Calculate the electric potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.

(N/A) Let us consider a point $P$ on the axis of the disc at a distance $x$ from the centre $O$ of the disc. The disc has a surface charge density $\sigma = \frac{Q}{\pi R^2}$.
Consider a thin elemental ring of radius $r$ and width $dr$ on the disc. The area of this ring is $dA = 2\pi r dr$.
The charge on this elemental ring is $dq = \sigma dA = \sigma (2\pi r dr)$.
The distance of every point on this ring from point $P$ is $\sqrt{r^2 + x^2}$.
The potential $dV$ at point $P$ due to this ring is $dV = \frac{k dq}{\sqrt{r^2 + x^2}} = \frac{1}{4\pi \epsilon_0} \frac{\sigma (2\pi r dr)}{\sqrt{r^2 + x^2}}$.
To find the total potential $V$,we integrate from $r = 0$ to $r = R$:
$V = \int_0^R \frac{\sigma 2\pi r dr}{4\pi \epsilon_0 \sqrt{r^2 + x^2}} = \frac{\sigma}{2\epsilon_0} \int_0^R \frac{r dr}{\sqrt{r^2 + x^2}}$.
Let $u = r^2 + x^2$,then $du = 2r dr$,or $r dr = \frac{du}{2}$.
$V = \frac{\sigma}{2\epsilon_0} \int_{x^2}^{R^2+x^2} \frac{du/2}{\sqrt{u}} = \frac{\sigma}{4\epsilon_0} [2\sqrt{u}]_{x^2}^{R^2+x^2} = \frac{\sigma}{2\epsilon_0} [\sqrt{R^2 + x^2} - x]$.
Substituting $\sigma = \frac{Q}{\pi R^2}$,we get:
$V = \frac{Q}{2\pi \epsilon_0 R^2} [\sqrt{R^2 + x^2} - x]$.