Final answer:
The electric field due to a uniformly charged infinitely large plane thin sheet is found using Gauss's law and symmetry considerations. A Gaussian surface is chosen, and by equating the electric flux through the surface to the enclosed charge divided by the permittivity of free space, the electric field magnitude E = σ/2ε0 is determined.
Step-by-step explanation:
To find the electric field due to a uniformly charged infinitely large plane thin sheet using Gauss's theorem, we consider symmetry arguments that dictate the field must be perpendicular to the plane of the sheet and have the same magnitude at all points equidistant from the plane. We then choose a Gaussian surface in the form of a cylinder with cross-sectional area A and height 2x, with the sheet bisecting the cylinder. According to Gauss's law, the total electric flux through the cylindrical surface is then the sum of the fluxes through both ends, given by 2EA (because there is no flux through the side of the cylinder), where E is the electric field magnitude and A is the area of each circular end of the cylinder. The enclosed charge by the Gaussian surface is the charge density σ times the area A of the sheet, σA. Applying Gauss's law: 2EA = σA/ε0, where ε0 is the permittivity of free space. Solving for E, we get E = σ/2ε0. Therefore, the electric field due to the sheet is uniform in magnitude and directed away from the sheet if the charge is positive or towards the sheet if the charge is negative.