BI I hi - Egloos

81

(i) ~ " BI " ( I hi ).-Eob cos0 + ~ bl+l ~(cosO) = ~ Alb + bl+l ~(cosO)j

" [ I 1 hi ] "BI€r ~ lAlb - - (l + 1)bl+2 ~(cosO) = -Eo cosO- ~(l + 1) bl+2 PI (cos 0);

A al + hi = 0 ~ h = -a2l+1 AI 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 ;

..) ( a3 At ) 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

There are four charges involved: (i) q, (ii) polarization charge surrounding q, (Hi) surface charge (CTb)onthe 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) isqt = q.+ qp = qf(1 + X~) = qf€~. As in Ex. 4.8,

[-1 qdf€~ CTb CT~

]

~

(a) CTb = €oXe _4 3 - _2 - - (here CTb= 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. ELECTROSTATIC FIELDS IN MATTER

Plug this into (a) and solve for 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 total bound surface charge is O"t=O"b + O"~= 417r qd ~ E' l +X(~- +x. , )/ 2 (which vanishes, as it should, when

(r2+d2) ~ ", x.

X~ = Xe)' The total bound charge is (compare Eq. 4.51):

qt = (X~ - Xe)q _I (

€~ - €r

)q

2€~ [1 + (Xe + X~)/2] - €~ + €r €~' Iand hence

V(r) = ~{

q/€~ + qt}

I(for z > 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(for z < 0).V(r) = 41T€0 ";X2 + y2 + (z d)2

\I

III

\Problem 4.26

From Ex. 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

Using Eq. 4.55: W = !f JE2 dr. From Ex. 4.2 and Eq. 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 (R3 P

)2

J1

( 2 . 2 )2 .

- - 6" 4cos B+ sm B r smBdrdBd<jJ2 3100 r

(R3 p)2 l1r

100 1 11'(R3p)2

8 211' (1 + 3 COS2B)sin BdB 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

This is the correct electrostatic energy of the configuration, but it is not the "total work necessary to assemblethe system," because it leaves out the mechanical energy involved i~ polarizing the molecules.

Using Eq. 4.58: W = ~ JD.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

now(-2) (~~p:~3) = - ~~ R::2, exactly cancelling the exterior term. Conclusion: I Wtot = 0.1 This is notsurprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problemthere is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted asthe "work necessary to assemble the configuration" -the latter would depend entirely on how you assemble it.

Problem 4.28

First find the capacitance, as a function of h:

Air part: E = -bL ==> V = ..1L In(bJa)

}

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), where f is the total height.

Q >"(Xeh+ f) (Xeh + f)C = V = 2>"ln(bJa) 411'100= 211'100In(bJa) .

Th t d l' .. b E 4 64' F - 1 V 2 dC - 1 V 2 21rfOXe

} I

V 2

e ne upwar LOrCeIS gIven y q. . . - 2" dh - 2" In(bfa) . h = fOXe .The gravitational force down is F = mg = p11'(b2 - a2)gh. p(b2 - a2)g In(bJa)


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 4 Z, or F2 = 4~z (upward).1rfO y Y 1rfoY 1rforz

y

To calculate F I, put P2 at the origin, pointing in the z direction; then PIis at -r z, and it points in the -y direction. So FI = (PI' V) E2 =-PI 8:2

1- - - ; we need E2 as a function of x, y, and z.y x-y-O, z--r

.I

I

E Ell

[

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)= - 3PIP2 Z.41rfO r5 41rfO r5 41rfor4

8E2 =8y

8E2

18y (0,0)=

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 = PIP2 (A

41rfor3 -X)jF (

-) (3PIP2 -) 3PIP2 A

I

N 2PIP2 A

r x 2 = ry x _4 4 Z =_4 3 x; so 2 =_4 3 X.1rfor 1rfor 1rfor

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 I to the right I (see diagram). Note that the field lines must bulge to the right, as shown, becauseE is perpendicular to the surface of each conductor.

E

85

Problem 4.31

P = kr = k(xx + yy + zz) =? Pb= -V.p = -k(l + 1+ 1) = 1-3k.1

Total volume bound charge: I Qvol = -3ka3.1

(Jb=P.il. At top surface, il = z, z = a/2j so O"b= ka/2. Clearly, I O"b = ka/21 on all six surfaces.

