Our professor told us that the Coulomb's law $F=k_e\frac{Qq}{r^2}$ includes more information than the Gauss's Law $\int\mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{1}{\varepsilon_0}\sum q$ (sorry that the editor seems don't support the symbol \$\oiint\$) in electrostatics because the Coulomb's law implies the Coulomb force is the conservative force while the Gauss's Law doesn't. However I think the two laws are equivalent in maths, and they should have the same meaning in physics (only in electrostatics).

  • $\begingroup$ Why do you think the two are mathematically equivalent? $\endgroup$ Aug 16, 2022 at 5:28

5 Answers 5


Technically he's right:

A force field that has zero curl is necessarily conservative. You could come up with a vector field that obey's Gauss' law but also has a non-zero curl, because the curl is basically ignored when taking the surface integral.

The laws aren't quite equivalent in math - Coulomb's law necessarily obey's Gauss' law but Gauss' law doesn't necessarily follow Coulomb's law (in reality it does, but in isolation the law doesn't require it).

  • $\begingroup$ Got it! Thank you! $\endgroup$
    – dcmpsr
    Aug 16, 2022 at 4:19
  • $\begingroup$ Does Gauss' law and Faraday's law without the presence of a changing magnetic field contain the same information as Coulomb's law? $\endgroup$
    – Habouz
    Aug 18, 2022 at 19:22
  • $\begingroup$ @Habouz almost.. Faraday's law says that the curl of E is will get a contribution from the changing B field, but it doesn't explicitly say the curl of E is only the changing B field. Basically all you need is something that tells you the curl of E is 0 for them to be equivalent. $\endgroup$
    – Señor O
    Aug 19, 2022 at 3:53

I would like to add a complement to @SeñorO's answer. The non-equivalence between Coulomb's law and Gauss' law may sound surprising if one refers to the argument frequently used to show how Coulomb's law can be obtained from Gauss' law ( see this other question and the excellent @EmilioPisanty's answer ). Actually, in that "proof", the symmetry argument implicitly excludes non-zero curl solutions. However, it is a theorem (Helmholtz's theorem) that a vector field is uniquely determined by its divergence and curl (plus a few additional conditions about its possible discontinuities and the way it vanishes at infinity).

Edit after answering another related question.

Helmholtz's theorem is not a purely mathematical gadget. It can be restated as saying that, in general, the sources of a vector field ${\bf E}$ are a scalar field equal to $\nabla \cdot {\bf E}$ and a vector field equal to $\nabla \times {\bf E}$. While Coulomb's Law contains the information that $\nabla \times {\bf E} = 0$ and then that there is no vector source of the electrostatic field, Gauss' Law alone is not enough to exclude such a physical possibility.

  • $\begingroup$ "Uniquely determined by its divergence and curl", assuming the field vanishes at infinity. This is important when considering the homogenous wave equation solution in addition to the specific solution obtained by jefimenkos equations. $\endgroup$ Aug 16, 2022 at 5:55
  • $\begingroup$ @jensenpaull Of course, there is the assumption on how the field vanishes at infinity and its possible discontinuities. I did not want to add too many details. However, you are making a good point. I'll modify my answer to add something in this direction. $\endgroup$ Aug 16, 2022 at 6:02

Adding onto others. The derivation from gauss law to coulombs law usually requires you to assume that:

$$\vec{E} = E(r) \hat r$$

This is just saying that the distribution is spherically symetric, and has zero curl [you do this instinctively when saying E and da are parrallel for a spherical gaussian surface]

Coulombs law is a specific case of gauss law, where the curl is zero and the distribution is spherically symettric.

