Question

A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the centre. What is the ratio of the magnitude of the electric field at a distance 2R from the centre to the magnitude of the electric field at a distance of $\frac{R}{2}$ from the centre $(i.e.,E_{r=2R}/E_{r=R/2})$?

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The correct option is B 2

$\rho =\alpha r^2$ [where $\alpha$ is a constant]
Charge in the shell [element]
$dq=\rho \left(4\pi r^{2}dr\right)=\alpha \left(4\pi \right)r^{4}dr$
$\therefore$ Charge enclosed in sphere of radius r,
$q=4\pi \alpha {\int }_0^r{r}^{4}dr=\frac{4\pi \alpha r^5}{5}$
By Gauss's theorem
at $r=\frac{R}{2}$
$E_{r=R/2}=\frac{1}{4\pi }\times \frac{4\pi \alpha {\left(\frac{R}{2}\right)}^5}{5{\in }_0}\times \frac{1}{{\left(\frac{R}{2}\right)}^2}$
$E_r=\frac{R}{2}=\frac{\alpha R^5}{32{\in }_0}\times \frac{4}{R^2\times 5}$
$E_{r=R/2}=\frac{\alpha R^3}{40{\in }_0}$

$E_{r=R/2}=\frac{\alpha R^3}{40{\in }_0}$
Total charge enclosed, $Q=\frac{4\pi \alpha R^5}{5}$
$\therefore E_{r=2R}=\frac{4\pi \alpha R^5}{5{\in }_0}\times \frac{1}{4\pi (2R{)}^2}=\frac{\alpha R^3}{20{\in }_0}$
$\frac{E_{r=2R}}{E_{r=R/2}}=\frac{\alpha R^3}{20{\in }_0}\times \frac{40{\in }_0}{\alpha R^3}$
$\frac{E_{r=2R}}{E_{r=R/2}}=\frac{40}{20}$
$\frac{E_{r=2R}}{E_{r=R/2}}=2$