Question

A spherically symmetric charge distribution is considered with charge density varying as $\rho (r)=\{\begin{array}{ll}{\rho }_0\left(\frac{3}{4}-\frac{r}{R}\right)& \text{ for }r\le R\\ \text{ Zero }& \text{ for }r>R\end{array}$ Where, $r(r<R)$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be :

• A. $\frac{{\rho }_{0}r}{4{\epsilon }_0}\left(\frac{3}{4}-\frac{r}{R}\right)$
• B. $\frac{{\rho }_{0}r}{3{\epsilon }_0}\left(\frac{3}{4}-\frac{r}{R}\right)$
• C. $\frac{{\rho }_{0}r}{4{\epsilon }_0}\left(1-\frac{r}{R}\right)$
• D. $\frac{{\rho }_{0}r}{5{\epsilon }_0}\left(1-\frac{r}{R}\right)$

Answer 1

Answer: (C)

$\oint \overrightarrow{E}\cdot d\overrightarrow{s}=\frac{Q_{in}}{{\epsilon }_o}$
$E\cdot 4\pi r^2=\frac{\underset{0}{\overset{r}{\int }}{\rho }_o\left(\frac{3}{4}-\frac{r}{R}\right)4\pi r^{2}dr}{{\epsilon }_0}$
$E4\pi r^2=\frac{{\rho }_{o}4\pi }{{\epsilon }_o}\left(\frac{3}{4}\frac{r^3}{3}-\frac{r^4}{4R}\right)$
$Er{r}^2=\frac{{\rho }_o{r}^3}{4{\epsilon }_o}\left\{1-\frac{r}{R}\right\}$
$E=\frac{{\rho }_{o}r}{4{\epsilon }_o}\left\{1-\frac{r}{R}\right\}$