Let us consider a ring of radius $\mathrm{R}=a$ having charge $\mathrm{Q}$ distributed uniformly over the ring. Also a point $\mathrm{P}$ at a distance $z$ on its axis passing through centre. Take $r$ is the distance of $\mathrm{P}$ from charge $d q$ on the ring.

$r=\sqrt{z^{2}+a^{2}} \quad$ where $a=\mathrm{R}$

$\therefore$ Electric potential at point $\mathrm{P}$,

$\mathrm{V}=\int \frac{k d q}{r}=k \int \frac{d q}{r}=k \int \frac{d q}{\sqrt{z^{2}+a^{2}}}$ $\therefore \mathrm{V}=\frac{k}{\sqrt{z^{2}+a^{2}}} \int d q=\frac{k \mathrm{Q}}{\sqrt{z^{2}+a^{2}}}\left[\because \int d q=\right.$ $\text { If }-q\text { is at } \mathrm{P} \text { then potential energy,}$

$\mathrm{U}=\mathrm{W}$

$=q \times \mathrm{V}$

$=-q \times \frac{k \mathrm{Q}}{\sqrt{z^{2}+a^{2}}}$

$U=-\frac{k Q q}{\left[\sqrt{\frac{z^{2}}{a^{2}}+1}\right]}$

Now $\frac{k Q q}{a}=\mathrm{S}$ supposing new constant,

$U=-\frac{S}{\left(1+\frac{z^{2}}{a^{2}}\right)^{1 / 2}}$

$z>>>a$, then $\frac{z^{2}}{a^{2}}>1$ but $z=0, \frac{z^{2}}{a^{2}}=0$ $\therefore \mathrm{U}=-\mathrm{S}$