Chapter 4
Electrostatic Fields in Matter
Problem 4.1
E = V/x = 500/10-3 = 5x 105. Table 4.1: a/47r€0 = 0.66x 10-30, so a = 47r(8.85x 10-12)(0.66x 10-30) =
7.34X10-41. p = aE = ed ~ d = aE/e = (7.34x 10-41)(5 x 105)/(1.6 x 10-19) = 2.29 X 10-16 m.
d/R = (2.29 x 10-16)/(0.5 x 10-10) = 14.6x 10-6.1 To ionize, say d = R. Then R = aE/e = aV/ex
Rex/a.= (0.5 x 10-10)(1.6 x 10-19)(10-3)/(7.34 x 10-41) = 1108v.1
~ V
=
Problem 4.2
First find the field, at radius r, using Gauss' law:
=
Qenc
=
l
r
pdT
0
2q
47rq
=-
l
r
7ra3 0
-
4q
e-2r/ar2dr = -
= E~Qenc,or E = 4;<0~Qenc.
a
-
--e-2r/a
a3 [ 2
a2
a2
)
- a2 [ e-2r/a (r2 + ar +"2
=
J E.da
r2 + ar +-
(
a2
2
r
)]l 0
r
r2
)] .
(
-"2 ] = q [ 1 - e-2r/a 1 + 2~ + 2 a2
~
[Note:Qenc(r --+ 00)
q.] So the field of the electron cloud is Ee = 4;<0 [1 - e-2r/a(1+ 2~+ 2~)].
protonwill be shifted from r
0 to the point d where Ee = E (the external field):
=
1 q
1-e47r€0d2 [
E=--
2d/ a
(
d ~
1+2-+2-.
a
a2
The
)]
Expandingin powers of (d/a):
e-2d/a
(
d
d2
)
1- e-2d/a 1 + 2- + 2a
a2
=
1-
2d
!2
2d
() ()
a
+
2
a
- .!. 2d
3! a
( ) +...
3
=1-
2~ + 2
a
(~a ) - ~3 (a~) +...
2
= 1- (1-2~+2(~r -~(~r +..-) (1+2~+2~)
d
cP.
d
cP.
d3
cP.
d3
4~
= r - r - 2ta - 2:+
d2 + 2ta + 4:+
d2 + 4:+
d3 - 2:+
d2 - 4:+
d3 + -3 -a3 + .. .
4
d
3
( ) + higher order terms.
= 3 ~
73
3
74
CHAPTER 4. ELECTROSTATIC
E
1
q
4 d3
( )
= --- = --(qd)
= -po
1 3a3
4
1.
471"€0
dl- 3 a3
471"€0
371"€oa3
I
a
= 311"!:oa3 .
[Not so different from the uniform sphere model of Ex. 4.1 (see Eq. 4.2).
FIELDS IN MATTER
I
Note that this result predicts
4;EOa = !a3 = ! (0.5 X 10-10)3 = 0.09 X 10-30 m3, compared with an experimental value (Table 4.1) of
0.66 x 10-30 m3. Ironically the "classical" formula (Eq. 4.2) is slightly closer to the empirical value.]
Problem 4.3
per)
~
= Ar.
471"r 471"
€oA r4
4
d: ad2/4€0
= Ar2
4€0 . This
=E
I
Electricfield (by Gauss's Law): §E.da = E (471"r2)
= -!oQenc
~
d
= EloJ;Ar471"r2
dr, or E =
"internal" field balances the external field E when nucleus is "off-center" an amount
= V4€oE/A.
So the induced dipole moment is p
= ed = 2ev€0/AVE. Evidently
p is proportional to El/2.1
For Eq. 4.1 to hold in the weak-field limit, E must be proportional to r, for small r, which means that p
must go to a constant (not zero) at the origin: I p(O) :/; 0 (nor infinite).
I
Problem
.
4.4
r
~
Field of q:
Q
A
411"E:r2
r.
0
q
~
f. Induced dipole moment of atom: P
1I"EO r
Field of this dipole, at location of q (0 = 71",
inEq. 3.103): E = _4 1 13
7I"€0r
Force on q due to this field: IF
Problem
2aq 2
471"€or
)
(to the right).
q
= 2a -47I"€0 2 r13 I(attractive).
( )
4.5
Field of PI at P2 (0
= 71"/2in
Field of P2 at PI (0
=
Torque on PI: N1
Problem
(
=aE =
= PI
71"
Eq. 3.103): E1
in Eq. 3.103): E2
X E2
= 47I"€or
PI 39
= 47I"€or
P2 3 (-2f)
.
2PIP2
= I-47I"€or3 I(pomts
(points down).
(points to the right).
.
mto the page).
4.6
(a)
Use image dipole as shown in Fig. (a). Redraw, placing Pi at the origin, Fig. (b).
E--
P
(
. - 471"€0(2z)32cosOf+sinO9);
P
= pcosOf
+ psinO9.
2
N
(b)
=
P
X Ei
= 471"€:(2Z )3 [(cos 0 f + sin 0 9) x (2cos0 f + sin0 9)]
p2
=
10 ~o
Pif/
Z
=
A
[cosOsinO4J + 2sinOcosO(-
4r.€0(2z)3
p2 sin 0 cos 0
A
471"€0(2z)3 (-4J)
(out of the page).
A
]
4J)
75
.
But sin 0 cos 0
= (1/2)
p2 sin 20
I = 4m:o(16z3)
sm 20, so N
(out of the page).
