Electrostatic Potential Energy question from MHT CET 2023 (11 May, Shift 1)

Answer: (d)

1. PE of a system: $U = \sum \frac{kq_iq_j}{r_{ij}}$ over every pair, $k = \frac{1}{4\pi\varepsilon_0}$.
2. Pair $q,q$ at $2r$: $\frac{kq^2}{2r}$. Each pair $q,-2q$ at $r$: $-\frac{2kq^2}{r}$, and there are two such pairs.
3. $U = \frac{kq^2}{r}\left(\frac{1}{2} - 4\right) = -\frac{7kq^2}{2r} = -\frac{7q^2}{8\pi\varepsilon_0 r}$
4. So the answer is (d) $-\frac{7q^2}{8\pi\varepsilon_0 r}$.