What is the proof for Gauss's law? Can you give me some explanation and details with the proof?

  • 6
    $\begingroup$ It should be noted that Gauss's law is one of the four Maxwell's equations which serve as axioms of Electromagnetism. Hence it's an experimental fact; There's no mathematical proof for it. However you can show that Gauss's law is logically equivalent to Coulomb's law in the electrostatic regime; You can derive Gauss's from Coulomb's and vice versa. $\endgroup$
    – Omar Nagib
    Mar 9, 2016 at 19:15
  • 6
    $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/38404/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 9, 2016 at 19:50

1 Answer 1


Gauss's law is the electrostatic equivalent of the divergence theorem. Charges are sources and sinks for electrostatic fields, so they are represented by the divergence of the field: $\nabla \cdot E = \frac{\rho}{\epsilon_0}$, where $\rho$ is charge density (this is the differential form of Gauss's law). You can derive this from Coulomb's law. For a charge density $\rho(\vec{r}')$,

$$E(\vec r) = \frac{1}{4\pi\epsilon_0} \int_\mathbb{R} \frac{\rho(\vec r') \widehat{(\vec r - \vec r')}}{\left|\vec r - \vec r'\right|^2}dV = \frac{1}{4\pi\epsilon_0} \int_\mathbb{R} \frac{\rho(\vec r') (\vec r - \vec r')}{\left|\vec r - \vec r'\right|^3}dV.$$

(This converges to $E=\frac{q}{4\pi\epsilon_0 r^2}\hat r$ for a point charge.) Taking the divergence of this,

$$\nabla \cdot E = \frac{4\pi}{4\pi \epsilon_0}\int_\mathbb{R}\rho(\vec r')\delta(\vec r - \vec r')dV = \frac{\rho(\vec r')}{\epsilon_0}.$$

Then to arrive at the integral form presented in most introductory EM textbooks, the divergence theorem gives:

$$\int_{\mathcal{V}} \left(\nabla \cdot \vec F \right) dV = \oint_{\partial\mathcal{V}} \vec F\cdot \vec{dS},$$

so in the case of electrostatics $\int_\mathcal{V} \rho(\vec r)$ is simply all of the charge enclosed in $\mathcal{V}$ so,

$$\int_{\mathcal{V}} \left(\nabla \cdot \vec E \right) dV = \int_{\mathcal{V}} \frac{\rho(\vec r)}{\epsilon_0} dV = \frac{Q_{\text{enclosed}}}{\epsilon_0} = \oint_{\partial\mathcal{V}} \vec E\cdot \vec{dA}.$$

  • 4
    $\begingroup$ You did not derive Gauss's here. What you did is that you derived the integral form form the differential one. This was not the original question. $\endgroup$
    – Omar Nagib
    Mar 9, 2016 at 19:25
  • $\begingroup$ The equivalence between Guass' Law and the divergence theorem is contingent on the electric field (a physical thing) being a vector field (mathematical thing) and on charge being a source for that vector field (a physical fact). So while one can prove a statement like "if the electric field is correctly modeled as a vector field and electric charges represent sources in that field, then the total flux through a closed surface is proportional to the enclosed charge" you still haven't proven Gauss' Law, because Gauss's Law claims that the precondition hold and you can only check that in the lab. $\endgroup$ Mar 9, 2016 at 19:29
  • $\begingroup$ Both of these comments are correct. I had originally linked to a brief proof deriving the differential form of Gauss's law, which is a bit more intuitively obvious, from Coulomb's law, but I've included that proof in the answer now. $\endgroup$ Mar 9, 2016 at 19:44
  • $\begingroup$ How did you performed the divergence inside the integral $\endgroup$
    – user471651
    Dec 7, 2017 at 14:39
  • $\begingroup$ It's worth noting that Gauss's law is more general than Coulomb's law. Gauss's law holds even if the charges enclosed by the surface are moving, so their fields, though radial and inverse square, are not spherically symmetrical. $\endgroup$ Mar 16, 2019 at 16:58

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