11.3 Solutions
499
Fig. 11.34
11.35 Let the charge q be taken from the centre of ring A to the centre of ring B,
Fig. 11.35
At A, U A =
q Q 1
4πε 0 R
+
q Q 2
4πε 0 (
√
2 R)
At B, U B =
q Q 2
4πε 0 R
+
q Q 1
4πε 0 (
√
2 R)
Work done, W = U B − U A =
(
√
2 − 1) q (Q 2 − Q 1 )
4
√
2πε 0 R
Fig. 11.35
11.36 (a) Consider an infinitesimal length of the rod, at distance y from the origin,
Fig. 11.36. The charge in dy will be λdy. The distance of P 1 from dy
will be 2a − y. The potential at P 1 is
V 1 =
1
4πε 0
a
−a
λ dy
(2a − y)
=
λ
4πε 0
ln 3
The potential at P 2 is
V 2 =
1
4πε 0
a
−a
λdy
y 2 + x 2
=
2λ
4πε 0
ln
a +
√
a 2 + x 2
x
499
Fig. 11.34
11.35 Let the charge q be taken from the centre of ring A to the centre of ring B,
Fig. 11.35
At A, U A =
q Q 1
4πε 0 R
+
q Q 2
4πε 0 (
√
2 R)
At B, U B =
q Q 2
4πε 0 R
+
q Q 1
4πε 0 (
√
2 R)
Work done, W = U B − U A =
(
√
2 − 1) q (Q 2 − Q 1 )
4
√
2πε 0 R
Fig. 11.35
11.36 (a) Consider an infinitesimal length of the rod, at distance y from the origin,
Fig. 11.36. The charge in dy will be λdy. The distance of P 1 from dy
will be 2a − y. The potential at P 1 is
V 1 =
1
4πε 0
a
−a
λ dy
(2a − y)
=
λ
4πε 0
ln 3
The potential at P 2 is
V 2 =
1
4πε 0
a
−a
λdy
y 2 + x 2
=
2λ
4πε 0
ln
a +
√
a 2 + x 2
x