1
-
z
[
VR2
Ifthe right-hand charge in (a) is-q, then V
+ Z2 ]
+ £2
Z A
.
I(agrees
WIt h
Ex... 2 1)
z I(agrees with Prob. 2.6).
= 0 I, which,naively,suggestsE =
- 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 I
hopeto compute Ex = - ~~ or Ey = - ~~. That was OK in part (a), because we knew from symmetry that Ex
= Ey = O. But
now E points
in the x direction,
so knowing
Problem 2.26 V(a)
41r€0 Jo
(where v'2h
1
V(b)
= uk In 4100
[
h
41r€0
a21rr
ch,
( 2 - V2
:.1V(a) - V(b)
)
4€0
(2 + V2)2
(
= ;€~ [1-ln(1
2
2€0
Vh2 + 1-2-
V2h1-.
Jh2 + 1-2- V2h1-ch
Vh2 + 1-2-
V2h1- + -. h In(2V h2 V2
v2
In
= ah.
v'2h
v2
V2 = ah
E.
1-
+ ':n In(2h+2V2h-V2h) - h - ':n In(2h-V2h)
2+
where ii =
1-
= 41r€0 V2 Jo
to determine
= 1-/V2)
rv'2h
1
2V2€0 [
~
r
-=.--
0
=~ = 2v2€0
1-
l ( )
=_ 4 1r€0 21ra
a
a21rr
( ) ch = 21ra ~(V2h) V2
rv'2h
=~
V on the z axis is insufficient
)
+ V2)] .
= ah
2€0
In(1 + V2).
]
=
~
':n[
2v2€0 v2
+ 1-2- V2h1-+ 21-- V2h) ] In(2h+V2h)
-In(2h-V2h)
0 ]
30
2. ELECTROSTATICS
CHAPTER
Problem
2.27
Cut the cylinder into slabs, as shown in the figure, and use result of Prob. 2.25c, with z -t x and C1-t Pdx:
1. -8do: 'r
z+L/2
V =!to
J (V R2 + X2 - x) dx z-L/2
= ...£
+
1. XVR2
2<0 2
[
x2
+ R2ln ( x + VR2 + X2) -
= I4fo{ (z+t h/ R2+(z+t)2 -(z( Note:
-
(z
L
2
+ 2") +
(z
L
- 2")
2
2
]
[::;: :::~:::] -2ZL}.
t h/R2+(z-t)2
= -z
x
Z+L/2 X2 I z-L/2
+R21n
~2
2
L2
L2
- zL - "'4 + z - zL + "'4
= -2zL.)
H_H_
E = -vv
= -z-
8V
z
L z + - 2+
{/ ( 4€0
= --E...
8z
-
R2 +
(z
) 2
/
+ L
,
t + V R2 + (z + t) 2
--~
~,;,O Problem
{2VR2 + (Z+~)'
+
2VRZ
--
[L-JR'+(Z+~)'
- 2L
Z - t +
(Z --~)'
(Z +
(z - t) 2] '
t) 2 - V R2
+ (Z -
t) 2
-2£}
+JR'+(Z-~)']Z
2.28
= ~41T<0 J 1-B..dT.
z
Here p is constant, dT = r2 sin e dr de d
r2 sin IJdr dIJdiP . V - -.1!. J vz2+r2-2rzcoslJ' - 41T<0 f1T
V R2 + 1
Orient axes so P is on z axis.
V
L
y'R2+(Z-t)2
1
V R2 +
sin IJ
Jo vz2+r2-2rzcoslJ
de =
f21T d'!"
JO
y x
- 27r
'f' -
.
..l. (vr2 + Z2 - 2rz cose) 1T= ..l. (vr2 + Z2+ 2rz - vr2 + Z2 - 2rz ) 0 rz rz 1
2/z,
rlz(r + z -IT
ifr < z,
zl) = { 2/r, if >
}
2
Z-"2
-
~
L ( z - 2" )
( 2 ) - VR2+(z-t)2
y'R2+(Z+t)2
z +
=
2
L z- --
1+
Z"2
2
[
E
- n__
R2 +
VR2+(Z+t)2
1+ +R
+ 2" L)
2
}
31
.211' . 2 ... V = --1!.411"<0
But P
=Fw'
Problem
R lr2dr + f z lr2dr z r
{f
so V (z)
0
z
=~~
(R2
-
= <0 .P...
