Gauss's Law from Coulomb's Law

Derivation

Gauss's Law from Coulomb's Law

D-284 Home PU-305 Threads fields · force Depends on Laplacian of 1/r as a Delta Function, divergence-theorem

Statement

Starting from the Coulomb electric field of a static charge distribution, \( \vec{E}(\vec{r}) = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \rho(\vec{r}\,')\,\dfrac{\vec{r}-\vec{r}\,'}{|\vec{r}-\vec{r}\,'|^{3}}\,d^{3}r' \), we derive the differential form of Gauss's law, \( \nabla\cdot\vec{E} = \rho/\varepsilon_0 \), valid at every point of space, using the divergence theorem and the distributional identity \( \nabla\cdot\!\left(\hat{r}/r^{2}\right) = 4\pi\,\delta^{3}(\vec{r}) \).

Why it matters

Gauss's law in differential form is one of the four Maxwell equations. Deriving it from Coulomb's law shows that the inverse-square law is not an independent postulate: the \( 1/r^{2} \) falloff is exactly what makes the divergence of \( \vec{E} \) collapse onto the local charge density and nothing else. Any other power law would leave a spurious source spread through empty space.

The result is also the practical bridge between the two ways physicists compute fields. The integral (Coulomb) form is a global recipe; the differential (Gauss) form is a local partial differential equation. Together with \( \nabla\times\vec{E}=0 \) for statics, it determines \( \vec{E} \) from \( \rho \) up to boundary conditions, which is the foundation of electrostatic boundary-value problems and of the potential formulation \( \nabla^{2}\phi = -\rho/\varepsilon_0 \).

Assumptions

Electrostatics (time-independent sources).If \( \rho \) and \( \vec{E} \) vary in time, the Coulomb integral is no longer the exact field; retardation and the induced magnetic term (\( \partial\vec{B}/\partial t \)) enter, and one must return to the full Maxwell equations. The divergence relation \( \nabla\cdot\vec{E}=\rho/\varepsilon_0 \) survives, but it can no longer be derived from the static Coulomb integral alone.

Linear superposition of point-charge fields.If fields did not add linearly (nonlinear vacuum electrodynamics), the integral representation of \( \vec{E} \) as a sum over source elements would be invalid and the derivation would not even start.

Vacuum (or a medium absorbed into \( \varepsilon_0 \)).In a polarizable medium the bound charge contributes to \( \rho \); dropping this, one gets \( \nabla\cdot\vec{D}=\rho_{\text{free}} \) instead, with \( \vec{D}=\varepsilon\vec{E} \). The vacuum constant \( \varepsilon_0 \) must then be replaced by the permittivity.

\( \rho \) is a locally integrable, sufficiently smooth density.The delta-function identity is a statement about distributions acting on test functions. If \( \rho \) is itself singular (e.g. a genuine point charge, a surface layer, or a dipole layer) the differential form must be read distributionally, and naive pointwise evaluation of \( \nabla\cdot\vec{E} \) fails at the singular support.

Sources are confined so the field falls off at infinity.Needed so surface terms in the divergence theorem vanish and the interchange of \( \nabla \) with the source integral is justified. For an infinite charged medium filling all space with nonzero mean density the Coulomb integral diverges and must be regularized.

Derivation

\[ \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^3} \rho(\vec{r}\,')\,\frac{\vec{r}-\vec{r}\,'}{|\vec{r}-\vec{r}\,'|^{3}}\;d^{3}r' \]

Coulomb's law for a point charge plus linear superposition: each element \( \rho\,d^{3}r' \) sources a \( 1/r^{2} \) radial field, and the total field is their integral. A

\[ \frac{\vec{r}-\vec{r}\,'}{|\vec{r}-\vec{r}\,'|^{3}} = -\,\nabla_{\vec r}\!\left(\frac{1}{|\vec{r}-\vec{r}\,'|}\right) \]

The Coulomb kernel is minus the gradient of the Newtonian potential; direct differentiation of \( 1/|\vec r-\vec r\,'| \) with respect to the field point \( \vec r \) gives exactly this. A

\[ \nabla\cdot\vec{E} = \frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^3}\rho(\vec{r}\,')\;\nabla_{\vec r}\cdot\!\left(\frac{\vec{r}-\vec{r}\,'}{|\vec{r}-\vec{r}\,'|^{3}}\right)d^{3}r' \]

Take the divergence with respect to \( \vec r \). The operator passes through the integral because it acts on \( \vec r \) while the integration variable is \( \vec r\,' \); the interchange is legitimate on test functions since \( \rho \) is confined. B

\[ \nabla_{\vec r}\cdot\!\left(\frac{\vec{r}-\vec{r}\,'}{|\vec{r}-\vec{r}\,'|^{3}}\right) = 4\pi\,\delta^{3}(\vec{r}-\vec{r}\,') \]

