The Continuity Equation for Electric Charge

The continuity equation  
\[\frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot \mathbf{J} =0\]
implies conservation of electric charge.
To prove it, use Maxwell's equations  
\[\mathbf{\nabla} \times \mathbf{H} =\epsilon \frac{\partial \mathbf{E}} {\partial t} + \mathbf{J}\]
\[\mathbf{\nabla} \cdot \mathbf{E} = \frac{\mathbf\rho}{\epsilon}\]

Take the divergence.
\[\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{H}) = \epsilon \mathbf{\nabla} \cdot \frac{\partial \mathbf{E}} {\partial t} + \mathbf{\nabla} \cdot \mathbf{J}\]
.  (1)
The left hand side of this equation is zero. Assuming that  
  has continuous derivatives, we can write  
\[\mathbf{\nabla} \cdot \frac{\partial \mathbf{E}}{\partial t} = \frac{\partial}{\partial t} (\mathbf{\nabla} \cdot \mathbf{E})\]
Equation (1) becomes
\[0 = \epsilon \frac{\partial } {\partial t} (\mathbf{\nabla} \cdot \mathbf{E}) + \mathbf{\nabla} \cdot \mathbf{J}\]
Use the second of Maxewell's equations above in the last equation to give
\[0 = \epsilon \frac{\partial } {\partial t} (\frac{\rho}{\epsilon}) + \mathbf{\nabla} \cdot \mathbf{J}\]
Then obviously  
\[ \frac{\partial \rho} {\partial t} + \mathbf{\nabla} \cdot \mathbf{J}=0\]

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