Question 4.15: A charge distribution with spherical symmetry has density ρv...

The D field has already been found in Section 4.6D using Gauss’s law.

(a) For r\geq R,  E=\frac{\rho_{o}R^{3}}{3\varepsilon_{o}r^{2}}a_{r}

Once E is known, V is determined as

V=-\int E\cdot dl=-\frac{\rho_{o}R^{3}}{3\varepsilon_{o}}\int\frac{1}{r^{2}}dr=\frac{\rho_{o}R^{3}}{3\varepsilon_{o}r}+C_{1},      r\geq R

Since V(r=\infty)=0,  C_{1}=0

(b) For r\leq R,  E=\frac{\rho_{o}r}{3\varepsilon_{o}}a_{r}

Hence

V=-\int E\cdot dl=-\frac{\rho_{o}}{3\varepsilon_{o}}\int rdr=-\frac{\rho_{o}r^{2}}{6\varepsilon_{o}}+C_{2}

From part (a) V\left(r=R\right) =\frac{\rho_{o}R^{2}}{3\varepsilon_{o}} Hence,

\frac{R^{2}\rho_{o}}{3\varepsilon_{o}}=-\frac{R^{2}\rho_{o}}{6\varepsilon_{o}}+C_{2}\rightarrow C_{2}=\frac{R^{2}\rho_{o}}{2\varepsilon_{o}}

and

V=\frac{\rho_{o}}{6\varepsilon_{o}}\left(3R^{2}-r^{2}\right)

Thus from parts (a) and (b)

V=\begin{cases}\frac{\rho_{o}R^{3}}{3\varepsilon_{o}r}, & r\geq R \\ \frac{\rho_{o}}{6\varepsilon_{o}}\left(3R^{2}-r^{2}\right), & r\leq R \end{cases}

(c) The energy stored is given by

W=\frac{1}{2}\int_{v} D\cdot Edv=\frac{1}{2}\varepsilon_{o}\int_{v}E^{2}dv

For r\leq R

E=\frac{\rho_{o}r}{3\varepsilon_{o}}a_{r}

Hence

W=\frac{1}{2}\varepsilon_{o}\frac{\rho^{2}_{o}}{9\varepsilon_{o}^{2}}\int_{r=0}^{R}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}r^{2}\cdot r^{2}\sin\theta d\phi d\theta dr

=\frac{\rho^{2}_{o}}{18\varepsilon_{o}}4\pi\cdot \frac{r^{5}}{5}|_{0}^{R}=\frac{2\pi\rho_{o}^{2}R^{5}}{45\varepsilon_{o}}J