For0 < 0 < '!r/2, N tends to rotate p counterclockwise;for '!r/2 < 0 < '!r,N rotates p clockwise. Thus the
stableorientation is perpendicular to the surface-either t or ..t..
Problem 4,7
Say the field is uniform and points in the y direction. First slide p
in from infinity along the x axis-this takes no work, since F is J.. dl.
(If E is not uniform, slide p in along a trajectory J.. the field.) Now
rotate (counterclockwise) into final position. The torque exerted by
pEsinOz. The torque we exert is N = pEsinO
E is N = pxE
clockwise, and dO is counterclockwise, so the net work done by us is
negative:
y
tE
O.
P
=
x
p
U = J:/2 pE sin OdO = pE (- cosO) 1~/2 = -pE (cosO- cos~)
Problem 4,8
U = -pI,E2, but E2 = ~-!:r
Problem
= (p . V)E
=
(
Px
(Eq. 4.5); E
8
8
8
8x
Y8y
8z
- + P - + pz1
q
= -p,E.
Qed
[3(p2,f) f - P2]. SOU = ~-!:r [PI'P2- 3 (pI,f) (p2,f)].
Qed
1 q ~
q
xx+yy+zz
=_
4'!rEOr~ r = _
4'!rEO(2x + y 2 +
)
z 2)3/2'
q
x
4'!rEO(X2 + y2 + Z2)3/2
3
2x
-
.
3
2y
4'!rEO{ Px [ (x2 + y2 + Z2)3/2 - 2x (X2 + y2 + Z2)5/2 ] + py [ -2x (X2 + y2 + Z2)5/2 ]
3
2z
q
p
3r(p.r)
q Px 3x
[
+ pz [ -2x (X2 + y2 + Z2)5/2]} = 4'!rEO[ r3 - ;:s(Pxx + Pyy + pzz) ] = 4'!rEOr3
r5
-
F
cos 0
4,9
(a) F
Fx
= -pE
.
_
]
x'
= I4 1
'!rEor~ [p - 3(p . f) f] .
(b) E
=_
1 r-;.{3[p.
4'!rEO
(-f)]( -f) - p}
1 r13[3(p.
=_
4'!rEO
f) f - p]. (This is from Eq. 3.104; the minus signs
are because r points toward p, in this problem.)
F = qE = 1-4'!rEO
1 rq3[3(p . f) f - p] ,
[Notethat the forces are equal and opposite, as you would expect from Newton's third law.]
Problem 4,10
~
(a) Ub
= P,n = §]
kR;
(b) For r
Pb = -V.p
1 8
= -3"r 8r (r
2
kr)
1
= -~3kr
r
2
~
=~
< R, E = 3~oprf (Prob. 2.12), so E = I-(k/EO) r.1
For r > R, same as if all charge at center; but Qtot
= (kR)(4'!rR2)
+ (-3k)(t'!rR3)
= 0, so IE = 0.1
CHAPTER 4. ELECTROSTATIC
76
Problem
FIELDS IN MATTER
4.11
Pb = 0; ab = P.il = :!:P (plus sign at one end-the
P points away from).
one P points toward; minus sign at the other-the
one
(i) L » a. Then the ends look like point charges, and the whole thing is like a physical dipole, Qflength Land
charge P-rra2. See Fig. (a).
(ii) L « a. Then it's like a circular parallel-plate capacitor. Field is nearly uniform inside; nonuniform "fringing
field" at the edges. See Fig. (b).
(iii) L ~ a. See Fig. (c).
(a) Like a dipole
Problem
p
p
p
(c)
(b) Like a parallel-plate capacitor
4.12
{ J
J
v = 4';EO I;jdT = p. 4';EO ~dT }. But the term in curly brackets is precisely the field of a uniformly
charged sphere, divided by p. The integral was done explicitly in Prob. 2.7 and 2.8:
R3
I
1
.t. dT -- 1
4W<O!
.'
Problem
(4/3);R3Pf,
411"€0
(r>R),
r
A
I
3€or2P.r=
I
...!...P.r
3€0
I
R3 P cos B
3€or2 '
I
( > R)
r
,
So V(r,B) =
- p{ ~
r, (r < R). }
411"€0(4/3)wR'p
R3
~
I" 'owoo,
1
(r < R).
4.13
Think of it as two cylinders of opposite uniform charge density :!:p. Inside, the field at a distance s from
the axis of a uniformly charge cylinder is given by Gauss's law: E211"se= -:OP1l"S2e
:::} E = (p/2€0)s. For
two such cylinders, one plus and one minus, the net field (inside) is E = E+ + E- = (p/2fO) (s+ - s_). But
s+ - s-
= -d,
so E
= l-pd/(2€0),where d is the vector from the negative axis to positive axis. In this case
I
the total dipole moment of a chunk of length e is P (1I"a2e)= (p7ra2e) d. So pd = P, and IE = -P /(2€0), I for
s < a.
77
Outside, Gauss's law gives E27r8£ = .1...p7ra2£
:::}E = 1!!£.
2a2!,
<0
<0 s for one cylinder. For the combination, E =
a2 !:t. - iL
, where
E+ + E- = 1!!£.
2<0
s:!:
(s+
=
-S:i: =
8?t
)
s-
d
S T -j
2
( d)(
(
ST-
2
1
=
8+ - L
(
s+
8-
)
82
= ~
82
a-
S :f: S--;2
[(
-1
12
2
)
8 +-Ts.d
4
(s . d)
d
~-
1
d
ST82
2
( )(
.
s+s~
a2
E(s)
-1
)
~-
1
d
ST82
2
( )(
1:f:-
s.d
82
)
(keepmg only 1st order terms in d).
