Question

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 center of the two rings is :

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

Answer 1

The correct option is D $\frac{Q}{2\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$

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 $V=\frac{Q}{4\pi {ϵ}_{0}R}+\frac{(-Q)}{4\pi {ϵ}_0\sqrt{d^2+R^2}}-\frac{(-Q)}{4\pi {ϵ}_{0}R}-\frac{Q}{4\pi {ϵ}_0\sqrt{d^2+R^2}}=\frac{Q}{2\pi {ϵ}_0}[\frac{1}{R}-\frac{1}{\sqrt{d^2+R^2}}]$