Electrostatic fields in matter

107

Chapter 4
Electrostatic Fields in Matter
Problem 4.1
−30
E = V / x = 500 /10−3 = 5 ×105 . Table 4.1: α / 4πε 0 = 0.66 ×10 , so
α = 4π (8.85 × 10−12 )(0.66 ×10 −30 ) = 7.34 ×10 −41.
p = α E = ed ⇒ d = α E / e = (7.34 × 10−41 )(5 ×105 ) /(1.6 ×10 −19 ) = 2.29 × 10−16 m.

d / R = (2.29 ×10−16 ) /(0.5 ×10 −10 ) = 4.6 ×10−6. To ionize, say d = R. Then R=

α E / e = αV / ex ⇒ V = R ex / α = (0.5x 10−10 )(1.6x10 −19 )( 10−3 )/(7.34x10 −41 ) = 108V .

Problem 4.2
1

First find the field, at radius r, using Gauss’ law: ∫ E.da = ε Qenc , or
0

1 1
E=
Qenc .
4πε 0 r 2
4π q r −2 r / a 2
4q  a −2 r / a  2
a2 
Qenc = ∫ ρ dr =
e
r dr = 3  − e
 r + ar + ÷

0
π a 3 ∫0
a  2
2 


2q  −2 r / a 2
a2 
r
r2 
−2 r / a
−
e
(
r
+
ar
+
)
1
−
e
(1
+
2
+
2
) .
=
= q

a 2 
2 
a
a 2 

r

r
0

[Note: Qenc (r → ∞) = q. ] So the field of the electron could is
Ee =

1 q 
r
r2 
−2 r / a
1
−
e
(1
+
2
+
2
) . The proton will be shifted from r = 0 to the
4πε 0 r 2 
a
a 2 

point d where E e = E (the external field):
1 q 
d
d 2 
−2 d / a 
E=
1 − e
1 + 2 + 2 2 ÷ .
4πε 0 d 2 
a
a 



108

Expanding in powers of (d/a):
2

e

3

2

3

d
d
d2 

 2d  1  2 d  1  2 d 
d  4d 
−2 d / a 
= 1− 
+
−
+
...
=
1
−
2
+
2
−
+
...1
−
e
1
+
2
+
2

÷
÷ 
÷

÷

 ÷
 ÷
a
a
a2 
 a  2  a  3!  a 
 a  3 a 

2
3


d
d
d2 
d  4d 
1 − 1 − 2 + 2  ÷ −  ÷ + ... ÷1 + 2 + 2 2 ÷

÷
a
a
a 
 a  3 a 


d
d2
d
d2
d3

d2
d3 4 d3
γ −γ − 2 − 2 2 + 2 + 4 2 + 4 3 − 2 2 − 4 3 +
+ ...
a
a
a
a
a
a
a
3 a3
3
4d 
 ÷ + higher order terms.
3 a 

−2 d / a

=
=
=

E=

1 q  4 d3 
1 4
1
=
(qd ) =

p. α = 3πε 0 a 3 .
2 
3 ÷
3
4πε 0 d  3 a  4πε 0 3a
3πε 0 a 3

[Not so different from the uniform sphere model of Ex.(see Eq. 4.2). Note that
1

3

3

this result predicts 4πε α = 4 a = 4 (0.5 ×10 ) = 0.09 ×10 m , compared with an
0
experimental value (table 4.1) of 0.66 x 10 −30 m 3 . Ironically the “classical”
formula (Eq. 4.2) is slightly closer to the empirical value.]
3

−10 3

30

3

Problem 4.3
ρ ( r ) = Ar . Electric field (by Gauss’s Law):
2
Ñ

∫ E.da = E (4π ) =

1
1
Qenc =
ε0
ε0

∫

r

0

r r2 r
1 4π A r 4 Ar 2
Ar 4π r d r , or E=
=
. This “internal”
4π r 2 ε 0 4 4ε 0

field balances the external field E when nucleus is “off-center” an amount d :
ad 2 4ε 0 = E ⇒ d = 4ε 0 E A .So the induced dipole moment is p = ed = 2e ε 0 A E
.Evidently p is proportional to E1 2 .
For Eq. 4.1 to hold in the weak – field limit , E must be proportional to r , for
small r , which mean that must go to a constant (not zero) at the origin : ρ (0) ≠ 0
(nor infinite)


109

Problem 4.4
1

αq

q

Field of q : 4π ∈ r 2 rˆ . Induced dipole moment of atom : p = α E = 4πε r 2 rˆ .
0
0
Field of this dipole , at location of q ( θ = π , in Eq.3.103 ) : E =

1 1  2α q 

÷
4πε 0 r 3  4πε 0 r 2 

( to the right ) .
Force on q due to this field : ( attractive ) .
Problem 4.5
p

1
Field of p1 at p2 ( θ = π 2 in Eq. 3.103 ) : E1 = 4πε r 3 θˆ ( points down ).
0

pp

o

1 2
Torque on p2 : N 2 = p2 × E1 = p2 E1 sin 90 = p2 E1 = 4πε r 3 ( points into the page).
0

2p p

1 2
Torque on p1 : N1 = p1 × E2 = 4πε r 3 (points into the page )
0

Problem 4.6
Use image dipole as shown in Fig .(a). Redraw , placing pi at the origin , Fig.
(b).
Ei =

p
4πε 0 ( 2 z )

3

( 2 cos θ rˆ + sin θθˆ ) ;

p = p cos θ rˆ + p sin θθˆ .

( out of the page ).
But sin θ cos θ = ( 1 2 ) sin 2θ , so N =

p 2 sin 2θ
(out of thae page )
4πε 0 ( 16 z 3 )

For 0 < θ < π 2, N tends to rotate p counterclockwise ; for π 2 < θ < π , N rotates p
clockwise . Thus
the stable orientation is perpendicular to the surface –either ↑ or ↓ .


