Question

There are two thin wire rings, each of radius $R$, whose axes coincide. The charges of the rings are $q$ and $-q$. Find the potential difference between the centres of the rings separated by a distance $a$.

Answer 1

The arrangement of the rings are as shown in the figure. Now, potential at the point $1,{\phi }_1=$ potential at $1$ due to the ring $1+$ potential at $1$ due to the ring $2$.

$=\frac{q}{4\pi {ϵ}_{0}R}+\frac{-q}{4\pi {ϵ}_0(R^2+a^2{)}^{1/2}}$

Similarly, the potential at point $2$,

$\phi =\frac{-q}{4\pi {ϵ}_{0}R}+\frac{q}{4\pi {ϵ}_0(R^2+a^2{)}^{1/2}}$

Hence, the sought potential difference,

${\phi }_1-{\phi }_2=△\phi =2(\frac{q}{4\pi {ϵ}_{0}R}+\frac{-q}{4\pi {ϵ}_0(R^2+a^2{)}^{1/2}})$

$=\frac{q}{2\pi {ϵ}_{0}R}(1-\frac{1}{\sqrt{1+(a/R{)}^2}})$.