JEE Main · 2005 · 08 May 2005 · Offline · Electrostatics

Two thin wire rings each having a radius

R

are placed at a distance

d

apart with their axes coinciding. The charges on the two rings are

+ q

and

- q

. The potential difference between the centres of the two rings is

Answer: (B)

Solution

Each centre feels the potential of its own ring plus that of the far ring. The near ring's charge sits at distance

R

from its centre and the far ring's charge at distance

R^{2} + d^{2}

, so centre

1

is at

V_{1} = \frac{k q}{R} - \frac{k q}{R ^{2} + d ^{2}}

and centre

2

V_{2} = - \frac{k q}{R} + \frac{k q}{R ^{2} + d ^{2}}

. Their difference is

V_{1} - V_{2} = \frac{2 k q}{R} - \frac{2 k q}{R ^{2} + d ^{2}} = \frac{q}{2 π ε _{0}} [\frac{1}{R} - \frac{1}{R ^{2} + d ^{2}}]

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option