110

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 ⊥ dl. (If E is not
uniform , slide p in along a trajectory ⊥ the field.) Now rotate (counterclockwise
) into final position . The torque exerted by E is N = p × E = pE sin θ zˆ. The torque
we exert is N = pE sin θ clockwise , and dθ is counterclockwise , so the net work
done by us is negative :
U=

θ

∫

π 2

pE sin θ dθ = pE ( − cos θ

)

θ
π 2

π

= − pE  cos θ − cos ÷ = − pE cos θ = − p.E.
2


qed

Problem 4.8
U = − p1.E2 , but E2 =

1 1
1 1
3( p2 .rˆ)rˆ − p2 ] . So U =
 p1 p2 − 3 ( p1rˆ ) ( p2 .rˆ )  .
3 [
4πε 0 r
4πε 0 r 3 

Problem 4.9

( a ) F = ( p.∇ ) E ( Eq.4.5) ; E =

qed

1 q
q
xxˆ + yyˆ + zzˆ
rˆ =
2

4πε 0 r
4πε 0 ( x 2 + y 2 + z 2 ) 3 2


∂
∂
∂ q
x
Fz =  px
+ py
+ pz
÷
∂x
∂y
∂z  4πε0 ( x 2 + y 2 + z 2 ) 3 2

=

 

1
3
2x
 
 + py
p
−
x

4πε0  ( x 2 + y 2 + z 2 ) 3 2 2 ( x 2 + y 2 + z 2 ) 5 2 


 

=

q
 pz 3 x

−
p
x
+
p
y
+
p
z
=
(
)
x
y
z
3
5

4πε0 
r
r
 4πε0

q

q



2y
− 3 x
 + pz
 2 ( x2 + y 2 + z 2 ) 5 2 



 p 3r ( p.r ) 
 3−
 .
5
r
r

x







111

F=

1 q
 p − 3 ( p.rˆ ) rˆ  .
4πε 0 r 3 

( b)

E=

{

}

1 1
1 1
3  p. ( −rˆ )  ( − rˆ ) − p =
3 ( p.rˆ ) rˆ − p  . (This is from
3
4πε 0 r
4πε 0 r 3 

Eq. 3.104; the minus signs are beause r points toward p , in this
problem .)
F = qE =

1 q
3 ( p.rˆ ) − p  .
4πε 0 r 3 

[Note that the forces are equal and opposite , as you would expect from
Newton’s third law .]
Problem 4.10
(a) σ b = P.nˆ = kR; ρb = −∇.P = −

1 ∂ 2
1
r kr ) = − 2 3kr 2 = −3k .
(
3
r ∂r
r

1

(b) For r < R , E= 3ε ρ rrˆ (Prob. 2.12 ) , so E = − ( k ε 0 ) r.
0

For r > R, same as if all charge at center ; but
4

Qtot = ( kR ) ( 4π R 2 ) + ( −3k )  π R 3 ÷ = 0 , so E = 0.
3


Problem 4.11
ρb = 0;σ = P.nˆ = ± P (plus sign at one end-the one P points toward ; minus

sign at the other-the one P points away from ).

112

(i)

L >> a. Then the ends look like point charges , and the whole
thing is like a physical dipole , of length L and charge Pπ a 2 .
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).

Problem 4.12
V=

1
4πε 0

P.rˆ

∫r

2

 1
rˆ 
dr = P. 
dr  . But the term in curly brackets is precisely the
2
∫
 4πε 0 r


field of a uniformly charged sphere , divided by p. The integral was done
esplicitly in Prob .2.7 and 2.8 :
 1 ( 4 3) π R 3 p
rˆ,

r2
1
rˆ
1  4πε 0
dr = 
4πε 0 ∫ r 2
p  1 ( 4 3) π R 3 p
rˆ,
 4πε
R3
0


( r > R) 

 R3
R 3 P cos θ
ˆ
P
.
r
=
,

( r > R ) ,
2
2
3
ε
r
3
ε
r

0
0

 So V ( r , θ ) = 
Pr cos θ
 1
( r < R ) , 
,
 3ε P.r = 3ε
0

 0



( r < R ) 

Problem 4.13
Think of it as two cylinders of opposite uniform charge density ± ρ . Inside the
field


113

At a distance s from the axis of a uniformly charge cylinder is given by Gauss’s
1

2
law : E 2π sl = ε ρπ s l ⇒ E = ( ρ 2ε 0 ) s. For two such cylinders , one plus and one
0

minus. The net field (inside) is E = E+ + E− = ( ρ 2ε 0 ) ( s+ − s− ) . But s+ − s− = −d , so
E = ρ d ( 2ε 0 ) , where d is the

Vector from the negative axis to positive axis . In this case the total dipole
2
2
moment of a chunk of length l is P ( π a l ) = ( ρπ a l ) d . So ρ d = P, and
E = − P ( 2ε 0 ) , for s < a .

Outside , Gauss s law give E 2π sl =

’

1
ρ a 2 sˆ
ρπ a 2l ⇒ E =
, For one cylinder . For
ε0
2ε 0 s

ρ a 2  sˆ+ sˆ− 
the combination , E = E+ + E− =
 − ÷, where
2ε 0  s+ s− 
d
s± = s m ;
2
−1

−1


s± 
d  2 d 2
1
d  s.d 
1
d  s.d 
=  s m ÷ s +
ms.d ÷ ≅ 2  s m ÷1 m 2 ÷ ≅ 2  s m ÷ 1 ± 2 ÷
2

s± 
2 
4
s 
2 
s 
s 
2 
s 

=

( s.d ) d 
1
s±s 2 m ÷
2 
s 
s
2

( keeping only 1st order terms in d )


 sˆ+ sˆ−  1 
( s.d ) d  
s   1  s ( s.d )
( s .d )
− d ÷.
 − ÷ = 2  s + s 2 − ÷−  s − s s 2 + ÷ = 2  2
2

s
2 
2   s 
s
 s+ s−  s 

E ( s) =

a2 1
 2 ( P.sˆ ) sˆ − P  , for s > a
2ε 0 s 2 

Problem 4.14


114

Total charge on the dielectric is Qtot = Ñ
∫ s σ b da + ∫ν ρ b dτ = Ñ
∫ s P.da − ∫ν ∇.Pdτ . But the
divergence theorem says

Ñ
∫ P.da = ∫ ∇.Pdτ ,
s

ν

so Qenc = 0. qed

Problem 4.15
1 ∂  2k
k
 r ÷= − 2 ;
2
r ∂r  r 
r
+ P.rˆ = k b
r = b, 
σ b = P.nˆ = 

− P.rˆ = − k a
r = a.

