Question

Inside an infinitely long circular cylinder charged uniformly with volume density $\rho$ there is a circular cylindrical cavity. The distance between the axes of the cylinder and the cavity is equal to $a$. Find the electric field strength $E$ inside the cavity. The permittivity is assumed to be equal to unity.

Answer 1

Let us consider a cylinderical Gaussian surface of radius $r$ and height $h$ inside an infinitely long charged cylinder with charge density $\rho$. Now from Gauss theorem:

$E_{r}2\pi rh=\frac{q_{inclosed}}{{ϵ}_0}$

(where $E_r$ is the field inside the cylinder at a distance $r$ from its axis.)

or, $E_{r}2\pi rh=\frac{\rho \pi r^{2}h}{{ϵ}_0}$ or $E_r=\frac{\rho r}{2{ϵ}_0}$

Now, field at a point $P$, inside the cavity, is

$\overrightarrow{E}={\overrightarrow{E}}_{+}+{\overrightarrow{E}}_{-}=\frac{\rho }{2{ϵ}_0}({\overrightarrow{r}}_{+}-{\overrightarrow{r}}_{-})=\frac{\rho }{2{ϵ}_0}\overrightarrow{a}$.