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Potential Energy Is Defined for a System

Two identical...

Question

Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If $Q_{1}$ coulomb and $Q_{2}$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is

Zero

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\frac{q (Q_{2} - Q_{1}) (\sqrt{2} - 1)}{\sqrt{2} . 4 π ϵ_{0} R}

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\frac{q \sqrt{2} (Q_{1} + Q_{2})}{4 π ϵ_{0} R}

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\frac{q (Q_{1} + Q_{2}) (\sqrt{2} + 1)}{\sqrt{2} .4 π ϵ_{0} R}

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Solution

The correct option is B $\frac{q (Q_{2} - Q_{1}) (\sqrt{2} - 1)}{\sqrt{2} . 4 π ϵ_{0} R}$
$W = q (V_{O_{2}} - V_{O_{1}})$

where $V_{O_{1}} = \frac{Q_{1}}{4 π ϵ_{0} R} + \frac{Q_{2}}{4 π ϵ_{0} R \sqrt{2}}$
and $V_{O_{2}} = \frac{Q_{2}}{4 π ϵ_{0} R} + \frac{Q_{1}}{4 π ϵ_{0} R \sqrt{2}}$
$\Rightarrow V_{O_{2}} - V_{O_{1}} = \frac{(Q_{2} - Q_{1})}{4 π ϵ_{0} R} [1 - \frac{1}{\sqrt{2}}]$
So, $W = \frac{q . (Q_{2} - Q_{1})}{4 π ϵ_{0} R} \frac{(\sqrt{2} - 1)}{\sqrt{2}}$

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