Obtain Coulomb's law from Gauss's law.

(N/A) Consider a point charge $+q$ placed at the origin $O$.
Construct a spherical Gaussian surface $S$ of radius $r$ centered at $O$ that encloses the charge $q$.
Consider an infinitesimal surface area element $d\vec{S}$ at a point $P$ on the surface. Due to spherical symmetry,the electric field $\vec{E}$ is directed radially outward and is parallel to the area vector $d\vec{S}$. Thus,the angle between them is $\theta = 0^{\circ}$.
According to Gauss's law,the total electric flux $\phi$ through the closed surface is given by:
$\phi = \oint \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_{0}}$
Since $\vec{E}$ is uniform over the spherical surface and $\vec{E} \parallel d\vec{S}$,we have:
$\oint E \, dS \cos 0^{\circ} = \frac{q}{\varepsilon_{0}}$
$E \oint dS = \frac{q}{\varepsilon_{0}}$
Since the surface area of the sphere is $\oint dS = 4\pi r^{2}$,we get:
$E(4\pi r^{2}) = \frac{q}{\varepsilon_{0}}$
$E = \frac{1}{4\pi \varepsilon_{0}} \frac{q}{r^{2}}$
If a test charge $q_{0}$ is placed at point $P$,the force experienced by it is $F = q_{0}E$. Substituting the expression for $E$:
$F = q_{0} \left( \frac{1}{4\pi \varepsilon_{0}} \frac{q}{r^{2}} \right) = \frac{1}{4\pi \varepsilon_{0}} \frac{q q_{0}}{r^{2}}$
This expression is Coulomb's law.