Question

Two identical thin rings, each of radius R in, are coaxially placed at a distance R m from each other. If $Q_1$ coulomb and $Q_2$ coulomb are the charges uniformly spread on the two rings, find the work done in moving a charge q from the center of one ring to that of the other.

Answer 1

If $r$ is the radius of each ring and $R$ is the distance of separation between two centers of two rings.

Potential at center of ring 2 $=V_2=\frac{k{Q}_1}{\sqrt{r^2+R^2}}+\frac{k{Q}_2}{r}$
Potential at center of ring 1 $=V_1=\frac{k{Q}_2}{\sqrt{r^2+R^2}}+\frac{k{Q}_1}{r}$
Work done in moving a unit charge $q$ from the center of ring 1 to the center of ring 2 will be:

$W=q(V_2-V_1)$

So, on substituting $V_1$ and $V_2$ we get:

$W=qk(Q_2-Q_1)[\frac{1}{\sqrt{R^2+r^2}}-\frac{1}{r}]$

putting $r=R=$ radius of ring;

$W=qk(Q_2-Q_1)[\frac{1}{\sqrt{2}R}-\frac{1}{R}]$

$=\frac{qk}{R}(Q_2-Q_1)\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{q}{\sqrt{2}R}(Q_1-Q_2)(\sqrt{2}-1)$