(a) ρb = −∇.P = −

1 Q

enc
Gauss’s law ⇒ E = 4πε r 2 rˆ. For r < a , Qenc = 0 , so E = 0 . For r > b , Qenc =0
0
(Prob. 4.14) , so E = 0
r  −k 
−k 
2
2
÷( 4π a ) + ∫a  2 ÷4π r dr = −4π ka − 4π k ( r − a ) = −4π kr ; so
a
r
 





For a < r< b , Qenc = 
E = − ( k ε 0 r ) rˆ.

(b)

Ñ
∫ D.da = Q

f enc

= 0 ⇒ D = 0 everywhere . D = ε 0 E + P = 0 ⇒ E = ( −1 ε 0 ) P, so

E = 0 (for r < a and r > b );

E = ( k ε 0 r ) rˆ (for a < r < b)

Problem 4.16
(a) Same as E0 minus the field at the center of sphere with uniform
1

polarization P. The latter (Eq. 4.14) is − P 3ε 0 . So E = E0 + 3ε P
0

1
1
2

D = ε 0 E = ε 0 E0 + P = D0 − p + p, So D = D0 − P.
3
3
3


115

(b) Same as E0 minus the field of ± charges at the two ends of the “needle” –
but these are small , and far away , so E=E0

D = ε0 E = ε0 E0 = D0 − P, so D = D0 − P.
(c) Same as E0 minus the field of a parallel-plate capacitor with upper plate at
σ = P . The later is E = E0 +

1
P − ( 1 ε 0 ) P, s D = ε 0 E = ε 0 E0 + P, so D = D0. o
ε0

Problem 4.18
(a) Apply ∫ D.da = Q f to the Gaussian surface shown . DA = σ A ⇒ D = σ . (Note :
D =0 inside the metal plate .) This is true in both slabs ; D points down .
(b) D = ε E ⇒ E = σ ε1 in slab 1 , E = σ ε 0 in slab 2 . But ε = ε 0ε r , so
enc

3
ε1 = 2ε 0 ; ε 2 = ε 0 . E1 = σ 2ε 0 , E2 = 2σ 3ε 0 .
2
−1
(c) p = ε 0 χ e E , so P = ε 0 χ e d ( ε 0 ε r ) = ( χ e ε r ) ; χ e = ε r − 1 ⇒ P = ( 1 − ε r ) σ .

P1 = σ 2, P2 = σ 3.

(d) V = E1a + E2 a = ( σ a 6ε 0 ) ( 3 + 4 ) = 7σ a 6ε 0 .
(e) ρb = 0; σ b = + P1 at bottom of slab ( 1) = σ 2,

( 2) = σ

( 2 ) = −σ

σ b = + P2 at bottom of slab

3,

σ b = − P1 at top of slab ( 1) = −σ 2,

σ b = + P2 at top of slab

3,

(f) In slab 1 : total surface charge above : σ − ( σ 2 ) = σ 2
total surface charge below : ( σ 2 ) − ( σ 3) + ( σ 3) − σ = −σ 2,
In slab 2 : total surface charge above : σ − ( σ 2 ) + ( σ 2 ) − ( σ 3) = 2σ 3,
total surface charge below : ( σ 3) − σ = −2σ 3,


116

problem 4.19
With no dielectric , C0 = Aε 0 d (Eq. 2.54).
In configuration (a) , with +σ on upper plate , −σ on lower , D = σ between the

plates .
E = σ ε 0 (in air ) and E = σ ε (in dielectric). So V =
Ca =

Q ε0 A  2
=

V
d  1+1 εr

σ d σ d
Qd  ε 0 
+
=
1+
.
ε 0 2 ε 2 2ε 0 A  ε ÷


 Ca
2ε 0
=
.
÷=
 C0 1 + ε r

In configuration (b) , with potential difference V : E = V d , so σ = ε 0 E = ε 0V d
( in air ).
P = ε 0 χ eV d (in dielectric) , so σ b = − ε 0 χ eV d (at top surface of dielectric ).
σ tot = ε 0V d = σ f + σ b = σ f − ε 0 χ eV d , so σ f = ε 0V ( 1 + χ e ) d = ε 0ε rV d (on top plate

above dielectric).
⇒ Cb =

Q 1 A
A A  V
V  Aε 0  1 + ε r
= σ + σ f ÷=
 ε0 + ε 0 ε r ÷=

V V 2
2  2V  d
d 
d  2

 Cb 1 + ε r
=
.
÷.
2
 C0

[Which is greater ?
1 + ε r ) − 4ε r 1 + 2ε r + 4ε r2 − 4ε r ( 1 − ε r )
Cb Ca 1 + ε r
(
2ε r
−
=
−
=

=
=
> 0. So Cb > Ca ]
C0 C0
2
1+ εr
2(1+ εr )
2( 1+ εr )
2( 1+ εr )
2

2

If the x axis points down :
Problem 4.20
4
1
⇒ D 4π r 2 = ρ π r 3 ⇒ D = ρ r ⇒ E = ( ρ r 3ε ) rˆ , for
3
3
4
r < R; D 4π r 2 = ρ π R 3 ⇒ D = ρ R 3 3r 2 ⇒ E = ( ρ R 3 3ε 0 r 2 ) rˆ, for r >R .
3

