Question : Use Gauss'theorem to find electric field intensity due to a Uniformly charged infinite plane sheet of charge .
Solution:
Consider a thin infinite plane sheet having uniform surface charge density (charge per unit area) $\sigma$
To calculate the electric field intensity $\vec{E}$ at a point P , distant r from the sheet , draw a Gaussian surface in the form of a closed cylinder of length r on each side of sheet with end caps of area S.
Electric field $\vec{E}$ is perpendicular to the sheet and hence it is perpendicular to the planes of ends caps I and II also.
Charge enclosed by Gaussian surface
$q = \sigma S$
According to Gauss'theorem
$\oint_{S} \vec{E} \cdot d\vec{S}= \frac{q}{\epsilon_{0}}$
$\oint_{S} \vec{E} \cdot d\vec{S}= \frac{\sigma S}{\epsilon_{0}}$ .....(i)
Gaussian surface is divided into three parts I, II, III i.e. end caps and the curved surface of the cylinder
From eqn. (i) becomes
$\oint_{I} \vec{E} \cdot d\vec{S} + \oint_{II} \vec{E} \cdot d\vec{S} + \oint_{III} \vec{E} \cdot d\vec{S}= \frac{\sigma S}{\epsilon_{0}}$ ...(ii)
Angle between $\vec{E}$ and $d\vec{S}$ is $90^\circ$ for the curved surface III. In this case, $\oint_{III} \vec{E} \cdot d\vec{S} = E \, dS \cos 90^\circ = 0$.
Hence eqn. (ii) becomes
$\oint_{I} \vec{E} \cdot d\vec{S} + \oint_{II} \vec{E} \cdot d\vec{S} = \frac{\sigma S}{\epsilon_{0}}$
Since angle between $\vec{E}$ and $d\vec{S}$ for surfaces I and II is zero, so,
$\oint_{I} E \, dS \cos 0^\circ + \oint_{II} E \, dS \cos 0^\circ = \frac{\sigma S}{\epsilon_{0}}$
$\oint_{I} E \, dS + \oint_{II} E \, dS = \frac{\sigma S}{\epsilon_{0}}$
Since electric field intensity $E$ is constant at every point of the Gaussian surface, so
$E \oint_{I} dS + E \oint_{II} dS = \frac{\sigma S}{\epsilon_{0}}$
$ES + ES = \frac{\sigma S}{\epsilon_{0}}$
or
$2ES = \frac{\sigma S}{\epsilon_{0}}$
or
$E= \frac{\sigma}{2 \epsilon_{0}}$ ... (iii)
In vector form, eqn. (iii) can be written as
$\vec{E} = \frac{\sigma}{2 \epsilon_{0}} \hat{n}$
where $\hat{n}$ is a unit vector perpendicular to plane of the sheet pointing away from it.
The electric field is directed away from the sheet if it is positively charged and it is directed towards the sheet if it is negativity charged.
$E \propto \sigma$
Electric Field intensity due to an infinite sheet of charge is independent of the distance of the point of observation.
Thus , electric field due to a charged infinite plane sheet is uniform.