I am aware of how to use both Gauss's and Ampere's Law but I am really interested in knowing and understanding how charge is conserved as a consequence of these laws, especially using these forms of the laws:

$$\vec{\nabla}\cdot\vec{\mathbf{D}}=\rho$$ $$\vec{\nabla}\times\vec{\mathbf{H}}= \vec{\mathbf{J}}+\frac{\partial\vec{\mathbf{D}}}{\partial t}$$

Can someone please help explain?

  • 5
    $\begingroup$ Take the time derivative of the first equation and add it to the divergence of the second. $\endgroup$
    – mike stone
    Dec 13, 2021 at 19:55
  • $\begingroup$ Use that the divergence of a rotation vanishes: $$\vec{0} = \vec{\nabla}\cdot(\vec{\nabla}\times\vec{\mathbf{H}})=\cdots$$ $\endgroup$
    – md2perpe
    Dec 13, 2021 at 21:22
  • $\begingroup$ @mikestone thanks, I used your comment and the answer provided below! $\endgroup$
    – harry
    Dec 14, 2021 at 6:10

1 Answer 1


You can prove it with Maxwell's Equations and some vector identities.

Conservation of charge is $\nabla \cdot \vec{J} + \partial \rho/\partial t = 0$

Given $\nabla \times \vec{B} = \mu_0 \vec{J}+(1/c^2)\partial \vec{E}/\partial t$.

Take the divergence of both sides. By a vector identity, the divergence of a curl is 0.By Gauss' Law $\nabla \cdot \vec{E} = \rho/\epsilon_0.$ So take the divergence of both sides:

$0 = \mu_0 (\nabla \cdot \vec{J})+(1/\epsilon_0c^2)\partial \rho/\partial t$

We know $\mu_0\epsilon_0=1/c^2$, so divide everything by $\mu_0$, you get the defining equation for charge conservation.

Something similar applies to $\vec{D}$ and $\vec{H}$.

  • $\begingroup$ Got it, thank you! $\endgroup$
    – harry
    Dec 14, 2021 at 6:09

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