Derivation of Maxwell's first equation

Maxwell's first equation is the differential form of Gauss's law of electrostatics.i.e

$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

Derivation:

According to Gauss's law for electrostatic-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{\epsilon_{0}} \qquad(1)$

For continuous charge distribution inside the surface-

$q=\oint_{v}\rho.dV$

Where
$\rho$→Charge density
dV→Small volume

Now substitute the value of $q$ in equation $(1)$ then

$\oint_{s}\overrightarrow{E}.\overrightarrow{dS}=\frac{1}{\epsilon_{0}} \oint_{v}\rho.dV \qquad(2)$

Now according to Gauss's divergence theorem-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV \qquad (3)$

From equation$(2)$ and equation$(3)$, we can write the above equation-

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV= \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV $

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV- \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV=0 $

$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{E}- \frac{\rho}{\epsilon_{0}})dV=0 $

On solving the above equation-

$\overrightarrow{\nabla}.\overrightarrow{E}- \frac{\rho}{\epsilon_{0}}=0 $

$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

This is Maxwell's first equation.