Question

A hollow charged conduct has a tiny hole cut into its surface Show that the electric field in the hole is $\left(\sigma /2{\epsilon }_{\circ }\right)$ $\hat{n}$, where $\hat{n}$ is he 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 electric charge, $\sigma$ is the charge density, and ${\epsilon }_0$ is the permittivity of free space.

Hence, the charge can be given as:

$|q|=\overrightarrow{\sigma }.\overrightarrow{dS}$

Thus according to Gauss' Law, the flux

$\varphi =\overrightarrow{E}.\overrightarrow{dS}=\frac{q}{{\epsilon }_0}$

$\Rightarrow \overrightarrow{E}.\overrightarrow{dS}=\frac{\overrightarrow{\sigma }.\overrightarrow{dS}}{{\epsilon }_0}$

$\Rightarrow \overrightarrow{E}=\frac{\sigma }{{\epsilon }_0}\hat{n}$

Therefore, electric field just outside the conductor would be $\frac{\sigma }{{\epsilon }_0}\hat{n}$. This field is a result of superposition of fields i.e, the field due to cavity ($E^{\prime }$) and field due to rest of the charged conductor ($E^{\prime }$). These fields are equal and oppositely directed inside the conductor, and are equal both in magnitude and direction outside the conductor.

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

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

Therefore, electric field due to rest of conductor is $\frac{E}{2}=\frac{\sigma }{2{\epsilon }_0}\hat{n}$; so hence proved.