Four equal charges, each Q, are placed at the four corners of a square of each side ‘a’. Work done in moving in a charge –Q from its centre to infinity is

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$\frac{\sqrt{2} Q^{2}}{4 {πε}_{0} a}$

$\frac{\sqrt{2} Q^{2}}{{πε}_{0} a}$

$\frac{Q^{2}}{2 {πε}_{0} a}$

answer is C.

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Detailed Solution

$\begin{array}{l} W = - Q [V_{\infty} - V_{0}] = - Q [0 - V_{0}] \\ = Q [4 \times \frac{1}{4 {πε}_{0}} \cdot \frac{Q}{r}] \\ = \frac{Q^{2}}{{πε}_{0} (a / \sqrt{2})} = \frac{\sqrt{2} Q^{2}}{{πε}_{0} a} \end{array}$

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