Physics - Electrostatics Question with Solution | TestHub

Charge density, $\rho ={\rho }_0\left(1-\frac{r}{R}\right)$ $$\begin{array}{rl}& dq=\rho dv\\ & q_{\text{in }}=\int dq=\rho dv\\ & ={\rho }_0\left(1-\frac{r}{R}\right)4\pi r^{2}dr\left(\because dv=4\pi {\mathrm{r}}^2\mathrm{dr}\right)\\ & =4\pi {\rho }_0{\int }_0^R\left(1-\frac{r}{R}\right)r^{2}dr\\ & =4\pi {\rho }_0{\int }_0^R{r}^{2}dr-\frac{r^2}{R}dr\\ & =4\pi {\rho }_0\left[{\left[\frac{r^3}{3}\right]}_0^R-{\left[\frac{r^4}{4R}\right]}_0^R\right]\\ & =4\pi {\rho }_0\left[\frac{R^3}{3}-\frac{R^4}{4R}\right]\end{array}$$ $=4\pi {\rho }_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=4\pi {\rho }_0\left[\frac{R^3}{12}\right]$ $q=\frac{\pi {\rho }_0{R}^3}{3}$ $$E.4\pi r^2=\left(\frac{\pi {\rho }_0{R}^3}{3{ϵ}_0}\right)$$ $\therefore$ Electric field outside the ball, $E=\frac{{\rho }_0{R}^3}{12{\in }_0{r}^2}$