Question

An annular disc of inner radius $a$ and outer radius '$2a$' is uniformly charged with uniform surface charge density $\sigma$. Find the potential at a distance '$a$' from the centre at a point '$P$' lying on the axis.

Answer 1

The charge contained in the elementary ring
$dq=\sigma \left(2\pi rdr\right)$
The potential due to the ring at $P$
$=dV=\frac{dq}{4\pi {\in }_0{r}^{\prime }}$
$⟹$ The total potential at $P$ due to the given disc $V$
$=\int dV=\frac{1}{4\pi {\in }_0}\int \frac{dq}{r}$
Since $dq=\sigma 2\pi rdr$ and
$r^{\prime }=\sqrt{a^2+r^2}$
$⟹V_p=\frac{\sigma }{2{\in }_0}\stackrel{r=2a}{\underset{r=a}{\int }}\frac{rdr}{\sqrt{a^2+r^2}}$
$⟹V_p=\frac{\sigma }{2{\in }_0}{\left[\sqrt{a^2+r^2}\right]}_{r=a}^{r=2a}=\frac{\sigma }{2{\in }_0}a(\sqrt{5}-\sqrt{2})$