Four charges +q, q, +q, and q are placed in order on the four consecutive corners of a square of side a. Find the work done in interchanging the positions of any two neighbouring charges of opposite sign.

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Solution

The diagonal of the square is $d = \sqrt{a^{2} + a^{2}} = \sqrt{2} a$
The potential energy in first case, $U_{i} = \frac{k (- q) q}{a} + \frac{k (- q) (- q)}{\sqrt{2} a} + \frac{k (- q) q}{a} = \frac{k q^{2}}{a} (\frac{1}{\sqrt{2}} - 2)$
The potential energy in second case, $U_{f} = \frac{k (- q) (- q)}{a} + \frac{k (- q) (q)}{\sqrt{2} a} + \frac{k (- q) q}{a} = - \frac{k q^{2}}{a} (\frac{1}{\sqrt{2}})$
As work done is the change in potential energy so
$W = U_{f} - U_{i} = \frac{k q^{2}}{a} (2 - \sqrt{2})$

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