1
= -2fO 82
=~
+ ~
2
82
s.d
82
.
)
- ~ - s-s~
) (
T "2
IT-
2 )]
82
[2(P. 8) § -
for 8
P] '
(2S(S.d)
82
-d
82
).
> a.
Problem 4.14
Total charge on the dielectric is Qtot = is O"b
da + Iv Pbdr = is P . da - Iv V.p dr. But the divergence
theorem says is p. da = Iv V.p dr, so Qenc = O. qed
Problem 4.15
(a)Pb=-v,p=-~~
r2 or
( r ) =-~; r2
r2~
O"b=P.ii=
+P.~=k/b
(atr=b),
(at r = a). }
{ -P . r = -k/a
Gauss's law:::} E = 4:<0Q;~cr. For r < a, Qenc = 0, so IE = 0.1 For r > b, Qenc = 0 (Prob. 4.14), so E
I
Fora < r < b, Qenc = (~k) (47ra2)+ I: (~) 47rf2dr
(b) fD.da
= Qfenc
= -47rka -
47rk(r- a)
= -47rkr;
= O:::}D = 0 everywhere. D = foE + P = O:::}E = (-l/fo)P,
IE = 0 (for r < a and r > b);j IE
Problem 4.16
= -(k/fOr)
r (for a < r < b).
so E
I
= -(k/for)
= 0.1
r.1
so
I
(a) Same as Eo minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14)
is-P/3fO. SoIE = Eo + ~p.1
D = foE= foEo+ ~P = Do- P + ~P, so D = Do - ~p.1
I
(b) Same as Eo minus the field of:f: charges at the two ends of the "needle"-but
away,so
!
E
= Eo.J
D
= foE = foEo = Do -
P,
SO
I
D
= Do -
these are small, and far
P .1
(c) Same as Eo minus the field of a parallel-plate capacitor with upper plate at 0" = P. The latter is
-(l/fo)P, so IE = Eo + !op.1 D = foE = foEo + P, so ID = Do.!
CHAPTER 4. ELECTROSTATIC
78
Problem
4.17
P
E
(uniform)
Problem
D
(same as E outside, but lines
continuous, since V.D = 0)
(field of two circular plates)
4.18
J
=
(a) Apply D . da
Q/enc to the gaussian surface shown. DA
metal plate.) This is true in both slabs; D points down.
~
(b) D
I
E2
FIELDS IN MATTER
= fE
::}
E
= a/fl
= aA
::}
I
D = a.1 (Note:D = 0 insidethe
2+u
in slab 1, E = a/f2 in slab 2. But f = tofT, so fl = 2fo; f2 = ~fO' lEI = a/2fo,
= 2a/3fo.1
(c) P = foXeE, so P = foXed/(fOfr) = (Xe/fr)a; Xe = fr -I::}
P = (1- f;l)a.
IPl=a/2,IIP2=a/3.1
(d) V = E1a + E2a = (aa/6fo)(3 + 4) = 17aa/6fo,!
ab = +P1 at bottom of slab (1)
(e)Pb=O;
ab = + P2 at bottom of slab (2) = a /3,
ab = -P2 at top of slab (2) = -a /3.
= a/2,
ab= -PI at top of slab (1) = -a/2;
total surface charge above: a - (a/2) = a/2,
(f) In slab 1: { total surface charge below: (a/2) - (a/3) + (a/3) - a = -a/2,
total surface charge above: a
-
+ (a/2) - (a/3) = 2a/3,
(a/2)
In slab 2: { total surface charge below:(a/3)
-
a
= -2a/3,
E -
~
..(
-
2a
..(
} ==> 1 - 2fO'
E
} ==>
]+u
-u/2
(!)
+u/2
-u/3
@
+u/3
]-u
Problem
4.19
With no dielectric, Co = Afo/d (Eq. 2.54).
In configuration (a), with +a on upper plate, -a on lower, D = a between the plates.
E = a/ fO(in air) and E = a / f (in dielectric).
2
- ~ - ~
Ca - V - d
1+1/(r
==> Ca
Co - 1 2fr
+ fr .
(
)
I
So V
= {;; ~ + 7~ = 2~~ (1 + ~)
.
I
In configuration (b), with potential difference V: E
= V /d, so a = foE = fO V / d (in air).
2 - 3fO'
I
79
P = EOXeE= EOXeV/d (in dielectric), so O'b= -EOXeV/d (at top surface of dielectric).
= EoV/d= O'f +
= O'f-
EoXeV/d,so O'f= EoV(l + Xe)/d = EOErV/d(on top plate above dielectric).
V
V
A
AEO
1 + Er
=? Cb = V = V
+ O'f2"
= 2V Eod + EOdEr = d ~
. Co
Cb = ~.
1 + Er
~l+<r = {l+<r)2-4<r
= 1+2<r+4<~-4<r
= 2(1+<r)
(1-<r)2> 0. So Cb > Ca.]
