Question

A solid ball of radius $R$ has a charge density $\rho$ given by $\rho ={\rho }_0(1-r/R)$ for $0\le r\le R$. The electric field outside the ball is:

• A. $\frac{{\rho }_0{R}^3}{{\in }_0{r}^2}$
• B. $\frac{{\rho }_0{R}^3}{12{\in }_0{r}^2}$
• C. $\frac{4rh{o}_0{R}^3}{3{\in }_0{r}^2}$
• D. $\frac{3{\rho }_0{R}^3}{4{\in }_0{r}^2}$

Answer 1

Answer: (B)

$\frac{{\rho }_0{R}^3}{12{\in }_0{r}^2}$

$\rho ={\rho }_0(1-\frac{r}{R})$

$dq=\rho dv$

$q_{in}=\int dq$

$=\rho dv$

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

$\because dv=4\pi r^{2}dr$

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

$=4\pi {\rho }_o{\int }_o^R{r}^{2}dr-\frac{r^2}{R}dr$

$=4\pi {\rho }_o[{\left[\frac{r^3}{3}\right]}_o^R-{\left[\frac{r^4}{4R}\right]}_o^R]$

$=4\pi {\rho }_o[\frac{R^3}{3}-\frac{R^4}{4R}]$

$=4\pi {\rho }_o[\frac{R^3}{3}-\frac{R^3}{4}]$

$=4\pi {\rho }_o\left[\frac{R^3}{12}\right]$

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

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

$\Rightarrow E=\frac{{\rho }_o{R}^3}{12{\in }_o{r}^2}$