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 . T - Physics MCQ AIEEE 2026 | ExamTest.live

hardMCQAIEEE2026Physics

1 mark

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:

Solution & Step-by-step Explanation

Potential at the centre of first ring (C_{1}): V_{1} = \frac{k q}{R} + \frac{k ( - q )}{R ^{2} + d ^{2}} Potential at the centre of second ring (C_{2}): V_{2} = \frac{k ( - q )}{R} + \frac{k q}{R ^{2} + d ^{2}} Potential difference Δ V = V_{1} - V_{2} : Δ V = (\frac{k q}{R} - \frac{k q}{R ^{2} + d ^{2}}) - (- \frac{k q}{R} + \frac{k q}{R ^{2} + d ^{2}}) Δ V = 2 k q [\frac{1}{R} - \frac{1}{R ^{2} + d ^{2}}] = \frac{2 q}{4 π ϵ _{0}} [\frac{1}{R} - \frac{1}{R ^{2} + d ^{2}}] Δ V = \frac{q}{2 π ϵ _{0}} [\frac{1}{R} - \frac{1}{R ^{2} + d ^{2}}]

Try it yourself before checking the explanation above.

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:

qR /4 π ϵ_{0} d^{2}

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

zero

\frac{q}{4 π ϵ _{0}} [\frac{1}{R} - \frac{1}{R ^{2} + d ^{2}}]

Practice with timed mock tests and track your performance across Physics.