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

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

$\frac{qR}{4 π \in_{0} d^{2}}$

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answer is A.

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Detailed Solution

At (1) using, potential $(v_{1}) = V_{self} + V_{due to (2)}$

$\Rightarrow V_{1} = \frac{1}{4 {πε}_{0}} [\frac{q}{R} - \frac{q}{\sqrt{R^{2} + d^{2}}}]$

At (2) using potential $(V_{2}) = V_{self} + V_{due to (1)}$

$\Rightarrow V_{2} = \frac{1}{4 {πε}_{0}} [\frac{- q}{R} + \frac{q}{\sqrt{R^{2} + d^{2}}}]$

$ΔV = V_{1} - V_{2}$

$= \frac{1}{4 {πε}_{0}} [\frac{q}{R} + \frac{q}{R} - \frac{q}{\sqrt{R^{2} + d^{2}}} - \frac{q}{\sqrt{R^{2} + d^{2}}}]$

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

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