[Which is greater?. ~Co - ~Co = l+<r.2
2(1+<r)
2(1+<r)
O"tot
Q
1
(
O'b
A
A
0'2"
)
)
(
(
)
I
I
If the x axis points down:
I
E
~
(a) air
ID.da
<r+l)
-
d
-
X
Er7x
(Er
-
~ O'f (top
X
I
Y
d
-
d
7
0
1)7
plate)
«r+l)
«r+l)
0
d x
surface)
0
-
x
d
] O'b (top
0
d X
«rH)
Yx
d
(b) dielectric
-
«rH)
-
2 v
«r+l) d X
Yx
d
(b) air
Problem
.2<r
«r+1) d x
(a) dielectric
p
D
-
<
-(Er - 1)7
(left)
Er7
(right)
4.20
= Qfenc => D41fr2
= p!1fr3 => D = lpr => E = (pr/3E) r, for r < R; D41fr2 = p!1fR3 => D =
pR3/3r2=>E = (pR3/3Eor2)r, for r > R.
fO
V
pR3 1
= - } E. dl =
00
R
P fo
3EO -:;. 00 - 3E
1
pR2
}R
=
rdr
3EO
+
3E
=
""2
1
pR2
P R2
I
3EO
( + ).
1
2Er
Problem 4.21
Let Q be the charge on a length £ of the inner conductor.
f D . da
V
=
D21fs£ = Q
a
-
-
f
-
Q
V£
D
b
E.dl-
C
C
£
=>
=
I
-
In(b/a)
_
Q
= 21fS~
0;
I(
a
-Q
21fEO£
Q
= 2-1fEOS~
E
0
C
(a
Q
< s < b), E = -21fES~0 (b < r < c).
) l( )
dS
-+
S
b
+ 21fEO
(l/Er) In(cjb)"
-Q
dS
21fd
S
---
b
Q
-
21fEO£
[
In -
(a )
EO +-In
(
C
E
b')]
I
Problem 4.22
Same method as Ex. 4.7: solve Laplace's equation for V;n(s, tj)) (s < a) and Vout(s, tj)) (s > a), subject to
the boundary conditions
x
(i)
V;n
=
Vout
(ii)
E8~n
=
EO8~~ut
{ (Hi) Vout
-+
- Eos cos tj)
at s = a,
at s = a,
for S
Eot
» a.
y
FromProb. 3.23 (invoking boundary condition (Hi)):
00
V;n(s, tj)) =
2::>k(ak cosktj)+ bksinktj)), Vout(s, tj))= -Eoscostj) +
k=l
00
L
k=l
s-k(Ck
cosk</> + dk sink</».
80
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
(1 eliminated the constant terms by setting V
= 0 on the
y z plane.) Condition (i) says
cosk</J+ bk sin k</J)= -Eoscos</J +
I::ak(ak
L
a-k(ck
cosk</J
+
dk sink</J),
while (ii) says
f.r
Evidently
bk
L kak-l(ak
cosk</J + bk sink</J)
= dk = 0 for all k,
ak
= -Eo
cos</J-
L ka-k-l(ck
cosk4>
+ dk sink</J).
= Ck = 0 unless k = 1, whereasfor k = 1,
aal = -Eoa + a-ICI,
f.ral = -Eo - a-2CI'
Solving for aI,
Eo
al =
and hence Ein(s, </J)=
Problem 4.23
Po
=
f.oXeEo;
Evidently En
EI
- (1 + Xe/2)'
-~
x =~2).1
= --31~
= (- ~e )n
so ~n(S, </J)=
Po
Eo
- (1 + Xe/2)
scos</J=
Eo
-x
(1 + Xe/2) ,
As in the spherical case (Ex. 4.7), the field inside is uniform.
= --Xe3 Eo;
PI
= f.oXeEI = --f.OX~
3 Eo;
E2
=
1
--PI
~
=
X~
-Eo;
9
Eo, so
E=Eo+EI+E2+"'=
[~(_~er]
Eo.
The geometric series can be summed explicitly:
00
'"' n ~x-l-x
n=O
-,
1
so
1
Eo,
E = (1 + Xe/3)
which agrees with Eq. 4.49. [Curiously, this method formally requires that Xe < 3 (else the infinite series
diverges), yet the result is subject to no such restriction, since we can also get it by the method of Ex. 4.7.]
Problem 4.24
Potentials:
Vout(r,O) = -EorcosO+): r~lPI(COSO), (r > b);
(a < r < b);
Vrned(r,O)= L(Alrl+r~l)1't(cosO),
= 0,
(r < a).
{ ~n(r,O)
Boundary Conditions:
{
(i) Vout
(ii) f.~ or
(Hi) Vrned
= Vrned,
=
=
f.0~ or '
0,
(r = b)j
(r = b)j
(r = a).
81
" BI
(i)
~
.-Eob cos0 + ~
(ii)
~
€r ~
(Hi)
~
A al + hi
"
= 0
al+l
Alb
hi
)
+ bl+l
~(cosO)j
]
lAlb -
I
I
(
hi
"BI
- (l + 1)bl+2 ~(cosO) = -Eo cosO- ~(l
I 1
[
"
bl+l ~(cosO) = ~
h =
~
A
-a2l+1
I
+ 1) bl+2 PI (cos 0);
I.
Fori =f.1 :
(i)
(ii)
BI
=
bl+1
(
A bl - a21+1AI
I
bl+l
1-1
[
a
2l+1
AI
bl+2
+ (l + 1)
€r lAlb
)
~
B = A
I
I
(b2l+1 -
BI
- -(l + 1)bl+2
a21+1
-
]
.
),
Bl -
~
l
-€rAI
[(
l
21+1
)
+1 b
21+1
]
+a
-~
Al
- Bl - O.
Fori = 1 :
B1
(i)
..