∫ D.da = Q
0

f enc

V = − ∫ E.dl =

∞

ρ R3 1
3ε 0 r

R
∞

−

ρ
3ε

∫

0

R

rdr =

ρ R2 ρ R2 ρ R2 
1 
+
=
1 +
÷.
3ε 0 3ε 2
3ε 0  2ε r 

117

Problem 4.21
Let Q be the charge on a length l of the inner conductor .
Q

Ñ
∫ D.da = D2π sl = Q ⇒ D = 2π sl ;

E=

Q
( a < s < b ) , E = Q ( b< r< c ).
2πε 0 sl
2πε sl

a
b
c  Q  ds
Q  ds
Q   b  ε 0  c 
V = − ∫ E.dl = ∫ 
+∫ 
=
÷
÷
ln  ÷+ ln  ÷ .
c
a 2πε l

b 2πε l

 s 2πε 0l   a  ε  b  
0  s

2πε 0
C Q
= =
.
l Vl ln ( b a ) + ( 1 ε r ) ln ( c b )

Problem 4.22
Same method as Ex. 4.7 : Solve Laplace’s equation for Vin ( s, φ ) ( s < a ) and
Vout ( s, φ ) (s > a ), subject the boundary condition .
(i) Vin = Vout
at s = a ,
∂Vout
∂s
(iii) Vout → − E 0 s cos φ

(ii) ε

at s = a ,

for s >> a .
From Prob. 3.23 (invoking boundary condition (iii)):
∞

Vin ( s, φ ) = ∑ s k ( ak cos kφ + bk sin kφ )
k =1

.

( I eliminated the constant terms by seting V = 0 on the y z plane .) Condition (i)
says

∑a ( a
k

k

cos kφ + bk sin kφ ) = − E0 s cos φ + ∑ a − k ( ck cos kφ + d k sin kφ ) ,

While (ii) says

ε r ∑ ka k −1 ( ak cos kφ + bk sin kφ ) = − E0 cos φ − ∑ ka − k −1 ( ck cos kφ + d k sin kφ ) .

Evidently bk = d k = 0 for all k , ak = ck = 0 unless k = 1 , whereas for k = 1 ,
aa1 = − E0 a + a −1c1 , ε r a1 = − E0 − a −2c1 .
Solving for a1 ,
a1 = −

E0
E0
E0
, so Vin ( s, φ ) = −
s cos φ = −
x,
( 1 + χe 2)
( 1 + χe 2)

( 1 + χe 2 )


118

∂V

E

in
0
and hence Ein ( s, φ ) = − ∂x xˆ = ( 1 + χ 2 ) . As in the spherical case ( Ex. 4.7 ) , the
e

field inside is uniform .
Problem 4.23
P0 = ε 0 χ e E0 ; E1 = −

χ
ε χ2
χ
1
1
p0 = − e E0 ; P1 = ε 0 χ e E1 = − 0 e E0 ; E2 = −
P1 = e E0 ; …
3ε 0
3
3
3ε 0
9

n

 χ 
Evidently En =  − e ÷ E0 , so
 3 
 ∞  χ e n 
E = E0 + E1 + E2 + ... =  ∑  − ÷  E0 .
 n = 0  3  

The geometric series can be summed explicitly :
∞

∑x
n =0

n

=

1
1
E
, so E =
( 1 + χ e 3) 0 ,
1− x

Which agrees with Eq. 4.49. [Curiously , this method formally requires that
χ e < 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 :
Bl

Vout ( r , θ ) = − E0 r cos θ + ∑ r l +1 Pl ( cos θ ) ,


 t Bl 
( a < r Vmed ( r , θ ) = ∑  At r + l +1 ÷Pl ( cos θ ) ,
r 


Vin ( r , θ ) = 0,



Boundary Conditions :


( i ) Vout = Vmed , (r = b);

∂Vmed
∂V
= ε 0 out , (r = b ) ;
(ii )
∂r
∂r
(iii )V = 0(r = a)
med


(i) ⇒ − E0b cos θ + ∑



Bl
B 
P cos θ ) = ∑  Al bl + l +l1 ÷Pl ( cos θ ) ;
l +1 l (
b
b 


l −1
(ii) ⇒ ε r ∑ lAl b − ( l + 1)



(iii) ⇒ A1al +

119

Bl 
B
P cos θ ) = − E0 cos θ − ∑ ( l + 1) l +l 2 Pc
1 ( cos θ ) ;
l+2  l (
b 
b

Bl
= 0 ⇒ − a 2l +1 Al .
a l +1

For l #1 :
B1
a 3 A1
=
A
b
−
⇒ B1 − E0b3 = A1 2 ( b3 − a 3 ) ;
1
2
2
b
b
3

a A
B
3
3
3
(ii) ε r  A1 + 2 3 1 ÷ = − E0 − 2 31 ⇒ −2 B1 − E0b = ε r A1 ( b + 2a ) .
b
b


−3E0

3
3
3
3
3
So −3E0b = A1  2 ( b − a ) + ε r ( b + 2a )  ; A1 = 2 1 − ( a b ) 3  + ε 1 + 2 ( a b ) 3  .

 r

3

−3E0
a 
Vmed =
r − 2 ÷cos θ ,
3
3 
r 
2 1 − ( a b )  + ε r 1 + 2 ( a b )  





(i) − E0b +

E ( r , θ ) = −∇Vmed =

 2a 3 
 a3 

ˆ 
ˆ
1
+
cos
θ
r
−
1
−
sin
θθ


÷

÷
3
3
.
3
3
r 

2 1 − ( a b )  + ε r 1 + 2 ( a b )  
 r 





3E0

Problem 4.25
There are four charges involved : (i) q , (ii) polarization charge surrounding q ,
(iii) surface charge ( σ b ) on the top surface of the lower dielectric , (iv) surface


charge ( σ

'
b

)

120

on the lower surface of the upper dielectric . In view of Eq. 4.39 ,

'
'
the bound charge (ii) is q p = −q ( χ e 1 + χ e ) , so the total ( point ) charge at ( 0 , 0 ,
'
'
d ) is ql = q + q p = q ( 1 + χ e ) = q ε r . As in Ex. 4.8,



