Eletrodinâmica griffiths resolução capitulo 4

Prévia do material em texto

Chapter 4
ElectrostaticFields in Matter
Problem4.1
E =V/x =500/10-3=5x 105.Table4.1:a/47r€0=0.66x10-30,soa =47r(8.85x10-12)(0.66x 10-30)=
7.34X10-41. p =aE =ed ~ d =aE/e =(7.34x 10-41)(5x 105)/(1.6x 10-19)=2.29X 10-16m.
d/R=(2.29x 10-16)/(0.5x 10-10)=14.6x 10-6.1Toionize,sayd=R. ThenR =aE/e =aV/ex ~ V =
Rex/a.=(0.5x 10-10)(1.6x 10-19)(10-3)/(7.34x 10-41)=1108v.1
Problem4.2
Firstfindthefield,at radiusr, usingGauss'law: J E.da = E~Qenc,orE = 4;<0~Qenc.
l r 47rqlr - 4q [ a - ( a2)]l
r
Qenc = pdT =- e-2r/ar2dr=- --e-2r/a r2+ar+-
0 7ra30 a3 2 2 0
2q
[ (
a2
)
a2
] [ (
r r2
)]= - a2 e-2r/a r2+ar+"2 -"2 =q 1- e-2r/a 1+2~+2a2 .
[Note:Qenc(r--+00) =q.]Sothefieldoftheelectroncloudis Ee=4;<0~[1- e-2r/a(1+2~+2~)].The
protonwillbeshiftedfromr =0 to thepointd whereEe=E (theexternalfield):
1 q
[
2d/a
(
d ~
)]E=-- 1-e- 1+2-+2-.47r€0d2 a a2
Expandingin powersof (d/a):
e-2d/a = 1- (
2d
)+! (
2d
)2 - .!.(
2d
)3 +...=1- 2~+2(~)2 - ~(~)3 +...a 2 a 3! a a a 3 a
= 1- (1-2~+2(~r-~(~r +..-)(1+2~+2~)
d cP. d cP. d3 cP. d3 4~= r- r - 2t - 2:++2t +4:++4:+- 2:+- 4:++- - +.. .a d2 a d2 d3 d2 d3 3a3
4
(
d
)
3
= 3 ~ + higherorderterms.
(
d d2
)1- e-2d/a 1+2- +2-a a2
73
74 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
1 q (
4d3
) 1 4 1. I 3 IE =-- -- =--(qd) =-po a =311"!:oa.471"€0dl- 3a3 471"€03a3 371"€oa3
[Not so differentfrom the uniformspheremodelof Ex. 4.1 (seeEq. 4.2). Note that this resultpredicts
4;EOa = !a3 = ! (0.5X 10-10)3= 0.09X 10-30m3,comparedwith an experimentalvalue(Table4.1)of
0.66x 10-30m3. Ironicallythe "classical"formula(Eq. 4.2)is slightlycloserto theempiricalvalue.]
Problem 4.3
per)=Ar. Electricfield(byGauss'sLaw):§E.da = E (471"r2)= -!oQenc=EloJ;Ar471"r2dr,orE =
~ 471"A r4 =Ar2 . This "internal"fieldbalancestheexternalfieldE whennucleusis "off-center"anamount471"r €o 4 4€0
d: ad2/4€0=E ~ d =V4€oE/A. So the induceddipolemomentis p =ed=2ev€0/AVE.Evidently
I p is proportionalto El/2.1
For Eq. 4.1to hold in theweak-fieldlimit, E mustbeproportionalto r, for smallr, whichmeansthatp
mustgotoa constant(notzero)at theorigin:I p(O):/;0 I (norinfinite).
Problem 4.4
r Field of q: ~ ~f. Induceddipolemomentof atom: P =aE =. 0 Q A 1I"EOr
q 411"E:r2 r.
Fieldofthisdipole,at locationofq(0=71",inEq. 3.103):E = _41 13( 2aq 2) (to theright).7I"€0r 471"€or
Forceon q dueto this field:IF =2a(-4q )2 13I(attractive).7I"€0 r
Problem 4.5
Field of PI at P2 (0=71"/2in Eq. 3.103): E1 =4 PI 39 (pointsdown).7I"€or
Field of P2 at PI (0 =71"in Eq. 3.103):E2 =4 P2 3 (-2f) (pointsto the right).7I"€or
I 2PIP2 I . .Torqueon PI: N1 =PI X E2 = -4 3 (pomtsmtothepage).7I"€or
Problem 4.6
(a)
Useimagedipoleasshownin Fig. (a). Redraw,placingPi at theorigin,Fig. (b).
E-- P (
. - 471"€0(2z)32cosOf+sinO9);
P =pcosOf+psinO9.
10~o
Pif/ Z
2
N = P X Ei =471"€:(2Z)3 [(cos0f +sin09)x (2cos0f +sin09)]
p2
[
A A
]= 4r.€0(2z)3cosOsinO4J+2sinOcosO(-4J)
p2sin0cos0 A
= 471"€0(2z)3(-4J) (out of the page).
(b)
75
. p2sin20
Butsin0cos0=(1/2)sm20,soIN =4m:o(16z3)(outof thepage).
