Question

A sphere of radius $R$ has a charge density $\rho \left(r\right)={\rho }_0\left(\frac{r}{R}\right)$ where ${\rho }_0$ is a constant and $r$ is the distance from the centre of the sphere. Find the electric field for $r>R$

• A. $E=\frac{{\rho }_0{r}^3}{4{ϵ}_0{R}^2}$
• B. $E=\frac{{\rho }_{0}R}{2{ϵ}_0}$
• C. $E=\frac{{\rho }_0{R}^3}{4{ϵ}_0{r}^2}$
• D. $E=\frac{4{\rho }_0{R}^3}{3{ϵ}_0{r}^2}$

Answer 1

The correct option is A $E=\frac{{\rho }_0{r}^3}{4{ϵ}_0{R}^2}$

$\rho \left(r\right)={\rho }_0\left(\frac{r}{R}\right)$

$\because \rho =\frac{Q}{V}$

$\therefore$ Total charge inside the sphere

$Q_{enc}={\int }_0^R{\rho }_0\left(\frac{r}{R}\right)\times 4\pi r^{2}dr$

$=\frac{{\rho }_{0}4\pi }{R}{\int }_0^R{r}^{3}dr$

$=\frac{S_{0}4\pi R^4}{4R}$

$\therefore Q_{enc}={\rho }_0\pi R^3$

Electric field outside the region $(\sigma >R)$

$\overrightarrow{E}.d\overrightarrow{A}=\frac{Q_{enc}}{{ϵ}_0}$

$⟹E.\left(4\pi r^2\right)=\frac{{\rho }_0\pi R^3}{{\epsilon }_0}$

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