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 $\overrightarrow{E}$ be the electric field in the hole due to rest of the charges conductor. In order to find $\overrightarrow{E}$ consider that the hole in the conductor is plugged. Let $\overrightarrow{E}$ be the electric field due to the charged conductor at a point outside it then,

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

After plugging the hole, the conductor becomes a closed one, the electric field inside it must be zero. It implies that inside the conductor, electric field due to the rest of the conductor must be equal and opposite to that due to small portion of the conductor, that the hole. However outside the conductor, the two fields are in the same direction i.e., along the outward normal to the surface of the conductor. It therefore,follows that the electric field in the hole due to the rest of the conductor,

$\overrightarrow{E}=\frac{1}{2}{\overrightarrow{E}}^{\prime }=\frac{\sigma }{2{\in }_0}\hat{n}$