For0<0<'!r/2,N tendstorotatep counterclockwise;for'!r/2 <0<'!r,N rotatesp clockwise.Thusthe
stableorientationisperpendiculartothesurface-eithert or ..t..
Problem4,7
Saythefieldis uniformandpointsin they direction.First slidep
in frominfinityalongthex axis-this takesnowork,sinceF is J..dl.
(If E is notuniform,slidep in alonga trajectoryJ.. thefield.) Now
rotate(counterclockwise)into finalposition.The torqueexertedby
E is N =pxE =pEsinOz. The torqueweexertis N =pEsinO
x clockwise,anddOis counterclockwise,so thenet workdoneby us is
negative:
U = J:/2 pE sinOdO=pE (- cosO)1~/2=-pE (cosO- cos~)=-pE cos0=-p,E. Qed
Problem4,8
U = -pI,E2, butE2 = ~-!:r [3(p2,f)f - P2].SOU =~-!:r[PI'P2- 3(pI,f) (p2,f)]. Qed
Problem 4,9
1 q ~ q xx+yy+zz
(a)F =(p .V)E (Eq. 4.5);E =_4 ~ r =_4 (2 2 2)3/2''!rEOr '!rEOx + y + z
y
tE
O.
P
p
(
8 8 8
) q xFx = Px - +P - +pz- - .8x Y8y 8z 4'!rEO(X2+y2+ Z2)3/2
q
{ [
1 3 2x
] [
3 2y
]- 4'!rEOPx (x2+ y2 + Z2)3/2- 2x (X2+y2+Z2)5/2 +py -2x (X2+y2+ Z2)5/2
[
3 2z
]}
q
[
Px 3x
]
q
[
p 3r(p.r)
]+ pz -2x (X2+y2+Z2)5/2 = 4'!rEOr3 - ;:s(Pxx+Pyy+pzz) = 4'!rEOr3 - r5 x'
.
F = I_
4 1 ~ [p- 3(p.f) f] .'!rEor
(b)E =_4 1 -;.{3[p. (-f)]( -f) - p}=_4 1 13[3(p. f) f - p]. (This is fromEq. 3.104;theminussigns'!rEOr '!rEOr
arebecauser pointstowardp, in thisproblem.)
F =qE=1-41 q3[3(p. f) f - p],'!rEOr
[Notethattheforcesareequalandopposite,asyouwouldexpectfromNewton'sthird law.]
Problem4,10
~
§] 1 8 2 1 2 ~(a) Ub=P,n = kR; Pb=-V.p =-3"-8 (r kr) =-~3kr =~r r r
(b) For r <R, E = 3~oprf(Prob.2.12),soE =I-(k/EO)r.1
For r > R, sameas if all chargeat center;but Qtot =(kR)(4'!rR2)+ (-3k)(t'!rR3) = 0, soIE = 0.1
76 CHAPTER4. ELECTROSTATIC FIELDS IN MATTER
Problem 4.11
Pb=0; ab=P.il =:!:P (plussignat oneend-the oneP pointstoward;minussignat theother-the one
P pointsawayfrom).
(i) L » a. Thentheendslooklikepointcharges,andthewholethingis likeaphysicaldipole,QflengthLand
chargeP-rra2.SeeFig. (a).
(ii) L « a. Thenit's likeacircularparallel-platecapacitor.Fieldisnearlyuniforminside;nonuniform"fringing
field"at theedges.SeeFig. (b).
(iii) L ~ a. SeeFig. (c).
p p p
(a)Likea dipole (b) Likea parallel-platecapacitor
(c)
Problem 4.12
v = 4';EOJ I;jdT =p.{4';EOJ ~dT}.But thetermin curlybracketsis preciselythefieldof a uniformly
chargedsphere,dividedby p. The integralwasdoneexplicitlyin Prob.2.7and2.8:
I
R3 A
I
R3P cosB I ( >R)
{
I (4/3);R3Pf, (r>R),
}
3€or2P.r= 3€or2 ' r ,
1 .t. 1 411"€0 r So V(r,B) =dT --
4W<O!.' - p ~ (4/3)wR'pr, (r<R). I ...!...P.r ~ I" 'owoo, 1 (r <R).411"€0 R3 3€0
Problem 4.13
Think of it as twocylindersof oppositeuniformchargedensity:!:p.Inside,thefieldat a distances from
the axis of a uniformlychargecylinderis givenby Gauss'slaw: E211"se= -:OP1l"S2e:::}E = (p/2€0)s.For
twosuchcylinders,oneplusandoneminus,thenet field(inside)is E =E+ +E- =(p/2fO)(s+- s_). But
s+- s- =-d,soE =l-pd/(2€0),I whered isthevectorfromthenegativeaxistopositiveaxis.In thiscase
thetotal dipolemomentof a chunkof lengthe is P (1I"a2e)=(p7ra2e)d. Sopd =P, andIE = -P /(2€0),I for
s <a.
77
Outside,Gauss'slaw givesE27r8£= .1...p7ra2£:::}E = 1!!£.2a2!, for onecylinder. For thecombination,E =<0 <0s
E++E- = 1!!£.2
a2 (!:t.- iL ), where<0 s+ s-
d
s:!: = S T -j2
( d)(
12
)
-1
( )( )
-1
( )( )S:i: 2 a- 1 d s.d 1 d s.d- = ST- 8 +-Ts.d ~- ST- IT- ~- ST- 1:f:-8?t 2 4 82 2 82 82 2 82
1
( (s.d) d) . .= 82 S :f:S--;2 T "2 (keepmgonly 1storder terms in d).
