Electrostatic Potential and Capacitance question for JEE Main (JEE Main 2019 (9 Apr, Shift 1)), with solution

Answer: (A) $\dfrac{1}{4\pi\varepsilon_0}\left[-\dfrac{q^2}{d} - \dfrac{qQd}{D^2}\right]$

Energy of the pair $+q$, $-q$: $-\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{d}$.

The pair is a dipole of moment $p = qd$ pointing from $-q$ to $+q$, i.e. away from $Q$. $Q$ lies on the dipole's axis at distance $D$ from its centre, on the side of $-q$, where the dipole's potential is
$$V = -\frac{1}{4\pi\varepsilon_0}\frac{qd}{D^2}$$
So the energy of $Q$ with the dipole is $QV$, and in total
$$U = \frac{1}{4\pi\varepsilon_0}\left[-\frac{q^2}{d} - \frac{qQd}{D^2}\right]$$

Solution by Sreeraj P, M.Sc Physics