Work And Energy In Electric Field

The total electrostatic potential energy of a system of charges is the sum of potential energies of all unique pairs: (U = \sum_{i<j} \frac{kq_i q_j}{r_{ij}}). For four charges at the corners of a square of side (a), the pairs and their separations are: four sides (separation (a)) contributing (4 \times \frac{kq^2}{a}), and two diagonals (separation (a\sqrt{2})) contributing (2 \times \frac{kq^2}{a\sqrt{2}}). Total pairs: (\binom{4}{2} = 6), confirmed. Summing: (U = 4 \cdot \frac{kq^2}{a} + 2 \cdot \frac{kq^2}{a\sqrt{2}} = \frac{kq^2}{a}\left(4 + \frac{2}{\sqrt{2}}\right) = \frac{kq^2}{a}\left(4 + \sqrt{2}\right) = \frac{2kq^2}{a}\left(2 + \frac{1}{\sqrt{2}}\right)). Both forms are equivalent since (4 + \sqrt{2} = 2(2 + 1/\sqrt{2})). Option (\frac{kq^2}{a}(4 + \sqrt{2})) is also correct in a different form — the given answer (\frac{2kq^2}{a}(2 + 1/\sqrt{2})) is the same value. Option (\frac{kq^2}{a}(4 + 1/\sqrt{2})) is incorrect because it uses (1/\sqrt{2}) instead of (2/\sqrt{2} = \sqrt{2}) for the diagonal contribution. Option (\frac{kq^2}{a}(2\sqrt{2}+1)) is numerically incorrect. This JEE Advanced problem tests systematic pairwise potential energy summation. Plausibility check: with 6 pairs all at positive (kq^2/r), the total must be positive and of order (kq^2/a).