Prior result laplacian-of-inverse-r-delta: \( \nabla^{2}(1/r)=-4\pi\delta^{3}(\vec r) \), i.e. \( \nabla\cdot(\hat r/r^{2})=4\pi\delta^{3}(\vec r) \). Naively the divergence is zero for \( \vec r\neq\vec r\,' \), but the flux of \( \hat r/r^{2} \) through any enclosing surface is \( 4\pi \) (divergence theorem, Step justification below), so all the divergence is concentrated at the origin as a delta. C

\[ \oint_{\partial V}\frac{\hat r}{r^{2}}\cdot d\vec a = \int_{0}^{2\pi}\!\!\int_{0}^{\pi}\frac{1}{r^{2}}\,r^{2}\sin\theta\,d\theta\,d\varphi = 4\pi = \int_{V}\nabla\cdot\!\left(\frac{\hat r}{r^{2}}\right)d^{3}r \]

Divergence theorem (prior result divergence-theorem) applied to a ball about \( \vec r\,' \): the surface flux is \( 4\pi \) independent of radius, forcing the volume integrand to be a unit-weight delta at \( \vec r\,' \). This fixes the coefficient in Step 4. C

\[ \nabla\cdot\vec{E} = \frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^3}\rho(\vec{r}\,')\;4\pi\,\delta^{3}(\vec{r}-\vec{r}\,')\;d^{3}r' \]

Substitute the identity of Step 4 into the integrand of Step 3. A

\[ \nabla\cdot\vec{E} = \frac{4\pi}{4\pi\varepsilon_0}\,\rho(\vec r) = \frac{\rho(\vec r)}{\varepsilon_0} \]

The sifting property of the Dirac delta, \( \int f(\vec r\,')\,\delta^{3}(\vec r-\vec r\,')\,d^{3}r' = f(\vec r) \), collapses the integral onto the local density; the factors of \( 4\pi \) cancel. A

Result

\[ \boxed{\;\nabla\cdot\vec{E}(\vec r) = \frac{\rho(\vec r)}{\varepsilon_0}\;} \]

Reading. The divergence of the electric field at a point equals the local charge density there divided by \( \varepsilon_0 \). Field lines begin only where positive charge sits and end only where negative charge sits; in charge-free regions \( \nabla\cdot\vec E=0 \) and field lines are continuous. The \( 4\pi \) of Coulomb's law has been "rationalized" into \( \varepsilon_0 \).

Units check. In SI, \( [\rho]=\mathrm{C\,m^{-3}} \) and \( [\varepsilon_0]=\mathrm{C^{2}\,N^{-1}\,m^{-2}}=\mathrm{F\,m^{-1}} \), so the right side has units \( \mathrm{C\,m^{-3}}\big/(\mathrm{C^{2}\,N^{-1}\,m^{-2}}) = \mathrm{N\,C^{-1}\,m^{-1}} \). The left side is \( [E]/[\text{length}] = (\mathrm{N\,C^{-1}})/\mathrm{m} = \mathrm{N\,C^{-1}\,m^{-1}} = \mathrm{V\,m^{-2}} \). The two sides match.

Limiting cases

Charge-free region (\( \rho=0 \)): \( \nabla\cdot\vec E=0 \); \( \vec E \) is solenoidal, and with \( \vec E=-\nabla\phi \) the potential obeys Laplace's equation \( \nabla^{2}\phi=0 \).
Single point charge \( q \) at the origin: \( \rho=q\,\delta^{3}(\vec r) \) gives \( \nabla\cdot\vec E=(q/\varepsilon_0)\delta^{3}(\vec r) \), reproducing \( \vec E=q\,\hat r/(4\pi\varepsilon_0 r^{2}) \) with all divergence at the point charge.
Integral (Gauss) form: integrating over any volume \( V \) and applying the divergence theorem gives \( \oint_{\partial V}\vec E\cdot d\vec a = Q_{\text{enc}}/\varepsilon_0 \), the flux law.
Uniform ball of charge, inside: constant \( \rho \) gives constant \( \nabla\cdot\vec E \), consistent with the linearly rising interior field \( E(r)=\rho r/(3\varepsilon_0) \).