(8+- L ) = ~[(s+s~ - ~)- (s-s~ +~)] =~(2S(S.d)-d ).s+ 8- 82 82 2 82 2 82 82
a2 1
E(s) =-- [2(P. 8)§ - P]
2fO 82 '
for 8 >a.
Problem4.14
Totalchargeon thedielectricis Qtot= is O"bda+Iv Pbdr = is P .da- Iv V.p dr. But thedivergence
theoremsaysis p. da=IvV.p dr, soQenc= O. qed
Problem4.15
(a)Pb=-v,p=-~~ (r2~)=-~; O"b=P.ii={ +P.~=k/b (atr=b), }r2or r r2 -P .r = -k/a (atr =a).
Gauss'slaw:::}E = 4:<0Q;~cr. For r < a, Qenc= 0,soIE = 0.1For r > b,Qenc= 0 (Prob.4.14),soI E =0.1
Fora<r <b,Qenc = (~k)(47ra2)+I: (~)47rf2dr=-47rka- 47rk(r- a)=-47rkr;soI E =-(k/for) r.1
(b)fD.da =Qfenc =O:::}D =0everywhere.D =foE+P =O:::}E = (-l/fo)P, so
IE =0 (forr <a andr > b);j IE =-(k/fOr) r (fora < r < b).I
Problem4.16
(a)SameasEo minusthefieldat thecenterofa spherewithuniformpolarizationP. The latter(Eq. 4.14)
is-P/3fO.SoIE =Eo+~p.1 D=foE=foEo+~P=Do- P +~P,soI D =Do- ~p.1
(b)SameasEo minusthefieldof:f: chargesat thetwoendsof the "needle"-but thesearesmall,andfar
away,so! E =Eo.J D =foE =foEo=Do - P, SOI D =Do- P .1
(c)Sameas Eo minusthe fieldof a parallel-platecapacitorwith upperplateat 0"=P. The latter is
-(l/fo)P, soIE =Eo+!op.1 D =foE=foEo+P, soID =Do.!
78 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
Problem 4.17
P
(uniform)
E
(fieldof twocircularplates)D
(sameasE outside,but lines
continuous,sinceV.D = 0)
Problem 4.18
(a) Apply J D .da=Q/encto thegaussiansurfaceshown.DA =aA ::} I D =a.1(Note:D =0insidethe
metalplate.)This is truein bothslabs;D pointsdown.
~ 2+u
(b) D =fE ::} E =a/fl in slab1,E = a/f2 in slab2. But f = tofT,so fl = 2fo;f2 = ~fO'lEI = a/2fo,I
I E2 =2a/3fo.1
(c)P =foXeE,soP = foXed/(fOfr)= (Xe/fr)a;Xe=fr -I::} P =(1- f;l)a.
(d)V =E1a+E2a= (aa/6fo)(3+4)=17aa/6fo,!
IPl=a/2,IIP2=a/3.1
(e)Pb=O; ab= +P1at bottomof slab(1)=a/2,ab= -PI attopofslab(1)=-a/2;
ab= +P2 at bottomof slab(2)=a/3,
ab= -P2 at topof slab(2) = -a /3.
(f) In slab1: {
In slab2: {
total surfacechargeabove:a - (a/2) =a/2,
}
E - ~ ..(
total surfacechargebelow:(a/2) - (a/3)+(a/3)- a = -a/2, ==>1- 2fO'
total surfacecharge above:a - (a/2) + (a/2) - (a/3) =2a/3,
}
E - 2a ..(
totalsurfacechargebelow:(a/3)- a =-2a/3, ==> 2 - 3fO'
]+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 onupperplate,-a on lower,D = a betweentheplates.
E =a/fO(inair)andE =a/f (in dielectric). So V ={;;~ +7~ =2~~(1+~) .
- ~ - ~ ( 2 ) I Ca - 2fr ICa - V - d 1+1/(r ==>Co - 1+fr .
In configuration(b), with potentialdifferenceV: E =V/d,soa =foE =fOV/d (in air).
79
P =EOXeE= EOXeV/d (in dielectric),soO'b= -EOXeV/d (at topsurfaceof dielectric).
O"tot=EoV/d=O'f+ O'b=O'f- EoXeV/d,soO'f=EoV(l+Xe)/d=EOErV/d(ontopplateabovedielectric).
Q 1 (
A A
)
A
(
V V
) AEO (1+Er )
I Cb 1+Er I=?Cb=V =V 0'2" + O'f2" =2V Eod +EOdEr =d ~ . Co=~.
[Whichis greater?~ - ~ = l+<r.- ~ = {l+<r)2-4<r= 1+2<r+4<~-4<r= (1-<r)2>0 SoC >C ]. Co Co 2 l+<r 2(1+<r) 2(1+<r) 2(1+<r) . b a.
