In the first configuration (1) as shown in th | vedclass

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(D) For configuration $(1)$,the potential energy $U_1$ is the sum of interaction energies of all pairs of charges. There are $4$ pairs at distance $a$

Vedclass · Accepted Answer

$\frac{Kq_0^2}{a}(3\sqrt{2}-2)$

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(D) For configuration $(1)$,the potential energy $U_1$ is the sum of interaction energies of all pairs of charges. There are $4$ pairs at distance $a$ and $2$ pairs at distance $a\sqrt{2}$.$U_1 = 4 \left( \frac{Kq_0^2}{a} \right) + 2 \left( \frac{Kq_0^2}{a\sqrt{2}} \right) = \frac{Kq_0^2}{a} (4 + \sqrt{2})$.For configuration $(2)$,the charges are at midpoints. The distances between pairs are: $4$ pairs at distance $a/\sqrt{2}$ (adjacent sides),$2$ pairs at distance $a$ (opposite sides),and $1$ pair at distance $a\sqrt{2}$ (diagonal). Wait,let's re-evaluate: The midpoints are $G, E, H, F$. The distance between adjacent midpoints is $a/\sqrt{2}$. There are $4$ such pairs. The distance between opposite midpoints is $a$. There are $2$ such pairs.$U_2 = 4 \left( \frac{Kq_0^2}{a/\sqrt{2}} \right) + 2 \left( \frac{Kq_0^2}{a} \right) = \frac{Kq_0^2}{a} (4\sqrt{2} + 2)$.The difference is $\Delta U = U_2 - U_1 = \frac{Kq_0^2}{a} (4\sqrt{2} + 2 - 4 - \sqrt{2}) = \frac{Kq_0^2}{a} (3\sqrt{2} - 2)$.

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