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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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$\large\frac{q^2}{4 \pi \in_0 a}$$ (4 -4 \sqrt 2)$
Hence A is the correct answer.
answered Jun 11, 2014 by meena.p
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