If thex axispointsdown:
I ~ E p ]O'b(top surface) ~ O'f (top plate) ID
Problem 4.20
ID.da =Qfenc=>D41fr2= p!1fr3=>D = lpr =>E = (pr/3E)r, for r < R; D41fr2=p!1fR3=>D =
pR3/3r2=>E = (pR3/3Eor2)r, forr >R.
fO pR3 1
1
R P fo pR2 P R2
I
pR2(
1
)V =- }00 E. dl=3EO -:;.00- 3E }R rdr =3EO +3E ""2 = 3EO 1+2Er .
Problem4.21
LetQ bethechargeon a length£of theinnerconductor.
V -
Q Q Q
D21fs£=Q =>D =_2 0; E =-2 0 (a <s <b), E =-2 0 (b<r <c).1fS~ 1fEOS~ 1fES~
f
a
I
b
( Q )
dS l
C
( Q )dS Q [ (
b
)
EO (
C
)]
- E.dl- - -+ - --- In - +-In -
C - a 21fEO£ S b 21fd S - 21fEO£ a E b'
Q I 21fEO IV£ = In(b/a)+(l/Er) In(cjb)"
f D .da =
C
£ -
Problem4.22
Samemethodas Ex. 4.7:solveLaplace'sequationfor V;n(s,tj))(s < a) andVout(s,tj))(s > a), subjectto
theboundaryconditions x
{
(i) V;n = Vout at s =a,
(ii) E8~n = EO8~~ut at s = a,
(Hi)Vout -+ -Eos costj) for S » a. Eot
y
FromProb.3.23(invokingboundarycondition(Hi)):
00
V;n(s,tj))= 2::>k(akcosktj)+bksinktj)),
k=l
00
Vout(s,tj))= -Eoscostj)+L s-k(Ck cosk</>+dksink</».
k=l
(a)air
< - .2<r - 0 0 Y«r+1)d x «rH) d X «r+l) d
(a)dielectric
2 v - x - -«r+l) d X «rH) d <r+l) d X «r+l) d -
(b) air Yx
-
0 0 7 (left)d d x
(b) dielectric Yx Er7x (Er - 1)7 X -(Er - 1)7 Er7 (right)d
80 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
(1eliminatedtheconstanttermsby settingV =0 ontheyz plane.)Condition(i) says
I::ak(ak cosk</J+ bksink</J)=-Eoscos</J+L a-k(ck cosk</J+dk sink</J),
while(ii) says
f.rL kak-l(ak cosk</J+bksink</J)=-Eo cos</J- L ka-k-l(ck cosk4>+dksink</J).
Evidently bk=dk =0 for all k, ak=Ck =0 unlessk =1,whereasfork=1,
aal= -Eoa +a-ICI, f.ral= -Eo - a-2CI'
Solvingfor aI,
Eo
al = - (1+Xe/2)' Eo scos</J=so ~n(S,</J)= - (1+ Xe/2)
Eo-x
(1+Xe/2) ,
andhenceEin(s,</J)= -~ x=~2).1 As in thesphericalcase(Ex. 4.7),thefieldinsideis uniform.
Problem 4.23
1 Xe f.OX~ 1 X~
Po =f.oXeEo; EI =--3 Po =--3 Eo; PI =f.oXeEI =-- 3 Eo; E2 =--PI =-Eo;~ ~ 9
EvidentlyEn =(- ~e)n Eo, so
E=Eo+EI+E2+"'= [~(_~er] Eo.
The geometricseriescanbesummedexplicitly:
00 1
'"' n - -,~x-l-xn=O
so 1 Eo,
E = (1+ Xe/3)
whichagreeswith Eq. 4.49. [Curiously,this methodformallyrequiresthat Xe < 3 (elsethe infiniteseries
diverges),yetthe resultis subjectto nosuchrestriction,sincewecanalsogetit by themethodof Ex. 4.7.]
Problem 4.24
Potentials:
{
Vout(r,O)= -EorcosO+):r~lPI(COSO),
Vrned(r,O)= L(Alrl+r~l)1't(cosO),
~n(r,O) = 0,
BoundaryConditions:
{
(i) Vout = Vrned,
(ii) f.~ = f. ~or 0 or '
(Hi)Vrned = 0,
(r > b);
(a < r < b);
(r < a).
(r = b)j
(r = b)j
(r = a).
81
(i) ~ " BI " ( I hi ).-Eobcos0+~ bl+l~(cosO) = ~ Alb +bl+l ~(cosO)j
" [ I 1 hi ] "BI€r~ lAlb - - (l + 1)bl+2 ~(cosO)= -Eo cosO- ~(l +1)bl+2PI (cos0);
A al + hi = 0 ~ h = -a2l+1 A
I al+l I I.
(ii) ~
(Hi) ~
Fori =f.1 :
(i) BI = (A bl - a21+1AI) ~ B = A (b2l+1 - a21+1) .bl+1 I bl+l I I ,
[
2l+1 A
]
B
[(
l
) ]1-1 a I - I - 21+1 21+1 --€r lAlb +(l +1) bl+2 - -(l +1)bl+2 ~ Bl - -€rAI l +1 b +a ~ Al - Bl - O.
Fori = 1:
(ii)
B1 a3Al 3
(
3 3
)-Eob+b2=A1b-~ ~ BI-Eob =A12 b -a ;
..
