11.3 Solutions
533
(d) V =
E(x)dx = E 0
d
0
dx
ax + b
(6)
Put y = ax + b, dx = dy/a. Then (6) becomes
V =
E 0
a
ad+b
b
dy
y
=
E 0
a
ln y|
ad+b
b
=
σ 0
ε 0 a
ln
1 +
ad
b
(7)
where we have used (5).
The capacitance C =
q
V
=
Aε 0 a
ln
1 +
ad
b
(8)
where we have used (7) and q = σ A.
(e) Vacuum polarization charge density
P(x) = ε 0 (k(x) − 1)E =
ε 0 E 0
k(x)
(k(x) − 1)
= ε 0 E 0
1 −
1
ax + b
(9)
11.98 Newton’s law of gravitation is
F = −
GMm
r 2 ˆ
e r = m g
The Gauss’ law for gravitation may be written as
g · ds = −
G M
r 2 4πr
2
= −4π G M
The divergence theorem gives
∇ · g d
3 r =
g · ds
∴ ∇ · g
4
3
πr
3
= −4π G M
∴ ∇ · g = −4π Gρm
This is analogous to the law for electric field
∇ · E =
ρ q
ε 0