The two thin co-axial rings, each of radius a and having charges + Q and − Q respectively, are separated by a distance of s. The potential difference between the centres of the two rings is

Given, charges on two rings are + Q and – Q, respectively as shown below.

Potential at centre of ring A due to charge on ring A and charge on ring B is given as

           ${\mathrm{V}}_{\mathrm{A}}=\frac{\mathrm{kQ}}{\mathrm{a}}-\frac{\mathrm{kQ}}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}$

Similarly, potential at centre of ring B due to charge on ring B and charge on ring A is given as

            ${\mathrm{V}}_{\mathrm{B}}=-\frac{\mathrm{kQ}}{\mathrm{a}}+\frac{\mathrm{kQ}}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}$

Potential difference between centre of both the rings is

          $\begin{array}{l}\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\\ =\frac{\mathrm{kQ}}{\mathrm{a}}-\frac{\mathrm{kQ}}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}-\left(-\frac{\mathrm{kQ}}{\mathrm{a}}+\frac{\mathrm{kQ}}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}\right)\\ =\frac{2\mathrm{kQ}}{\mathrm{a}}-\frac{2\mathrm{kQ}}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}\end{array}$

                  $=\frac{\mathrm{Q}}{2{\mathrm{\pi \epsilon }}_0} \left(\frac{1}{\mathrm{a}}-\frac{1}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}\right) \left(\because \mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0} \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}\right)$

Thus, potential difference between centre of both the rings is

             $\frac{\mathrm{Q}}{2{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\mathrm{a}}-\frac{1}{\sqrt{{\mathrm{s}}^2+{\mathrm{a}}^2}}\right)$