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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.

1 Answer

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)$
answered Jun 7, 2014 by meena.p