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(C) The potential $V$ at the center of the square due to the four charges $Q$ at the corners is given by $V = 4 	imes \frac{kQ}{r}$,where $r$ is the

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$\sqrt{2} Q^2 / \pi \varepsilon_0 a$

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(C) The potential $V$ at the center of the square due to the four charges $Q$ at the corners is given by $V = 4 	imes \frac{kQ}{r}$,where $r$ is the distance from the corner to the center. For a square of side $a$,the diagonal is $a\sqrt{2}$,so $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Thus,$V = 4 	imes \frac{kQ}{a/\sqrt{2}} = \frac{4\sqrt{2}kQ}{a}$.The potential energy $U_i$ of the charge $-Q$ at the center is $U_i = (-Q)V = -\frac{4\sqrt{2}kQ^2}{a}$.The potential energy $U_f$ at infinity is $0$.The work done $W$ is $W = U_f - U_i = 0 - (-\frac{4\sqrt{2}kQ^2}{a}) = \frac{4\sqrt{2}kQ^2}{a}$.Substituting $k = \frac{1}{4\pi\varepsilon_0}$,we get $W = \frac{4\sqrt{2}Q^2}{4\pi\varepsilon_0 a} = \frac{\sqrt{2}Q^2}{\pi\varepsilon_0 a}$.

Four equal charges $Q$ are placed at the four corners of a square of side $a$. The work done in moving a charge $-Q$ from its centre to infinity is

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