An annular disc of inner radius a and outer radius ‘2a’ is uniformly charged with 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 ring $=dq = \sigma (2 \pi r dr)$

The potential due to the ring at $P = dV =\large\frac{dq}{4 \pi \in_0r'}$

=> the total potential at P due to the given disc $V = \int dV $

$\quad= \large\frac{1}{4 \pi \in_0} \int \large\frac{dq}{r'}$

Since $dq = \sigma 2 \pi r dr$ and

$r' =\sqrt {a^2+r^2}$

=> $V_p =\large\frac{\sigma}{2 \in_0 } \int \limits_{r=a} ^{r=2a} \large\frac{rdr}{\sqrt {a^2+r^2}}$

=> $V_p =\large\frac{\sigma}{2 \in_0} \bigg[ \sqrt {a^2+r^2} \bigg]_{=a}^{=2a}$

$V_p =\large\frac{\sigma}{2 \in_0}$$a (\sqrt {5} -\sqrt 2)$