Cylindrical Capacitor

Problem

A cylindrical capacitor consists of a cylindrical wire of radius $a$, and a coaxial cylindrical shell of radius $b$. Both the wire and cylindrical shell have length $L$. The wire has a total charge of $Q$ distributed on its surface. The cylindrical shell has a total charge of $-Q$ distributed on its surface.

Calculate the following:

Part 1: Electric field in the region $a < r < b$

Part 2: Electric potential in the region $a < r < b$

Part 3: Capacitance of the spherical capacitor.

Solution

Part 1: Electric field

The first thing to calculate is the electric field between the spherical shells. Use Gauss' law

$\oint \boldsymbol{E} \cdot d\boldsymbol{A} = \frac{Q_{en}}{\epsilon_0}$.

The total charge enclosed in a Gaussian surface between the wire and the cylindrical shell is $Q$. For cylindrical geometry

$\oint \boldsymbol{E} \cdot d\boldsymbol{A} = |\boldsymbol{E}| (2\pi rL)$

$|\boldsymbol{E}| (2\pi rL) = \frac{Q}{\epsilon_0}$

$|\boldsymbol{E}| = \frac{Q}{2\pi \epsilon_0 Lr}$.

The electric field from the inner spherical shell emanates radially outward, so

$\boldsymbol{E} = \frac{Q}{2\pi \epsilon_0 Lr} \boldsymbol{r}$.

Part 2: Electric potential

$V = -\int_{b}^{a} \boldsymbol{E} \cdot d\boldsymbol{l}$

$V = -\int_{b}^{a} \frac{Q}{2\pi \epsilon_0 Lr} dr$

$V = -\frac{Q}{2\pi \epsilon_0 L} \left(\ln(a) - \ln(b) \right)$

$V = \frac{Q}{2\pi \epsilon_0 L} \ln\left(\frac{b}{a}\right)$

Part 3: Capacitance

$C = \frac{Q}{V}$

$C = \frac{2\pi \epsilon_0 L}{\ln\left(\frac{b}{a}\right)}$