Question

Let there be a spherically symmetric charge distribution with charge density varying as $\rho\left(r\right) = \rho_{0} \left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r = R$, and $\rho \left(r\right) = 0$ for $r > R$, where r is the distance from the origin. The electric field at a distance $r\left(r < R\right)$ from the origin is given by

• A. $\frac{4\pi\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right) $
• B. $\frac{\rho_{0}r}{4\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right) $
• C. $\frac{4\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right) $
• D. $\frac{\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right) $

Answer 1

Answer: (B)

Apply shell theorem the total charge upto distance r can be calculated as followed
$dq = 4\pi r^{2}.dr.\rho$
$ = 4\pi r^{2}.dr.\rho_{0} \left[\frac{5}{4}-\frac{r}{R}\right] $
$= 4\pi\rho_{0}\left[\frac{5}{4}r^{2}dr-\frac{r^{3}}{R}dr\right]$
$\int dq = q = 4\pi \rho _{0}\int\limits^{r}_{0}\left(\frac{5}{4}r^{2}dr-\frac{r^{3}}{R}dr\right) $
$4\pi \rho _{0} = \left[\frac{5}{4} \frac{r^{3}}{3}-\frac{1}{R} \frac{r^{4}}{4}\right]$
$E = \frac{kq}{r^{2}}$
$= \frac{1}{4\pi\varepsilon_{0}} \frac{1}{r^{2}}. \left[\frac{5}{4} \left(\frac{r^{3}}{3}\right)- \frac{r^{4}}{4R}\right]$
$ E = \frac{\rho_{0}r}{4\varepsilon_{0}} \left[\frac{5}{4}-\frac{r}{R}\right] $