Final answer:
The electric field along the x-axis outside an infinitely long sheet of charge is derived using Gauss's law. By considering a Gaussian surface, we determine the perpendicular electric field component to be E = (η / (2ε_0)) &hat;i, with the direction of the field along the x-axis.
Step-by-step explanation:
To derive the expression for the electric field E along the x-axis for points outside the sheet of charge, we will use Gauss's law, which relates the electric field to the charge enclosed by a Gaussian surface. For an infinitely long sheet of width L with surface charge density η, we consider a Gaussian pillbox with height h that extends symmetrically about the sheet and with faces perpendicular to the sheet. Using symmetry, the electric field components parallel to the sheet cancel out, and only the perpendicular component (Ex) contributes to the net flux.
The total flux through the Gaussian surface is ΦE = Ex A, where A is the area of one end of the Gaussian pillbox. Now, since the total charge enclosed is σ·A, where σ is the surface charge density of the plate (η), we can apply Gauss's law ΦE = Qenc / ε0, yielding Ex = η·A / (2ε0·A) = η / (2ε0).
Thus, the electric field at a point along the x-axis outside the sheet, for x > L/2 and x > 0, is E = (η / (2ε0)) &hat;i, pointing in the &hat;i direction.