[12] A solid ball of radius $\mathrm{R}$ has a charge density $\rho$ given by…
[12]
A solid ball of radius $\mathrm{R}$ has a charge density $\rho$ given by $\rho=\rho_0\left(1-\frac{r}{R}\right)$ for $0 \leq r \leq R$. The electric field outside the ball is:
$\frac{\rho_0 R^3}{\varepsilon_0 r^2}$
$\frac{4 \rho_0 R^3}{3 \varepsilon_0 r^2}$
$\frac{3 \rho_0 R^3}{4 \varepsilon_0 r^2}$
$\frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$
Solution
Charge density, $\rho=\rho_0\left(1-\frac{r}{R}\right)$
$
\begin{aligned}
&d q=\rho d v \\
&q_{\text {in }}=\int d q=\rho d v \\
&=\rho_0\left(1-\frac{r}{R}\right) 4 \pi r^2 d r \quad\left(\because d v=4 \pi \mathrm{r}^2 \mathrm{dr}\right) \\
&=4 \pi \rho_0 \int_0^R\left(1-\frac{r}{R}\right) r^2 d r \\
&=4 \pi \rho_0 \int_0^R r^2 d r-\frac{r^2}{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]
\end{aligned}
$
$=4 \pi \rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=4 \pi \rho_0\left[\frac{R^3}{12}\right]$
$q=\frac{\pi \rho_0 R^3}{3}$
$
E .4 \pi r^2=\left(\frac{\pi \rho_0 R^3}{3 \epsilon_0}\right)
$
$\therefore \quad$ Electric field outside the ball, $E=\frac{\rho_0 R^3}{12 \in_0 r^2}$