Home exercise 3.2.1: Uniformly charged sphere

The charge distribution is spherically symmetric, so $E=E_r(r)a_r$. For the left-hand side of Eq. (3.2.13) we take a concentric sphere of radius $r$. As in Problem 3.2.5,

to give

The charge is uniform, so the enclosed fraction of $Q$ is the volume fraction:

Equating both sides, $4\pi r^2{\epsilon }_0{E}_r=Q{r}^3/b^3$, which gives

The Gaussian sphere encloses the whole charge,

which gives

$|E|$ grows linearly from zero at the centre to its largest value $Q/(4\pi {\epsilon }_0{b}^2)$ at the surface $r=b$, where the two expressions agree, and then decreases as $1/r^2$ (Figure 3.2.23).

Both expressions give $Q/(4\pi {\epsilon }_0{b}^2)$ at $r=b$ (no surface charge, no jump), and $E=0$ at the centre, where by symmetry no direction is preferred. Outside, the sphere acts like a point charge $Q$ at its centre. The result is Problem 3.2.5 with $a\to 0$ and ${\rho }_v=3Q/(4\pi b^3)$: indeed ${\rho }_{v}r/(3{\epsilon }_0)=Qr/(4\pi {\epsilon }_0{b}^3)$. In vector form the field inside is $E={\rho }_{v}r/(3{\epsilon }_0)$, a form used in Exercise 3.2.9.