A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\frac{\sigma}{2 \epsilon_{0}} \hat{n}$, where $\mathrm{n}^{\hat{ }}$ is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.

Home » A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is , where is the unit vector in the outward normal direction and is the surface charge density near the hole.

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\frac{\sigma}{2 \epsilon_{0}} \hat{n}$ , where $\mathrm{n}^{\hat{ }}$ is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.

CBSE | Class 12 | Electric Charges and Fields | NCERT | Physics

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\frac{\sigma}{2 \epsilon_{0}} \hat{n}$ , where $\mathrm{n}^{\hat{ }}$ is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.

Solution:

If we assume that the hole has been filled, then Gauss’s formula may be used to calculate the electric field intensity at a location near to the surface of the conductor, where

Flux, $\phi=E . d s=\frac{q}{\epsilon_{0}}$

$q=\sigma x d s$

here, $\sigma$ is the surface charge density

ds is the area of the Gaussian surface

$\mathrm{Eds}=(\sigma \mathrm{x} \mathrm{ds}) / \mathrm{\epsilon}_{0}$

$\mathrm{E}=\sigma / \mathrm{E}_{0}$

$=\frac{\sigma}{\epsilon_{0}} \hat{n}$

Let $E_{1}$ be the electric field due to the hole and $E_{2}$ is the electric field due to the rest of the conductor. $\mathrm{E}{1}$ and $\mathrm{E}{2}$ are opposite in direction and equal in magnitude since the total electric field inside the conductor is zero.