Q. Two thin wire rings each having radius R are placed at a distance *d *apart with their axes coinciding. The charges on the two rings are +Q and -Q The potential difference between the centres of the two rings is

(a)Zero

(b) $ \displaystyle \frac{Q}{4\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]$

(c) $ \displaystyle \frac{Q}{4\pi\epsilon_0 d^2} $

(d) $ \displaystyle \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]$

Ans: (d)

Sol:

$ \displaystyle V_1 = \frac{-Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{Q}{4\pi \epsilon_0 R} $

$ \displaystyle V_2 = \frac{Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{-Q}{4\pi \epsilon_0 R} $

$ \displaystyle V_1 -V_2 = \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}] $