Question

Define electric flux. Apply Gauss' law to obtain an expression for the electric field intensity at a point due to an infinitely long uniformly charged straight wire. Draw the necessary diagram.

Answer 1

Electric flux is the measure of flow of the electric field through a given area. It is proportional to the number of electric field lines going through a normally perpendicular surface.

For electric field $\overrightarrow{E}$ passing through an area $\overrightarrow{A}$, the flux passing through it is $\overrightarrow{E}.\overrightarrow{A}$.

According to Gauss' law, flux passing through a closed surface is proportional to the amount of charge in the enclosed volume.

$\overrightarrow{E}.\overrightarrow{A}=\frac{q_{enc}}{{ϵ}_0}$

Let us take a cylindrical surface co-axial with the charge carrying straight wire as Gaussian surface.

Enclosed charge in this volume as shown in figure, assuming linear charge density to be $\lambda$, is

$q_{enc}=\lambda L$

Hence according to the Gauss' law,

$E\left(2\pi RL\right)=\frac{\lambda L}{{ϵ}_0}$

$⟹E=\frac{\lambda }{2\pi {ϵ}_{0}R}$