'
σ b σ b' 
 −1 qd ε r

−
−
(a) σ b = ε 0 χ e 
( here σ b = P.nˆ = + Pz = ε 0 χ e Ez );
3
4πε 0 2
2ε 0 2ε 0 
2 2


(r +d )




σ b σ b' 
qd ε r'
'
'  1
−
−
(b) σ b = ε 0 χ e 
( here σ b = − Pz = −ε 0 χ e' Ez );
3

4πε 0 2

( r + d 2 ) 2 2ε 0 2ε 0 

'

Solve for ( σ b ) , ( σ b ) : first divide by χ e and χ e' ( respectively ) and subtract :


'
 σ b 1 qd ε r 
⇒σ = χ  +
3
3  .
χ e 2π 2
2
2 2
2 2

r
+
d
r
+
d
(
)
(
) 

Plug this into (a) and solve for σ b , using ε r' = 1 + χ e' :

σ b' σ b
1
−
=

'
χ e χ e 2π

σb =

−1
4π

qd ε r'

qd ε r'

(r

2

+d

3
2 2

)

'
b

χ e ( 1 + χ e' ) −

'
e

σb
χ e + χ e' )
(
2

, so

σb =

1
σ
=
σ
+
σ
=
t
b
The total bound surface chage is
4π
'
b

−1
4π

ε r χ e' ε r'

qd

(r

2

+d

3
2 2

2

+d

(χ −χ )
1 + ( χ + χ )

'
e

qd

(r

)

1 + ( χ e + χ e' ) 2 



3
2 2

)

ε r'

e

e

'
e

2 

( Which vanishies , as it should , when χ e' = χ e ) . The total bound charge is
( compare Eq. 4.51 ) :
qt =

(χ

'
e

− χe ) q

2ε r' 1 + ( χ e + χ e' )

 ε ' − εr  q

=  r'
÷ ' , and hence
2   ε r + ε r  ε r

.


121


qt
1 
q ε r'
V ( r) =
+

2
2
2
2
4πε 0  x 2 + y 2 + ( z − d )
x
+
y
+
z
+
d
(
)








(for z > 0 )

1
q
q  ε r' − ε r 
2q
Meanwhile , since ' + qt = ' 1 + '
, V ( r ) = 4θε
= '
εr
εr  εr + εr  εr + εr
0

 2q ( ε r' + ε r ) 


x2 + y 2 + ( z − d )

(for z < 0 ).
Problem 4.26
From Ex. 4.5:



0, ( r < a )



0, ( r < a )



 Q

D= Q
rˆ, ( a < r < b ) 
 , E=
2
rˆ, ( r > a ) 

 4πε r

 4π r 2

 Q

 4πε r 2 rˆ, ( r > b ) 
0


W=
=

1 b 1 1

1
1 Q
1
D.Edr =
4π  ∫ 2 2 r 2 dr +
2
∫
2
2 ( 4π )
ε0
ε a r r

Q2
8πε 0

∫

∞

b

1  Q2
dr  =
r 2  8π

 1  1 1  1 
Q2
 1 χe
−
+

=


÷ 
 +
 ( 1 + χ e )  a b  b  8πε 0 ( 1 + χ e )  a b

 1  −1  b 1  −1  ∞
  ÷ +  ÷ 
ε  r  a ε 0  r  b 


÷.


Problem 4.27
Using Eq. 4.55 : W =

ε0
E 2 dr . From Ex. 4.2 and Eq. 3.103 ,
2 ∫

2


122

 −1
 3ε Pzˆ,
 0

E= 3
 R P 2cos θ rˆ + sin θθˆ ,
 3ε 0 r 3

(

)

( r < R) 


 , so
( r > R ) 

2

Wr < R

ε  P  4 3 2π P 2 R 3
= 0
.
÷ πR =
2  3ε 0  3
27 ε 0

Wtot =

2π R 3 P 2
9ε 0

This is the crrect electricstic energy of the configuration , but it is not the “ total
work necessary to assemble the system ,” because it leaves out the mechanical
energy involved in polarizing the molecules .
1
1
D.Edr. For r < R, D = ε 0 E + D = − P + P = −2ε 0 E , so
∫
2
3
 2π P 2 R 3 
4π R 3 P 2
ε
1
=
−
D.E = −2 0 E 2 , and this contribution is now ( −2 ) 
, exactly
÷
27 ε 0
2
2
 27 ε 0 

Using Eq. 4.58 : W =

canceling the exterior term . Conclusion : Wtot = 0 . 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
First find the capacitance , as a function of h :


123

2λ
2λ
E=
⇒V =
ln ( b a )
4πε 0 s
4πε 0



λ λ' ' ε

⇒
= ; λ = λ = ε r λ.

Air part :
ε
ε
ε0
2λ '
2λ '
2λ '
0
D=

⇒E=
⇒V =
ln ( b a ) , 

4π s
4πε s
4πε


Oil part :

Q = λ ' h + λ ( l − h ) = ε r λ h − λ h + λl = λ ( ε r − 1) h + l  = λ ( χ e h + l ) , where l is the total

height .
C=

( χ h + l)
Q λ ( χeh + l )
=
4πε 0 = 2πε 0 e
V 2λ ln ( b a )
ln ( b a )
1

dC

1

2πε χ

2
2
0 e
The net upward force is given by Eq. 4.64 : F = 2 V dh = 2 V ln ( b a )
2
2
The gravitational force down is F = mg = ρπ ( b − a ) gh
Problem 4.29

∂

(a) Eq. 4.5 ⇒ F2 = ( p2∇ ) E1 = p2 ∂y ( E1 ) ;
p

p

1
1
Eq. 3.103 E1 = 4πε r 3 θˆ = − 4πε y 3 zˆ . Therefore
0

E2 = −

0

3 p1 p2
p1 p2  d  1  
3 p1 p2
F2 =
zˆ ( upward )

