Q. Let there be a spherically symmetric charge distribution with charge density varying as$\ \rho(r) = \rho_{0}\left( \frac{5}{4} - \frac{r}{R} \right)$ upto $r = R,$ and $\rho(r) = 0$ for $r > R,$ where $r$ is the distance from the origin. The electric field at a distance $r(r < R)$ from the origin is given by
A) $\frac{\rho_{0}r}{4\varepsilon_{0}}\left( \frac{5}{4} - \frac{r}{R} \right)$
B) $\frac{{4\pi\rho}_{0}r}{3\varepsilon_{0}}\left( \frac{5}{3} - \frac{r}{R} \right)$
C) $\frac{\rho_{0}r}{4\varepsilon_{0}}\left( \frac{5}{3} - \frac{r}{R} \right)$
D) $\frac{{4\rho}_{0}r}{3\varepsilon_{0}}\left( \frac{5}{4} - \frac{r}{R} \right)$
Explanation
(c) Total charge $Q = \int_{0}^{r}{\rho\ dv =}\int_{0}^{r}{\rho_{0}\left( \frac{5}{4} - \frac{r}{R} \right)4\pi r^{2}dr}$
$= 4\pi\rho_{0}\int_{0}^{r}\left( \frac{5r^{2}}{4} - \frac{r^{3}}{R} \right)dr = 4\pi\rho_{0}\left\lbrack \frac{5r^{3}}{12} - \frac{r^{4}}{4R} \right\rbrack$
$E = \frac{KQ}{r^{2}} = \frac{1}{4\pi\varepsilon_{0}r^{2}}4\pi\rho_{0}\left\lbrack \frac{5}{12}r^{3} - \frac{r^{4}}{4R} \right\rbrack = \frac{\rho_{0}r}{4\varepsilon_{0}}\left\lbrack \frac{5}{3} - \frac{r}{R} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.