}
{l
z3
z 3
+
R2-z2
2
}= L ( R2 - ) z2
2<0
~) = -d!oR (3 - ~) ;
V (r)
I
3'
=
= 4;<0 f p(r') ('V2k)dr (since p is a function air', -L- f p(r') (-4mP (r - r') ] dr = -1...p (r ) . ,f 411"<0 <0 .
Problem
(3-
~)..I
,f
2.29
\72V = 4;<0'V2 f(*)dr
=
~
not r)
2.30.
(a) Ex. 2.4: Eabove =
-2u<0 ft; Ebelow= --2u<0 ft
(ft always pointing up); Eabove- Ebelow= .£.ft. ,f <0
Ex. 2.5: At each surface, E = 0 one side and E = .£. <0 other side, so 6.E = .£.. <0 ,f Prob.
0
(b)
2.11:
Eout
:
:
'IF"
'. .~ ../
--'
= .£.f; Ein = 0; so 6.E = <0 .£.f. ,f <0
= ~f
5 R
§E. da = E(211's ) l
Outside:
Inside:
= 0, so E
Qenc
l
(c)
Vout
~
ar
Problem
= = R2u
= O.:. 6.E = .£.§. <0
,f
at surface ) ; \In = Ru ; so Vout = \In- ,f Ru <0 ( <0
= _R2u = _.£. (at surface ) - 8V;n= O' so ~ 8r ~ <0 ' 8r '
-
8V;n
8r
= _.£..,f <0
2.31
( ) V - -L".!Ji.. - -L {=.!1. a - 411"<0 L" rij - 411"<0 a q2
(
+
.'. W4 = qV = 141I'foa -2 +
(b) WI = 0, W2 = 1 Wtot
= <0 1...Qenc = <0 = .£.§ (at surface 1...(211'R)l=} E = .£.B.§ <0 s <0
2
4;<0(=f);
7{
- -L-
-L V2a. + =.!1. a } -
1 V
M
2
)
(-2 + ...1V2 ) -
411"
(1).
-
W3 = 4;<0(:;a - ~ ; W4 = (see(a)). 1
II
= 411"<0 - 1 + ' V i - 1 - 2 + V 2}
=
411'fO
2q2
(
1
)
-;;: - 2+ V2 -
(2)
.(4)
-
+
.+
-
.
(3)
).
CHAPTER 2. ELECTROSTATICS
32
Problem
2.32
(a) W =
! J pVdT.
From Frob. 2.21 (or Frob. 2.28): V = f
1 1 q {R W = 2P 411"100 2R Jo
=
(3 - R2 ) 411"rdr = 4€oR qR2 ~= ~ ~q2 ( )
qp R2= 5100
5100 ~11" R3
J E2dT.
(b) W = !f .
-
Outside (r > R) E
100
1
2
= = !f { is
(c) W
1
q2
{(
"2 411"100
.
= 4;€O ~r
j
roo 1
.. W - 2" (411"€O)2q{ JR 1
5 R 411"100
-;:
{R
r4 (r 411"dr)+ Jo 1
)1 R
+ R6
"5
W
=
100
2
-
1 ~ 4 11"100 r
=a
100
1
-
)(
q2
1 ~ 4 11"100 r
(
R3
-
5
)
2
(411"rdr) } 1
1
q2
1
(
+ 5R
R
)=
13q2 411"100:5R"/
q2
14
)
= 4;€O~'
Jo
411"
1
4
2
{ (411"100)2 5R + (411"100)21I"q ~ 11" + (411"100)2 q2
1
1
As a --t 00, the contribution Problem
2
(R3 )
r2 sinOdOd4>+ {R E2dT + {a
1
1
1
= 411"100"2~ { + 5R - ~ + R } =
(~f.(~
I
R3
VE. da + Jv E2dT} , where V is large enough to enclose all the charge, but otherwise
{/ (
= 2"
= 4€oR
- R2"5 ] 0
)I 0 } = 411"100"2
Let's use a sphere of radius a > R. Here V
arbitrary.
r
R
r5
(
[ 3"3
qp
R
Inside (r < R) E = 4;€OWrr.
