# A hollow charged conductor has a tiny hole cut into its surface.

Question:

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.

Solution:

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.