(ll
)
3
So -3Eob
€r
[2 (b
Vrned(r,O) =
E(r,O)
=
a3 At
(
+ 2~
Al
3
= Al
a3Al
-Eob+b2=A1b-~
3
- a
)
+ €r
(
3
)
BI-Eob
3
=A12
B1
(b
3
-a
);
3
3
3
= -Eo - 2b3 ~ -2B1 - Eob = €rAI (b + 2a ) .
3
+ 2a )] ;
b
3
~
-3Eo
2[1- (afb)3] + €r[1 + 2(afb)3]
-3Eo
Al = 2[1 - (afb)3] + €r[1 + 2(afb)3].
a3
( -;:2 )
r
cosO,
-VVrned = 12[1- (afb)3] ~E:r[1 + 2(afb)3]
{ (1+ 2r~3)cosOi'-
(1- ;:) sinoo}.
Problem 4.25
There are four charges involved: (i) q, (ii) polarization charge surrounding q, (Hi) surface charge (CTb)on
the top surface of the lower dielectric, (iv) surface charge (CT~)on the lower surface of the upper dielectric.
In view of Eq. 4.39, the bound charge (ii) is qp = -q(X~f(1 + X~), so the total (point) charge at (0,0, d) is
qt = q.+ qp = qf(1 + X~) = qf€~. As in Ex. 4.8,
(a) CTb
(b)
I
CTb
=
=
-1
qdf€~
3
€oXe _
[ 47I"€o(r2 + cF):2
I
_
€oXe 4
[
1
CTb
-_
2 -
€o
CT~
~
2€o]
(here CTb
= P.n = +Pz = €oXeEz)j
CTb - _
CT~
- _
2
2
+ cF)2 €o €o]
I
qdf€~
( here CTb= -Pz = -€OXeEz )
3
7I"€o(r2
Solvefor CTb,CT~:first divide by Xe and X~ (respectively) and subtract:
CT~
CTb
1
qdf€~
X~ - Xe = 271"
(r2 + cF)~
I
~
CTb=
I
CTb
1
qdf€~
Xe [ Xe + 271"
(r2 + cF)~ ] .
.
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
82
Plug this into (a) and solve for O"b,using €~ = 1 + X~:
' O"b (
'
qd/€~
)
(
)
";!Xel+Xe--Xe+Xe,soO"b=41T (r2 + d2)2
2
O"b -
-1
-
, O"b-
,
-I
qd
{ 41T (r2
Xe
1
-1
Xe
3
, ;
(
[
41T(r2 + d2)2 1 + Xe + Xe)/]2
1
qd
,
qd/€~
1
qd
€rX~/€~
+ d2) ~ [1 + (Xe + X~)/2] + 21T(r2 + d2) ~ } ' so O"b= 41T(r2 + d2) ~ [1 + (Xe + X~)/2]'
X x. ,
~- + )/ 2 (which vanishes, as it should, when
The total bound surface charge is O"t O"b + O"~ = 417r
qd ~ E'~ l + ( ",
(r2+d2)
x.
X~ = Xe)' The total bound charge is (compare Eq. 4.51):
=
qt
V(r)
=
(X~ - Xe )q
€~
_
2€~ [1 + (Xe + X~)/2] -
I
(
-
€r
€~ + €r
)
q
€~' I and hence
I
=~
q/€~
+
qt
(for z
41T€0{ ";X2 + y2 + (z - d)2
";x2 + y2 + (z + d)2 }
\
> 0).
I
I
I
q
since €r
~ + qt
Meanwhile,
Problem
= 7"r
€~
-
€r
[ 1 + €'r + €r ]
-
~
€~
+ €r '
V(r)
=
41T€0
I
I
1
";X2
[2q/(€~ + €r)]
(for z < 0).
+ y2 + (z d)2
\
4.26
From Ex. 4.5:
0,
D
w = -1
2
I
=
(r
Q
O'Q
{ 41Tr2r,
(r < a)
(r > a) } ,
I
1 Q2
D.EdT=--41T
-
2 (41T)2
{€
l
b
1 1
E=
--r2dr+-
a r2 r2
-r,
41Ttt
(a < r < b)
~
00
1
€o b
1
=-Q2
-dr
r2
}
1
1
Q2
1 Xe
= 81T€0{ (1 + Xe) ( ~ - b) + b } = 81T€0(1
+ Xe) ( ~ + b ) .
Q2
1
1
.
(r > b) }
{ ~r,
41T€or
1
< a)
~
I
-1
- -
81T { €
( )
r
I
b
a
1
+--
€o
00
-1
( )
r
1
b
}
83
Problem 4.27
Using Eq. 4.55: W
= !f J E2 dr.
From Ex. 4.2 and Eq. 3.103,
-I
E =
{
Wr<R =
fa
2
, so
(r > R) }
( ) =
.
.
(
)
6"
4cos
B+
sm
B
r
smBdrdBd<jJ
( )J
.£-. 2 ~11'R3
3100 3
R3 P
3100
fa
Wr>R = 2
=
(r < R)
3102
P z,
R
~(2cosBf+sinBO),
3for
2
211'P2R3.
27 fa
1
2
2
2
r
l
1r
00
1
(R3 p)2
B) sin BdB R 4"
1 8fa 211' a (1 + 3 COS2
r dr
11'(R3P)2
9100
1
(~ ) =
11'(R3 p)2
=-
(-COSB-COS3B)I~(-3~3)1:
9100
411'R3p2.
