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

Answer: 2

Using Gauss's theorem for spherical surface of radius $r$ outside the sphere.
$\oint E\cdot ds=\frac{Q_{max}}{{ϵ}_0}$

Here, $Q_{enc}=q+q^{{}^{\prime }}=q+\underset{R}{\overset{r}{\int }}\rho \left(4\pi r^2\right)dr$
$\therefore E\left(4\pi r^2\right)=\frac{1}{{ϵ}_0}\left(q+\underset{R}{\overset{r}{\int }}\frac{\alpha }{r}\left(4\pi r^2\right)dr\right)$
$E\left(4\pi r^2\right)=\frac{1}{{ϵ}_0}\left(q+\underset{R}{\overset{r}{\int }}(4\pi \alpha r)dr\right)$ $=\frac{1}{{ϵ}_0}\left(q+2\pi \alpha r^2-2\pi \alpha R^2\right)$
$\therefore E\left(4\pi r^2\right)=\frac{\left(q-2\pi \alpha R^2\right)}{{ϵ}_0}+\frac{2\pi \alpha r^2}{{ϵ}_0}$
For $q-2\pi \alpha R^2=0$
$E=\frac{2\pi \alpha r^2}{4\pi r^2{ϵ}_0}=\frac{\alpha }{2{ϵ}_0}$
i,e., $E$ becomes independent of $r.$
$\therefore$ The intensity $E$ does not depend on $r$ if
$q-2\pi \alpha R^2=0$
$\therefore q=2\pi \alpha R^2=6.28\alpha R^2$
$\therefore N=2$