Past exam of the physics 2 paper of the University of Oxford / 2014 resit / 8

In matter, Gauss's law for free electric charge isThe area normal points out of $\mathcal{V}$. In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$, so these become $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enclosed}}/{ϵ}_0$.

Put charges $+Q$ and $-Q$ on the concentric conductors. Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ between them, $a<r<2a$. The potential difference isThe capacitance of this vacuum spherical capacitor is thereforeThe conductors are equipotentials; a uniform dielectric filling would replace ${ϵ}_0$ by its permittivity.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.