) ( a3At ) B1 3 ( 3 3)(ll €r Al +2~ = -Eo - 2b3 ~ -2B1 - Eob = €rAI b +2a .
3 [ ( 3 3) ( 3 3)]
-3Eo
So -3Eob = Al 2 b - a +€r b +2a ; Al = 2[1- (afb)3]+€r[1+ 2(afb)3].
-3Eo (
a3
)Vrned(r,O)= 2[1- (afb)3]+€r[1+ 2(afb)3] r -;:2 cosO,
E(r,O) = -VVrned= 12[1- (afb)3]~E:r[1+2(afb)3]{(1+2r~3)cosOi'- (1- ;:) sinoo}.
(i)
Problem 4.25
Thereare four chargesinvolved: (i) q, (ii) polarization charge surrounding q, (Hi) surfacecharge (CTb)on
thetop surfaceof the lower dielectric, (iv) surfacecharge (CT~)on the lower surface of the upper dielectric.
In viewof 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,
[
-1 qdf€~ CTb CT~
]
~
(a) CTb = €oXe _4 3 - _2 - - (hereCTb= P.n = +Pz = €oXeEz)j7I"€o(r2+cF):2 €o 2€o
( )
I I
[
1 qdf€~ CTb CT~
]
(
I
)b CTb = €oXe_4 3 - _2 - _2 here CTb= -Pz = -€OXeEz .
7I"€o(r2 +cF)2 €o €o
Solvefor CTb,CT~:first divide by Xe and X~(respectively)and subtract:
CT~ CTb 1 qdf€~ I I
[
CTb 1 qdf€~
]X~- Xe = 271"(r2+cF)~ ~ CTb= Xe Xe+ 271"(r2+cF)~ .
82 CHAPTER 4. ELECTROSTATICFIELDS IN MATTER
Plug this into (a) andsolvefor O"b,using€~= 1+X~:
O"b - -1 qd/€~ (
'
)
O"b
(
' )
-1 qd Xe
- ";!Xel+Xe--Xe+Xe,soO"b=- 3[ ( ,)/]
;
41T(r2 +d2)2 2 41T(r2+d2)2 1+ Xe+Xe 2
,
{
-I qd 1 1 qd/€~
}
, 1 qd €rX~/€~
Xe 41T(r2 + d2)~[1+ (Xe+X~)/2]+ 21T(r2+ d2)~ ' so O"b= 41T(r2+d2)~[1+ (Xe+X~)/2]'
, -O"b-
The totalboundsurfacechargeis O"t=O"b+O"~= 417r qd ~E' l +X(~- +x., )/2 (whichvanishes,asit should,when(r2+d2) ~ ", x.
X~ =Xe)'The total boundchargeis (compareEq. 4.51):
qt= (X~- Xe)q _
I (€~ - €r )
q
2€~[1+ (Xe+ X~)/2] - €~+ €r €~'Iand hence
V(r) =~
{
q/€~ + qt
}
I(forz >0).
41T€0";X2+y2+ (z - d)2 ";x2 +y2+ (z + d)2
q
[
€~- €r
]
- ~
Meanwhile, since ~ +qt=7" 1+ €' + €r - €~ +€r'€r r r 1 [2q/(€~+€r)] I(forz <0).V(r) =41T€0 ";X2 +y2+ (z d)2
\
I
I
I
I
\Problem 4.26
FromEx. 4.5:
D =
{
O'Q (r <a)
41Tr2r, (r >a)
{
0,
Q ~-r,
} , E = 41Ttt ~
~r,41T€or
(r <a)
}
(a<r <b) .
(r>b)
1I 1 Q2 {I lb 1 1 1 100 1 } Q2 { I (-1)I b 1 (-1)1
00
}
w = - D.EdT=--41T - --r2dr+- -dr =- - - +--
2 2 (41T)2 € a r2 r2 €o b r2 81T € r a €o r b
Q2
{
1
(
1 1
)
1
}
Q2
(
1 Xe
)= 81T€0(1+Xe) ~- b +b = 81T€0(1+Xe) ~+b .
Problem 4.27
83
UsingEq. 4.55:W =!f J E2 dr. FromEx. 4.2andEq. 3.103,
E =
Wr<R =
Wr>R=
Wtot=
=
{
-I
3102P z,
3R ~(2cosBf+sinBO),for
fa (.£-.)2 ~11'R3=211'P2R3.2 3100 3 27 fa
fa (
R3P
)
2
J
1
( 2 . 2 )
2 .
- - 6" 4cosB+sm B r smBdrdBd<jJ2 3100 r
(R3p)2 l
1r
1
00 1 11'(R3p)2
8 211' (1+ 3COS2B)sinBdB 4"dr =-1 fa a R r 9100
11'(R3P)2(~ )=411'R3p2.9100 3R3 27100
211'R3 p2
9100
(-COSB-COS3B)I~(-3~3)1:
(r <R)
}
,
(r>R)
so
Thisisthecorrectelectrostaticenergyof theconfiguration,but it is not the "totalworknecessaryto assemble
thesystem,"becauseit leavesoutthemechanicalenergyinvolvedi~polarizingthemolecules.
UsingEq. 4.58: W =~J D.E dr. For r <:R, D = foE, sothiscontributionis thesameasbefore.
