Question

use Gauss theorem to find the electric field due to a uniformly charged infinite plane sheet ​

Answer 1

Answer:

Consider an infinite thin plane sheet of positive charge with a uniform surface charge density σ on both sides of the sheet. Let P be the point at a distance a from the sheet at which the electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.

Explanation:

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

​

.

Answer 2

Answer:

sigma/(2 epsilon0)

Explanation:

Using Gauss theorem derive an expression for electric field due to a uniformly charge infinite plane sheet.?

please see the enclosed attachment for detailed explanation and step by step working.

take an infinitely large charged sheet MNOP of charge density sigma.

Take a point P at the centre of EFGH. To find E at P, construct a Gaussian surface - a cuboid IJKL - EFGH . The electric field and flux lines flow out of the sheet perpendicular to it.

Gauuss's law:

closed Integral over cuboid of E • dA = q/epsilon0.

2 E = sigma/ epsilon0

E = sigma/(2 epsilon0)