Question

Two thin rings each having a radius $R$ are placed at distance $d$ apart with their axes coinciding. The charges on the two rings are $+q,-q$. The potential difference between the rings is:

• A. $\frac{qR}{4\pi {\epsilon }_0{d}^2}$
• 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

Answer: (B)

$\frac{q}{2\pi {\epsilon }_0}(\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}})$

Potential due to a ring of charge $q$ at any point d distance away from the center of the ring on its axis is given by $kq/\sqrt{R^2+d^2}$ where R is the radius of the ring.
From the diagram:
the potential on the center of ring 1$=kq(1/R-1/\sqrt{R^2+d^2})$
and the potential on the center of ring 2$=-kq(1/R-1/\sqrt{R^2+d^2})$
Difference in the potentials is thus,
$=2kq(1/R-1/\sqrt{R^2+d^2})$