Forr < R, D = foE+P = -!p +P = jp = -2fOE,so~D.E= -2!fE2, andthiscontributionis
now(-2) (~~p:~3)=- ~~R::2, exactlycancellingthe exteriorterm. Conclusion:I Wtot =0.1This is not
surprising,sincethederivationin Sect.4.4.3calculatestheworkdoneonthefreecharge,andin this problem
thereisnofreechargein sight.Sincethisis anonlineardielectric,however,theresultcannotbeinterpretedas
the"worknecessaryto assembletheconfiguration"-the latterwoulddependentirelyon howyouassembleit.
Problem4.28
Firstfindthecapacitance,asa functionof h:
Air part:E =-bL ==>V = ..1L In(bJ a)
}
41rfOS 41rfO' >.. >'" 10
==>- =-j >..'=->..=lOr>".
OilPart:D = ~ ==>E = 2>"==>V = 2>"In(bJa)
fa 10 fa
41rs 41rfS 41rf '
Q=>..'h+>"(f- h) =fr>"h - >"h+>"f=>..[(fr- l)h +f)=>"(Xeh+ f), wheref is thetotalheight.
Q >"(Xeh+ f) (Xeh+ f)
C =V =2>"ln(bJa) 411'100=211'100In(bJa) .
Th t d l' .. b E 4 64' F - 1V2 dC - 1V2 21rfOXe
} I
V2
ene upwar LOrCeIS gIven y q. . . - 2" dh - 2" In(bfa). h = fOXe .
Thegravitationalforcedownis F =mg=p11'(b2- a2)gh. p(b2- a2)gIn(bJa)
84 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
Problem 4.29
8
(a) Eq. 4.5:::} F2 =(P2.V) EI =P2~ (Ed;
. uy
. PI A PI A
Eq. 3.1O3:::}EI = ~ ()= --4 3 z. Therefore4m:or 1rfoY
~y?jr
PIP2
[
d
(
1
)] A 3PIP2 A I 3PIP2 A IF2 =_4- -d 3" z =_4 4Z, or F2 = 4~z (upward).1rfO y Y 1rfoY 1rfor
z
y
To calculateF I, put P2 at the origin, pointingin the z direction;then PI
is at -r z, and it pointsin the -y direction. So FI = (PI' V) E2 =
-PI 8:2
1
- - - ; weneedE2 asa functionof x, y, andz.y x-y-O,z--r
.
I
I
E E
ll
[
3(P2' r)r
]
h A A A A d hFrom q. 3.104: 2 =-3" ? - P , were r = xx +yy +ZZ,P2 = -P2Y, an ence41rfOr r-
P2 .r = -P2Y'
E2 = ~
[
-3Y(XX+yy +zz) +(x2+y2+Z2)y
]
=~
[
-3XYX+ (x2- 2y2+z2)y- 3YZZ
]41rfO (x2 +y2+z2)5/2 41rfO (x2+y2+z2)5/2
~
{-~~2Y[-3XYX+(x2- 2y2+Z2)y- 3yzz]+~(-3xx - 4yy- 3ZZ)}
;
~~ 2~ ~
~ -3z z; FI =-PI (~ 3r Z)=- 3PIP2Z.41rfO r5 41rfOr5 41rfor4
8E2 =
8y
8E2
18y (0,0)
=
TheseresultsareconsistentwithNewton'sthirdlaw:FI = -F2.
(b)Frompage165,N2= (P2x EI) +(r x F2). ThefirsttermwascalculatedinFrob.4.5;thesecondwe
getfrom(a), usingr =ry:
P2 X EI = PIP2 ( A
41rfor3 -X)j
F (
-
) (3PIP2 -) 3PIP2 A I N 2PIP2 Ar x 2 = ry x _4 4 Z =_4 3 x; so 2=_4 3 X.1rfor 1rfor 1rfor
This is equalandoppositeto thetorqueonPI dueto P2,with respectto thecenterof PI (seeFrob. 4.5).
Problem 4.30
Netforceis I to therightI (seediagram).Notethatthefieldlinesmustbulgeto theright,asshown,because
E is perpendicularto thesurfaceofeachconductor.
E
85
Problem4.31
P =kr =k(xx +yy +zz) =? Pb= -V.p = -k(l +1+1)=1-3k.1
Totalvolumeboundcharge:I Qvol =-3ka3.1
(Jb=P.il. At topsurface,il =z, z =a/2jsoO"b=ka/2.Clearly,I O"b=ka/21onall sixsurfaces.
T~talsurfacebound charge: I Qsurf=6(ka/2)a2=3ka3.1 Total bound chargeis zero. if
Problem4.32
f q ~ 1 q f qXe fD.da=Qfonc::}D = -4 2 rj E =-D =4 (1 ) 2"; P =fOXeE = 4 (1 ) 2"'7rr 10 7rfO +Xe r 7r +Xe r
Pb=-V.p = 47r(~~Xe)(V. ~)=-q1 ~eXe83(r) (Eq. 1.99)jO"b=P.f = 47r(1~X~e)R2;
Qsurf=(Jb(47rR2)=I q Xe .1 The compensatingnegativechargeis at thecenter:1+Xe
j PbdT =- l qXe j 83(r)dT=-q-1Xe .+Xe +Xe
Problem4.33
Ell is continuous(Eq. 4.29);Dl. is continuous(Eq. 4.26,with O"f= 0). So EXl =-EX2' DYl = DY2 ::}
E1EYI =f2EY2'andhence
tan02= EX2/EY2= EYl = E2. Qed
tan01 EXl/EYl EY2 El
If 1isairand2 is dielectric,tanO2/tan01= E2/EO> 1,andthefieldlinesbendawayfromthenormal.This is
theoppositeof lightrays,soa convex"lens"woulddefocusthefieldlines.