2
00
1 r5
r3
qp
2
- r;) = 4;€O m (3- ~)
JR
1
-
(
1 ~ 411"100 r
)
( -;: ) R } l
.
3 q2
411"100:5R ../
from the surface integral (4;€O~)
1 = dqV = dq -411"100 ~, r
( )
4
goes to zero, while the volume integral
dq
0~~ W =
=
(q = total charge on sphere).
2
= 411"r 1
dW
(q = charge on sphere of radius r).
r3
q = 311"r3 p = q R3
411"r2
1
3q 2 R3r dr
( ) ;: (
R3 r dr.
)=
1 3q2 {R 1 3q2 R5 R65 411"100 R6 Jo r4dr = 411"100
2
3q
=
dr p = 43qdr 311"R
qr3 411"100R3
}
a
- 1)) picks up the slack. 2.33 dW
2 (411"r2dr)
1
3q2 4
411"100 R6 r dr
1
3q2
( )
= 411"100:5R
../
33 Problem 2.34 (a) W
= ~ f E2 dr.
E
=
2
W - ~ -
2
b 1 ~ 2d J. ( ) 24 a;:2 1rr r 41TfO
W2 = 8;fOT,
E1.E2 = (4;fO)2~,
(a) aR I
. E2
dr
= 4;R2; aa = ~;-q q
f::O
-I L (~- ~b ) .
J. b 1 81TfO a;:2 -
a
81rfO
El = 4;fO!-rr (r > a), E2 = 4;fO::;ir (r > b). So
(r > b), and hence f E1.E2 dr = - (4;fO)2 = - 4::0b' q2go ;!r471T2dr 2
+ 100 fEI
(b) V(O)= -
zero elsewhere.
2
= 8;fO7'
Wtot = WI + W2 Problem 2.35
< b),
r
- L
( )
2
(b) WI
<
4;fO!-r (a
= ~q2 (~+! -~) = ~ (~-!).,(
ab = 4;b2' q
I
E. dl = - f::O(4;fO!-r)dr - fba(O)dr - faR (4;fO !-r)dr - f~(O)dr
(c) ab ~ 0 (the charge "drains off"); V(O) = - f::O(O)dr - faR(~!-r Problem 2.36 I
( ) a
I
I
aa
= -~; qa
I
(b) Eout
II ab
= -4 1
aR
I
~ (t ~ -~) ~ (~ - ~)
)dr - f~(O)dr =
+
I
= qa41rR2 + qb .
qa + qb ~ r, where r = vector f rom center of large sphere. 1rfO r
1
A
I
Eb = _ 1 q~rb, Iwhere ra (rb) is the vector from center of cavity a (b). 41rfO rb
qa A
-ra, (c) lEa = 41rfOr~ (d) IZera.
= -4;b2; qb I
=
I
(e) aR changes (but not aa or ab); Eoutside changes (but not Ea or Eb); force on qa and qb still zero. Problem 2.37
Betweenthe plates, E = 0; outside the plates E = a / 100 = Q / foA. So p
= 100 E2 = 100~ 2
-1Q2l -~
2 f5A2
Problem 2.38 Inside, E = 0; outside, E = -41TfOr 1 ~r; so Eave = ~ 4;fO
~ r;
fz
= a(Eave)z;a = 41T~2'
Fz = J fzda = f( 41T~2) ~ (4;fO ~ ) cas
-
1 -9- 2 ( )
2<6 41TR
f1T/2.
21rJo
cME
sm () cas
d -
() ()
() R2 sin () d() dq;
1 9... 2 ( ) 1TfO4R
(
1 . 2 2"sm
())
1
1T/2
-
0
-
1
2
(9... ) 211:fO4R
_ -
I
Q2
321rR2fO .
.1
.1
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: _1 s. Potential difference between the C linders is . L = .!..Qenc y E. da = E . 21TS = .!..Q => E = -92 "'EO L 8 EO EO
I
-
V(b)
=-
V(a)
l
b
a
E.d1= --
Problem
~ ==~...t~)' so capacitance
1
b
-
b
Q
( ) .
-dB= --In
21T€oL a S
As set up here, a is at the higher potential, so V = V(a)
C=
l
Q
V(b)
-
21T€oL
=~
a
In (~).
