Question

A ball of radius $R$ carries a positive charge whose volume density depends only on a separation $r$ from the ball's centre as $\rho ={\rho }_0(1-r/R)$, where ${\rho }_0$ is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find:
(A) the magnitude of the electric field strength as a function of the distance $r$ both inside and outside the ball;
(b) the maximum intensity $E_{max}$ and the corresponding distance $r_m$.

Answer 1

(a) We assume the ball to be divided into infinite number of concentric thin shell of thickness $dr$. Let us assume such a shell at a radial distance $r$.
Volume of this shell $=pdV=4\pi r^2\rho dr=4\pi r^2{\rho }_0(1-\frac{r}{R})dr$
Net charge enclosed in the sphere of radius $r$
$q=\int pdv=\stackrel{r}{\underset{0}{\int }}4\pi {\rho }_0{r}^2(1-\frac{r}{R})dr=4\pi {\rho }_0(\frac{r^3}{3}-\frac{r^4}{4R})$
Electric field at radial distance $r$
$E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{1}{4\pi {\epsilon }_0{r}^2}\times 4\pi {\rho }_0(\frac{r^3}{3}-\frac{r^4}{4R})=\frac{{\rho }_{0}r}{3{\epsilon }_0}(1-\frac{3r}{4R})$
Now for those points outside the sphere, we have to take into account the total charge contained in the sphere of radius $R$.
Total charge $Q=\stackrel{R}{\underset{0}{\int }}{\rho }_0(1-\frac{r}{R})4\pi r^{2}dr=\frac{\pi {\rho }_0{R}^3}{3}$
Electric field, using Gauss's laws, at a point which is at distance $r(>R)$ from the centre;
$\int E.ds=\frac{Q}{{\epsilon }_0}⟹E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{{\rho }_0{R}^3}{12{\epsilon }_0{r}^2}$
(b) Again let us consider the expression for electric field within the sphere of radius $R$:
$E=\frac{{\rho }_{0}r}{3{\epsilon }_0}(1-\frac{3r}{4R})$
For maximum electric field:
$\frac{dE}{dr}=0⟹1-\frac{6r}{4R}=0⟹r=\frac{2R}{3}$
Putting, $r=\frac{2R}{3}$, value of maximum electric field, $E_{max}=\frac{{\rho }_{0}R}{9{\in }_0}$