  3 ÷ zˆ =
4 or
4πε 0 r 4
4πε 0  dy  y  
4πε 0 y

To calculate F1 , put p2 at the origin , poiting in the z direction ; then p1 is at −rzˆ ,
and it points in the − yˆ direction . So F1 = ( p1∇ ) E2 = − p1
need E2 as a function of x , y , and z .
From Eq. 3.104 : E2 =
and hence p2 .r = − p2 y

∂E2
∂y

; we
x = y = 0, z = − r


1 1  3 ( p2 .r ) r
− p  , where r = xxˆ + yyˆ + zzˆ , p2 = − p2 yˆ ,
3 
2
4πε 0 r  r



124

2

2
2
2
2
2




p2  −3 y ( xxˆ + yyˆ + zzˆ ) + ( x + y + z ) yˆ 
p2  −3xyxˆ + ( x − 2 y + z ) yˆ − 3 yzzˆ 
E2 =
=
52
2
2
2 52
 4πε 0 

4∞ε 0 
x
+
y
+
z
(
)
( x2 + y 2 + z 2 )





∂E2
p  5 1
1

= 2 − 7 2 y  −3 xyxˆ + ( x 2 − 2 y 2 + z 2 ) yˆ − 3 yzzˆ  + 5 ( −3xxˆ − 4 yyˆ = 3 zzˆ ) 
∂y 4πε 0  2 r
r


∂E2
∂y

=

( 0, 0 )

 p 3r 
p 2 −3 z
3p p
zˆ; F1 = − p1  2 5 zˆ ÷ = − 1 24 zˆ.
5
4πε 0 r
4πε 0 r
 4πε 0 r 

There results are consistent with Newton’s third law : F1=-F2 .
(c) From page 165 , N 2 = ( p2 × E1 ) + ( r × F2 ) . The first term was calculated in
Prob . 4.5 ; the second we get from (a) , using r = ryˆ :

p2 × E1 =

p1 p2
 3p p
− xˆ ) ; r × F2 = ( ryˆ ) ×  1 24
3 (
4πε 0 r
 4πε 0 r

2 p1 p2
 3 p1 p2
xˆ
zˆ ÷ =
xˆ ; so N 2 =
3
4πε 0 r 3
 4πε 0 r

This is equal and opposite to the torque on p1 due to p2 , with respect to the
center of p1 ( see Prob. 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 , beause E is perpendicular to the surface of each conductor .
Problem 4.31
P = kr = k ( xxˆ + yyˆ + zzˆ ) ⇒ ρb = −∇.P = − k ( 1 + 1 + 1) = −3k .
Total volume bound charge : Qvol = −3ka3
σ b = P.nˆ . At top surface , nˆ = zˆ, z = a 2 ; so σ b = ka 2 . Clearly , σ b = ka 2 on all

six surface .
2

3
Total surface bound charge : Qsorf = 6 ( ka 2 ) a = 3ka . Total bound charge is
zero .


125

Problem 4.32

Ñ
∫ D.da = Q

f enc

ρb = −∇.P = −

⇒D=

qχe
q
1
q
rˆ
rˆ
rˆ; E = D =
; P = ε 0 χe E =
.
2
2
2

4π r
ε
4πε 0 ( 1 + χ e ) r r
4π ( 1 + χ e ) r 2

qχe
4π ( 1 + χ e )

Qsurf = σ b ( 4π R 2 ) = q

χe 3
qχe
 rˆ 
δ ( r ) ; ( Eq.1.99 ) ; σ b = P.rˆ =
 ∇. 2 ÷ = − q1
1 + χe
4π ( 1 + χ e ) R 2
 r 

χe
. The compensating negative charge is at the center :
1 + χe

qχ
qχ
∫ ρ dr = − 1 + χ ∫ δ ( r ) dr = − 1 + χ
3

e

e

.

b

e

e

Problem 4.33
E is continuous ( Eq. 4.29 ) ; D ⊥ is continuous ( Eq. 4.26 , with σ f = 0 ) . So
Ex = Ex ; Dy = Dy ⇒ ε1 E y = ε 2 E y , and hence
1

2

1

2

1

2

tan θ 2
ε
=
=
= 2 . qed

tan θ1 Ex1 E y1 E y2 ε1
Ex2 E y2

E y1

tan θ 2

ε2

If 1 is air 2 is dielectric , tan θ = ε > 1 , and the field lines bend away from the
1
1
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
p' = p −

1
1
1
p=
p = p .We want the potential produced by p’ ( at the
1 + χe
1 + χe
εr


center ) and σ b ( at R ). Use separation of variables :
∞


Bl

( Eq.3.72 )
Outside
:
V
r
,
θ
=
P cos θ )
(
)
∑

l +1 l (

l =0 r






∞
1 p cos θ
l
 Inside : V ( r , θ ) =
+

A
r
P
cos
θ
(
)
∑
( Eqs.3.66,3.102 ) 
l
l
4πε 0 ε r r 2
l =0


126




or
 .

p
B1 =
+ A1R 3 
4πε 0ε r

B
∂V

∂V
1 2 p cos θ
1
−
= −∑ ( l + 1) l +l 2 Pl ( cosθ ) +
− ∑ lAl R l −1 Pl ( cos θ ) = − σ b
3
∂r R + ∂r R −
R
4πε 0 ε r R
ε0

 Bl
l
 l +1 = Al R
 R
R
⇒
V continuous at R

B
1
p
 12 =
+ A1 B,
4πε 0 ε r R 2
 R

=−

Bl = R 2l +1 Al

 1 2 p cos θ

1
1
∂V
P.rˆ = − ( ε 0 χ e Erˆ ) = χ e
= χ e −
+ ∑ lAl R l −1Pl ( cos θ )  .
3
ε0
ε0
∂r R −
 4πε 0 ε r R


− ( l + 1)

