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

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$\frac{q R}{4 π \in_{0} d^{2}}$

Zero

$\frac{q}{2 π \in_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

$\frac{q R}{4 π \in_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

answer is A.

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Detailed Solution

$V_{1} = \frac{K q}{R} - \frac{K q}{\sqrt{d^{2} + R^{2}}}$

$V_{2} = - \frac{K q}{R} + \frac{K q}{\sqrt{d^{2} + R^{2}}}$

$Δ V = V_{1} - V_{2} = \frac{q}{2 {πε}_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

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