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}