question_answer

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 \[+\text{ }q\]and\[-\text{ }q\]. The potential difference between the centres of the two rings is     AIEEE  Solved  Paper-2005

A) \[\frac{qR}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]

B) \[\frac{q}{2\pi {{\varepsilon }_{0}}}\left[ \frac{1}{R}-\frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]\]

C) zero

D) \[\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{R}-\frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]\]

Correct Answer: B

Solution :

                To determine the potential at a point due to lots of charges, we determine the potentials of each pair of charges belonging to system and after that add it, to get the net potential at point due to all the charges. The net potential at the centre, A of first ring Potential due to charge + q on ring A + Potential due to charge - q on ring B                                ?..(i) Similarly, the net potential at the centre, b of second ring Potential difference