Question

The region between two concentric spheres of radii $a$ and $b$ $(>a)$ contains charge for which the volume charge density is $\rho \left(r\right)=\frac{c}{r}$, where $c$ is a constant and $r$ is the radial distance from centre as shown in figure. A point charge $q$ is placed at the origin, $r=0$. The value of $c$ for which electric field in the region between the spheres is constant (i.e. independent of $r$ ) will be

• A. $\frac{q}{2\pi a^2}$
  
• B. $\frac{q}{\pi a^2}$
  
• C. $\frac{q}{3\pi a^2}$
  
• D. $\frac{q}{4\pi a^2}$
  

Answer 1

The correct option is A $\frac{q}{2\pi a^2}$


On taking a Gaussian sphere of radius $r$ inside the region between two concentric spheres.


By Gauss's law,
$\oint \overrightarrow{E}\cdot \overrightarrow{dA}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}...\left(1\right)$
Here, $\oint \overrightarrow{dA}=4\pi r^2$( surface integration)
The net enclosed charge within gaussian surface is charge of point charge placed at centre + charge enclosed ;
$\Rightarrow q_{\text{enclosed}}={\int }_a^r\rho \left(4\pi r^2\right)dr+q={\int }_a^r\frac{c}{r}\left(4\pi r^2\right)dr+q$
Or, $q_{\text{enclosed}}=2\pi c(r^2-a^2)+q$
Now, from Eq. $\left(1\right)$
$E\times 4\pi r^2=\frac{2\pi c(r^2-a^2)+q}{{\epsilon }_0}$
$\Rightarrow E=\frac{k[2\pi c(r^2-a^2)+q]}{r^2}$
$\Rightarrow E=\frac{k[2\pi c{r}^2-2\pi c{a}^2+q]}{r^2}$
$\Rightarrow E=2\pi kc-\frac{2\pi c{a}^{2}k}{r^2}+\frac{kq}{r^2}$
For $\text{E}$ to be constant (i.e. independent of $r$),
$\Rightarrow \frac{-2\pi c{a}^{2}k}{r^2}+\frac{kq}{r^2}=0$
$\therefore c=\frac{q}{2\pi a^2}$