# Derivation of electric field intensity due to uniformly charged thin infinite plane sheet

Let σ be the charge density. That is charge per unit area

Let us imagine a cylindrical portion being perpendicular to the plane sheet

Let A be the area of cross section.

Hence σA is the charge enclosed within that closed surface

By Gauss idea the flux coming out has to be 1/εo * (σA)

Now let us consider the two extreme flat faces of area A

Then by manipulation we have E A as the flux from each flat face

Hence 2 E A from both the faces.

Note there is no chance of getting a electrical lines of force crossing the curved surface area

Hence 2 E A = 1/εo * (σA)

==>

**E = σ/2ε**o

It does not have any distance of the point right from the charged sheet

Hence field has to be uniform on both sides of the sheet

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