energy conservation and charge conservation

1. Charge conservation
Do the divergence of ampere law(2nd maxwell equation). Since curl has no divergence(another one-gradient has no curl), we have

\begin{align*}
0 &=\nabla \cdot (\nabla \times \pmb{H})=\nabla \cdot (J+\frac{\partial \pmb{D}}{\partial t}) ~\Rightarrow \\
\nabla & \cdot J +\frac{\partial \rho}{\partial t}=0 \\
o &r, in~ integral~ form \\
I &=\int_V\nabla \cdot Jdv=\oint_SJda=\frac{\partial} {\partial t}\int_V \rho dv=\frac{\partial Q} {\partial t}
\end{align*}

Current flow out of the volume V equals to the charge depletion per unit time.
For conductors, J=sigmaE,

\begin{align*}
\nabla & \cdot J= \nabla  \cdot \sigma E=\frac{\sigma \rho }{\epsilon} ~ \Rightarrow \\
\nabla & \cdot J+\frac{\partial \rho}{\partial t}=\frac{\sigma \rho }{\epsilon}+\frac{\partial \rho}{\partial t}=0  ~ \Rightarrow \\
\rho &(r,t)  = \rho_0  e^{-\sigma t / \epsilon } \\
d &ue ~ to ~ \sigma >> \epsilon ~ in ~ conductor, \rho -> 0 ~ in~ a ~short~ time. 
\end{align*}

e.g. Copper, sigma=5.7x10^7, epsilon=8.85x10^-12, epsilon/sigma=1.6x10^-19 sec.


2. Energy conservation.
* Energy per unit volume, \inline w=\frac{1}{2}\epsilon |E|^2+\frac{1}{2}\mu |H|^2
* Energy flux (Poynting vector): \inline \pmb{P}=\pmb{E} \times \pmb{H}

\begin{align*}
 \nabla  &\cdot (\pmb{E} \times \pmb{H})=\pmb{H}\cdot (\nabla \times \pmb{E})-\pmb{E}\cdot (\nabla \times \pmb{H})= \\
-\mu & \pmb{H}\cdot \frac{\partial \pmb{H}}{\partial t} -\mu \pmb{E}\cdot \frac{\partial \pmb{E}}{\partial t}-\pmb{E}\cdot \pmb{J}  ~\Rightarrow\\
\nabla  &\cdot \pmb{P}=  -\frac{\partial w}{\partial t}- \pmb{E}\cdot \pmb{J}\\
N&ote: \epsilon \pmb{E} \cdot \frac{\partial \pmb{E} }{\partial t}=\frac{1}{2}\epsilon \frac{\partial |E|^2}{\partial t}, \mu \pmb{H} \cdot \frac{\partial \pmb{H} }{\partial t}=\frac{1}{2}\mu \frac{\partial |H|^2}{\partial t}
\end{align*}

the rate of energy flow out of volume equals to the depletion rate of stored energy in the volume and ohmic loss in the volume.