Question

State Guass law in electrostatics. Derive an expression of electric field due to an infinitely long straight uniformly charged wire. Draw necessary diagram.

Answer 1

The Gauss law states that electric flux passing through any closed surface is equal to the charge enclosed by that surface divided by permittivity of vacuum.

By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. $\overrightarrow{E}$ and $\overrightarrow{ds}$ are along the same direction.

Now here we have the two surfaces, one curved and other the plane caps,

First, the flux through the curved surface,

$\oint \overrightarrow{E}\cdot \overrightarrow{ds}=\frac{q_{in}}{{ϵ}_0}$

$E\left(2\pi rl\right)=\lambda l/{ϵ}_0$

$E=\frac{\lambda }{2\pi r{ϵ}_0}$

Now due to the plane caps,

The angle between $\overrightarrow{E}$ and $\overrightarrow{ds}$ is 90,

so the flux through that part is zero

so, Total flux through the closed surface is,

$E=\frac{\lambda }{2\pi r{ϵ}_0}$