Question

A solid ball of radius $R$ has a charge density $\rho$ given by $\rho ={\rho }_{o}1-/R$ for $0\le r\le$.the electric field outside the ball is

• A. $\frac{{\rho }_o{R}^3}{{ϵ}_o{r}^2}$
• B. $\frac{{\rho }_o{R}^3}{12{ϵ}_o{r}^2}$
• C. $\frac{4{\rho }_o{R}^3}{3{ϵ}_o{r}^2}$
• D. $\frac{3{\rho }_o{R}^3}{4{ϵ}_o{r}^2}$

Answer 1

The correct option is B $\frac{{\rho }_o{R}^3}{12{ϵ}_o{r}^2}$

If P lies outside

$E_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

$E_{out}=\frac{\rho R^3}{3{\epsilon }_0{r}^2}$

$V_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$

$V_{out}=\frac{\rho R^3}{3{\epsilon }_{0}r}$

Total charge on sphere:

$q={\int }_0^R\rho dv={\int }_0^R{\rho }_0(1-\frac{r}{R})4\pi r^{2}dr$

$q={\rho }_{0}4\pi [{\int }_0^R{r}^{2}dr-{\int }_0^R\frac{r^3}{R}dr]$

$q={\rho }_{0}4\pi (\frac{R^3}{3}-\frac{R^3}{4})={\rho }_0\left(4\pi \right)\frac{R^3}{12}$

$q=\frac{{\rho }_0\pi R^3}{3}$

$E_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}=\frac{1}{4\pi {\epsilon }_0}\frac{{\rho }_0\pi r^3}{3r^2}$

$E=\frac{{\rho }_0{R}^3}{12{\epsilon }_0{r}^2}$