Electric Intensity Due To An Infinite Sheet Of Charge. Assume we have a plane sheet of infinite extent on which the positive charges are uniformly distributed.

Now imagine a cylinder passing through this sheet.

The charge enclosed by the surface is A. 

The first face of the infinite sheet

1=E.Acos

cos=1

1=E.A

Second face of the infinite sheet

2=E.Acos

cos=1

2=E.A

Third face of the infinite sheet

3=E.Acos

cos=0

3=0

The total net flux will be determined by

=1+2+3

=E.A+E.A+0

=2E.A

According to the Gauss’s law

=1Qt

Since is the charge density, Q can be considered as Q=.A

1A

2E.A=1A

A and A will be canceled out.

This leads us to 

2E=1

2E=

E=2