11.3 Solutions
513
(b) q 1 = 4πε 0 aV 1 , q 2 = 4πε 0 aV 2
U 1 =
1
2
q 1 V 1 =
1
2
q 2
1
4πε 0 a
, U 2 =
1
2
q 2
2
4πε 0 a
Initially total energy U = U 1 + U 2 =
q 2
1 + q 2
2
8πε 0 a
(1)
When the spheres are connected they reach a common potential and
when disconnected let the charges be q
1 and q
2 .
q
1 = 4πε 0 aV, q
2 = 4πε 0 aV
∴ q
2 = q
1
Final energy in the spheres
U
1 =
1
2
q
1 V =
q
2
1
8πε 0 a
, U
2 =
q
2
2
8πε 0 a
Total final energy U
= U
1 + U
2 =
q
2
1 + q
2
2
8πε 0 a
(2)
q 1 + q 2 = q
1 + q
2 = 2q
1 (charge conservation)
(3)
U = U 1 − U 2 =
1
8πε 0 a
q
2
1 + q
2
2 − 2q
2
1
(4)
Eliminating q
1 between (3) and (4) and simplifying
U =
1
16πε 0 a
(q 1 − q 2 )
2
(5)
This energy is dissipated in Joule heating of the wire.
11.61 When the spheres are brought into contact they reach a common potential,
say V . If the charges on them are now Q
1 and Q
2
V =
Q
1
4πε 0 R 1
=
Q
2
4πε 0 R 2
(1)
σ
1 =
Q
1
4π R 2
1
=
ε 0 V
R 1
(2)
where we have used (1).