Question

A nollow charged conductor has a tiny hole cut into its surface. show that the electric fleld in the hole is $\left(\sigma /2{\epsilon }_0\right)\hat{n},$ where $\dot{n}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.

Answer 1

We can calculate three electric field due to hollow charge sphere by using gauss theorem. Let, the surface charge density near the hole $=6$. Let, $\hat{n}$ be the normal vector which is directed outwards. Let, $P$ be the point on the hole.

According of Gauss theorem,

$\int E.ds=\frac{9}{{ϵ}_0}$

Where $9$ is the charge hear the hole

$6=\frac{9}{ds}$

$\Rightarrow 9=6ds$

$Eds\cos \theta =\frac{6ds}{{ϵ}_0}$

But the angle between the electric field and area vector is $0^0$.

$Eds=\frac{6ds}{{ϵ}_0}$

$E=\frac{6}{{ϵ}_0}$

$E=\frac{6}{{ϵ}_0}\hat{n}$

The electric field is due to the filed up hole and the field due to the root of the change conductor.

Hence, there is no electric field inside the conductor. But outside the conductor, the electric fields are equal and are in the same direction.

Hence the electric field at $P$ due to each point $=\frac{1}{2}E=\frac{6}{2{ϵ}_0}\hat{n}$