Q. Two thin wire rings each having 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
A) Zero
B) $\frac{Q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
C) $QR/4\pi\varepsilon_{0}d^{2}$
D) $\frac{Q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
Explanation
(d) Potential at the centre of rings are
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$V_{O_{1}} = \frac{k.q}{R} + \frac{k( - q)}{\sqrt{R^{2} + d^{2}}},\ V_{O_{2}} = \frac{k( - q)}{R} + \frac{kq}{\sqrt{R^{2} + d^{2}}}$
$\Rightarrow V_{O_{1}} - V_{O_{2}} = 2kq\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack = \frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.