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v3	v4
52	52	$ \mathbf{E} = -\mathbf{v} \times \mathbf{B} \color{green}{+ \mathbf{E}_{\scriptscriptstyle R}}$
53	53
54		We then need to add this contribution to the induction equation $\color{green}{-\nabla \times \mathbf{E'}}$ and the energy equation where it enters both the Poynting flux $-\color{green}{\nabla \cdot \left ( \mathbf{E'} \times \mathbf{B} \right)}$ and the work done on the the charge distribution $\color{green}{ - \mathbf{J} \cdot \mathbf{E'}}$.
	54	We then need to add this contribution to the induction equation $\color{green}{-\nabla \times \mathbf{E}_{\scriptscriptstyle R}}$ and the energy equation where it enters both the Poynting flux $-\color{green}{\nabla \cdot \left ( \mathbf{E}_{\scriptscriptstyle R} \times \mathbf{B} \right)}$ and the work done on the the charge distribution $\color{green}{ - \mathbf{J} \cdot \mathbf{E}_{\scriptscriptstyle R}}$.
55	55
56	56	However the work done on the charge distribution would need to also be added to the thermal energy cancelling any changes in the total energy due to the $\color{green}{\mathbf{J} \cdot \mathbf{E}_{\scriptscriptstyle R}}$ term. This leaves us with.