Deriving Coulomb's Law from Gauss's Law

I've been thinking about this for the past couple of days. I apologize if my explanation isn't very clear.

I have already seen derivations of this, but I'm still not satisfied.

In the derivations of Coulomb's law from Gauss's law that I've seen, we take a spherical shell of radius $r$ around a point charge and calculate the electric flux through it.

$$\oint\limits_A \vec{E}\cdot d\vec{A} = E\cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \Rightarrow E = \frac{Q}{4\pi\epsilon_0 r^2}$$

What is assumed here, however, is that the electric field is perpendicular to the surface and has the same magnitude at every point on the spherical shell. Gauss's law does not state that explicitly, though, but Coulomb's law explicitly gives us the magnitude and direction of the force between two charges (and thus the direction of the electric field of a single charge).

Am I correct in thinking that, in addition to Gauss's law, we also need to state (as another law) that the electric field of a point charge points radially outwards (or inwards), and that its magnitude only depends on the distance from the point charge?

Another way to formulate my question would be: is this "other law" somehow hidden in Gauss's law?

• I think your broad question is: how to know that a contained charge will create a particular E. field. Dec 20, 2016 at 0:22
• Not really a law, but more of a principle: your derivation implicitly assumes that the field is radially symmetric because the charge density in question is. Similarly, the field would have an asymuthal symmetry if you were to calculate the field of an infinite charged wire. There is no a priori reason for the field to not have the same symmetry as the configuration itself. I agree that this is in a way cheating, but that's the luxury we as physicists have over mathematicians. Also, since you got a solution, it must be unique, and that's all we need. Dec 20, 2016 at 0:32
• Yes, it makes 100% sense to assume that the field must be spherically symmetric for a point charge. My question was whether this assumption must be explicitly added to Gauss's law or whether it is somehow a consequence of Gauss's law itself. Dec 20, 2016 at 0:49
• For the opposite question: physics.stackexchange.com/q/38404/2451 May 18, 2020 at 11:30

You are correct that Gauss's law alone cannot be used to derive Coulomb's law. Instead, you need to supplement it with the hypothesis that space is isotropic, but nothing more.

The thing is really in the language. You start with Gauss's law, and then you say "consider a point charge...", without actually saying much about what you mean by that. In particular, it's not enough to say 'it's this thing that has charge but zero volume', because point electric dipoles also take up zero volume and they are very different beasts than point charges (and indeed you could combine the two to make a non-point-charge object that "has charge but zero volume").

What you mean by a point charge, then, is a bit stronger, and in particular you mean that it is a spherically symmetric object: it is meaningless to speak about "rotating" a point charge.

In addition to this, you require the assumption that electromagnetism itself is also rotationally symmetric: that if you rotate a set of charges, then you'll get a rotated set of fields. This is not included in Gauss's law, but it is a plenty reasonable assumption to add.

If you have both of those things, then the result follows:

• The electric field produced at $\mathbf r$ by a point charge at $\mathbf r_0$ must point along $\mathbf r-\mathbf r_0$, because if you rotate the world around that axis, the charges don't change, so therefore the fields cannot change.
• The electric field produced at $\mathbf r_1$ and $\mathbf r_2$ produced by a point charge at $\mathbf r_0$, where both points are at equal distances $|\mathbf r_1-\mathbf r_0|=|\mathbf r_2 - \mathbf r_0|$ from the point charge, must have equal magnitude, because both points are related by a rotation about $\mathbf r_0$ which preserves the charge configuration.

However, it is important to remember that it isn't Gauss's law alone that gets you there - you do need the isotropy assumption.