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(D) The work done by an external agent is given by $W_{ext} = U_f - U_i$.In Figure $(1)$,the charges are at the corners of a square of side $a$. The p

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$\frac{-kq^2}{a}(3 - \sqrt{2})$

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(D) The work done by an external agent is given by $W_{ext} = U_f - U_i$.In Figure $(1)$,the charges are at the corners of a square of side $a$. The potential energy $U_i$ is the sum of the potential energies of all pairs:$U_i = 4 	imes \frac{kq^2}{a} + 2 	imes \frac{kq^2}{a\sqrt{2}} = \frac{kq^2}{a}(4 + \sqrt{2})$.In Figure $(2)$,the charges are at the corners of a square inscribed in a circle of radius $a$. The distance between adjacent charges is $a\sqrt{2}$ and the distance between opposite charges is $2a$. The potential energy $U_f$ is:$U_f = 4 	imes \frac{kq^2}{a\sqrt{2}} + 2 	imes \frac{kq^2}{2a} = \frac{kq^2}{a}(2\sqrt{2} + 1)$.Now,calculate the work done:$W_{ext} = U_f - U_i = \frac{kq^2}{a} [(2\sqrt{2} + 1) - (4 + \sqrt{2})]$$W_{ext} = \frac{kq^2}{a} (\sqrt{2} - 3) = -\frac{kq^2}{a} (3 - \sqrt{2})$.

Consider the configuration of a system of four charges each of value $+q$. The work done by an external agent in changing the configuration of the system from Figure $(1)$ to Figure $(2)$ is:

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