Let us consider a ring of radius $(R=a)$ having charge $Q$, distributed uniformly.
Now take point $\mathrm{P}$ to be at a distance $\mathrm{x}$ from the centre of the ring or perpendicular to plane of ring as shown in figure. The charge element dq is at a distance $\mathrm{x}$ from point P. Therefore, $\mathrm{V}$ can be written as
$$
\begin{aligned}
V &=k_e \int \frac{d q}{r}=k_e \int \frac{d q}{\sqrt{x^2+a^2}} \\
(\because r&\left.=\sqrt{\left(x^2+a^2\right)}\right)
\end{aligned}
$$
Where, $\mathrm{k}_{\mathrm{e}}=\frac{1}{4 \pi \varepsilon_0}$, thus each element $\mathrm{dq}$ is at the same distance from point $P$, so we get the net potential is :
$$
\mathrm{V}=\frac{\mathrm{k}_{\mathrm{e}}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}} \int \mathrm{dq}=\frac{\mathrm{k}_{\mathrm{c}} \mathrm{Q}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}
$$


Hence, the net electric potential
$$
\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}
$$