~O
per unit length is In 21T (Ii) .1
n "
I
2.40
(a) W
= (force)x(distance) = (pressure)
(b) W
= (energy per unit
= ~ E2 A€.I
x (area) x (distance)
I
volume)x (decrease in volume) = (€O~2) (A€). 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 4Aaz
1
E--- 41T€o(z2 + ~2)
J
Z2
Z.
+
-~- da 2
a22
- a--..
Here A -+ ad2a (see figure), and we integrate over a from 0 to ii:
E
= ~2az
l
a
.
-2
1
. Let u = "4' so a da =
41T€O
1
J
a2
{ tan-l
E
»
2a
+
2 Z
(
=-
a -+ 00 (infinite Plane ) : E z
du
tan -1
1T€O[
= k "'EO [tan-l
a (point charge): Let f(x)
az
=-
(u + z2)v2u +
2
2 duo .
a /4
= -4az = 1T:
-a+aa-
a2
alia
0 (z2 + a;) /z2 + a22
41T€O
z2
1T€O
v2u
2
-tan-l [Z
(
g
- tan-l(l)
Z
0.2/4
Z2
)] 0
1+
-
2
2z2
}; 1T
- - Z. A
4]
= k "'EO (oo) - 1!:. 4]
(1!:.2
,f ) = ...fL. 2EO
1!:. 4
= tan-l vI +x - 7' and expand
as a Taylor series:
1 f(x)
+
z2
)
-~- da 2
= f(O) + x!,(O) + "2x21"(0)+ ...
35
0) = tan
f( Here
-1
1< 1< 1< . ' 12 v'I+X 1- -- 2(2+x).,!1+x' 1 1 SO (1) - 4" = 4"- 4"-, O f (x ) -- 1+(1+x)
1' (0) = 1.4'
SO
1
f(x) = -x 4 + ( )x2 + ( )x3 + . . .
. a2 Thus (Since W = x« I)E ,
1 a2
20"
(
)
~ 1rlo 4W
1
1
O"a2
/
q
= 41
Problem 2.42
p - fOV . E -
fO
I 8 -r2 -8r {
1
2A
BSinocos
8( ( -r ) + --r sin0 r
1 1 B sin 0 . -( sm
= fO
8
=
-
2(A r fO
I
}
r B sin
Problem 2.43
= 4;EO~r.
From Prob. 2.12, the field inside a uniformly charged sphere is: E is f=pE = (~~3)(41<~R3)r
So the force per unit volume
= ~(41<~3)2r, and the force in the z direction on dr is:
3 dFz = fz dr = fO
2
Q 471-R3 r cas O(r2sin 0 dr dOd
( )
The total force on the "northern" hemisphere is:
Fz =
3
( 3 = fO (
f fzdr
= fO
2
Q
R
)1 1 ) ( )(
r3dr 0
471'R30
2
Q
R4
Sin2°
"4
~
471'R3
21f
1f/2
1
cas 0 sin 0 dO 0
1
)
0
d
13Q2
1
(271')=1 ~OR2'
Problem 2.44
Vcenter=
v.pole
-
1
47I'fO
=
1-
7I'fO
f
-
1 a(271'R2)
471'fO RV2 aR
:::: Problem
M
v 2fO
(1- 0)
da
10
f
= 271'R2
{ 1-2=
1f/2
a R
da
= -4 1
7I'fO
a
R
2 (271'R)
= -aR 2 fO
sinO dO,
R2 + R2 - 2R2 cosO = 2R2(1 - cosO).
sinO dO
VI - cas 0
= - aR
2V2fO
(2Vl
-
.
1f/2
cas 0) 1
0
aR In :. Vpole- Vcenter= -2 (v2 - 1). fO
aR In'
= V 2fO
2.45
First let's determine
fO
f
a ' 1 - - d a, With 471'fO 1-
-
= -4 i
a
-da
f E .
da
=
the electric field inside and outside
f0471'r2 E
=
Q
enc =
f
P dr
=
the sphere,
using Gauss's
p k1')1'2sin 0 d1'dOd = 471'k lr
law:
1'3d1' = { : ~~4
(r < R), (r > R).