T~tal surface bound charge: I Qsurf= 6(ka/2)a2 = 3ka3.1 Total bound charge is zero. ifProblem 4.32

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. ~ )= -q 1 ~eXe 83(r) (Eq. 1.99)j O"b=P.f = 47r(1~X~e)R2;

Qsurf= (Jb(47rR2) = I q Xe .1 The compensating negative charge is at the center:1 + Xe

j PbdT = - lqXe j 83(r)dT = -q-1 Xe .+ 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. Qedtan 01 EXl/EYl 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 istheopposite 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 p = tp. We want thepotentialproduced by pi (at the center) and O"b(at R). Use separation of 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- 3R2 - 47rEoErR2 + AIR, or B1 - 47rfOf~+ AIR

8V

I

avi8r 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


-(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 AIR3

Forl=l: -2-+---AI=Xe ---+A1 -BI+---=---+Xe-;R3 411"fO frR3 411"fOfrR3 411"fOfr 2 411"fo fr 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, for r::; R, V(r, 0) = ~P cosO + 1 prcos(} 2(fr -1)411"fo frr2 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

Given two solutions, VI (and EI = -VVI, DI = fEd and V2 (E2 = -VV2, D2 = fE2), defineV3==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.

But V.D3 = 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 I f(E3)2 dr = 0. But f > 0, so E3 = 0, so V2 - VI = constant. Butat surface, V2 = VI, so V2 = VI everywhere. qedProblem 4.36

(a) Proposed potential: I VCr) = Vo~.1 If so, then IE = - VV = VO~ f, I in which case P = 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 outof 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 sphericalsymmetry, the net charge must be uniform:

CTtot411"R2 = Qtot = 411"fORVo(since Vo = Qtot/411"foR), so CTtot= foVo/R. Therefore

-{

(fOVo/ R), on northern hemisphere}CTI - (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 = CTtot(411"R2) = fOVo R2 = VoR. ./411"for 411"fOr R for r

(c) Since everything is consistent, and the boundary conditions (V = Vo at r = R, V -+ 0 at 00) are met,Prob. 4.35 guarantees that this is the solution.

87

(d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to ft,sowe'd get bound charge on this surface, spoiling the symmetry.Problem 4.37

Eext = ~ 8. Since the sphere is tiny, this is essentially constant, and hence P = €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 - - (~ ) )..2R3 81 + Xe/3 47r2€0 833 - 3 + Xe 7r€083 .

=

Problem 4.38

The density of atoms is N = (4/3)7rR3'The macroscopic field E is Eself + Eelse, where Eself is the averagefieldover 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 infact equal, as we found in Prob. 3.41d.] Now

1 pEself = - 47r€0R3

(Eq. 3.105), so1 0:

(0:

) (NO:

)E = - 47r€0 R3 Eelse + Eelse = 1 - 47r€oR3 Eelse = 1 - 3€0 Eelse.

So

P= No:(1 - N o:/3€0) E = €oXeE,

and henceNo:/€o

Xe = (1 - No:/3€0)'

Solving for a:No: No: No:

( Xe)Xe - -3 Xe= - =} - 1+ _3 = Xe,

€o €o €oor

€o Xe 3€0 Xe 3€0 (€r - 1

)a = N (1 + Xe!3) = Ii (3 + Xe' But Xe = €r - 1, so 0:= 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. No:/€o =(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.

H: (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 ..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, N 0:/ €o = (3.4 X 10-4)(1.64) = 5.6 x 10-4, Xe = 5.2 X 10-4


Problem 4.40

(a) (u) -J~:E ue-u/kT du - (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, so the graph starts at the origin, with an initial slope of 1/3. As x --+00, y --+coth(oo) = 1, so the graphgoes asymptotically to y = 1 (see Figure).

.E...np'

11 """""""""""""'" .

pe/kT

(b) For small x, y :::::::kx, so;; :::::::-f!-r, or P :::::::~E = €oXeE =>P is proportional to E, and Xe = Np2 .p ~~

For water at 20° = 293 K p = 6.1 X 10-30 em' N = molecules= moleculesX molesX !\rams., 'volume mole gram volume

N -(6 0 1023

) (1

) (106)

- 0 33 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 . givesanexperimental 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.

N = 6.0 X 10232.85 X 10-2 = 2.11 X 1025.

,(2.11 x 1025)(6.1x 10-30)2 -

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.

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