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Deriving Gauss's law from Coulomb's law

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 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).

Coulomb's law states that the electric field due to a stationary point charge is: E (r) = q 4 π ε 0 e r r 2 {\displaystyle \mathbf {E} (\mathbf {r})={\frac {q}{4\pi \varepsilon _{0}}}{\frac {\mathbf {e} _{r}}{r^{2}}}} where

Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give E (r) = 1 4 π ε 0 ∫ ρ (s) (r − s) | r − s | 3 d 3 s {\displaystyle \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} } where 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

∇ ⋅ (r | r | 3) = 4 π δ (r) {\displaystyle \nabla \cdot \left({\frac {\mathbf {r} }{|\mathbf {r} |^{3}}}\right)=4\pi \delta (\mathbf {r})} where δ (r) is the Dirac delta function, the result is ∇ ⋅ E (r) = 1 ε 0 ∫ ρ (s) δ (r − s) d 3 s {\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r})={\frac {1}{\varepsilon _{0}}}\int \rho (\mathbf {s})\,\delta (\mathbf {r} -\mathbf {s})\,\mathrm {d} ^{3}\mathbf {s} }

Using the " sifting property " of the Dirac delta function, we arrive at ∇ ⋅ E (r) = ρ (r) ε 0, {\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r})={\frac {\rho (\mathbf {r})}{\varepsilon _{0}}},} 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.

Let Ω ⊆ R 3 {\displaystyle \Omega \subseteq R^{3}} be a bounded open set, and E 0 (r) = 1 4 π ε 0 ∫ Ω ρ (r ′) r − r ′ ‖ r − r ′ ‖ 3 d r ′ ≡ 1 4 π ε 0 ∫ Ω e (r, r ′) d r ′ {\displaystyle \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} '}} be the electric field, with ρ (r ′) {\displaystyle \rho (\mathbf {r} ')} a continuous function (density of charge).

It is true for all r ≠ r ′ {\displaystyle \mathbf {r} \neq \mathbf {r'} } that ∇ r ⋅ e (r, r ′) = 0 {\displaystyle \nabla _{\mathbf {r} }\cdot \mathbf {e} (\mathbf {r,r'})=0}.

Consider now a compact set V ⊆ R 3 {\displaystyle V\subseteq R^{3}} having a piecewise smooth boundary ∂ V {\displaystyle \partial V} such that Ω ∩ V = ∅ {\displaystyle \Omega \cap V=\emptyset }. It follows that e (r, r ′) ∈ C 1 (V × Ω) {\displaystyle e(\mathbf {r,\mathbf {r} '})\in C^{1}(V\times \Omega)} and so, for the divergence theorem:

∮ ∂ V E 0 ⋅ d S = ∫ V ∇ ⋅ E 0 d V {\displaystyle \oint _{\partial V}\mathbf {E} _{0}\cdot d\mathbf {S} =\int _{V}\mathbf {\nabla } \cdot \mathbf {E} _{0}\,dV}

But because e (r, r ′) ∈ C 1 (V × Ω) {\displaystyle e(\mathbf {r,\mathbf {r} '})\in C^{1}(V\times \Omega)},

∇ ⋅ E 0 (r) = 1 4 π ε 0 ∫ Ω ∇ r ⋅ e (r, r ′) d r ′ = 0 {\displaystyle \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} for the argument above (Ω ∩ V = ∅ ⟹ ∀ r ∈ V ∀ r ′ ∈ Ω r ≠ r ′ {\displaystyle \Omega \cap V=\emptyset \implies \forall \mathbf {r} \in V\ \ \forall \mathbf {r'} \in \Omega \ \ \ \mathbf {r} \neq \mathbf {r'} } and then ∇ r ⋅ e (r, r ′) = 0 {\displaystyle \nabla _{\mathbf {r} }\cdot \mathbf {e} (\mathbf {r,r'})=0})

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

Now consider r 0 ∈ Ω {\displaystyle \mathbf {r} _{0}\in \Omega }, and B R (r 0) ⊆ Ω {\displaystyle B_{R}(\mathbf {r} _{0})\subseteq \Omega } as the sphere centered in r 0 {\displaystyle \mathbf {r} _{0}} having R {\displaystyle R} as radius (it exists because Ω {\displaystyle \Omega } is an open set).

Let E B R {\displaystyle \mathbf {E} _{B_{R}}} and E C {\displaystyle \mathbf {E} _{C}} be the electric field created inside and outside the sphere respectively. Then,

Φ (R) = ∮ ∂ B R (r 0) E 0 ⋅ d S = ∮ ∂ B R (r 0) E B R ⋅ d S + ∮ ∂ B R (r 0) E C ⋅ d S = ∮ ∂ B R (r 0) E B R ⋅ d S {\displaystyle \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} }

The last equality follows by observing that (Ω ∖ B R (r 0)) ∩ B R (r 0) = ∅ {\displaystyle (\Omega \setminus B_{R}(\mathbf {r} _{0}))\cap B_{R}(\mathbf {r} _{0})=\emptyset }, and the argument above.

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

Φ (R) = Q (R) ε 0 = 1 ε 0 ∫ B R (r 0) ρ (r ′) d r ′ = 1 ε 0 ρ (r c ′) | B R (r 0) | {\displaystyle \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})|} with r c ′ ∈ B R (r 0) {\displaystyle r'_{c}\in \ B_{R}(\mathbf {r} _{0})}

Where the last equality follows by the mean value theorem for integrals. Using the squeeze theorem and the continuity of ρ {\displaystyle \rho }, one arrives at:

∇ ⋅ E 0 (r 0) = lim R → 0 1 | B R (r 0) | Φ (R) = 1 ε 0 ρ (r 0) {\displaystyle \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})}