JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Electrostatics

The two thin coaxial rings, each of radius '

a

' and having charges

+ Q

and

- Q

respectively are separated by a distance of '

s

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

Answer: (A)

Solution

Every element of a ring of radius

a

is the same distance from its own centre, so a ring of charge

q

gives

\frac{q}{4 π ε _{0} a}

at its centre, and

\frac{q}{4 π ε _{0} s ^{2} + a ^{2}}

at a point on its axis a distance

s

away.
At the centre of the ring carrying

+ Q

V_{1} = \frac{Q}{4 π ε _{0} a} - \frac{Q}{4 π ε _{0} s ^{2} + a ^{2}}

At the centre of the ring carrying

- Q

both contributions change sign, so

V_{2} = - V_{1}

. The potential difference is therefore

V_{1} - V_{2} = 2 V_{1} = \frac{Q}{2 π ε _{0}} [\frac{1}{a} - \frac{1}{s ^{2} + a ^{2}}]

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option