Editing Gauss's law (section)

===Equivalence of integral and differential forms===
{{Main article|Divergence theorem}}

The integral and differential forms are mathematically equivalent, by the divergence theorem. Here is the argument more specifically.

{{math proof|title=Outline of proof 
|proof=The integral form of Gauss's law is:
:{{oiint|intsubscpt=<math>{\scriptstyle _S}</math>|integrand=<math>\mathbf{E} \cdot \mathrm{d}\mathbf{A}</math>}}<math> = \frac{Q}{\varepsilon_0}</math>
for any closed surface {{mvar|S}} containing charge {{mvar|Q}}. By the divergence theorem, this equation is equivalent to:

<math display="block">\iiint_V \nabla \cdot \mathbf{E} \, \mathrm{d}V = \frac{Q}{\varepsilon_0}</math>

for any volume {{mvar|V}} containing charge {{mvar|Q}}. By the relation between charge and charge density, this equation is equivalent to:
<math display="block">\iiint_V \nabla \cdot \mathbf{E} \, \mathrm{d}V = \iiint_V \frac{\rho}{\varepsilon_0} \, \mathrm{d}V</math>
for any volume {{mvar|V}}. In order for this equation to be ''simultaneously true'' for ''every'' possible volume {{mvar|V}}, it is necessary (and sufficient) for the integrands to be equal everywhere. Therefore, this equation is equivalent to:

<math display="block">\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}.</math>
Thus the integral and differential forms are equivalent.
}}