27100
3R3
211'R3 p2
Wtot =
9100
This is the correct electrostatic energy of the configuration, but it is not the "total work necessary to assemble
the system," because it leaves out the mechanical energy involved i~ polarizing the molecules.
Using Eq. 4.58: W
= ~ J D.E dr.
For r <: R, D = foE, so this contribution is the same as before.
Forr < R, D = foE + P = -!p + P = jp = -2fOE, so ~D.E = -2!fE2, and this contributionis
=-
~~ R::2, exactly cancelling the exterior term. Conclusion: Wtot = 0.1 This is not
surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem
there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as
the "work necessary to assemble the configuration" -the latter would depend entirely on how you assemble it.
Problem 4.28
now(-2) (~~ p:~3)
I
First find the capacitance, as a function of h:
Air part: E
= -bL
==> V
41rfOS
= ..1L
In(bJa)
41rfO'
>..
==>
OilPart: D = 41rs
~ ==>E = 41rfS
2>" ==>V = 41rf
2>"In(bJa) ' }
Q = >..'h + >"(f - h)
= fr>"h -
>"h+ >"f
= >..[(fr -
Q
C
Th
t
e ne upwar
d
l'
..
LOrCeIS gIven
y
The gravitational force down is F
q.
2>"ln(bJa)
4 64 ' F .
.
= -j>'"10
= >"(Xeh+ f),
>"(Xeh+ f)
=V =
b E
l)h + f)
-fa
10
>..'
where f is the total height.
(Xeh + f)
= 211'100In(bJa)
.
411'100
- 1 V 2 21rfOXe
dh - 2" In(bfa) .
1 V 2 dC
- 2"
= mg = p11'(b2 -
= fa->..= lOr>".
a2)gh.
h =
}
I
V2
fOXe
p(b2 - a2)g In(bJa)
.
84
CHAPTER 4. ELECTROSTATIC
Problem
4.29
(a) Eq. 4.5
:::}
.
.
FIELDS IN MATTER
F2
8
= (P2 . V)
PI
Eq. 3.1O3:::} EI = 4m:or
~
F2
= P2~uy
EI
PI
A
(Ed;
~y
?jr
A
() = -- 41rfoY3 z. Therefore
= _ -PIP2
d
4
d
1
1rfO [ y
z
41rfoY4 Z, or
( Y )] z = _
3PIP2
A
3PIP2
A
A
F2 = ~z
41rfor
3"
I
I(upward).
.
I
To calculate F I, put P2 at the origin, pointing in the z direction; then PI
is at -r z, and it points in the -y direction.
y
=
So FI
(PI'
V) E2
=
I
-PI 8:2y x-y-O,
- - z--r
- ; we need E2 as a function of x, y, and z.
1
From
E q. 3.104:
E2
ll
-3"
41rfOr
=
P2 . r = -P2Y'
~
E2 =
8y
8E2
8y
1
(0,0)
= ~
~~
[
?
r-
h
- P] , were
r = xx + yy + ZZ, P2 = -P2Y, and hence
A
-3Y(XX + yy + zz) + (x2 +y2 + Z2)y
41rfO [
8E2
3(P2' r)r
(x2
+ y2 + z2)5/2
A
A
-3XYX + (x2 - 2y2 + z2)y - 3YZZ
=~
41rfO [
]
A
(x2 + y2 + z2)5/2
]
-~~2Y[-3XYX + (x2- 2y2+ Z2)y - 3yzz] + ~(-3xx - 4yy - 3ZZ) ;
~
{ 2~
}
= ~ -3z
41rfO r5
z;
FI
Z =
= -PI ~41rfO 3r
r5
(
)
- 3PIP2 Z.
41rfor4
These results are consistent with Newton's third law: FI = -F2.
(b) From page 165, N2 = (P2 x EI) + (r x F2). The first term was calculated in Frob. 4.5; the second we
get from (a), using r = r y:
P2 X EI
= 41rfor3
PIP2
(
A
-X)j
r x F2
= (ry-) x
-
(_ ) = _
3PIP2
41rfor 4
Z
3PIP2
N2
41rfor 3 x; so
A
I
_
2PIP2
= 41rfor 3 X.
A
This is equal and opposite to the torque on PI due to P2, with respect to the center of PI (see Frob. 4.5).
Problem 4.30
Net force is to the right (see diagram). Note that the field lines must bulge to the right, as shown, because
E is perpendicular to the surface of each conductor.
I
I
E
85
Problem 4.31
P
= kr = k(xx
+ yy + zz) =? Pb= -V.p
Total volume bound charge:
= P.il.
I
= -k(l + 1 + 1) = 1-3k.1
= -3ka3.1
Qvol
At top surface, il = z, z = a/2j so O"b
= ka/2. Clearly,
(Jb
T~tal surface bound charge:
I
I
O"b
= ka/21
on all six surfaces.
Qsurf= 6(ka/2)a2 = 3ka3.1 Total bound charge is zero.
if
Problem 4.32
f D.da = Qfonc::}D = 4- q
~
2 rj
7rr
Pb= -V.p
=
47r(~~Xe) (V.
E
1
q
f
4 7rfO(1 + Xe) 2";
r
P
= -D
=
10
)
qXe
= fOXeE =
=
~ = -q 1 ~eXe 83(r)
=
(Eq. 1.99)j O"b P.f
f
4 7r(1 + Xe) 2"'
r
47r(1~X~e)R2;
Qsurf= (Jb(47rR2) = q 1 +XeXe .1 The compensating negative charge is at the center:
I
j
= - l qXe
PbdT
j
83(r)dT
+ Xe
= -q-1 Xe .
