Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 10 / b

Choose $z=0$ at the sheet, positive $z$ upwards, so the uniformly charged slab occupies $-d<z<0$. Use the usual isolated-plane superposition convention, with no imposed uniform background field. The slab alone produces $+\rho d/(2{ϵ}_0)\hat{\mathbf{z}}$ above it, while the negatively charged sheet produces $-\sigma /(2{ϵ}_0)\hat{\mathbf{z}}$ there. ConsequentlySince $\rho d>\sigma$, the field is upward, away from the net positive charge distribution, with magnitude $(\rho d-\sigma )/(2{ϵ}_0)$. The equal halves in the separate infinite-sheet fields follow from their reflection symmetry before superposition.

Below the slab, its own field is downward, $-\rho d/(2{ϵ}_0)\hat{\mathbf{z}}$. The negative sheet attracts a positive test charge upwards and contributes $+\sigma /(2{ϵ}_0)\hat{\mathbf{z}}$. Adding the fields givesThe magnitude equals the above-sheet value, but the direction is downward. This equality of outside magnitudes describes the zero-background solution for the given infinite charge planes; adding an externally imposed uniform field would constitute extra boundary data not specified here.

Inside the uniformly charged slab, its own electric field is $\rho (z+d/2)\hat{\mathbf{z}}/{ϵ}_0$: it vanishes at the slab's midplane and has derivative $\rho /{ϵ}_0$. Everywhere below the negative sheet, that sheet adds $+\sigma \hat{\mathbf{z}}/(2{ϵ}_0)$. ThereforeIts magnitude is the absolute value of this scalar coefficient. The field changes direction atFor $z<z_0$ it points downward; for $z>z_0$ it points upward, towards the negative sheet. Thus it is neither uniform nor directed upwards throughout the slab. The derivative and the field at the lower slab boundary also agree with Gauss's law and the previously found outside field.