Editing Gauss's law (section)

===Deriving Gauss's law from Coulomb's law===
{{Citation needed|date=June 2024|reason=A reliable source is needed for the entire derivation.}}
Strictly speaking, Gauss's law cannot be derived from Coulomb's law alone, since Coulomb's law gives the electric field due to an individual, [[Electrostatic charge|electrostatic]] [[point charge]] only. However, Gauss's law ''can'' be proven from Coulomb's law if it is assumed, in addition, that the electric field obeys the [[superposition principle]]. The superposition principle states that the resulting field is the vector sum of fields generated by each particle (or the integral, if the charges are distributed smoothly in space).

{{math proof|title=Outline of proof
|proof=
Coulomb's law states that the electric field due to a stationary [[point charge]] is:
<math display="block">\mathbf{E}(\mathbf{r}) = \frac{q}{4\pi \varepsilon_0} \frac{\mathbf{e}_r}{r^2}</math>
where
*{{math|'''e'''<sub>''r''</sub>}} is the radial [[unit vector]],
*{{mvar|r}} is the radius, {{math|{{abs|'''r'''}}}},
*{{math|''ε''<sub>0</sub>}} is the [[electric constant]],
*{{mvar|q}} is the charge of the particle, which is assumed to be located at the [[origin (mathematics)|origin]].

Using the expression from Coulomb's law, we get the total field at {{math|'''r'''}} by using an integral to sum the field at {{math|'''r'''}} due to the infinitesimal charge at each other point {{math|'''s'''}} in space, to give
<math display="block">\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{s})(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} \, \mathrm{d}^3 \mathbf{s}</math>
where {{mvar|ρ}} is the charge density. If we take the divergence of both sides of this equation with respect to '''r''', and use the known theorem<ref>See, for example, {{cite book | last=Griffiths | first=David J. | title=Introduction to Electrodynamics | edition=4th | publisher=Prentice Hall | year=2013 | page=50 }} or {{cite book | last=Jackson | first=John David | title=Classical Electrodynamics | edition=3rd | publisher=John Wiley & Sons | year=1999 | page=35}}</ref>

<math display="block">\nabla \cdot \left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right) = 4\pi \delta(\mathbf{r})</math>
where {{math|''δ''('''r''')}} is the [[Dirac delta function]], the result is
<math display="block">\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{1}{\varepsilon_0} \int \rho(\mathbf{s})\, \delta(\mathbf{r}-\mathbf{s})\, \mathrm{d}^3 \mathbf{s}</math>

Using the "[[Dirac delta function#Translation|sifting property]]" of the Dirac delta function, we arrive at
<math display="block">\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{\rho(\mathbf{r})}{\varepsilon_0},</math>
which is the differential form of Gauss's law, as desired.
}}

Since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and, in this respect, Gauss's law is more general than Coulomb's law.

{{math proof|title=Proof (without Dirac Delta)
|proof=
Let <math>\Omega \subseteq R^3 </math> be a bounded open set, and  <math display="block">\mathbf E_0(\mathbf r) = \frac {1}{4 \pi \varepsilon_0} \int_{\Omega} \rho(\mathbf r')\frac {\mathbf r - \mathbf r'} {\left \| \mathbf r - \mathbf r' \right \|^3} \mathrm{d}\mathbf{r}'
\equiv \frac {1}{4 \pi \varepsilon_0} \int_{\Omega} e(\mathbf{r, \mathbf{r}'}){\mathrm{d}\mathbf{r}'}</math> be the electric field, with <math>\rho(\mathbf r')</math> a continuous function (density of charge).

It is true for all <math>\mathbf{r} \neq \mathbf{r'}</math> that <math>\nabla_\mathbf{r}  \cdot \mathbf{e}(\mathbf{r, r'}) = 0</math>.

