Let us consider a ring of radius $(\mathrm{R}=\mathrm{a})$ having charge $\mathrm{Q}$, distributed uniformly over a ring.
Now take a point $P$ to be at a distance $z$ from the centre of the ring or perpendicular to plane of ring as shown in figure. The charge element dq is at a distance $z$ from point P. So, V can be written as
$$
\mathrm{V}=\mathrm{k}_{\mathrm{e}} \int \frac{\mathrm{dq}}{\mathrm{r}}=\mathrm{k}_{\mathrm{e}} \int \frac{\mathrm{dq}}{\sqrt{\mathrm{z}^2+\mathrm{a}^2}} \quad\left(\because \mathrm{r}=\sqrt{\mathrm{z}^2+\mathrm{a}^2}\right)
$$
where, $\mathrm{k}=\frac{1}{4 \pi \varepsilon_0}$, since each element $\mathrm{dq}$ is at the same distance from point $\mathrm{P}$, so we have net potential
$$
\mathrm{V}=\frac{\mathrm{k}_{\mathrm{e}}}{\sqrt{\mathrm{z}^2+\mathrm{a}^2}} \int \mathrm{dq}=\frac{\mathrm{k}_{\mathrm{e}} \mathrm{Q}}{\sqrt{\mathrm{z}^2+\mathrm{a}^2}}
$$


Considering $-\mathrm{q}$ charge at $\mathrm{P}$, the potential energy is
$\mathrm{U}=\mathrm{W}=\mathrm{q} \times$ potential difference
or,
$$
\begin{aligned}
&=\frac{\mathrm{k}_{\mathrm{e}} \mathrm{Q}(-\mathrm{q})}{\sqrt{\mathrm{z}^2+\mathrm{a}^2}} \\
\mathrm{U} &=\frac{1}{4 \pi \varepsilon_0} \frac{-\mathrm{Qq}}{\sqrt{\mathrm{z}^2+\mathrm{a}^2}} \\
&=\frac{1}{4 \pi \varepsilon_0 \mathrm{a}} \frac{-\mathrm{Qq}}{\sqrt{1+\left(\frac{\mathrm{z}}{\mathrm{a}}\right)^2}}
\end{aligned}
$$
Let, $\quad \frac{\mathrm{Qq}}{4 \pi \in_0 \mathrm{a}}=\mathrm{S}$, a new constant
Then,
$$
\begin{aligned}
U &=\frac{-S}{\left[1+\left[\frac{z}{a}\right]^2\right]^{1 / 2}} \\
& \because z \gg \gg a ; z^2 \gg>a^2 \text { at } z=0(U=-S)
\end{aligned}
$$

Hence, the variation of potential energy with $\mathrm{z}$ is shown in the figure. The charge $-\mathrm{q}$ displaced would perform oscillations.