Past exam of the physics 2 paper of the University of Oxford / 2016 resit / 2 / Solution

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.