Question

Define Gauss theorem by using this theorem find the relation for electric field intensity (e) due to charged infinite thin sheet

Answer 1

Answer:

Consider an infinite plane which carries the uniform charge per unit area σ

Let the plane coincides with the y−z plane.

Let us draw a cylindrical gaussian surface, whose axis is normal to the plane, and which is cut in half by the plane.

Let the cylinder run from x=−a to x=+a, and let its cross-sectional area be A. According to Gauss' law,

2E(a)A=  

ϵ  

0

​  

 

σA

​  

,

where E(a)=−E(−a) is the electric field strength at x=+a. Here, the left-hand side represents the electric flux out of the surface.

The only contributions to this flux come from the flat surfaces at the two ends of the cylinder. The right-hand side represents the charge enclosed by the cylindrical surface, divided by ϵ  

0

​  

. It follows that

E=  

2ϵ  

0

​  

 

σ

​  

.

Explanation:

Answer 2

Gauss Theorem :

According to Gauss Theorem , the total flux through a closed surface is $\frac{1}{\epsilon_0}$ times the net charge enclosed by the closed surface .

Numerically , it can be expressed as :

$\phi_{\tiny{E}} = \oint \vec{E} . \vec{dS} = \dfrac{q}{\epsilon_0}$

For electric field due to charge infinitely thin plane sheet , refer to attachment .

The angle between E and ds in 1 and 2 is 0° (from the figure) and in 3 is 90° (from the figure) . Hence , the effect of flux in 3 will become 0 .

If the plane sheet is thick , the formula gets reduced to $\dfrac{\sigma}{\epsilon_0}$ because the charge accumulated on the Gaussian surface will be twice as charge will be distributed on both side of sheet .

Where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space .