Consider a non-conducting solid sphere of radius R and centre O has uniform volume density of charge \rho .
Let us calculate electric field intensity \vec E at any point P, where OP = r. With O as centre and r as radius, a sphere is imagined, which acts as a Gaussian surface. At every point of this surface, magnitude of \vec E is same and directed radially outwards.
If q’ is the charge enclosed by the imagined sphere, then following Gauss’s law,
\oint_s\vec{E}.|vec{dS}= \oint_s\vec{E}. \widehat{n} ds = E\oint_sds=\frac{q’}{\epsilon}
where \epsilon is the electrical permittivity of the material of the sphere. or,
or, E\times4\pi r^2=\frac{q’}{\epsilon}
\therefore E=\frac{q’}{4\pi\epsilon r^2}
Now, q’ =\frac43\pi r^3\times\rho
\therefore E=\frac{4/3\pi r^2\rho}{4\pi\epsilon r^2}
or, E=\frac{\rho}{3\epsilon}r
At the centre of the sphere, r=0, i.e., E=0
At the surface of the sphere, r=R, i.e., E=\frac{\rho}{3\epsilon} R- maximum
At r >> R E\propto \frac1{r^2}