36
CHAPTER
So E
= 4~0 r2
f (r < R);
2. ELECTROSTATICS
E = 4~~:2f (r > R).
Method I: €Q
W
j
="2
€
=47r"'£ 2
E2dr (Eq. 2A5) 2
k
4€Q
r6dr+R8
Q
2
4€Q
2
47rr dr
1
00
-dr
1R
r2
=-
{'X>
€Q
+
7rk2
8€Q
}
1R
"2
(
4€Qr2
R7
{
2
kR4
(
-+R8
--
7
)
47rr2dr 00
1
)1 R }
r
=-
R7
7rk2
8€Q
(-+R7 ) 7
.1
II:
j
w=~
pV dr
l
For r < R, V(r)=-
R
~
4€Q R
l
:. W
(Eq. 2A3).
r
00
=-
=- 1 2 0 = 27rk2 3€0
Problem
( )
= "2 lQ
R
( ) {l -
= l7rk27€QR7 Method
kr2
€Q {R
r
kR4
l( )
kr2
k
4€Q
4€Q
(- )dr=-(- + - ) = ~ ( - ) . (kr) (R3 - - )] 47rr2dr = - 1 ( ~ = = . .( _! () {
E.d1=-
--"2
00
R3
r3 3
4€Qr
1R
R3
R3 3
k
dr-
R7
4
3€0
7rk2R7
4 7 }
R
27rk2
4
R3 R4
2. 3€0
( )
R4 --
r
R
I
00
+-
r
r3 3
1
R}
r3 4
3€0
r3
[ 3€0
{
1
1
R3r3
0
)
- -r6 dr 4
7rk2 R7
7
7€0
2.46 ~
E = - VV -- - A -v
~
-e
'
-Ar
(r )
r A
-Ar
-- - A r (-A )e
{
~
- e
-Ar
-- Ae-Ar + Ar ' -.r ) (1 A
r A
}
I
~
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).
p
= €o V.
So ~. V (e-Ar(1 + Ar)) = -Ar2e-Ar, and Ip = €oA [47r83(r)- ~2e-Ar} .
Q= jPdr But
It re-Ardr = ~,
Problem
= €OA{47r /83(r)dr-A2 so Q
je~Ar
47rr2dr} =€OA(47r-A247r 100 re-Ardr).
= 47r€oA (1- ~) = Izero. I
2.47
(a) Potential of +A is V+ = - 2:'0 In (~),
Potential of -A is V-
= +-
A In 2 ?TfO
where s+ is distance from A+ (Prob. 2.22).
(!=.) , where s- is distance from A-. a
37
.'.TotallV = ~ln
21T1:0
Nows+
= yI(y
( )
(X,Y,Z)
.
s+
- a)2 + Z2, and s- = yI(y + a)2 + z2, so
Vexy z) = ~ "
z
s_
In
21T<0
y'(y+a)2+z2
(y'(y-a)2+z2 )
(b) Equip otentials are given by y2 + 2ay + a2 + Z2 = k(y2
~
(y + a)2 + z2 In 471"Eo [ (y - a)2 + z2 ]
=
-
.
= e(41T<0 Va/A)= k = constant
(y+a)2+z2 (y-a)2+z2
2ay + a2 + Z2) => y2(k
y
).
-).
.,
That is'
- 1) + z2(k - 1 ) + a2(k - 1 ) - 2ay(k + 1) = 0, or
y2 + Z2+ a2 - 2ay (fI:t) = O. The equation for a circle, with center at (Yo,0) and radius R, is (y - yo? + z2 = R2, or y2 + z2 + (Y5 - R2) - 2yyo = O.
Evidentlythe equipotentials are circles,with Yo = a (fI:t) a2-- Y 02 - R2 => R2 -- Y 02 - a2 -- a2 !:tl k-l
( )
2
-
a2
-
-
and
a2 (k2+2kH-k2+2k-l)
(k-l)2
a2
-
-
4k
(k-l)2 ,
Or
R = 2av'k . or in terms of Vr. Ik-ll' , o.
-
Yo R
= 2a
e41T
e21T
e211"
=a
-1 e41T
2
=
271"EOVO I
a
a coth
(-
=
- e-21T
).
a csch I
).
