Question

The charge per unit volume inside an insulating sphere of radius $R$ is given as a function of distance from centre $\rho \left(r\right)=cr$ , Where $0<r<R$. The magnitude of electric field at a distance ${}^{\prime }{b}^{\prime }$ from the centre of sphere will be

• A. $\frac{c}{\pi b^4}$
• B. $\frac{c{b}^2}{4{ϵ}_0}$
• C. $\frac{cb}{\pi {ϵ}_0}$
• D. $\frac{c{b}^4}{{ϵ}_0}$

Answer 1

The correct option is B $\frac{c{b}^2}{4{ϵ}_0}$


Let us consider a shell of thickness $dr$ at a distance ${}^{\prime }{r}^{\prime }$ from centre.
Volume of the shell,
$dv=\left(4\pi r^2\right)dr$
($dv=$ surface area $\times$ thickness)
The total charge enclosed in the shell will be
$d{q}_{in}=\rho dv$
$\Rightarrow d{q}_{in}=crdv(\because \rho =cr)$
Let us construct a Gaussian sphere at a distance $b$ from centre, thus net charge enclosed by this sphere:
$q_{in}=\int d{q}_{in}=\int crdv$
$\Rightarrow q_{in}={\int }_0^{b}cr\left(4\pi r^2\right)dr$
or, $q_{in}=4\pi c{\int }_0^b{r}^{3}dr$
or, $q_{in}=4\pi c{\left[\frac{r^4}{4}\right]}_0^b$
$\therefore q_{in}=\pi c{b}^4$
Applying Gauss's law at spherical surface.
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{q_{in}}{{ϵ}_0}$
or $EA=\frac{\pi c{b}^4}{{ϵ}_0}$
$\Rightarrow E\left(4\pi b^2\right)=\frac{\pi c{b}^4}{{ϵ}_0}$
$(\because A=4\pi b^2)$
$\Rightarrow E=\frac{c{b}^2}{4{ϵ}_0}$

$\begin{array}{c}\text{Why this question}?\\ \text{It intends to test your understanding in finding}{q}_{in}\\ \text{for a nonuniformly charged sphere and applying Gauss law}\\ \text{Tip: Electric field can be found by considering a gaussian}\\ \text{surface (sphere), if we can obtain net charge enclosed by it.}\end{array}$