Problem4.34
In viewof Eq. 4.39, the net dipole momentat the center is pi =P - 1~~ep =I';Xo p =tp. We want the
potentialproducedby pi (at thecenter)andO"b(at R). Useseparationof variables:
1
00 B
)
Outside:V(r,O)=L rl:l Pz(cosO) (Eq.3.72)
1=0
. 1 pcosO 00 .lnstde: V(r,O)=_4 ~ +LAlrIPI(cosO) (Eqs.3.66,3.102)
7rEO Err 1=0
{
R~L=AIRI, or BI =R2/HAI (l ¥' 1)
}
V continuousat R ::} .
B1 - 1 P - p- 3
R2 - 47rEoErR2+ AIR, or B1 - 47rfOf~+AIR
8V
I
av
i8r R+ ar R-
""' BI 1 2pcosO ""' I 1 1= - L)l +1)Rl+2Pz(cosO)+_4 R3 - LlAIR - PI(cosO)=--O"b7rEO Er EO
1 ~ 1
(
~
)
av
I {
I 2pcosO ""' I 1
}
= --p. r =-- EoXeE.r =Xe _a =Xe --4 R3 + ~lAIR - PI(cosO) .EO EO r R- 7rEO lOr
86 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
-(I +1)R~~2- lAIRI-I = XelAIRI-I (l ¥-1); or - (2l+ l)AIRI-I = XelAIRI-I =>Al = 0 (£¥-1).
BI 1 2p ( 1 2P ) P AIR3 1 XeP AIR3Forl=l: -2-+---AI=Xe ---+A1 -BI+---=---+Xe-;R3 411"fO frR3 411"fOfrR3 411"fOfr 2 411"fofr 2
-~ - AIR3+ ~ - AIR3 = _~XeP +XeAIR3 =>AIR3 (3+Xe)=~XeP.
411"fOfr 411"fOfr 2 411"fOfr 2 2 411"fOfr
=>Al = ~ 2XeP =~ 2(fr -l)p; BI = ~
[
1+ 2(fr - 1)
]
=~~.
411"foR3fr(3+Xe) 411"foRafr(fr + 2) 411"fOfr (fr + 2) 411"fOfrfr + 2
V(r,O)=(p COSO)(~ )(r ~R).411"for2 fr +2
Meanwhile,forr::; R, V(r,0)= ~P cosO+ 1 prcos(}2(fr-1)
411"fofrr2 411"fO R3 freEr +2)
P cos()
[ (
fr - 1)
r3
]= I 4 2 1+2 - 2 R3 (r::;R).1I"for fr fr +
Problem 4.35
Giventwosolutions,VI (andEI =-VVI, DI =fEd andV2(E2=-VV2, D2=fE2),defineV3==V2- VI
(E3=E2- EI' Da=D2- DI).
Iv V.(VaD3)dr = Is V3Da'da=0, (Va=° onS), soI(VV3) .D3dr +I V3(V.D3)dr =0.
But V.D3 =V.D2 - V.DI =PI- PI = 0,andVV3 = VV2 - VVI = -E2 +EI = -E3, so IE3' D3dr =O.
But D3 = D2 - DI = fE2 - EEl = fE3, soI f(E3)2 dr =0. But f > 0, soE3 = 0,soV2- VI = constant.But
at surface,V2= VI, so V2= VI everywhere.qed
Problem 4.36
(a) Proposedpotential:I VCr)=Vo~.1If so,thenIE = - VV =VO~f, I in whichcase P =fOXeVo~f,
in theregionz < 0. (P = o for z > 0,of course.)ThenCTb= fOXeVo; (f.ft) =1-fO~Vo.1(Note: ft points out
of dielectric=> ft=-f.) This CTbis on thesurfaceat r = R. Theflatsurfacez =0carriesnoboundcharge,
sinceft = z 1..f. Noris thereanyvolumeboundcharge(Eq. 4.39). If V is to havethe requiredspherical
symmetry,the netchargemustbeuniform:
CTtot411"R2 =Qtot = 411"fORVo(sinceVo = Qtot/411"foR),so CTtot= foVo/R. Therefore
-
{
(fOVo/R), onnorthernhemisphere
}CTI - (fOVolR)(l +Xe), onsouthernhemisphere'
(b) By construction, CTtot=CTb+CTI =foVo/R is uniform(onthenorthernhemisphereCTb=0,CTI =foVo/R;
on the southern hemisphereCTb= -foXeVo/R, soCTI= fVo/R). Thepotentialof a uniformlychargedsphereis
Vo= Qtot =CTtot(411"R2)=fOVo R2 =VoR. ./
411"for 411"fOr R for r
(c) Sinceeverythingis consistent,andtheboundaryconditions(V =Voat r = R, V -+0at00)aremet,
Prob.4.35guaranteesthat this is thesolution.
