[NOTES/EM-03008]-Maxwell's Second Equations from Coulomb's Law

$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$

Maxwell's Second Equations from Coulomb's Law
Consider a small volume $\Delta V$, at position $\vec{r}^\prime$,
of a continuous charge distribution in a  volume \(V\) and charge density $\rho$. The charge inside the volume $\Delta V$  is $ \Delta q=\Delta V\rho(\vec{r}^\prime)$.
Electric field at \(\vec r\),due to this charge at $\vec{r}^\prime$, is given by
\[\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\big(\rho(\vec{r})\Delta V\big)  \frac{(\vec{r}-\vec{r^\prime})}{{\mid\vec{r}-\vec{r^\prime}\mid}^3}.\]
With $\Delta V \rightarrow d^3r^\prime$, the total electric field is
\begin{eqnarray}
\vec{E}&=&\frac{1}{4\pi\epsilon_0}\int_V  \rho (\vec{r}^\prime)\frac{(\vec{r}-\vec{r}^\prime)}{{\mid\vec{r}-\vec{r}^\prime\mid}^3}d^3r^\prime .
\end{eqnarray}

 The proof of Maxwell's second equation is straight forward. We have to show that for arbitrary charge distribution $\nabla \times \vec{E} = 0.$
Using superposition principle, it is sufficient to prove this result
$\vec{E}$ for a single point charge. Therefore,  taking the electric field to be that for a point charge \(q\) at \(\vec r_0=(x_0, y_0, z_0)\)
\begin{equation}
 \vec E =\frac{q}{4\pi\epsilon_0}  \frac{(\vec{r}-\vec{r}_0)}{{\mid\vec{r}-\vec{r}_0\mid}^3}
\end{equation}
we compute
\begin{eqnarray}
\nabla \times \vec{E} =
\left | \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\[2mm]
\dfrac{\partial{}}{\partial{x}} &  \dfrac{\partial{}}{\partial{y}} &  \dfrac{\partial{}}{\partial{z}} \\[3mm]
E_x & E_y & E_z
\end{array} \right|.
\end{eqnarray}
To compute the derivatives involved, it is useful to introduce \(\vec R=(X,Y,Z)\equiv(\vec r-\vec r_0)\) and
\[\pp{x}=\pp{X},\quad\pp{y}=\pp{Y},\quad\pp{z}=\pp{Z}.\]
Skipping details of a straight forward computation, we get  
\begin{eqnarray}\nonumber
(\nabla \times \vec{E})_x &=& \Big(\frac{\partial{E_z}}{\partial{y}} - \frac{\partial{E_y}}{\partial{z}}\Big) = 0,
\\\nonumber
\big(\nabla \times \vec{E}\big)_y &=& \Big(\frac{\partial{E_z}}{\partial x} - \frac{\partial{E_x}}{\partial z}\Big) = 0, \\\nonumber
\big(\nabla \times \vec{E}\big)_z &=& \Big(\frac{\partial{E_y}}{\partial x} - \frac{\partial{E_x}}{\partial y}\Big) = 0.
\end{eqnarray}
%
Hence  
\begin{equation}
\nabla \times \vec{E} = (\nabla \times \vec{E})_x \hat{i}+ ( \nabla \times \vec{E})_y \hat{j} + ( \nabla \times \vec{E})_z \hat{k}= 0.
\end{equation}