Bl
− lAl R l −1 = χ e lAl R l −1 ( l #1) ; or − ( 2l + 1) Al R l −1 = χ elAl R l −1 ⇒ Al = 0 ( l #1) .
l +2
R

For l = 1
−2



B1

1 2p
1 2p
p
A1 R 3
1 χe p
A1 R 3
+
−
A
=
χ
−
+
A
−
B
+
−
=
−
+
χ
l
e
1÷
1
e
3
R 3 4πε 0 ε r R 3
4πε 0ε r

2
4πε 0 ε r
2
 4πε 0 ε r R


p
p
A1 R3
1 χe p
A1 R3
A1 R3
1 χe p
3
−
− A1 R +
−
=−
+ χe
⇒
( 3 + χe ) =
4πε 0ε r
4πε 0ε r
2
4πε 0 ε r
2
2
4πε 0 ε r
⇒ A1 =

2χe p
1
1 2 ( ε r − 1) p
p
=
; B1 =
3
3
4πε 0 R ε r ( 3 + χ e ) 4πε 0 R ε r ( ε r + 2 )
4πε 0ε r
 q cos θ  3 
V ( r ,θ ) = 
÷( r ≥ R ) .
2 ÷
 4πε 0 r   ε r + 2 

 2 ( ε r − 1) 
p
3ε r
.
1 +
=
ε
+
2
4
πε
ε
ε
+

2
(
)
r
0 r
r




Meanwhile , for

1 p cos θ
1 pr cos θ 2 ( ε r − 1)
r ≤ R, V ( r , θ ) =
+
2
4πε 0 ε r r
4πε 0
R3
εr + 2

127

 ε r −1  r3 
p cos θ 
1
+
2



÷ 3  ( r ≤ R) .
4πε 0 r 2ε r 
 εr + 2  R 

Problem 4.35
Given two solution ,V1 ( and E1 = −∇V1 , D1 = ε E1 ) V2 ( E2 = −∇V2 , D2 = ε E2 ) ,
define V3 ≡ V2 − V1 ( E3 = E2 − E1 , D3 = D2 − D1 ) .

∫ ∇. ( V D ) dr = ∫ V D .da = 0, (V = 0 on S ) , so ∫ ( ∇V ) .D dr + ∫ V ( ∇.D ) dr = 0.
3

3

ν

3

3

3

s

3

3

3

3

But ∇.D3 = ∇.D2 − ∇D1 = ρ f − ρ f = 0 ,and ∇.V3 = ∇.V2 − ∇V1 = − E2 + E1 = − E3 , so

∫ E .D dr = 0 .
3

3

But D3 = D2 − D1 = ε E2 − ε E1 = ε E3 ,so ∫ ε ( E3 ) dr = 0 . But ε > 0 , so E3 = 0, so V2V1=constant . But at surface , V2=V1 , so V2=V1 everywhere . qed
2

Problem 4.36
(a) Proposed potential : V ( r ) = V0
case P = ε 0 χ eV0

R
R
If so , then E = −∇V = V0 2 rˆ in which
r
r

R
rˆ, in the region z < 0 . ( P = 0 for z > 0 , of course ) .
r2

Then σ b = ε 0 χ eV0

ε χV
R

rˆ.nˆ ) = − 0 e 0 . (Note : points out of dielectric ⇒ nˆ = − rˆ
2 (
r
R

) . This is on the surface at r = R . The flat surface z = 0 carries no hound
charge , since nˆ = zˆ ⊥ rˆ . Nor is there any volume bound charge ( Eq. 4.39 )
. If V is to have the required spherical symmetry , the net charge must be
unliform :
σ tot 4π R 2 = Qtot = 4πε 0 RV0 (since V0 = Qtot 4πε 0 R , so σ tot = ε 0V0 R . Therefore


128

( ε 0V0 R )
on northern hemisphere 
σf =
.
( ε 0V0 R ) ( 1 + χ e ) on southern hemisphere 

(b) By construction , σ tot = σ b + σ f = ε 0V0 R is uniform (on the northern
hemisphere σ b = 0, σ f = ε 0V0 R ; on the southern hemisphere σ b = − ε 0 χ eV0 R
, so σ f = εV0 R ) . The potential of a uniform charge sphere is
σ tot ( 4π R 2 ) ε 0V0 R 2
Qtot
R
V0 =
=
=
= V0 .

4πε 0 r
4πε 0 r
R ε 0r
r

(c) Since everything is consistent , and the boundary conditions (
V = V0 at r = R at ∞ ) are met , Prob. 4.35 guaratees that this the solution .
(d) Figure (b) works the way , but Fig. (a) does not : on the flat surface , P is
not perpendicular to , so we’d get bound charge on this surface , spoiling
the symmetry .
Problem 4.37
λ
sˆ Since the sphere is tiny , this is essentially constant , and hence
2πε 0 s
ε χ
P = 0 e Eext ( Ex. 4.7 )
1 + χe 3
Eext =

2

 ε χ  λ  d  λ 
 ε χ 

ˆ =  0 e ÷ λ ÷
F = ∫  0 e ÷
÷ 
÷sdr
 1 + χ e 3   2πε 0 s  ds  2πε 0 s 
 1 + χ e 3  2πε 0 s 

 1  −1  ˆ
 ÷ 2 ÷s ∫ dr
 s  s 

 − χ e  λ 2  1 4
 χ e  λ 2 R3
3
ˆ
π
R
s
=
−
sˆ.
=
÷ 2 ÷ 3

÷
3
 1 + χ e 3  4π ε 0  s 3
 3 + χ e  πε 0 s

Problem 4.38
1

The density of atoms is N = ( 4 3) π R 3 . The macroscopic field E is , where Eself is
the average field over the sphere due to the atom itself .

