Question 88 of 311

A spherically symmetric charge distribution is considered with charge density varying as
$\rho \left(r\right)=\left\{\begin{matrix}{\rho }_{0}\left(\dfrac{3}{4}-\dfrac{r}{R}\right) & \text{for}r\leq R \\ Zero & & & forr\gt R\end{matrix}\right.$

Where, $r\left(r\lt R\right)$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be :

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

Show answer and solution▾

Answer: (C) $\dfrac{{\rho }_{0}r}{4{\epsilon }_{0}}\left(1-\dfrac{r}{R}\right)$

According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ${\epsilon }_{0}$.
$\Rightarrow \oint \vec{E\cdot }d\vec{s}=\dfrac{{Q}_{in}}{{\epsilon }_{0}}$
$\Rightarrow E4\pi {r}^{2}=\dfrac{{\int }_{0}^{r}{\rho }_{0}\left(\dfrac{3}{4}-\dfrac{r}{R}\right)4\pi {r}^{2}dr}{{\epsilon }_{0}}$
$\Rightarrow E4\pi {r}^{2}=\dfrac{{\rho }_{o}4\pi }{{\epsilon }_{o}}\left(\dfrac{3}{4}\dfrac{{r}^{3}}{3}-\dfrac{{r}^{4}}{4R}\right)$
$\Rightarrow E=\dfrac{{\rho }_{o}r}{4{\epsilon }_{o}}\left(1-\dfrac{r}{R}\right)$