Q 12-02-141JEE MainJEE Main 2021 (26 Aug, Shift 2)Medium
The two thin coaxial rings, each of radius $a$ and having charges $+Q$ and $-Q$ respectively are separated by a distance of $s$. The potential difference between the centres of the two rings is:
Answer: (A) $\frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{a} - \frac{1}{\sqrt{s^2+a^2}}\right]$
Every point of a ring is at distance $a$ from its own centre and $\sqrt{s^2 + a^2}$ from the other ring's centre. With $k = \frac{1}{4\pi\varepsilon_0}$:
$V_1 = \dfrac{kQ}{a} - \dfrac{kQ}{\sqrt{s^2+a^2}}$, $V_2 = -\dfrac{kQ}{a} + \dfrac{kQ}{\sqrt{s^2+a^2}}$
$$V_1 - V_2 = \frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{a} - \frac{1}{\sqrt{s^2+a^2}}\right]$$
Solution by Sreeraj P, M.Sc Physics