Exercise 3.2.1: Two coaxial charged cylinders

Gauss's law, Eq. (3.2.13), on a coaxial cylinder of radius $\rho$ and length $L$. By symmetry $E=E_{\rho }(\rho )a_{\rho }$, the caps contribute nothing, and

For the right-hand side, the enclosed charge is $0$ for $\rho <a$, $2\pi aL{\rho }_{S1}$ for $a<\rho <b$, and $2\pi aL{\rho }_{S1}+2\pi bL{\rho }_{S2}$ for $\rho >b$. Equating both sides in each region:

The field outside vanishes when the total charge per unit length, $2\pi (a{\rho }_{S1}+b{\rho }_{S2})$, is zero:

At each surface the field jumps by ${\rho }_S/{\epsilon }_0$: at $\rho =a$ from $0$ to ${\rho }_{S1}/{\epsilon }_0$, at $\rho =b$ by $b{\rho }_{S2}/({\epsilon }_{0}b)={\rho }_{S2}/{\epsilon }_0$. This is the charge distribution of a coaxial cable: equal and opposite charges per unit length on the two conductors confine the field to the space between them (Application 3.2.2, Figure 3.2.26). The outer surface charge density is smaller in magnitude than the inner one by the factor $a/b$, because the same charge per unit length is spread over a larger circumference.