Physics - Elecrostatics Question with Solution | TestHub

Let us consider a spherical shell of radius $x$ and thickness $\mathrm{dx}$.
Charge on this shell $$dq=\rho .4\pi x^{2}dx={\rho }_0\left(1-\frac{x}{R}\right)\cdot 4\pi x^{2}dx$$ $\therefore$ Total charge in the spherical region from centre to $r(r<R)$ is $$\begin{array}{rl}& q=\int dq=4\pi {\rho }_0{\int }_0^r\left(1-\frac{x}{R}\right)x^{2}dx\\ & =4\pi {\rho }_0{\left[\frac{x^3}{3}-\frac{x^4}{4R}\right]}_0^r\\ & =4\pi {\rho }_0\left[\frac{r^3}{3}-\frac{r^4}{4R}\right]\\ & =4\pi {\rho }_0{r}^3\left[\frac{1}{3}-\frac{r}{4R}\right]\end{array}$$ $\therefore$ Electric field at $\mathrm{r},\mathrm{E}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\mathrm{q}}{{\mathrm{r}}^2}$ $$\begin{array}{rl}& =\frac{1}{4\pi {\epsilon }_0}\cdot \frac{4\pi {\rho }_0{r}^3}{r^2}\left[\frac{1}{3}-\frac{r}{4R}\right]\\ & =\frac{{\rho }_0}{{\epsilon }_0}\left[\frac{r}{3}-\frac{r^2}{4R}\right]\end{array}$$