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. $\frac{Q.R}{4\pi {\epsilon }_0{d}^2}$
• B. $\frac{Q}{2\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$
• C. $\frac{Q}{{4\pi \epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$
• D. 0

Answer 1

The correct option is A $\frac{Q.R}{4\pi {\epsilon }_0{d}^2}$

Two thin wire ring each having a radius $=R$

distance $=d$

two rings are $+q$ and $-q$.

$PD=\frac{1}{4\pi {ϵ}_0}\frac{QR}{d^2}$

Now in this question, we get the solution.