  • $\begingroup$ In the second paragraph, I assume that you were meaning that the field (not the charge distribution) is spherically symmetric. Then, zero-curl is a consequence of the symmetry and not an additional property. $\endgroup$ Aug 16, 2022 at 6:09
  • $\begingroup$ Yes, I meant that a spherically symetric field implies zero curl and is a consequence of the field symmetry, and is not an additional prescribed condition. However I am also saying that a spherically symetric field also implies a spherically symmetric distribution. That is not to say that a spherically symmetric distribution must always have a spherically symetric field[as seen from the addition of a homogenous solution]. $\endgroup$ Aug 16, 2022 at 6:46

You don't need to introduce divergence and rotational to realize Gauss's law is not equivalent to Coulomb's law. The former is a scalar relationship (1 equation) while the later is vector relation (3 equations). It could be the three equations are not independent, but it is not the case and Coulomb's law contains more information than Gauss's law. As a mater of fact, Maxwell's equations are redundant, they somewhat repeat themselves. There is a formulation of electromagnetism in terms of normal variables that eliminates these redundancies. Gauss's law is not part of this framework.


Other answers have hinted at the following but not (IMO) made it really clear and explicit:

Coulomb's law can be derived, in full, from Gauss' law in electrostatics, with no other assumptions.

One can also derive the Gauss law from Coulomb's law. Hence the two are equivalent.

In order to derive Coulomb's law one sets up the question: what is the electric field of a point charge? By a point charge here we mean a charged entity of negligible physical size and which has no sense of direction (such as a spin). Therefore the situation has spherical symmetry about the location of the charge. It follows that the size of the electric field can only depend on radius, not angles in a spherical coordinate system centred on the charge. It also follows that the direction of the electric field must be radial. For, if it had a non-radial component then, by symmetry, that component would have to be independent of location throughout any given spherical shell. But this requirement is only met by a zero non-radial component. It is important to note that this fact is derived not assumed in this proof.

We have now established that $\bf E$ has the form ${\bf E} = E(r) \hat{\bf r}$. It remains to substitute this into Gauss' law, perform an integral over the volume of a sphere of radius $r$, convert one side to a surface integral, and hence obtain $E(r) = (4 \pi \epsilon_0 r^2)^{-1}$ in SI units.

I think the original statement from a professor, asserting that Coulomb's law contains information not contained in Gauss' law, is coming from a mathematical hunch that a statement about an integral (or about a divergence) will not in general describe a vector field as fully as a Green's function will. This is correct in general but the spherical symmetry makes the field of a point charge a special case, and it can be derived from the divergence with no further information. (One also also needs to assume the field is well-behaved, i.e. no infinite derivatives, away from the charge.)

  • 1
    $\begingroup$ You say "with no other assumptions", but then you make further assumptions (spherical symmetry of the solution). This assumptions boils down (in this particular case) to the requirement that the curl is zero. While Gauss law is still valid for a moving particle, coulombs law in general is not. I disagree with the assertion that both are equivalent. $\endgroup$ Aug 16, 2022 at 13:30
  • $\begingroup$ To use Gauss 'law you must calculate the electric flux on an orientable surface. Not all 3D surfaces are orientable and have two faces, and you would be hard pressed to calculate the electric field on a non orientable surface by using Gauss' law. $\endgroup$
    – Shaktyai
    Aug 16, 2022 at 14:27
  • $\begingroup$ If it were true that Coulomb's law can be derived from Gauss' law without any other assumption, there would be no room for the Heaviside–Feynman formula en.wikipedia.org/wiki/Jefimenko%27s_equations . The symmetry assumption is an additional hypothesis. One effect of which is the elimination of the rotational component of the field. $\endgroup$ Aug 16, 2022 at 17:28
  • $\begingroup$ @GiorgioP I claim that a symmetric source (the particle) must give an effect (the field) with that same symmetry (here invar. under rotations). Are you saying otherwise? $\endgroup$ Aug 16, 2022 at 18:04
  • $\begingroup$ @Quantumwhisp the issue is whether the spherical sym of solution must hold, irrespective of the Maxwell Eqns. I claim that a symmetric source must give a field with that same symmetry. Are you saying otherwise? $\endgroup$ Aug 16, 2022 at 18:05

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