Breaks when

Time-dependent fields. With changing sources the electrostatic Coulomb integral is no longer the field (retardation), so this derivation breaks even though the Maxwell equation \( \nabla\cdot\vec E=\rho/\varepsilon_0 \) still holds. One must instead take it as a field equation, not a consequence of the static integral.
Non-inverse-square force law. If the field went as \( 1/r^{n} \) with \( n\neq 2 \), then \( \nabla\cdot(\hat r/r^{n})\neq 4\pi\delta^{3}(\vec r) \): for \( n\neq 2 \) the divergence is nonzero throughout empty space (\( \propto (2-n)/r^{\,n+1} \)), producing spurious volume sources and destroying the clean local relation. Gauss's law is equivalent to exact inverse-square (used in photon-mass bounds).
Polarizable/magnetizable media treated as vacuum. Inside a dielectric the bound charge \( \rho_b=-\nabla\cdot\vec P \) is real charge; omitting it makes \( \varepsilon_0 \) wrong. The correct free-charge statement is \( \nabla\cdot\vec D=\rho_{\text{free}} \).
Singular sources evaluated pointwise. At a genuine point, line, or surface charge the field diverges and \( \nabla\cdot\vec E \) is not an ordinary function; using it as a classical derivative at the singularity gives nonsense. The relation is exact only in the distributional sense.

Failure modes

"The divergence of \( \hat r/r^{2} \) is zero everywhere." True for \( r\neq 0 \), but this ignores the delta at the origin; the flux argument shows the missing \( 4\pi\delta^{3} \). Concluding \( \nabla\cdot\vec E=0 \) for a point charge is the classic error.
Differentiating the integrand with respect to \( \vec r\,' \). The divergence in \( \nabla\cdot\vec E \) is with respect to the field point \( \vec r \), not the source point \( \vec r\,' \). Mixing the two flips a sign or kills the delta.
Losing the \( 4\pi \). Forgetting the \( 4\pi \) in \( \nabla\cdot(\hat r/r^{2})=4\pi\delta^{3} \) leaves a stray \( 1/(4\pi) \) and gives \( \nabla\cdot\vec E=\rho/(4\pi\varepsilon_0) \), which is dimensionally consistent but numerically wrong.
Treating \( \varepsilon_0 \) and \( 1/(4\pi\varepsilon_0) \) interchangeably. Gaussian vs SI unit confusion; in Gaussian units the result is \( \nabla\cdot\vec E=4\pi\rho \), and swapping systems midway corrupts factors.
Using \( \vec D \) and \( \vec E \) forms together. Writing \( \nabla\cdot\vec E=\rho_{\text{free}}/\varepsilon_0 \) inside a dielectric double-counts or omits bound charge.

Discussion

The derivation makes precise a statement that is often asserted: Gauss's law and Coulomb's law are the same physics viewed globally versus locally. The only nontrivial ingredient is the delta-function identity, and that identity is itself a consequence of two facts — the field is radial with an inverse-square magnitude, and the surface area of a sphere grows as \( r^{2} \). The two \( r^{2} \) factors cancel in the flux integral, leaving a radius-independent \( 4\pi \). This geometric cancellation is why inverse-square laws in three dimensions produce clean local source equations; in \( d \) dimensions the "Coulomb" field goes as \( 1/r^{d-1} \) precisely so that \( \nabla\cdot\vec E \) still localizes.

Combined with the irrotationality of the electrostatic field, \( \nabla\times\vec E=0 \), the divergence relation determines \( \vec E \) completely (Helmholtz's theorem) from \( \rho \) and boundary data. Introducing the potential \( \vec E=-\nabla\phi \) converts it into Poisson's equation \( \nabla^{2}\phi=-\rho/\varepsilon_0 \), whose Green's function is exactly the \( 1/(4\pi|\vec r-\vec r\,'|) \) kernel we started from. The logic is therefore a closed loop: the Coulomb kernel is the Green's function of the very Laplacian whose action on it produces the delta.

Physically, the differential form is a statement of local charge accounting: field lines are conserved objects that can be created or destroyed only at charge. This is what allows the field to be visualized as flux and is the electrostatic ancestor of the continuity equation. It also underlies why the interior of a conductor in equilibrium has \( \vec E=0 \) and all net charge on the surface — a direct reading of \( \nabla\cdot\vec E=\rho/\varepsilon_0 \).

At the deepest level the derivation is a theorem about the Laplacian's fundamental solution on \( \mathbb{R}^3 \), and its robustness is what makes Gauss's law survive into regimes where Coulomb's static integral does not. In quantum electrodynamics Gauss's law reappears as a constraint (the generator of gauge transformations) that must be imposed on physical states, \( (\nabla\cdot\hat{\vec E}-\hat\rho)\lvert\psi\rangle=0 \); its classical content is exactly the relation derived here. Experimental tests of the inverse-square law — null searches for deviations, which would signal a nonzero photon mass via the Proca equation \( \nabla\cdot\vec E=\rho/\varepsilon_0-\mu^{2}\phi \) — are precisely tests of whether \( \nabla\cdot(\hat r/r^{2}) \) is a pure delta. Current bounds place the photon mass below about \( 10^{-18}\,\mathrm{eV}/c^{2} \).

