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

Answer 1

Answer: (B)

Apply shell theorem the total charge upto distance r can be calculated as followed
$dq=4\pi r^2.dr.\rho$
$=4\pi r^2.dr.{\rho }_0\left[\frac{5}{4}-\frac{r}{R}\right]$
$=4\pi {\rho }_0\left[\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr\right]$
$\int dq=q=4\pi {\rho }_0\underset{0}{\overset{r}{\int }}\left(\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr\right)$
$4\pi {\rho }_0=\left[\frac{5}{4}\frac{r^3}{3}-\frac{1}{R}\frac{r^4}{4}\right]$
$E=\frac{kq}{r^2}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{1}{r^2}.\left[\frac{5}{4}\left(\frac{r^3}{3}\right)-\frac{r^4}{4R}\right]$
$E=\frac{{\rho }_{0}r}{4{\epsilon }_0}\left[\frac{5}{4}-\frac{r}{R}\right]$