Obtain Gauss's law from the flux associated with a sphere of radius $r$ and charge $q$ at its centre.

(N/A) 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 the figure.
The flux through an area element $\Delta \overrightarrow{S}$ is,
$\Delta \phi = \overrightarrow{E} \cdot \Delta \overrightarrow{S} = E \Delta S \cos(0^{\circ}) = E \Delta S$
Using Coulomb's law for the electric field due to a point charge $q$ at distance $r$:
$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
Since the electric field $\overrightarrow{E}$ and the area vector $\Delta \overrightarrow{S}$ are in the same direction (radially outward),
$\Delta \phi = \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \right) \Delta S$
The total flux $\phi$ through the sphere is the sum of fluxes through all such area elements:
$\phi = \sum \Delta \phi = \sum \left( \frac{q}{4 \pi \varepsilon_{0} r^{2}} \Delta S \right)$
$\phi = \frac{q}{4 \pi \varepsilon_{0} r^{2}} \sum \Delta S$
Since $\sum \Delta S = S = 4 \pi r^{2}$ (total surface area of the sphere),
$\phi = \frac{q}{4 \pi \varepsilon_{0} r^{2}} \times 4 \pi r^{2} = \frac{q}{\varepsilon_{0}}$
This is Gauss's law for a point charge,which states that the total electric flux through any closed surface is equal to the net charge enclosed by the surface divided by $\varepsilon_{0}$.