Question

A system consists of a uniformly charged sphere of radius $R$ and a surrounding by a medium with volume charge density $\rho =\frac{\alpha }{r},$ where $\alpha $ is a positive constant and $r$ is the distance from the centre of the sphere. If $3.14N\alpha R^{2}$ is the charge on the sphere when the electric field intensity outside the sphere is independent of $r$ . What is the value of $N?$

Answer 1

Using Gauss's theorem for spherical surface of radius $r$ outside the sphere.
$\oint E\cdot ds=\frac{Q_{max}}{\epsilon _{0}}$

Here, $Q_{enc}=q+q^{'}=q+\int\limits _{R}^{r}\rho \left(4 \pi r^{2}\right)dr$
$\therefore E\left(4 \pi r^{2}\right)=\frac{1}{\epsilon _{0}}\left(q + \int \limits_{R}^{r} \frac{\alpha }{r} \left(4 \pi r^{2}\right) dr\right)$
$E\left(4 \pi r^{2}\right)=\frac{1}{\epsilon _{0}}\left(q + \int\limits _{R}^{r} \left(\right. 4 \pi \alpha r \left.\right) dr\right)$ $=\frac{1}{\epsilon _{0}}\left(q + 2 \pi \alpha r^{2} - 2 \pi \alpha R^{2}\right)$
$\therefore E\left(4 \pi r^{2}\right)=\frac{\left(q - 2 \pi \alpha R^{2}\right)}{\epsilon _{0}}+\frac{2 \pi \alpha r^{2}}{\epsilon _{0}}$
For $q-2\pi \alpha R^{2}=0$
$E=\frac{2 \pi \alpha r^{2}}{4 \pi r^{2} \epsilon _{0}}=\frac{\alpha }{2 \epsilon _{0}}$
i,e., $E$ becomes independent of $r.$
$\therefore $ The intensity $E$ does not depend on $r$ if
$q-2\pi \alpha R^{2}=0$
$\therefore q=2\pi \alpha R^{2}=6.28\alpha R^{2}$
$\therefore N=2$