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. $\frac{qR}{4\pi {\epsilon }_0{d}^2}$
• B. $\frac{q}{2\pi {\epsilon }_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$
• C. zero
• D. $\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

Answer 1

Answer: (B)

The electric potential at a distance $x$ from charged ring is $V=\frac{q}{4\pi {ϵ}_0\sqrt{x^2+R^2}}$
Thus, the potential at the center of ring of charge $+Q=$ potential due to itself $+$ potential due to other ring of charge $-Q$.
or $V_1=\frac{Q}{4\pi {ϵ}_0\sqrt{0^2+R^2}}+\frac{(-Q)}{4\pi {ϵ}_0\sqrt{d^2+R^2}}$
also the potential at the center of ring of charge $-Q=$ potential due to itself $+$ potential due to other ring of charge $+Q$.
or $V_2=\frac{(-Q)}{4\pi {ϵ}_0\sqrt{0^2+R^2}}+\frac{Q}{4\pi {ϵ}_0\sqrt{d^2+R^2}}$
Thus, potential difference the center of the two rings is $V=V_1-V_2$
or $V=\frac{Q}{4\pi {ϵ}_{0}R}+\frac{(-Q)}{4\pi {ϵ}_0\sqrt{d^2+R^2}}-\frac{(-Q)}{4\pi {ϵ}_{0}R}-\frac{Q}{4\pi {ϵ}_0\sqrt{d^2+R^2}}=\frac{Q}{2\pi {ϵ}_0}\left[\frac{1}{R}-\frac{1}{\sqrt{d^2+R^2}}\right]$