Q. 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 two rings is
A) $\frac{qR}{4\pi\varepsilon_{0}d^{2}}$
B) $\frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
C) Zero
D) $\frac{q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
Explanation
The correct answer is Option B: $\frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.