Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 8

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

Take the grounded electrical conductor to occupy $z\le 0$ and approximate the negligible-radius charged sphere by a point charge $Q$ at ${\mathbf{r}}_{+}=D\hat{\mathbf{z}}$. Put a fictitious image charge $-Q$ at ${\mathbf{r}}_{-}=-D\hat{\mathbf{z}}$. The constructed electric potential is zero on $z=0$, has the correct point charge singularity, and vanishes at infinity. The uniqueness theorem for the electrostatic potential makes it the physical electric potential for $z>0$:Taking its negative gradient gives the requested electric field outside the small sphere:On the plane at cylindrical radius $r$, the tangential components cancel and the normal component isThe electrostatic boundary conditions at a conductor, with outward unit normal $+\hat{\mathbf{z}}$ and zero interior electric field, give the induced surface charge densityFinally integrate over annuli of area $2\pi rdr$:The image charge is a computational device, not a real charge below the plane. For a finite conducting sphere, its induced polarization would change the point-source approximation; the stated negligible radius is essential.