Question

Use Gauss's law to obtain the expression for the electric field due to a uniformly charged infinite plane sheet of charge.

Answer 1

Let us consider an infinitely thin plane sheet of positive charge having a uniform surface charge density σ. Let P be the point where the electric field E is to be found. Let us imagine a cylindrical Gaussian surface of length 2r and containing P, as shown. The net flux is through the cylindrical Gaussian surface.

${\displaystyle \varphi =\text{∮}\overrightarrow{E}\cdot \overrightarrow{dA}}$

= ${\displaystyle \underset{RCF}{\int }\overrightarrow{E}\cdot \overrightarrow{dA}+\underset{LCF}{\int }\overrightarrow{E}\cdot \overrightarrow{dA}+\underset{CS}{\int }\overrightarrow{E}\cdot \overrightarrow{dA}}$

= ${\displaystyle \underset{RCF}{\int }E\cdot dA\cos 0\text{°}+\underset{LCF}{\int }E\cdot dA\cos 0\text{°}+\underset{CS}{\int }E\cdot dA\cos 90\text{°}}$

= EA + EA + 0

= 2EA   ...(1)

Here, A is the area of the cross-section of each circular face, i.e., LCF and RCF.

The total charge enclosed by the Gaussian cylinder

= σA   ...(2)

Using Gauss’s theorem, from (1) and (2),

2EA = ${\displaystyle \frac{\sigma A}{{\epsilon }_0}}$

E = ${\displaystyle \frac{\sigma }{2{\epsilon }_0}}$