Electric Potential & Equipotentials question from JEE Main 2020 (2 Sep, Shift 2)

Answer: (a)

1. Equal $\sigma$: $q \propto$ radius$^2$, so $q_r = \frac{Qr^2}{r^2 + R^2}$, $q_R = \frac{QR^2}{r^2 + R^2}$.
2. $V_{centre} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_r}{r} + \frac{q_R}{R}\right) = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q(r + R)}{r^2 + R^2}$
3. So the answer is (a).