87
(d)Figure(b) worksthesameway,but Fig. (a)doesnot: ontheflat surface,P is notperpendicularto ft,
sowe'dgetboundchargeonthissurface,spoilingthesymmetry.
Problem4.37
Eext= ~ 8. Sincethe sphereis tiny, this is essentiallyconstant,and henceP = €oXe/ Eext (Ex. 4.7).27r€08 1+ Xe 3
( )( ) ( ) ( )( )
2
( )( )F - €oXe ~ ~ ~ 8dr - €oXe ~ ! -1 8 dr- J 1 + Xe!3 27r€08 d8 27r€08 - 1+ Xe/3 2no 8 82 J
-Xe (~ )2-~7rR38- - (~ ) )..2R381+Xe/3 47r2€0 833 - 3 + Xe 7r€083.=
Problem4.38
Thedensityof atomsis N = (4/3)7rR3'The macroscopicfieldE is Eself+ Eelse,whereEselfis theaverage
fieldoverthespheredueto theatomitself.
p =o:Eelse =}P =No:Eelse.
[Actually,it is thefieldat the center,not theaverageoverthesphere,that belongshere,but thetwoarein
factequal,aswefoundin Prob.3.41d.]Now
1 p
Eself= - 47r€0R3
(Eq.3.105),so
1 0:
(
0:
) (
NO:
)E =- 47r€0R3 Eelse +Eelse = 1- 47r€oR3 Eelse = 1 - 3€0 Eelse.
So
P= No:
(1 - N o:/3€0)E = €oXeE,
andhence
No:/€o
Xe= (1- No:/3€0)'
Solvingfor a:
No: No: No:
( Xe)Xe - -3 Xe=- =}- 1+_3 =Xe,€o €o €o
or
€o Xe 3€0 Xe 3€0 (€r - 1)a = N (1+Xe!3)= Ii (3+Xe' ButXe=€r- 1, so0:= Ii z+2 . qed
Problem4.39
Foranidealgas,N =Avagadro'snumber/22.4liters=(6.02x 1023)/(22.4x 10-3)=2.7X 1025.No:/€o =
(2.7x 1O25)(47r€0x 1O-3O),8/€0=3.4X 10-4,8,where,8is thenumberlistedinTable4.1.
H: (3=0.667,No:/€o=(3.4x 10-4)(0.67)=2.3x 10-4, Xe=2.5X 10-4
}
He: (3=0.205,No:/€o= (3.4x 10-4)(0.21)=7.1x 10-5, Xe =6.5X 10-5 ..
Ne: (3=0.396,No:/€o=(3.4x 10-4)(0.40)=1.4x 10-4, Xe=1.3X10-4 agreementISqUItegood.
Ar: (3=1.64, N0:/€o =(3.4X 10-4)(1.64)=5.6x 10-4, Xe= 5.2X 10-4
88 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
Problem 4.40
(a) (u) - J~:E ue-u/kTdu - (kT)2e-u/kT[-(u/kT) - l]I~~E
JPE e-u/kT du - -kTe-U/kT lpE-pE -pE
{
[e-pE/kT- ePE/kT]+ [(pE/kT)e-pE/kT + (pE/kT)ePE/kT]
}
= kT
e-pE/kT - epE/kT
[
ePE/kT +e-PE/kT
] (
PE
)= kT - pE epE/kT- e-pE/kT =kT - pE coth kT .
A - - -(u)
I { (
PE )
kT
}P =N(p); p =(pcosO)E=(p. E)(E/E) =-(u)(E/E); P =Np pE =Np coth kT - pE .
Lety ==P/Np, x ==pE/kT.Theny =cothx-1/x. Asx --+0,y= (~+f - ~;+...)-~ =f-~;+...~
0,sothegraphstartsat theorigin,withaninitialslopeof 1/3. Asx --+00,y --+coth(oo)= 1,sothegraph
goesasymptoticallyto y =1 (seeFigure).
.E...
np'
11 """""""""""""'" .
pe/kT
(b) For smallx, y :::::::kx, so;; :::::::-f!-r,or P :::::::~E = €oXeE=>P is proportionaltoE, and Xe= Np2 .p ~~
For waterat 20°= 293K p = 6.1X 10-30em' N = molecules= moleculesX molesX !\rams., 'volume mole gram volume
N - (60 10
23
) (
1
) (10
6
)
- 033 1029. - (O.33Xl029)(6.1Xl0-30)2- j"1;)l12 T bl 4 2
.
- . X X 18 x-. X , Xe- (3)(8.85xl0-12)(1.38XlO-23)(293)- ~ a e . givesan
experimentalvalueof 79,soit's prettyfar off.
For watervaporat 100°= 373K, treatedasanidealgas,v~~r::e= (22.4X 10-3)X (~~~)= 2.85X 10-2 m3.
N = 6.0X 1023
2.85X 10-2 = 2.11X 1025.,
(2.11x 1025)(6.1x 10-30)2 -
Xe= (3)(8.85x 10-12)(1.38x 10-23)(373)=15.7x 10 3.1
Table4.2gives5.9x 10-3,sothis timetheagreementis quitegood.