A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\frac{\sigma}{2 \in_{0}}\right) \hat{n}$, where $\hat{n}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
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 $|q|=\vec{\sigma} \times \overrightarrow{d s}$
According to Gauss’s law,
Flux, $\phi=\vec{E} \cdot \overrightarrow{d s}=\frac{|q|}{\epsilon_{0}}$
$E d s=\frac{\vec{\sigma} \times \overrightarrow{d s}}{\epsilon_{0}}$
$\therefore E=\frac{\sigma}{\epsilon_{0}} \hat{n}$
Therefore, the electric field just outside the conductor is $\frac{\sigma}{\epsilon_{0}} \hat{n}$. This field is a superposition of field due to the cavity ( $E$ ) 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.
$\therefore E^{\prime}+E^{\prime}=E$
$E^{\prime}=\frac{E}{2}$
$=\frac{\sigma}{2 \in_{0}} \hat{n}$
Therefore, the field due to the rest of the conductor is $\frac{\sigma}{\epsilon_{0}} \hat{n}$.
Hence, proved.