+ Xe
Problem 4.33
Ell is continuous (Eq. 4.29); Dl. is continuous (Eq. 4.26, with O"f = 0). So EXl =-EX2' DYl = DY2 ::}
E1EYI
= f2EY2'and hence
tan02 = EX2/EY2= EYl = E2. Qed
EXl / EYl
tan 01
EY2
El
If 1 is air and 2 is dielectric, tan O2/ tan 01 = E2/ EO> 1, and the field lines bend away from the normal. This is
the opposite of light rays, so a convex "lens" would defocus the field lines.
Problem 4.34
In view of Eq. 4.39, the net dipole moment
at the center is pi
=P -
=
1~~e p
I';Xo
= tp.
p
We want the
potentialproduced by pi (at the center) and O"b(at R). Use separation of variables:
B
00
Outside: V(r,O)=
L rl:l
1=0
_
.
lnstde:
V(r,O)= 4
1
Pz(cosO)
1 pcosO
~
(Eq.3.72)
7rEO Err
R~L = AIRI,
av
8V
8r R+
I
ar Ri
= -
or
B1 1
P
R2 - 47rEoErR2 + AIR,
""'
L)l
1
= --p.
BI
EO
or
_
1
BI
~
= R2/HAI
2pcosO
Er R3
av
= --1 (EoXeE.r) = Xe _
a
~
EO
)
(l ¥' 1)
.
p3
B1 - 47rfOf~+ AIR
+ 1)R l+2Pz(cosO)+ 47rEO
r
(Eqs.3.66,3.102)
1=0
V continuous at R ::}
{
.
00
+ LAlrIPI(cosO)
r
I
R-
-
""'
LlAIR
}
I 1
-
1
PI(cosO)
I 2pcosO
= Xe { --47rEO
lOrR3
= --O"b
EO
""'
+ ~lAIR
I 1
-
PI(cosO) }
.
86
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
-(I
+ 1) R~~2 -
Forl=l:
lAIRI-I
BI
1
-2-+---AI=Xe
R3
411"fO
= XelAIRI-I (l ¥- 1); or - (2l + l)AIRI-I
1
---+A1
2p
(
frR3
2P
-~
- AIR3 + ~
=> Al =
~
2XeP = ~ 2(fr -l)p;
411"foR3fr(3 + Xe) 411"foRafr(fr + 2)
411"fOfr
- AIR3 = _~XeP
411"fOfr
2
(
cosO +
411"fo frr2
p
P
-BI+---=---+Xe-;
AIR3
2
BI =
COSO
411"for2
~
2
411"fOfr [
)(~ + 2)
fr
(r
=>Al = 0 (£ ¥- 1).
AIR3
411"fOfr
+ XeAIR3 =>
411"fOfr
V(r,O)=
Meanwhile, for r::; R, V(r, 0) = ~P
)
411"fOfrR3
= XelAIRI-I
2
1
XeP
AIR3
411"fo fr
2
(3 + Xe) = ~XeP.
411"fOfr
1 + 2(fr - 1)
(fr + 2) ]
= ~~.
411"fOfrfr + 2
~ R).
1 prcos(} 2(fr -1)
R3
freEr + 2)
411"fO
fr - 1
P cos ()
= I 41I"for 2 fr [ 1 + 2 -fr + 2
(
r3
) ] (r::;
R).
R3
Problem 4.35
Given two solutions, VI (and EI = -VVI, DI = fEd and V2 (E2 = -VV2, D2 = fE2), define V3 ==V2- VI
(E3 = E2 - EI' Da = D2 - DI).
Iv V.(VaD3) dr = Is V3Da' da = 0, (Va= ° on S), so I(VV3) . D3 dr + I V3(V.D3) dr = 0.
= V.D2 - V.DI = PI- PI = 0, and VV3 = VV2 - VVI = -E2 + EI = -E3, so IE3' D3 dr = O.
But D3 = D2 - DI = fE2 - EEl = fE3, so f(E3)2 dr = 0. But f > 0, so E3 = 0, so V2 - VI = constant. But
at surface, V2 = VI, so V2 = VI everywhere. qed
Problem 4.36
But V.D3
I
(a) Proposed potential:
I
VCr) = Vo~.1 If so, then IE = - VV = VO~ f, in which case P
I
= fOXe Vo ~ f,
;
in the region z < 0. (P = o for z > 0, of course.) Then CTb= fOXeVo (f.ft) = 1- fO~ Vo .1 (Note: ft points out
of dielectric => ft = -f.) This CTbis on the surface at r = R. The flat surface z = 0 carries no bound charge,
since ft = z 1..f. Nor is there any volume bound charge (Eq. 4.39). If V is to have the required spherical
symmetry, the net charge must be uniform:
CTtot411"R2
=
Qtot = 411"fORVo(since Vo = Qtot/411"foR), so CTtot= foVo/R.
CT
I
- {
=
Therefore
(fOVo/ R), on northern hemisphere
(fOVol R)(l + Xe), on southern hemisphere' }
=
=
(b) By construction,
CTtot CTb+CTI
foVo/R is uniform (on the northern hemisphere CTb= 0, CTI foVo/R;
on the southern hemisphere CTb= -foXeVo/R, so CTI= fVo/R). The potential of a uniformly charged sphere is
Vo =
Qtot
411"for
= CTtot(411"R2)
= fOVo
R2 = VoR.
411"fOr
R for
r
(c) Since everything is consistent, and the boundary conditions (V
Prob. 4.35 guarantees that this is the solution.
