Cylindrical Capacitance

Use Gauss' Law to determine the electric field between the cylinders. On a cylindrical Gaussian surface of length $L$ , have:

$E(2\pi r L) = \frac{\lambda L}{ \epsilon_0}.$

where $\lambda$ is the charge density per unit length. So the electric field between the cylinders is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}.$

Integrating from $a$ to $b$ , find the potential:

$V = \frac{\lambda}{2\pi \epsilon_0} \int_a^b \frac{dr}{r} = \frac{\lambda}{2\pi \epsilon_0} \log (b/a).$

From $Q=CV$ , since $Q = \lambda L$ , have:

$Q = \frac{2\pi \epsilon_0 L}{\log(b/a)} V.$

which gives the capacitance as claimed.