Common misconceptions. Gauss's law is not "more general" than Coulomb's law in electrostatics — they are equivalent given the field's radial symmetry and irrotationality. Its greater generality appears only when the field is not static: Gauss's law remains a Maxwell equation while the Coulomb integral does not. Also, \( \nabla\cdot\vec E=0 \) in a region does not mean \( \vec E=0 \) there; it means no net charge is enclosed, e.g. a uniform field or the exterior of a neutral distribution.

Worked examples

Uniformly charged ball: recover the interior field from the local law.

A ball of radius \( R \) carries uniform density \( \rho_0 \). Use \( \nabla\cdot\vec E=\rho/\varepsilon_0 \) with spherical symmetry to find \( E(r) \) inside.

\[ \nabla\cdot\vec E=\frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(r^{2}E_r\right)=\frac{\rho_0}{\varepsilon_0}\quad(r<R) \]

Spherical divergence of a purely radial field \( \vec E=E_r(r)\hat r \). B

\[ \frac{\partial}{\partial r}\!\left(r^{2}E_r\right)=\frac{\rho_0}{\varepsilon_0}r^{2}\;\Rightarrow\; r^{2}E_r=\frac{\rho_0}{3\varepsilon_0}r^{3}+C \]

Integrate once in \( r \); regularity at the center forces \( C=0 \). A

\[ E_r(r)=\frac{\rho_0\,r}{3\varepsilon_0} \]

Solve for \( E_r \). Now insert numbers: take \( \rho_0=1.0\times10^{-6}\,\mathrm{C\,m^{-3}} \) and \( r=0.20\,\mathrm{m} \). A

\[ E_r=\frac{(1.0\times10^{-6})(0.20)}{3(8.854\times10^{-12})}\,\mathrm{V\,m^{-1}} \]

Substitute \( \varepsilon_0=8.854\times10^{-12}\,\mathrm{F\,m^{-1}} \), keeping units explicit. A

\[ E_r \approx 7.5\times10^{3}\ \mathrm{V\,m^{-1}} \ \ (\text{radially outward}) \]

Reading. The interior field rises linearly from the center; the local law with constant \( \rho \) forces constant divergence, which integrates to \( E\propto r \). This matches the Gauss-flux answer, confirming consistency.

Units check. \( \mathrm{(C\,m^{-3})(m)}/\mathrm{(F\,m^{-1})}=\mathrm{C\,m^{-2}}/\mathrm{(C\,V^{-1}\,m^{-1})}=\mathrm{V\,m^{-1}} \). Correct.

Divergence of a given field: read off the charge density.

Measurements fit \( \vec E(\vec r)=A\,r^{2}\,\hat r \) with \( A=3.0\times10^{4}\,\mathrm{V\,m^{-3}} \) in some region. Find \( \rho(\vec r) \) at \( r=0.50\,\mathrm{m} \).

\[ \rho=\varepsilon_0\,\nabla\cdot\vec E=\varepsilon_0\,\frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(r^{2}\cdot A r^{2}\right) \]

Rearrange the derived law for \( \rho \), then apply the spherical divergence. A

\[ =\varepsilon_0\,A\,\frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(r^{4}\right)=\varepsilon_0 A\,\frac{4r^{3}}{r^{2}}=4\varepsilon_0 A\,r \]

Differentiate \( r^{4} \) and simplify; symbolic result before numbers. A

\[ \rho=4(8.854\times10^{-12})(3.0\times10^{4})(0.50)\ \mathrm{C\,m^{-3}} \]

Insert \( \varepsilon_0 \), \( A \), and \( r=0.50\,\mathrm{m} \). A

\[ \rho \approx 5.3\times10^{-7}\ \mathrm{C\,m^{-3}} \]

Reading. A field growing as \( r^{2} \) requires a charge density growing linearly with \( r \); empty space (\( \rho=0 \)) would demand \( r^{2}E_r=\text{const} \), i.e. the inverse-square field, not this one.

Units check. \( \mathrm{(F\,m^{-1})(V\,m^{-3})(m)}=\mathrm{(C\,V^{-1}\,m^{-1})(V\,m^{-3})(m)}=\mathrm{C\,m^{-3}} \). Correct.

