Question

A solid ball of radius $R$ has a charge density $\rho$ given by $\rho ={\rho }_o(1-r/R)$ for $0\le r\le R$. 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

Answer: (B)

Given: Charge density $\rho ={\rho }_0\left(1-\frac{r}{R}\right)$. By Gauss's Law, we have
$\int E\cdot ds=\frac{Q_{cnc}}{{\epsilon }_0}\Rightarrow E\times 4\pi r^2=\frac{Q_{enc}}{{\epsilon }_0}$
Now $Q_{\text{enc }}=\int \rho dV=\int {\rho }_0\left(1-\frac{r}{R}\right)dV=\underset{0}{\overset{R}{\int }}{\rho }_0\left(1-\frac{r}{R}\right)4\pi r^{2}dr$
$\Rightarrow Q_{\text{enc }}=4\pi {\rho }_0\underset{0}{\overset{R}{\int }}\left(1-\frac{r}{R}\right)r^{2}dr=4\pi {\rho }_0\underset{0}{\overset{R}{\int }}{r}^{2}dr-\frac{r^3}{R}dr=4\pi {\rho }_0\left\{{\left[\frac{r^3}{3}\right]}_0^R-{\left[\frac{r^4}{4R}\right]}_0^R\right\}=4\pi {\rho }_0\left(\frac{R^3}{3}-\frac{R^4}{4R}\right)$
Therefore, $Q_{\text{anc }}=4\pi {\rho }_0\frac{R^3}{12}$
Substitute $Q_{\text{enc }}$ in Eq. (1), we get
$E\times 4\pi r^2=4\pi {\rho }_0\frac{R^3}{12}\times \frac{1}{{\epsilon }_0}$
$\Rightarrow E=\frac{{\rho }_0{R}^3}{12{\epsilon }_0{r}^2}$