Does the Coulomb's law include more information than the Gauss's Law in electrostatics?

Question

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).

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.)