(-
z
y
Problem 2.48 d2V (a) V'2V = _L (Eq. 2.24), so d 2 X <0
-
I
()b qV - 2mv -+ v -, , I-tqv --:;;;. (c) dq = Apdx j ¥i
= ap~~ =
I
1
= -'--po EO
I
Apv
= I
I
(constant).
(Note: p, hence also I, is negative.)
271"EOVO
).
).
38
(d )
CHAPTER 2.
= -.1.. p
d2V
-;rxr
dx2 = {3V-1/21 ' where {3= tPV (Note: [ is negative, so {3is positive; q is positive.) <0
(e) Multiply by V'
=
I
~~ :
I
= {3V-1/2 dV v' dV' ~ ~ But V(O)
m2qV:::}
ELECTROSTATICS
= V'(O) = 0
V'dV'
:::}
I
= {3
V-1/2
dV
:::}
~V'2 = 2{3V1/2+ constant.
2
(cathode is at potential zero, and field at cathode is zero), so the constant is zero, and
V'2 = 4{3V1/2:::}dV dx = 2VP V1/4 :::}V-1/4dV = 2vp~' '
I
V-1/4
I
= 2VP
dV
dx
~ V3/4
:::}
= 2VP
x + constant.
But V(oy = 0, so this constant is also zero. 3
= 2VPx,
V3/4
4/3
3
(2VP )
so Vex) =
X
Interms of Vo (instead of I):
I
Vex)
9
( )
X4/3, or Vex) ="4{3
= -€o
(f ) V (d ) [
-
dX2
Problem
= Vo (d )
I
(see graph).
= Vo(~).
v.
2/3
X
= V m:vv =
I
v2qVo/m (d )
811 m = Vr 0 = (32<5A2q )
4
2
d4/3 :::} 11,3= 81md [2. 0 ~,
- K TT3/2
9v'1nd2vo
.
1/3
2
vo , were
-
K
h
I
x
d
2
[2 = 32
-
4€oA f2q - 9d2 V m:'
2.49
1
(a)
x4/3.
I
4V2
-
(32€~A~q)
x4/3 =
4€oVo 1 4 1 -2/3 _ = -€o Vod4/3 3 . 3x - - 9(d2x)2/3 .
d2V
f2q
V
1/3
81I2
4/3
Without space-charge, V would increase linearly: Vex)
p
2/3
I (
IE = ~47f€o
A
1 + ~ e-1J-/Adr.
fJ'I-. 1-2
A
)
(b) Yes.! The field of a point charge at the origin is radial and symmetric, so V XE = 0, and hence this is also true (by superposition) for any collection of charges. r r 1 1 r (c) V = "2 1 + - e-r/Adr E . dl = --q I
i
00
1
= -q
47f€o
47f€o
OO
{r
1
-
r2
i
00 r
(
A
)
q r e-r/Adr = 1 + ( ) A
OO
{{r
47f€o
1
1
r2
A r
-e-r/Adr + -
00
1
.' -e-r/)..dr
1
r
.
}
.'
39 1
Now I ~e-r
/ A
dr = -~
-r/).
- X I ~dr
-r/).
1
t-- exactly right to kill the last term. Therefore q
V(r)=-
41rfO{
--
l
Js
!
Vdr=-
l
q MfO
V
2
R e-r/A
-r
0
:.
1 E.
Js
da +
q
#dr=-
r
fO
= ,X2f~ { -e-R/A
~ r V dr = ,X2 Jv
(1 !J...
r } -
I
I)?
= !J...
l
-
R
q
e-r/A
41rfO r
.
(1 + R )
fO'x
e-R/\ e-r/\
q
re r/Adr=-
-
R
r ---1
(
)] 0
fO [ (1/ ,X)2,X
0
~) + I } .
+
fO'x{
---
r
1 E .da = ~q~ (1 + ~) e-R/A# MfO l)!l-,X
(d)
oo
e-r/A
I+ R
(
)
e-R/A -
(
1+ R
)
e-R/A
,X
+I
}
= !J.... fO
qed
(e)Doesthe result in (d) hold for a nonspherical surface? Suppose we makea "dent" in the sphere-pushing a patch (area R2 sin BdBd
D. fE. da = =
4:fO {;2
4:fO
(1 + ~) e-S/A(S2 sin BdBd
[(1 +~)
I
-1
1 - -~
D. ,X2
V dr - ,X241rfO
I
=-
~-rIA r
2
1 - -~
.
