514
11 Electrostatics
Similarly
σ
2 =
ε 0 V
R 2
(3)
Thus σ α
1
R
.
11.62 (a) V = Er = 5 × 10 6 × 0.1 = 5 × 10 5 V
(b) Pressure p =
1
2
σ 2
ε 0
=
1
2ε 0
(Eε 0 ) 2 =
1
2
ε 0 E 2
=
1
2
× 8.85 × 10
−12
× (5 × 10
6
)
2
= 110.6 N/m
2
(c) If q is the charge and C the capacitance then the electrostatic energy
U =
1
2
q 2
C
=
1
2
q 2
4πε 0 r
Now q = 4πε 0 r 2 E =
(0.1) 2 × 5 × 10 6
9 × 10 9
= 5.55 × 10 −6 C
∴
U =
1
2
× 9 ×
10 9
0.1
× (5.55 × 10
−6
)
2
= 1.386 J
11.63 Work done in the isobaric expansion (constant pressure) is
W = Pv =
4π
3
R
3
2 − R
3
1
p
(1)
where we have written υ for volume.
Increase in electrostatic energy
U =
1
2
q 2
C 1
−
1
2
q 2
C 2
=
q 2
2
1
4πε 0
1
R 1
−
1
R 2
(2)
where C is the capacitance of the spherical bubble. Equating (1) and (2) and
simplifying, we obtain
q =
32
3
π
2
ε 0 p R 1 R 2
R
2
1 + R 1 R 2 + R
2
2
1/2
(3)