Editing Gauss's law (section)

==Equation involving the {{math|E}} field==

Gauss's law can be stated using either the electric field {{math|'''E'''}} or the electric displacement field {{math|'''D'''}}. This section shows some of the forms with {{math|'''E'''}}; the form with {{math|'''D'''}} is below, as are other forms with {{math|'''E'''}}.

===Integral form===
[[File:Electric-flux-surface-example.svg|thumb|Electric flux through an arbitrary surface is proportional to the total charge enclosed by the surface.]]
[[File:Electric-flux-no-charge-inside.svg|thumb|No charge is enclosed by the sphere. Electric flux through its surface is zero.]]
Gauss's law may be expressed as:<ref name="GrantPhillips"/>

<math display="block">\Phi_E = \frac{Q}{\varepsilon_0}</math>

where {{math|Φ<sub>''E''</sub>}} is the [[electric flux]] through a closed surface {{mvar|S}} enclosing any volume {{mvar|V}}, {{mvar|Q}} is the total charge enclosed within {{mvar|V}}, and {{math|''ε''<sub>0</sub>}} is the [[electric constant]]. The electric flux {{math|Φ<sub>''E''</sub>}} is defined as a [[surface integral]] of the electric field:

:{{oiint|preintegral=<math>\Phi_E = </math>|intsubscpt=<math>\scriptstyle _S</math>|integrand=<math>\mathbf{E} \cdot \mathrm{d}\mathbf{A}</math>}}

where {{math|'''E'''}} is the electric field, {{math|d'''A'''}} is a vector representing an [[infinitesimal]] element of [[area]] of the surface,{{refn|More specifically, the infinitesimal area is thought of as [[Plane (mathematics)|planar]] and with area {{math|d''N''}}. The vector {{math|d'''R'''}} is [[Normal (geometry)|normal]] to this area element and has [[magnitude (vector)|magnitude]] {{math|d''A''}}.<ref>{{cite book|last=Matthews|first=Paul|title=Vector Calculus|publisher=Springer|year=1998|isbn=3-540-76180-2}}</ref>|group=note}} and {{math|·}} represents the [[dot product]] of two vectors.

In a curved spacetime, the flux of an electromagnetic field through a closed surface is expressed as

:{{oiint|preintegral=<math>\Phi_E = c </math> |intsubscpt=<math> \scriptstyle _S</math>|integrand=<math> F^{\kappa 0} \sqrt {-g} \, \mathrm{d} S_\kappa </math>}}

where <math>c</math> is the [[speed of light]]; <math>F^{\kappa 0}</math> denotes the time components of the [[electromagnetic tensor]]; <math>g</math> is the determinant of [[metric tensor]]; <math> \mathrm{d} S_\kappa = \mathrm{d} S^{ij} = \mathrm{d}x^i \mathrm{d}x^j </math> is an orthonormal element of the two-dimensional surface surrounding the charge <math>Q</math>; indices <math> i,j,\kappa = 1,2,3</math> and do not match each other.<ref>{{cite journal | last1 = Fedosin | first1 = Sergey G. | title = On the Covariant Representation of Integral Equations of the Electromagnetic Field | journal = Progress in Electromagnetics Research C | volume = 96 | pages = 109–122| year = 2019 | url = https://rdcu.be/ccV9o| doi = 10.2528/PIERC19062902| arxiv = 1911.11138 | bibcode=2019arXiv191111138F| s2cid = 208095922 }}</ref>

Since the flux is defined as an ''integral'' of the electric field, this expression of Gauss's law is called the ''integral form''.
[[File:Gauss's law - surface charge - boundary condition on D.svg|thumb|A tiny Gauss's box whose sides are perpendicular to a conductor's surface is used to find the local surface charge once the electric potential and the electric field are calculated by solving Laplace's equation. The electric field is perpendicular, locally, to the equipotential surface of the conductor, and zero inside; its flux ''πa''<sup>2</sup>·''E'', by Gauss's law equals ''πa''<sup>2</sup>·''σ''/''ε''<sub>0</sub>. Thus, {{nowrap|1=''σ'' = ''ε''<sub>0</sub>''E''}}.]]
In problems involving conductors set at known potentials, the potential away from them is obtained by solving [[Laplace's equation]], either analytically or numerically. The electric field is then  calculated as the potential's negative gradient. Gauss's law makes it possible to find the distribution of electric charge: The charge in any given region of the conductor can be deduced by integrating the electric field to find the flux through a small box whose sides are perpendicular to the conductor's surface and by noting that the electric field is perpendicular to the surface, and zero inside the conductor.

The reverse problem, when the electric charge distribution is known and the electric field must be computed, is much more difficult.  The total flux through a given surface gives little information about the electric field, and can go in and out of the surface in arbitrarily complicated patterns.

An exception is if there is some [[symmetry]] in the problem, which mandates that the electric field passes through the surface in a uniform way. Then, if the total flux is known, the field itself can be deduced at every point. Common examples of symmetries which lend themselves to Gauss's law include: cylindrical symmetry, planar symmetry, and spherical symmetry. See the article [[Gaussian surface]] for examples where these symmetries are exploited to compute electric fields.

===Differential form===

By the [[divergence theorem]], Gauss's law can alternatively be written in the ''differential form'':
<math display="block">\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}</math>

where {{math|∇ · '''E'''}} is the [[divergence]] of the electric field, {{math|''ε''<sub>0</sub>}} is the [[vacuum permittivity]] and {{mvar|ρ}} is the total volume [[charge density]] (charge per unit volume).

===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.
}}