.
s
I
r sm B , dr dBd - ,X241rfOsm BdB dR re
4:fO sin B dB d(e-r/A
= - 4:fO [(1 +~)
e-R/A] sin BdBd.
(1
+
-rIA dr
i))I:
e-S/A - (1 + ~) e-R/A] sin BdBd.
.
Sothe changein -b IV dr exactly compensates for the change in §E da, and we get t;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 distortionsof the sphere, so the result holds for all shapes. By superposition, if there are many charges inside, the total is .!..Qenc. Charges outside do not contribute EO
(in the argument above we found that
0
for this
volume fE . da + -b IV dr = a-and, again, the sum is not changed by distortions of the surface, as long as q remainsoutside). So the new "Gauss's Law" holds for any charge configuration.
(f) In differential form, "Gauss's
VE -
I
law" reads: IV.E
+
~ V =
~p, lor, putting it all in terms of E:
A12 E. d1= ~p. SinceE = -VV, this alsoyields"Poisson'sequation": _\72V + A12V = ~p.
CHAPTER 2. ELECTROSTATICS
40
~
,,~.. .~"
-;$i"
~
-'!;
""co -;, "'A
E=-VV
V=- JE.dl
Problem
p
2.50
= 100V. E = 100tx (ax) =
The same
charge
density
foa (constant everywhere). would be compatible (as far as Gauss's I
I
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 ("E 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 =
~ J p~dT )-the
Problem
integral is hopelessly indefinite, in this case.)
2.51
Compare Newton's law of universal gravitation to Coulomb's law:
F = _Gmlm2 ~r;
A
F
=- 1
qlq2
471"100-;:2
A
r.
Evidently 4;<0 -t G and q -t m. The gravitational energy of a sphere (translating Prob. 2.32) is therefore
IWgrav=~G~.1 Now, G = 6.67 X 10-11 N m2/kg2, and for the sun M = 1.99 X 1030kg, R = 6.96 X 108 m, so the sun's gravitational energy is W = 2.28 X 1041J. At the current rate, this energy wouldbe dissipated in a time t --
W 2.28 X X 1041 P -- 3.86 1026
= 5.90 X 1014s = 11.87 X 107 years.!
41 Problem 2.52 First eliminate
z, using the formula for the ellipsoid: Q
a(x,y)=_m
1
47rab yI C2 (X2 f a4)
+ C2(y2 fb4) + 1 - (x2 f a2) - (y2fb2)
.
Now(for parts (a) and (b)) set c -+ 0, "squashing" the ellipsoid down to an ellipse in the xy plane: a(x,y)=-
1
Q 27rabyl1-
.
(xfa)2 - (yfb)2
(1multiplied by 2 to count both surfaces.) Q 27r R ..; R2 - r2 1 .1
=
(a) For the circular disk, set a = b = R and let r == ylx2 + y2.1 a(r)
= 2A7r ..; a2 1-
(b) For the ribbon, let Qfb ==A, and then take the limit b -400:1 a(x) (c) Let b =
x2
.\
c, r == yly2 + z2, making an ellipsoid of revolution:
. Q r2 - + - = 1, wIth a = --,--
1
X2
47rac2 yI x2 f a4
C2
a2
+
r2
f c4
.
The charge on a ring of width dx is
2xdx 2rdr Now ~ a + ~c A(X)
=0
dq
= a27rr
dB,
c2x
~
dr -d x
where ds = yldx2 + dr2 = dxJ1 + (drfdx)2.
= -~, ar
so ds
~ Q = -d = 27rr-4 2
= dx V 1 + 1
C4X2
42 ar
2 -ylx2fa4
7rac ylx2fa4 + r2fc4 r
x
c2
r = dx-Jx2fa4 +r2fc4
=
+ r2fc4. Thus
I
Q -2a '
(Constant!)
0'(r)
0'(r)
r
R
(b)
(a)
I
II (c)
I
' -a
X
J -a
(d)
I
1 A(X)
..
a
Lx
a
r
