Past exam of the physics 2 paper of the University of Oxford / 2016 / 3

Translational and reflection symmetry parallel to the slabs imply $\mathbf{E}=E_x(x)\hat{\mathbf{x}}$. The differential Gauss's law is $d{E}_x/dx=\rho /{ϵ}_0$. Starting at the left boundary with $E_x(-a)=0$ givesThere is no singular surface charge at $x=0$, so the normal electric field is continuous there: $E_x(0)=-C/{ϵ}_0$. Integrating through the positive slab givesThe second expression indeed returns to zero at $x=b$, because the two total charges per unit area are $-C$ and $+C$. For $C>0$, the electric field throughout the interior points from the positive slab towards the negative slab, along $-x$; reversing $C$ reverses the directions. The piecewise linear profile is a useful consequence of the uniform volume charge density.

Define the requested electric potential difference as $\Delta V=V(b)-V(-a)$. Since $E_x=-dV/dx$,soGeometrically, the integral is the area of two triangles under the signed electric field profile. For $C>0$, the right-hand plane is at higher electric potential. Choosing $V(-a)=0$, a continuous interior electric potential is $V=C(x+a{)}^2/(2{ϵ}_{0}a)$ in the first slab and $V=Ca/(2{ϵ}_0)+C(x-x^2/(2b))/{ϵ}_0$ in the second. These expressions also verify the boundary values directly.