Question

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

Answer 1

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let E is the electric field just outside the conductor, q is the electric charge, $\sigma$ is the charge density, and ${\in }_0$ is the permittivity of free space.

charge $|q|=\overline{\sigma }\times \overline{ds}$

According to Gauss's law,

Flux, $\varphi =\overline{E}.\overline{ds}=\frac{|q|}{{\in }_0}$

$Eds=\frac{\overline{\sigma }\times \overline{ds}}{{\in }_0}$

$\therefore E=\frac{\sigma }{{\in }_0}\hat{n}$

Therefore, the electric field just outside the conductors is $\frac{\sigma }{{\in }_0}\hat{n}$. This field is a superposition of field due to the cavity $\left(E^{\prime }\right)$ and the field due to the rest of the charged conductor $\left(E^{\prime }\right)$. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

$\therefore E^{\prime }+E^{\prime }=E$

$E^{\prime }=\frac{E}{2}$

$E^{\prime }=\frac{\sigma }{2{\in }_0}\hat{n}$

Therefore, the field due to the rest of the conductor is $\frac{\sigma }{{\in }_0}\hat{n}$. Hence, proved.