A uniformly charged non-conducting sphere of radius $r_0$ and charge density $\rho C/m^3$, has electric field on its surface given by

Using, $\oint E·dA=\frac{q_{\text{eaclosure }}}{{\epsilon }_0}$

For an outside point $E \Vert dA$

$E·dA=EdA$, as $\cos \theta =1$

and $q_{enclosure}$ $=$ charge density $\times$ volume

$=\rho \times \frac{4}{3}\pi r_0^3$

So, $E\oint dA=\frac{\rho \times \frac{4}{3}\pi r_0^3}{{\epsilon }_0}$

$E·4\pi r^2=\frac{4}{3}\frac{\pi \rho r_0^3}{{\epsilon }_0}$

$E=\frac{\rho r_0^3}{3{\epsilon }_0{r}^2}\left(E\right.$ at $\left.r>r_0\right)$

lor a point inside the sphere, $r<r_0$

I of any elemental patch $dA$ of Gaussian sphere,

$$\begin{gathered} E\Vert dA \\ \theta =0° \end{gathered}$$

$E·dA=EdA.$ Also, by symmetry $E$ is constant.

$$\begin{gathered} \therefore \oint E·dA=\frac{q_{enclosure}}{{\epsilon }_0} \\ \Rightarrow E\oint dA=\frac{\rho ·\frac{4}{3}\pi r^3}{{\epsilon }_0} \\ \Rightarrow E-4\pi r^2=\frac{\rho \frac{4}{3}\pi r^3}{{\epsilon }_0} \\ \Rightarrow E_{\text{inide }}=\frac{\rho r}{3E_0}\text{ and at }r=r_0 \\ E_{\text{surtace }}=\frac{\rho r_0}{3{\epsilon }_0} \end{gathered}$$