129

p = α Eelse ⇒ P = N α 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. 3.41. ] Now
1

α

(Eq. 3.105) , so Eself = − 4πε R 3 Eesle
0
E=−

So
P=


 Nα 
1 α
α 
Eesle + Eelse = 1 −
E = 1 −
÷Eesle .
3
3 ÷ esle
4πε 0 R
 4πε 0 R 
 3ε 0 

Nα
E = ε 0 χe E,
( 1 − Nα 3ε 0 )

And hence
χe =

Nα ε 0
.
( 1 − Nα 3ε 0 )
Nα

Nα  χ e 
1 + ÷ = χ e ,
3 


Nα

Solving for α : χ e − 3ε χ e = ε ⇒ ε
0
0
0
ε

χ

3ε

χ

3ε  ε − 1 
0
e
0
e
Or α = N ( 1 + χ 3) = N ( 3 + χ ) . But χ e = ε r − 1 , so α = 0  r ÷. qed
N  εr + 2 
e
e

Problem 4.39
For an ideal gas , N = Avagadro’s number / 22.4 liters
= ( 6.02 ×1023 )

( 22.4 ×10 ) = 2.7 ×10
−3

. Nα ε 0 = ( 2.7 × 1025 ) ( 4πε 0 ×10 −30 ) β ε 0 = 3.4 ×10 −4 β

25

, where is the number listed in Table 4.1 .

H : β = 0.667, Nα ε 0 = ( 3.4 ×10−4 ) ( 0.67 ) = 2.3 ×10 −4 , χ e = 2.5 × 10 −4 

He : β = 0.205, Na ε 0 = ( 3.4 ×10−4 ) ( 0.21) = 7.1×10−5 , χ e = 6.5 ×10 −4 
 agreement is
Ne :β = 0.396, Nα ε 0 = ( 3.4 ×10−4 ) ( 0.40 ) = 1.4 ×10−4 , χ e = 1.3 × 10−4 


Ar :β = 1.64, Nα ε 0 = ( 3.4 ×10−4 ) ( 1.64 ) = 5.6 ×10−4 , χ e = 5.2 ×10−4 

quite good .


130

Problem 4.40
(a)
u =

∫

pE

ue

− pE
pE

∫

− pE

− u kT

du

e − u kT

=

( kT )

2

pE
e − u kT  − ( u kT ) − 1
− pE
pE
−kTe −u kT
− pE

 e − pE kT − e pE kT  + ( pE kT ) e − pE kT + ( pE kT ) e pE kT  
 

= kT  

− pE kT
pE kT
e
−e


 e pE kT + e − pE kT 
 pE 
= kT − pE  pE kT
= kT − pE coth 
÷.
− pE kT 

−e
 kT 
e

− u

 pE  kT 
P = N ( p ) ; p = ( p cos θ ) Eˆ = ( P.E ) Eˆ E = − u Eˆ E ; P = N p
= N p coth 
.
÷−
pE
 kT  pE 


(

)

(

)

Let y ≡ P N ( p ) ; x ≡ pE kT . Then y = coth x − 1 x . As
 1 x x3
 1 x x3
x → 0, y =  + − + ... ÷− = − + ... → 0 , so the graph starts at the origin ,
 x 3 45
 x 3 45
with an initial slope of 1/3 . As x → ∞, y → coth ( ∞ ) = 1 , so the graph goes

asymptotically to y = 1 (see Figure ) .
P
pE
1
Np 2
≈
x
,
y
≈
x
,
P
≈
E = ε 0 χ e E ⇒ P is proportional to
(b) For small
so N 3kT , or
3
3kT
p

E , and χ e =

Np 2
3ε 0 kT
molecules

0
−30

For water at 20 = 293K , p = 6.1×10 Cm; N = volume =

molecules moles gram
×
×
.
mole
gram volume

0.33 ×1029 ) ( 6.1×10−30 )
(
1
23
6
29
N = ( 6.0 ×10 ) ×  ÷× ( 10 ) = 0.33 × 10 ; χ e =
= 12.
 18 
( 3) ( 8.85 ×10−12 ) ( 1.38 ×10−23 ) ( 293)
2


131

Table 4.2 gives an experimental value of 79 , so it’s pretty far off .
For water vapor at 100o = 373K , treated as an ideal gas ,
volume
 373 
−2
3

= ( 22.4 ×10−3 ) × 
÷ = 2.85 ×10 m
mole
 293 

( 2.11×1025 ) ( 6.1×10−30 )
6.0 ×1023
25
N=
=
2.11
×
10
;
χ
=
= 5.7 ×10−3.
e
−2
−12
−23
2.85 × 10
( 3) ( 8.85 ×10 ) ( 1.38 ×10 ) ( 373)
2

Table 4.2 gives 5.9 ×10−3 , so this time the agreement is quite good .
Chapter 5
Manetostatics
Problem 5.1
Since v × B points upward , and that is also thr direction of the force , q must the

positive . To find R , in terms of a and d , use the Pythagorean theorem :

( R−d)

2

+ a 2 = R 2 ⇒ R 2 − 2 Rd + d 2 + a 2 = R 2 ⇒ R =

a2 + d 2
2d

The cyclotron formula then gives

(a
p = qBR = qB

2

+ d2)

2d

.

Problem 5.2
The general solution is ( Eq. 5.6 ) :
E
t + C3 ; z ( t ) = C2 cos ( ωt ) − C1 sin ( ωt ) + C4
B
( a ) y ( 0 ) = z ( 0 ) = 0; y&( 0 ) = E B; z&( 0 ) . Use these to determine C1 , C2 , C3 , and

y ( t ) = C1 cos ( ωt ) + C2 sin ( ωt ) +

C4 .

y ( 0 ) = 0 ⇒ C1 + C3 = 0; y&( 0 ) = ωC2 + E B ⇒ C2 = 0; z ( 0 ) = 0 ⇒ C2 + C4 = 0 ⇒ C4 = 0 ,

z&( 0 ) = 0 ⇒ C1 = 0 and hence also C3=0 . So y ( t ) = Et B; z ( t ) = 0 . Does this

make sense ? The magnetic force is q ( v × B ) = −q ( E B ) Bzˆ = −qE , which