./
= Vo at
r = R, V -+ 0 at 00) are met,
87
(d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to ft,
so we'd get bound charge on this surface, spoiling the symmetry.
Problem 4.37
~
Eext = 27r€08 8. Since the sphere is tiny, this is essentially constant,
F
J (1 + Xe!3
-
-Xe
=
1 + Xe/3
) (~
2
~ ~
-
) ( ) (
(~ ) 2-~7rR38 - - (~ )
€oXe
27r€08
47r2€0
d8
833
Xe / 3 Eext (Ex. 4.7).
and hence P = 1 +€oXe
8dr
27r€08
-
€oXe
1 + Xe/3
~
)( ) (! ) ( ) J
-1
2no
8
8
dr
82
)..2R3 8
7r€083 .
3 + Xe
Problem 4.38
The density of atoms is N = (4/3)7rR3'The macroscopic field E is Eself + Eelse, where Eself is the average
fieldover the sphere due to the atom itself.
p
= o:Eelse
=} P = No:Eelse.
[Actually,it is the field at the center, not the average over the sphere, that belongs here, but the two are in
fact equal, as we found in Prob. 3.41d.] Now
1 p
Eself = - 47r€0R3
(Eq. 3.105), so
1
E=-
0:
47r€0 R3 Eelse
+ Eelse
=
0:
(
1-
47r€oR3
)
Eelse
=
(
1-
NO:
3€0
)
€r
-
Eelse.
So
P=
No:
(1 - N o:/3€0) E = €oXeE,
and hence
No:/€o
Xe = (1 - No:/3€0)'
Solving for a:
Xe
No:
No:
- -3 Xe= €o
€o
No:
=} -
€o
Xe
(1+ _3 ) = Xe,
or
€o
Xe
3€0
a = N (1 + Xe!3) = Ii
Xe
(3 + Xe' But Xe = €r - 1, so 0:=
3€0
1
Ii (z+2 ) .
qed
Problem 4.39
Foran ideal gas, N = Avagadro's number/22.4liters = (6.02 x 1023)/(22.4x 10-3)
= 2.7 X 1025.
(2.7x 1O25)(47r€0x 1O-3O),8/€0= 3.4 X 10-4,8, where,8 is the number listed in Table 4.1.
(3 = 0.667, No:/€o = (3.4 x 10-4)(0.67) = 2.3 x 10-4, Xe = 2.5 X 10-4
He: (3= 0.205, No:/€o = (3.4 x 10-4)(0.21) = 7.1 x 10-5, Xe = 6.5 X 10-5
No:/€o
=
H:
Ne: (3 = 0.396, No:/€o= (3.4x 10-4)(0.40)= 1.4x 10-4, Xe= 1.3X10-4
Ar: (3 = 1.64, N 0:/ €o = (3.4 X 10-4)(1.64) = 5.6 x 10-4, Xe = 5.2 X 10-4
..
agreementISqUItegood.
}
CHAPTER 4. ELECTROSTATIC
88
Problem
4.40
(a)
(u)
-
=
J~:E ue-u/kT du - (kT)2e-u/kT
J-pE
PE e-u/kT
du
-
= N(p);
p
- l]I~~E
pE
-pE
l
[e-pE/kT - ePE/kT] + [(pE/kT)e-pE/kT
kT
e-pE/kT
kT - pE
[ epE/kT
A
= (pcosO)E = (p.
-
+ (pE/kT)ePE/kT]
epE/kT
}
PE
+ e-PE/kT
ePE/kT
=
[-(u/kT)
-kTe-U/kT
{
P
FIELDS IN MATTER
- e-pE/kT
E)(E/E)
]
= kT
- pE coth
-
( kT ) .
-(u)
= -(u)(E/E);
P
= Np
=
pE
I
Np { coth
PE
kT
(
kT
) - pE } .
Asx --+0, y = (~+ f - ~; +... )-~ = f-~; +... ~
0, so the graph starts at the origin, with an initial slope of 1/3. As x --+00, y --+coth(oo) = 1, so the graph
Lety ==P/Np,
x ==pE/kT. Theny
goes asymptotically to y
= cothx-1/x.
= 1 (see Figure).
.E...
np'
11
"""""""""""""'"
.
pe/kT
(b) For small x, y :::::::
kx, so;; p :::::::
-f!-r, or P :::::::
~E
= €oXeE => P is proportional to E, and Xe
= ~~Np2
.
For water at 20° = 293 K , p = 6.1 X 10-30 em' 'volume
N = molecules= molecules
X molesX
!\rams.
mole
gram
volume
.
(O.33Xl029)(6.1Xl0-30)2
- j"1;)l
- 0 33 1029.
12 T bl 4 2
X
, Xe - (3)(8.85xl0-12)(1.38XlO-23)(293)
- ~
a e . gives an
6
1
N -- (6 .0 X 1023) X ( 18
) x-. (10 )
experimental value of 79, so it's pretty far off.
For water vapor at 100° = 373 K, treated as an ideal gas, v~~r::e= (22.4 X 10-3) X (~~~) = 2.85 X 10-2 m3.
(2.11 x 1025)(6.1x 10-30)2
-
N = 6.0 X 1023
,
2.85 X 10-2 = 2.11 X 1025. Xe = (3)(8.85 x 10-12)(1.38x 10-23)(373)= 15.7x 10 3.1
Table 4.2 gives 5.9 x 10-3, so this time the agreement is quite good.