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The Gauss law of electrostatics follows from the Coulomb’s law for a point charge and superposition principle. The Gauss law along with the Gauss divergence theorem of vector calculus imply Maxwell’s first equation \(\nabla\cdot\bar{E}=\rho/\epsilon_0\) for electrostatics
1. \(\nabla\cdot\bar{E}=\rho/\epsilon_0\) from Gauss law
Let \(S\) be a surface which encloses a volume V. Flux of $\bar{E}$ through \(S\) is \(\int_S\bar{E}\cdot\hat{n}dS. \) The total charge enclosed is \(Q=\int_V\rho dV.\) The Gauss law for the flux of the electric field implies, \begin{equation} \int_S\bar{E}\cdot\hat{n}dS=\frac{1}{\epsilon_0}\int_V\rho d^3x \end{equation} Using the Gauss divergence theorem of vector calculus we get
\begin{eqnarray} \int_V\big(\nabla\cdot\bar{E}\big)\, dV=\frac{1}{\epsilon_0}\int_V\rho dV . \end{eqnarray} Rearranging we get that \begin{eqnarray} \int_V\Big(\nabla\cdot\bar{E}-\frac{\rho}{\epsilon_0}\Big)\,dV = 0 \label{Eq3} \end{eqnarray} is true for arbitrary \(V\). Considering \(V\) to be a small volume \(dxdydz\) at \((x,y,z)\), the left hand side of the above equation becomes \begin{eqnarray} \Big(\nabla\cdot\bar{E}-\frac{\rho}{\epsilon_0}\Big)\,dxdydz =0 \end{eqnarray} and we get Maxwell's first equation \begin{equation} \nabla\cdot\bar{E}=\frac{\rho}{\epsilon_0}. \label{Eq4} \end{equation} This equation is just the differential form of the Gauss law.
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4727:Diamond Point
