Electric Potential & Equipotentials question from JEE Main 2021 (26 Aug, Shift 2)

Answer: (d)

1. Every point of a ring is at distance $a$ from its own centre and $\sqrt{a^2 + s^2}$ from the other centre.
2. $V_1 = \dfrac{kQ}{a} - \dfrac{kQ}{\sqrt{s^2 + a^2}}$, $V_2 = -\dfrac{kQ}{a} + \dfrac{kQ}{\sqrt{s^2 + a^2}}$
3. $V_1 - V_2 = 2kQ\left[\dfrac1a - \dfrac{1}{\sqrt{s^2 + a^2}}\right] = \dfrac{Q}{2\pi\varepsilon_0}\left[\dfrac1a - \dfrac{1}{\sqrt{s^2 + a^2}}\right]$
4. So the answer is (d).