Problems

(A) Charge-free check. Show that \( \vec E=E_0\,\hat z \) (uniform) satisfies \( \nabla\cdot\vec E=0 \), and state the implied charge density.
Solution
In Cartesian components \( \vec E=(0,0,E_0) \) with \( E_0 \) constant, so \( \nabla\cdot\vec E=\partial_x 0+\partial_y 0+\partial_z E_0=0 \). By the derived law \( \rho=\varepsilon_0\nabla\cdot\vec E=0 \): a uniform field is sourced by no local charge (the charges sit on distant plates).
(A) Point charge as a delta source. Given \( \vec E=\dfrac{q}{4\pi\varepsilon_0}\dfrac{\hat r}{r^{2}} \), write \( \rho(\vec r) \).
Solution
\( \rho=\varepsilon_0\nabla\cdot\vec E=\dfrac{q}{4\pi}\nabla\cdot\!\left(\dfrac{\hat r}{r^{2}}\right)=\dfrac{q}{4\pi}\,4\pi\delta^{3}(\vec r)=q\,\delta^{3}(\vec r). \) All the charge is concentrated at the origin, as expected; \( \int\rho\,d^{3}r=q \).
(B) Shell. A spherical shell of radius \( R \) carries surface density \( \sigma \). Using the local law (distributionally) and symmetry, find \( E_r(r) \) for \( r<R \) and \( r>R \), and evaluate for \( \sigma=2.0\times10^{-6}\,\mathrm{C\,m^{-2}} \), \( R=0.10\,\mathrm{m} \), at \( r=0.20\,\mathrm{m} \).
Solution
For \( r\neq R \), \( \rho=0 \) so \( \tfrac{1}{r^{2}}\partial_r(r^{2}E_r)=0\Rightarrow r^{2}E_r=\text{const} \). Regularity inside gives \( E_r=0 \) for \( r<R \). Outside, \( r^{2}E_r=k \); the jump at \( r=R \) supplies total charge \( Q=4\pi R^{2}\sigma \), and matching to a point-like exterior field gives \( E_r=Q/(4\pi\varepsilon_0 r^{2})=\sigma R^{2}/(\varepsilon_0 r^{2}) \). Numerically \( Q=4\pi(0.10)^2(2.0\times10^{-6})=2.51\times10^{-7}\,\mathrm{C} \); at \( r=0.20\,\mathrm{m} \), \( E_r=\dfrac{2.51\times10^{-7}}{4\pi(8.854\times10^{-12})(0.20)^2}\approx 5.7\times10^{4}\,\mathrm{V\,m^{-1}} \).
(B) Non-inverse-square counterexample. Suppose a hypothetical field were \( \vec E=B\,\hat r/r \). Compute \( \nabla\cdot\vec E \) for \( r\neq 0 \) and comment on the implied \( \rho \).
Solution
\( \nabla\cdot\vec E=\tfrac{1}{r^{2}}\partial_r(r^{2}\cdot B/r)=\tfrac{1}{r^{2}}\partial_r(Br)=\tfrac{B}{r^{2}} \). This is nonzero everywhere \( r\neq0 \), so \( \rho=\varepsilon_0 B/r^{2}\neq0 \) throughout empty space. A \( 1/r \) field cannot be sourced by a localized charge alone — it demands charge spread through all space, illustrating why only the inverse-square field gives a pure delta at the source.
(C) Poisson consistency. A potential is measured as \( \phi(\vec r)=\alpha\,r^{2} \) (with \( r^{2}=x^2+y^2+z^2 \)). Find \( \vec E \), then \( \rho \) via the derived law, and confirm agreement with \( \nabla^{2}\phi=-\rho/\varepsilon_0 \). Take \( \alpha=100\,\mathrm{V\,m^{-2}} \).
Solution
\( \vec E=-\nabla\phi=-2\alpha\,\vec r \), i.e. \( E_r=-2\alpha r \). Then \( \nabla\cdot\vec E=\tfrac{1}{r^{2}}\partial_r(r^{2}(-2\alpha r))=\tfrac{1}{r^{2}}(-6\alpha r^{2})=-6\alpha \), so \( \rho=\varepsilon_0\nabla\cdot\vec E=-6\alpha\varepsilon_0 \). Cross-check: \( \nabla^{2}\phi=\tfrac{1}{r^{2}}\partial_r(r^{2}\cdot2\alpha r)=6\alpha \), and \( -\rho/\varepsilon_0=6\alpha \); the two agree. Numerically \( \rho=-6(100)(8.854\times10^{-12})=-5.3\times10^{-9}\,\mathrm{C\,m^{-3}} \), a uniform negative density.