Question

Use Gauss' law to find the E field of an infinite solid cylinder of charge of radius R and charge density lambda per unit length. As usual, draw everything on the diagrams and label them.

Answer 1

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

As we know that electric field due to long cylinder on a cylindrical Gaussian surface must be constant

so on the Gaussian surface we will have

$\int E. dA = \frac{q_{en}}{\epsilon_0}$

now the electric field is passing normally through curved surface area of the cylinder

so we will have

$E (2\pi rL) = \frac{q_{en}}{\epsilon_0}$

here enclosed charge in the cylinder is given as

$q_{en} = \lambda L$

from above equation

$E(2\pi rL) = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$