Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively.

$\frac{1}{4 \pi \varepsilon_0} \frac{q r}{R^3}$

The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.

Now, apply Gauss’s law to a spherical Gaussian surface of radius r (r > R for point A)

$\varepsilon_0 \oint \vec{E} . \vec{ds}=q_{i n} \quad \Rightarrow \quad \varepsilon_0 E\left(4 \pi r^2\right)=q$

$E=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$ where q is the total charge

For point B (r < R)

$\varepsilon_0 \oint \vec{E} . \vec{ds}=\varepsilon_0 E\left(4 \pi r^2\right)=q$

$q'=\frac{q \frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}=q\left(\frac{r}{R}\right)^3$

$E=\frac{1}{4 \pi \varepsilon_0} \frac{q\left(\frac{r}{R}\right)^3}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q r}{R^3}$