Home exercise 3.2.3: Uniformly charged cylinder

The cylinder has the symmetry of Problem 3.2.1, so $E=E_{\rho }(\rho )a_{\rho }$. For the left-hand side we take a coaxial cylindrical surface of radius $\rho$ from $z=-L/2$ to $z=L/2$. As in Problem 3.2.1,

The Gaussian cylinder encloses a solid cylinder of radius $\rho$ and length $L$:

and equating both sides, $2\pi L{\epsilon }_0\rho E_{\rho }={\rho }_v\pi {\rho }^{2}L$:

The Gaussian cylinder encloses the whole cross-section $\pi b^2$:

which gives

Outside, the cylinder acts like a line charge ${\rho }_L=\pi b^2{\rho }_v$ on its axis (Problem 3.2.1), just as the hollow cylinder of Problem 3.2.2 did with ${\rho }_L=2\pi b{\rho }_S$: only the charge per unit length matters outside. Inside, the field grows linearly from zero on the axis (Figure 3.2.25), like the field inside the uniformly charged sphere, ${\rho }_{v}r/(3{\epsilon }_0)$, but with $2{\epsilon }_0$ instead of $3{\epsilon }_0$. Both expressions give ${\rho }_{v}b/(2{\epsilon }_0)$ at $\rho =b$: no surface charge, no jump.