Consider now a compact set <math>V \subseteq R^3</math> having a [[piecewise]] [[Smooth surface|smooth boundary]] <math>\partial V</math> such that <math>\Omega  \cap V = \emptyset</math>. It follows that <math>e(\mathbf{r, \mathbf{r}'}) \in C^1(V \times \Omega)</math> and so, for the divergence theorem:

<math display="block">\oint_{\partial V} \mathbf{E}_0 \cdot d\mathbf{S} = \int_V \mathbf{\nabla} \cdot \mathbf{E}_0 \, dV</math>

But because <math>e(\mathbf{r, \mathbf{r}'}) \in C^1(V \times \Omega)</math>,

<math display="block">\mathbf{\nabla} \cdot \mathbf{E}_0(\mathbf{r}) = \frac {1}{4 \pi \varepsilon_0} \int_{\Omega} \nabla_\mathbf{r} \cdot e(\mathbf{r, \mathbf{r}'}){\mathrm{d}\mathbf{r}'} = 0 </math>  for the argument above (<math>\Omega  \cap V = \emptyset \implies \forall \mathbf{r} \in V \ \  \forall \mathbf{r'} \in \Omega \ \ \  \mathbf{r} \neq \mathbf{r'}  </math> and then <math>\nabla_\mathbf{r}  \cdot \mathbf{e}(\mathbf{r, r'}) = 0</math>)

Therefore the flux through a closed surface generated by some charge density outside (the surface) is null.

Now consider <math>\mathbf{r}_0 \in \Omega</math>, and <math>B_R(\mathbf{r}_0)\subseteq \Omega</math>  as the sphere centered in <math>\mathbf{r}_0</math> having <math>R</math> as radius (it exists because <math>\Omega</math> is an open set).

Let <math>\mathbf{E}_{B_R}</math> and <math>\mathbf{E}_C</math> be the electric field created inside and outside the sphere respectively. Then,

:<math>\mathbf{E}_{B_R} = \frac {1}{4 \pi \varepsilon_0} \int_{B_R(\mathbf{r}_0)} e(\mathbf{r, \mathbf{r}'}){\mathrm{d}\mathbf{r}'}</math>,    <math>\mathbf{E}_C = \frac {1}{4 \pi \varepsilon_0} \int_{\Omega \setminus B_R(\mathbf{r}_0)} e(\mathbf{r, \mathbf{r}'}){\mathrm{d}\mathbf{r}'}</math> and <math>\mathbf{E}_{B_R} + \mathbf{E}_C = \mathbf{E}_0 </math>

<math display="block">\Phi(R) =
\oint_{\partial B_R(\mathbf{r}_0)} \mathbf{E}_0 \cdot d\mathbf{S} =
\oint_{\partial B_R(\mathbf{r}_0)} \mathbf{E}_{B_R} \cdot d\mathbf{S} +
\oint_{\partial B_R(\mathbf{r}_0)} \mathbf{E}_C \cdot d\mathbf{S} =
\oint_{\partial B_R(\mathbf{r}_0)} \mathbf{E}_{B_R} \cdot d\mathbf{S} </math>

The last equality follows by observing that <math>(\Omega \setminus B_R(\mathbf{r}_0)) \cap B_R(\mathbf{r}_0) = \emptyset</math>, and the argument above.

The RHS is the electric flux generated by a charged sphere, and so:

<math display="block">\Phi(R) =\frac {Q(R)}{\varepsilon_0}  = \frac {1}{\varepsilon_0} \int_{B_R(\mathbf{r}_0)} \rho(\mathbf r'){\mathrm{d}\mathbf{r}'} =
\frac {1}{\varepsilon_0} \rho(\mathbf r'_c)|B_R(\mathbf{r}_0)| </math> with <math> r'_c \in \ B_R(\mathbf{r}_0)</math>

Where the last equality follows by the mean value theorem for integrals. Using the [[squeeze theorem]] and the continuity of <math> \rho  </math>, one arrives at:

<math display="block">\mathbf{\nabla} \cdot \mathbf{E}_0(\mathbf{r}_0) =
\lim_{R \to 0} \frac{1}{|B_R(\mathbf{r}_0)|}\Phi(R) = 
\frac {1}{\varepsilon_0} \rho(\mathbf r_0)  </math>
}}