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Starting with the Gauss law and using divergence theorem of vector calculus we derive Maxwell's first equation $\nabla\cdot \vec{E}= \rho/\epsilon_0$.
1. $\nabla\cdot\bar{E}=\rho/\epsilon_0$ from Gauss law.
\begin{FileText}{ Let S be a surface which encloses a volume \(V\). Flux of $\bar{E}$ through S is $\phi_S$ \begin{eqnarray} \phi_S=\int_S\bar{E}\cdot\hat{n}dS \label{eq1} \end{eqnarray}
The total charge enclosed is Q.
\begin{eqnarray} Q=\int_V\rho dV \label{eq2} \end{eqnarray}
Now using Gauss law,
\begin{eqnarray} \nonumber\int_S\bar{E}\cdot\hat{n}dS&=&\frac{1}{\epsilon_0}\int_V\rho dV \\ \nonumber\int_V\nabla\cdot\bar{E}dV&=&\frac{1}{\epsilon_0}\int_V\rho dV \\ \int_V(\nabla\cdot\bar{E}-\frac{1}{\epsilon_0}\rho)dV&=&0 .\label{eq3} \end{eqnarray}
This is true for arbitrary \(V\). Consider \(V\) to be a small volume between \(x\) and \(x+dx, y\) and \(y+dy, z\) and \(z+dz\). Then \(dV\) is equivalent to \(dx\,dyd\,z\) Therefore
\begin{eqnarray} (\nabla\cdot\bar{E}-\frac{1}{\epsilon_0}\rho)dxdydz&=&0 \label{eq4} \\
\nabla\cdot\bar{E}&=&\rho/\epsilon_0 \label{eq5} \end{eqnarray}
\eqref{eq5} holds at point (x,y,z) and \( (x,y,z)\) is an arbitrary point.
Therefore, the equation holds true everywhere. Hence we proved the first Maxwell's equation.
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