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. $\frac{Q R}{4 \pi \in _{0} d^{2}}$
• 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 $V_{A}=V_{+ Q}+V_{- Q}=\frac{K Q}{R}+\frac{K \left(\right. - Q \left.\right)}{\sqrt{R^{2} + d^{2}}}$
$\Delta V=V_{A}-V_{B}=\frac{K Q}{R}-\frac{K Q}{\sqrt{R^{2} + d^{2}}}-\frac{K Q}{\sqrt{R^{2} + d^{2}}}+\frac{K Q}{R}$
on solving we get $\Delta V=\frac{Q}{2 \pi \left(\epsilon \right)_{0}}\left(\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}\right)$