Question

In solving for the magnitude of the electric field E⃗(z) produced by a sheet charge with charge density σ, use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane parallel to the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A which will cancel out of the expression for E(z) in the end. The result of applying Gauss's law to this situation then gives an expression for E(z) for both z>0 and z<0. (Figure 3) Express E(z) for z>0 in terms of some or all of the variables/constants σ, z, and ϵ0.

Answer 1

AX 7oCcharge density Et A (area Let the charge sheet passes through the middle of the cylinder's length, along Z-axis, so the cylinder is perpendicular to the surface. Hence the flux through each end is dE=E ZEA GS 0° OE = EA Due to symmetry, the total flux through Gaussicin cylinder = 2EA. Hence by Gauss's theorem, E - VAT (TE) therefore, 2EA = DA : E = or Ez = 7 (i) If z>o (along positive z-axis), (ii) If zko (along negative z-axis)