Question

Prove that the electric field on the surface of a charged
conducting shell is $\sigma /{\epsilon }_0$.

Answer 1

$\oint \overrightarrow{E}\overrightarrow{ds}=\frac{Q_{enc}}{{\epsilon }_0}$ [Gauss Law]

For a charged conducting shell;

$Q_{enc}=\sigma .s=\sigma .4\pi r^2$

So, Gauss Law becomes;

$\mid \overrightarrow{E}\mid \left(4\pi r^2\right)=\frac{\sigma }{{\epsilon }_0}\left(4\pi r^2\right)$

$\mid \overrightarrow{E}\mid =\frac{\sigma }{{\epsilon }_0}$

$E=\frac{\sigma }{{\epsilon }_0}$