Obtain Gauss's law from Coulomb's law.

(N/A) Consider a point charge $q$ placed at the origin. The electric field $\overrightarrow{E}$ at a distance $r$ from the charge $q$ is given by Coulomb's law as:
$\overrightarrow{E} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}} \hat{r}$
Now,consider a spherical Gaussian surface of radius $r$ centered at the charge $q$. The electric flux $\phi_E$ through this surface is given by the surface integral:
$\phi_E = \oint \overrightarrow{E} \cdot d\overrightarrow{s}$
Since the electric field $\overrightarrow{E}$ is radial and the area vector $d\overrightarrow{s}$ is also radial (outward normal),the angle between them is $0^\circ$. Thus,$\overrightarrow{E} \cdot d\overrightarrow{s} = E ds \cos(0^\circ) = E ds$.
$\phi_E = \oint E ds = E \oint ds$
Since $E$ is constant at all points on the spherical surface,and $\oint ds = 4 \pi r^{2}$ (the surface area of the sphere):
$\phi_E = \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}} \right) \cdot (4 \pi r^{2})$
$\phi_E = \frac{q}{\varepsilon_{0}}$
This is Gauss's law: $\oint \overrightarrow{E} \cdot d\overrightarrow{s} = \frac{q}{\varepsilon_{0}}$.