Question

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

• A. $\frac{\rho_o \, R^3}{\epsilon_o \, r^2}$
• B. $\frac{\rho_o \, R^3}{ 12 \, \epsilon_o \, r^2}$
• C. $\frac{4 \, \rho_o \, R^3}{3 \, \epsilon_o \, r^2}$
• D. $\frac{ 3 \, \rho_o \, R^3}{4 \, \epsilon_o \, r^2}$

Answer 1

Answer: (B)

Given: Charge density $\rho=\rho_{0}\left(1-\frac{r}{R}\right)$. By Gauss's Law, we have
$\int E \cdot d s=\frac{Q_{ cnc }}{\varepsilon_{0}} \Rightarrow E \times 4 \pi r^{2}=\frac{Q_{ enc }}{\varepsilon_{0}}$
Now $\,\,\,Q_{\text {enc }}=\int \rho d V=\int \rho_{0}\left(1-\frac{r}{R}\right) d V=\int\limits_{0}^{R} \rho_{0}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r$
$\Rightarrow Q_{\text {enc }}=4 \,\pi\, \rho_{0} \int\limits_{0}^{R}\left(1-\frac{r}{R}\right) r^{2} d r=4\,\pi\, \rho_{0} \int\limits_{0}^{R} r^{2} d r-\frac{r^{3}}{R} d r=4 \,\pi\, \rho_{0}\left\{\left[\frac{r^{3}}{3}\right]_{0}^{R}-\left[\frac{r^{4}}{4 R}\right]_{0}^{R}\right\}=4 \pi \rho_{0}\left(\frac{R^{3}}{3}-\frac{R^{4}}{4 R}\right)$
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}{\varepsilon_{0}} $
$\Rightarrow E=\frac{\rho_{0} R^{3}}{12 \varepsilon_{0} r^{2}}$