Question

Volume charge density of a non conducting solid sphere of radius $R=30cm$ is given as $\rho ={\rho }_o(1-\frac{r}{R}).$ Electric field inside sphere is minimum at

• A. $r=15cm$
• B. $r=30cm$
• C. $r=10cm$
• D. $r=20cm$

Answer 1

The correct option is D $r=20cm$

By Gauss law, $\oint \overrightarrow{E}.\overrightarrow{dr}=\frac{q_{enclosed}}{{ϵ}_o}$
So, $E\times 4\pi r^2=\frac{q_{enclosed}}{{ϵ}_0}$ ,
where$q_{enclosed}=\int \rho dV$
$q_{enclosed}={\int }_0^r{\rho }_o(1-\frac{r}{R})4\pi r^{2}dr$

$q_{enclosed}=4\pi {\rho }_o{\int }_o^r(r^2-\frac{r^3}{R})dr$
$=4\pi {\rho }_o{[\frac{r^3}{3}-\frac{r^4}{4R}]}_o^r$

$q_{enclosed}=4\pi {\rho }_o(\frac{r^3}{3}-\frac{r^4}{4R})E\times 4\pi r^2=\frac{4\pi {\rho }_o{r}^2}{{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$
$\therefore E=\frac{{\rho }_o}{{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$
For maxima or minima, $\frac{dE}{dr}=0$
$\frac{1}{3}-\frac{2r}{4R}=0$
$r=\frac{2R}{3}=\frac{2\times 30}{3}cm=20cm$