Answer:
Explanation:
Consider an endless sheet of uniform charge thickness per unit area [tex]\sigma[/tex]
For a boundless sheet of charge, the electric field will be opposite to the surface. In this way, just the closures of a round and hollow Gaussian surface will add to the electric transition. For this situation, around and hollow Gaussian surface opposite to the charge sheet is utilized. The subsequent field is a large portion of that of a conductor at harmony with this surface charge thickness.
By balance, we expect the electric field on one or the other side of a plane to be an element of x just to be guided typical to the plane and to point away from/towards the plane contingent upon whether, [tex]\sigma[/tex] is positive/negative.
According to the law;
[tex]2EA = \dfrac{q_{enc}}{\varepsilon_o}[/tex]
[tex]where; \ q_{enc} = total \ enclosed charge = \sigma A \\ \\ thus; \\ \\ 2EA = \dfrac{\sigma A}{\varepsilon_o} \\ \\ E = \dfrac{\sigma}{2 \varepsilon _o}[/tex]
