Obtain Coulomb’s law from Gauss’s law.

As shown in figure, consider a point charge $+q$ kept at $\mathrm{O}$.

A Gaussian surface ' $\mathrm{S}$ ' is shown in figure includes charge $q$.

Consider surface area $\overrightarrow{d s}$ at point $\mathrm{P}$. Here, $\overrightarrow{\mathrm{E}} \| \overrightarrow{d s}$ and so that $\theta=0^{\circ}$.

According to Gauss's law,

$\phi=\frac{q}{\varepsilon_{0}}$ $\therefore \int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_{0}}$ $\therefore \int \mathrm{E} \cdot d s \cos 0^{\circ}=\frac{q}{\varepsilon_{0}}[\because \overrightarrow{\mathrm{E}} \| \overrightarrow{d s}]$ $\therefore \mathrm{E} \int d s=\frac{q}{\varepsilon_{0}}\left[\therefore \cos 0^{\circ}=1\right]$ $\therefore \mathrm{E} \times 4 \pi r^{2}=\frac{q}{\varepsilon_{0}}\left[\therefore \int d s=4 \pi r^{2}\right]$ $\therefore \mathrm{E}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}$ $\therefore \frac{\mathrm{F}}{q}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}$ $\therefore \mathrm{F}=\frac{\mathrm{Kq} q_{0}}{r^{2}}$ $\mathrm{This}$ is Coulomb's law.