Obtain Gauss’s law from the flux associated with a sphere of radius $\mathrm{'r'}$ and charge $\mathrm{'q'}$ at centre.

Let us consider the total flux through a sphere of radius $r$, which encloses a point charge $q$ at its centre.

Divide the sphere into small area elements, as shown in figure.

The flux through an area element $\Delta \overrightarrow{\mathrm{S}}$ is,

$\Delta \phi=\overrightarrow{\mathrm{E}} \cdot \Delta \overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{E}} \cdot \hat{r} \Delta \mathrm{S}$

where we have used Coulomb's law for the electric field due to a single charge $q$.

The unit vector $\hat{r}$ is along the radius vector from the centre to the area element.

The area element $\Delta \overrightarrow{\mathrm{S}}$ and $\hat{r}$ have the same direction,

$\therefore \Delta \phi=\frac{q}{4 \pi \varepsilon_{0} r^{2}} \Delta S=\frac{k q}{r^{2}} \Delta S\left(\because k=\frac{1}{4 \pi \varepsilon_{0}}\right)$

The total flux through the sphere is obtained by adding up flux through all the different area elements.

$\therefore \phi=\sum_{\Delta S} \frac{k q}{r^{2}} \cdot \Delta S$

$\therefore \phi=\frac{k q}{r^{2}} \sum_{\Delta S} \Delta S=\frac{q}{4 \pi \varepsilon_{0} r^{2}} S(\because \Sigma \Delta S=S)$

$\therefore \phi=\frac{q}{4 \pi \epsilon_{0} r^{2}} \times 4 \pi r^{2}=\frac{q}{\epsilon_{0}}$

$\left(\because S=4 \pi r^{2}\right. \text { for sphere) }$

Gauss's law : Electric flux through a closed surface $\mathrm{S}$,

$\phi=\Sigma \frac{q}{\varepsilon_{0}}$

$\Sigma q=$ total charge enclosed by $\mathrm{S}$.

"The electric flux associated with any closed surface is equal to the ratio of net charge enclosed by surface to $\epsilon_{0}$ ".