\[\frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot \mathbf{J} =0\]
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
\[ \frac{\partial \rho} {\partial t} + \mathbf{\nabla} \cdot \mathbf{J}=0\]