1. Potential energy of a system of charges

For a system of point charges, $U = \sum_{i<j} \frac{K q_i q_j}{r_{ij}}$

Here all four charges are identical and equal to $q_0$, so each pair contributes $\frac{K q_0^2}{r}$

Since there are $\binom{4}{2}=6$ pairs, we just need the pairwise distances in each configuration.

2. Configuration (1): charges at the corners of a square

Let the square side be $a$.

Among the 6 pairs:

• 4 pairs are along the sides, each at distance $a$
• 2 pairs are along the diagonals, each at distance $a\sqrt{2}$

So, $U_1 = 4\left(\frac{Kq_0^2}{a}\right) + 2\left(\frac{Kq_0^2}{a\sqrt{2}}\right)$

$U_1 = \frac{Kq_0^2}{a}\left(4 + \frac{2}{\sqrt{2}}\right)$

Since $\frac{2}{\sqrt{2}}=\sqrt{2}$, $U_1 = \frac{Kq_0^2}{a}(4+\sqrt{2})$

3. Configuration (2): charges at the midpoints of the sides

The midpoints $E,F,G,H$ form another square inside the original square.

Now find distances among these 4 midpoint charges:

• Adjacent midpoints are separated by $\sqrt{\left(\frac{a}{2}\right)^2+\left(\frac{a}{2}\right)^2} = \frac{a}{\sqrt{2}}$ There are 4 such pairs.

• Opposite midpoints are separated by distance $a$. There are 2 such pairs.

Hence, $U_2 = 4\left(\frac{Kq_0^2}{a/\sqrt{2}}\right) + 2\left(\frac{Kq_0^2}{a}\right)$

$U_2 = 4\left(\frac{\sqrt{2}Kq_0^2}{a}\right) + 2\left(\frac{Kq_0^2}{a}\right)$

$U_2 = \frac{Kq_0^2}{a}(4\sqrt{2}+2)$

4. Required difference: configuration (2) minus configuration (1)

$U_2-U_1 = \frac{Kq_0^2}{a}(4\sqrt{2}+2) - \frac{Kq_0^2}{a}(4+\sqrt{2})$

$U_2-U_1 = \frac{Kq_0^2}{a}\big(4\sqrt{2}+2-4-\sqrt{2}\big)$

$U_2-U_1 = \frac{Kq_0^2}{a}(3\sqrt{2}-2)$

5. Option matching

This matches Option D: $\boxed{\frac{Kq_0^2}{a}(3\sqrt{2}-2)}$