Question

A solid ball of radius R has a charge density p given by $p=p_0(1-r/R)$ for $0\le r\le R.$ The electric field outside the ball is:

• A. $\frac{p_0{R}^3}{{ϵ}_0{r}^2}$
• B. $\frac{p_0{R}^3}{12{ϵ}_0{r}^2}$
• C. $\frac{4p_0{R}^3}{3{ϵ}_0{r}^2}$
• D. $\frac{3p_0{R}^3}{4{ϵ}_0{r}^2}$

Answer 1

The correct option is B $\frac{p_0{R}^3}{12{ϵ}_0{r}^2}$

$Accordingtoquestion.....................q={\int }_0^{R}pdv={\int }_0^{R}po\left(1-\frac{r}{R}\right).4\pi r^{2}drr\Rightarrow q=po.4\pi \left[{\int }_0^R{r}^{2}dr-{\int }_0^R\frac{r^3}{R}dr\right]\Rightarrow q=po.4\pi \left(\frac{R^3}{3}-\frac{R^3}{4}\right)=po\left(4\pi \right)\frac{R^3}{12}\therefore q=\frac{po.\pi R^3}{3}E=\frac{1}{4\pi {\in }_0}\times \frac{q}{r^2}=\frac{1}{4\pi {\in }_0}\times \frac{po.\pi R^3}{3r^2}=\frac{po.R^3}{12{\in }_0{r}^2}\therefore E=\frac{po.R^3}{12{\in }_0{r}^2}thereforthecorrectoptionisB.$