## 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
$\mathbf{\nabla} \times \mathbf{H} =\epsilon \frac{\partial \mathbf{E}} {\partial t} + \mathbf{J}$
and
$\mathbf{\nabla} \cdot \mathbf{E} = \frac{\mathbf\rho}{\epsilon}$

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
$\mathbf{E}$
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
$\frac{\partial \rho} {\partial t} + \mathbf{\nabla} \cdot \mathbf{J}=0$