# Four charges q , -q , q , -q are placed in order on the four consecutive corners of a square of side a . The work done in interchanging the positions of any two neighbouring charges of the opposite sign is

$(a)\;\large\frac{k\;q^2}{a}\;(-4+\sqrt{2})\qquad(b)\;\large\frac{k\;q^2}{a}\;(4+2 \sqrt{2})\qquad(c)\;\large\frac{k\;q^2}{a}\;(4-2 \sqrt{2})\qquad(d)\;\large\frac{k\;q^2}{a}\;(4+\sqrt{2})$

## 1 Answer

Answer : (c) $\;\large\frac{k\;q^2}{a}\;(4-2 \sqrt{2})$
Explanation :
initial P .E = $\;\large\frac{-4\;k\;q^2}{a} + \large\frac{2\;k\;q^2}{\sqrt{2}\;a}$
$=(\sqrt{2}-4)\;\large\frac{k\;q^2}{a}$
final P .E = $\;\large\frac{k\;q^2}{a}-\large\frac{k\;q^2}{a}-\large\frac{k\;q^2}{a}+\large\frac{k\;q^2}{a}-\large\frac{2\;k\;q^2}{\sqrt{2}\;a}$
$work done = \large\frac{- \sqrt{2}\;K\;q^2}{a}-\large\frac{(\sqrt{2}-4)\;k\;q^2}{a}$
$=\large\frac{k\;q^2}{a}\;(4-2\;\sqrt{2})\;.$
answered Feb 5, 2014 by
edited Aug 21, 2014 by meena.p
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