Question

Derive the electric field intensity due to line charge?

Answer 1

The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.

Answer 2

By applying Gauss’ law, the field created due to the line charge can be found.

For simplifying the concept, let’s assume that the charge density on the line is linearly uniform throughout.

Furthermore, assume an imaginary surface around the line that forms into a cylindrical surface with r as a radius. Such an imaginary surface is termed to be Gaussian surface.

At all points on the surface, the field would possess equal magnitude.

The direction of electric field is normally outward.  

Then, we know electric flux is the electric field times the area of the Gaussian surface.

The electric flux is given as,

$\phi = E. 2\pi rL=\frac{\Lambda L}{\varepsilon_0}$  

The above equation can be rewritten as,

$E=\frac{\lambda}{2\pi \varepsilon_0r}$  

This gives the electric field intensity due to a line charge.