When we computed the electrostatic force on a surface with surface charge density σ, a conceptual problem arose because the electric field at the charged surface was discontinuous. One way to avoid such a problem is to consider that the surface charge is actually spread out over a layer of finite thickness d, so that there is a volume charge density ρ = σ/d. The electric field in this situation is then continuous and one can without any problems write that the force per unit volume is f = ρE.
(a) Consider an infinite flat slab of finite thickness d with uniform charge density ρ, that is oriented with normal in the x direction. Compute E(r) for this configuration. Do not assume that the field is necessarily antisymmetric about the center of the slab.
(b) Now compute the force per unit area on the slab, f = ∫dx ρ(x)E(x) and show that you get the same result that we derived in class for the charged surface, i.e. f = (1/2)σ [Eabove + Ebelow], where σ = ρd, and Eabove is the electric field just on one side of the slab and Ebelow is the electric field just on the other side of the slab.
(c) Now suppose that the charge density in the slab is not necessarily uniform, but could be some arbitrary function of x such that ∫ dx ρ(x) = σ, with σ a constant. Prove mathematically that the electrostatic force per unit area on the slab remains the same as in part (b) [Hint: Consider Gauss' Law in differential form as applied to this problem.]