Question

Find the electric potential at the axis of a uniformly charged disc and use potential to find the electric field at same point.

Answer 1

Potential due to ring element of radius $r$ and thickness $dr$ is $dv=\stackrel{R}{\underset{0}{\int }}\frac{1}{4\pi {\epsilon }_0}\frac{\sigma \left(2\pi rdr\right)}{\sqrt{r^2+x^2}}$
$⟹V=\frac{2\sigma }{4\pi {\epsilon }_0}[\sqrt{r^2+x^2}{]}_0^R=\frac{\sigma }{2{\epsilon }_0}(\sqrt{R^2+x^2}-x)$
$E=-\frac{dV}{dx}=-\frac{\sigma }{2{\epsilon }_0}(\frac{2x}{2\sqrt{R^2+x^2}}-1)=\frac{\sigma }{2{\epsilon }_0}(1-\frac{x}{\sqrt{R^2+x^2}})$