Editing Gaussian units (section)

=== Maxwell's equations ===
{{main|Maxwell's equations}}

Here are Maxwell's equations, both in macroscopic and microscopic forms. Only the "differential form" of the equations is given, not the "integral form"; to get the integral forms apply the [[divergence theorem]] or [[Stokes' theorem|Kelvin–Stokes theorem]].

{| class="wikitable plainrowheaders"
|+ Maxwell's equations in Gaussian system and ISQ
|-
! scope="col" | Name
! scope="col" | Gaussian system
! scope="col" | {{abbr|ISQ|International System of Quantities}}
|-
! scope="row" | [[Gauss's law]]{{br}}(macroscopic)
| <math>\nabla \cdot \mathbf{D}^{_\mathrm{G}} = 4\pi\rho_\mathrm{f}^{_\mathrm{G}}</math>
| <math>\nabla \cdot \mathbf{D}^{_\mathrm{I}} = \rho_\mathrm{f}^{_\mathrm{I}}</math>
|-
! scope="row" | [[Gauss's law]]<br />(microscopic)
| <math>\nabla \cdot \mathbf{E}^{_\mathrm{G}} = 4\pi\rho^{_\mathrm{G}}</math>
| <math>\nabla \cdot \mathbf{E}^{_\mathrm{I}} = \frac{1}{\varepsilon_0} \rho^{_\mathrm{I}}</math>
|-
! scope="row" | [[Gauss's law for magnetism]]
|<math>\nabla \cdot \mathbf{B}^{_\mathrm{G}} = 0</math>
|<math>\nabla \cdot \mathbf{B}^{_\mathrm{I}} = 0</math>
|-
! scope="row" | Maxwell–Faraday equation<br />([[Faraday's law of induction]])
| <math>\nabla \times \mathbf{E}^{_\mathrm{G}} + \frac{1}{c}\frac{\partial \mathbf{B}^{_\mathrm{G}}} {\partial t} = 0</math>
| <math>\nabla \times \mathbf{E}^{_\mathrm{I}} + \frac{\partial \mathbf{B}^{_\mathrm{I}}} {\partial t} = 0</math>
|-
! scope="row" | [[Ampère–Maxwell equation]]<br /> (macroscopic)
| <math>\nabla \times \mathbf{H}^{_\mathrm{G}} - \frac{1}{c} \frac{\partial \mathbf{D}^{_\mathrm{G}}} {\partial t} = \frac{4\pi}{c}\mathbf{J}_\mathrm{f}^{_\mathrm{G}}</math>
| <math>\nabla \times \mathbf{H}^{_\mathrm{I}} - \frac{\partial \mathbf{D}^{_\mathrm{I}}} {\partial t}= \mathbf{J}_\mathrm{f}^{_\mathrm{I}}</math>
|-
! scope="row" | [[Ampère–Maxwell equation]]<br /> (microscopic)
| <math>\nabla \times \mathbf{B}^{_\mathrm{G}} - \frac{1}{c}\frac{\partial \mathbf{E}^{_\mathrm{G}}} {\partial t} = \frac{4\pi}{c}\mathbf{J}^{_\mathrm{G}}</math>
| <math>\nabla \times \mathbf{B}^{_\mathrm{I}} - \frac{1}{c^2}\frac{\partial \mathbf{E}^{_\mathrm{I}}} {\partial t} = \mu_0\mathbf{J}^{_\mathrm{I}}</math>
|}