Question

Two thin wire rings each having a 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. $\frac{Q}{4\pi {\in }_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$
• C. $\frac{QR}{4\pi {\in }_0{d}^2}$
• D. $\frac{Q}{2\pi {\in }_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

Answer 1

Answer: (D)

$\Delta V=V_A-V_B$
where $V_A=V_{+Q}+V_{-Q}=\frac{KQ}{R}+\frac{K(-Q)}{\sqrt{R^2+d^2}}$
$\Delta V=V_A-V_B=\frac{KQ}{R}-\frac{KQ}{\sqrt{R^2+d^2}}-\frac{KQ}{\sqrt{R^2+d^2}}+\frac{KQ}{R}$
on solving we get $\Delta V=\frac{Q}{2\pi {\left(ϵ\right)}_0}\left(\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right)$