Question

A ball of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball's centre as: $\rho ={\rho }_0(1-\frac{r}{R})$ where ${\rho }_0$ is constant. Assume $ϵ$ as the permittivity of the ball.

Then the magnitude of the electric field as a function of the distance r outside the ball is given by :

• A. $E=\frac{{\rho }_0{R}^3}{8ϵr^2}$
• B. $E=\frac{{\rho }_0{R}^3}{12ϵr^2}$
• C. $E=\frac{{\rho }_0{R}^3}{16ϵr^2}$
• D. $E=\frac{{\rho }_0{R}^3}{24ϵr^2}$

Answer 1

The correct option is B $E=\frac{{\rho }_0{R}^3}{12ϵr^2}$

Using Gauss's law,$\int \overrightarrow{E}.\hat{n}ds=\frac{Q_{enclosed}}{ϵ}$

For field at $r$ outside the sphere,

$Q_{enclosed}={\int }_0^R\rho 4\pi r^{2}dr=4\pi {\rho }_0{\int }_0^R(1-\frac{r}{R})r^{2}dr=4\pi {\rho }_0[\frac{R^3}{3}-\frac{R^4}{4R}]=4\pi {\rho }_0\frac{R^3}{12}=\frac{\pi {\rho }_0{R}^3}{3}$

thus, $E.4\pi r^2=\frac{\pi {\rho }_0{R}^3}{3ϵ}$

$E=\frac{{\rho }_0{R}^3}{12ϵr^2}$