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

A)            Zero                                         

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

C)                                      \[QR/4\pi {{\varepsilon }_{0}}{{d}^{2}}\]

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

Correct Answer: D

Solution :

                      Potential at the centre of rings are \[{{V}_{{{O}_{1}}}}=\frac{k.q}{R}+\frac{k(-q)}{\sqrt{{{R}^{2}}+{{d}^{2}}}}\], \[{{V}_{{{O}_{2}}}}=\frac{k(-q)}{R}+\frac{kq}{\sqrt{{{R}^{2}}+{{d}^{2}}}}\] Þ \[{{V}_{{{O}_{1}}}}-{{V}_{{{O}_{2}}}}=2kq\,\left[ \frac{1}{R}-\frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]\]\[=\frac{q}{2\pi {{\varepsilon }_{0}}}\left[ \frac{1}{R}-\frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]\]