\begin{equation}
\bar{\nabla} . \bar{D}=\rho \text{Coulomb's law}\\
\bar{\nabla} .\bar{H} = \bar{J} \text{Ampere's law}\\
\bar{\nabla}\times \bar{E}+\dfrac{\partial \bar{B}}{\partial t}=0 \text{Faraday's law}\\
\bar{\nabla} . \bar{B} =0
\end{equation}
Where $\bar{D}=\epsilon \bar{E}$ and $\bar{B}=\mu \bar{H}$
\begin{equation}
\bar{\nabla}.(\bar{\nabla} \times \bar{B})=\mu_0 \left( \bar{\nabla} .\bar{J}+\epsilon \dfrac{\partial}{\partial t}\bar{\nabla}.E\right)\\
0=\mu_0 \left( \bar{\nabla} .\bar{J}+\epsilon_0 \dfrac{\partial}{\partial t} \dfrac{\rho}{\epsilon_0}\right)\\
\left( \bar{\nabla} .\bar{J}+ \dfrac{\partial \rho}{\partial t}\right)=0
\end{equation}
\begin{equation}
\bar{\nabla} \times \bar{H}=\bar{J}\\
\bar{\nabla}.(\bar{\nabla} \times\bar{H})=\bar{\nabla}.\bar{J}\\
0=\bar{\nabla}.\bar{J}
\end{equation}
\begin{equation}
\bar{\nabla} .\bar{J}+ \dfrac{\partial \rho}{\partial t}=0\\
\bar{\nabla} .\bar{J}+ \dfrac{\partial }{\partial t} (\bar{\nabla}.\bar{D})=0\\
\bar{\nabla} .\bar{J}+ \bar{\nabla}. \dfrac{\partial \bar{D}}{\partial t}=0\\
\bar{\nabla} . \left( \bar{J}+ \dfrac{\partial \bar{D}}{\partial t}\right)=0\\
\end{equation}
\begin{equation}
\bar{J}\rightarrow \bar{J}+\dfrac{\partial \bar{D}}{\partial t}
\end{equation}
\begin{equation}
\bar{\nabla} . \bar{D}=\rho \text{Coulomb's law}\\
\bar{\nabla} .\bar{H} = \bar{J}+\dfrac{\partial \bar{D}}{\partial t} \text{Ampere's law}\\
\bar{\nabla}\times \bar{E}+\dfrac{\partial \bar{B}}{\partial t}=0 \text{Faraday's law}\\
\bar{\nabla} . \bar{B} =0
\end{equation}
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