Question 4.9: A charge distribution with spherical symmetry has density ρv...

The charge distribution is similar to that in Figure 4.16. Since symmetry exists, we can apply Gauss’s law to find E.

\varepsilon _{o}\oint_{S}E\cdot dS=Q_{enc}=\int_{v}\rho_{v}dv

(a) For r \lt R

\varepsilon _{o}E_{r}4\pi r^{2}=Q_{enc}=\int_{0}^{r}\int_{0}^{\pi}\int_{0}^{2\pi}\rho_{v}r^{2}\sin\theta d\phi d\theta dr

=\int_{0}^{r}4\pi r^{2}\frac{\rho _{o}r}{R}dr= \frac{\rho _{o}\pi r^{4}}{R}

or

E=\frac{\rho _{o}r^{2}}{4\varepsilon _{o} R}a_{r}

(b) For r\gt R

\varepsilon _{o}E_{r}4\pi r^{2}=Q_{enc}=\int_{0}^{r}\int_{0}^{\pi}\int_{0}^{2\pi}\rho_{v}r^{2}\sin\theta d\phi d\theta dr

=\int_{0}^{R}\frac{\rho _{o}r}{R}4\pi r^{2}dr+\int_{R}^{r}0\cdot4\pi r^{2}dr= \pi\rho _{o}R^{3}

or

E=\frac{\rho _{o}R^{3}}{4\varepsilon _{o} r^{2}}a_{r}