Use gauss' law to obtain the electric field of an infinite sheet of charge with uniform charge density ps. take the sheet as lying in the xy plane of the cartesian coordinate system.

Draw a cylinder with height H and having circular faces each with area A such that the infinite sheet is parallel to the cylinder's circular faces and the sheet cuts the cylinder in half.

Because the sheet is infinitely large and lies in the x-y plane, the electric field only points in the +z or -z direction.

Calculate the electric flux through the cylinder. The cylinder's body runs parallel to the electric field while its circular faces are perpendicular to the field, so there is only electric flux through the cylinder's circular faces. The electric field E is the same along the circular faces, so the flux through one circular face is equal to EA. Therefore the total electric flux through the cylinder is Φ = 2EA.

Remember that the electric flux is proportional to the enclosed charge:

Φ = Q_{enclosed}/ε₀

The cylinder encloses a portion of the sheet having an area equal to one of its circular faces, A. Therefore the cylinder encloses an amount of charge given by:

Q_{enclosed} = σA

Now we have:

2EA = σA/ε₀

Solve for E:

E = σ/(2ε₀)

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