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What is the smallest and biggest distance in which Coulomb's law is valid?

Please provide a reference to a scientific journal or book. Just saying that this law is valid from this range to that range is not enough. Which experiments did result to these ranges?

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How do you know there's a limit on the distance of interaction? –  jhobbie Jul 17 at 19:28
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I don't know. That's why I am asking. However, now I am thinking that if our experimental accuracy at detecting coulombs interaction is limited, at least for large distances, by technology or in principle, then why shouldn't it be finite while we cannot detect the effect at arbitrary large distnces. –  yashar Jul 17 at 19:34
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I believe it is valid in all cases. But it might just not be as useful in some since it would be near zero or dwarfed by more dominant forces. –  jkeuhlen Jul 17 at 19:35
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Or does "valid" mean "experimentally measurable"? –  jkeuhlen Jul 17 at 19:36
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possible duplicate of What are the limits of applicability of Coulomb's Law? –  alemi Jul 18 at 6:35

6 Answers 6

Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by:

$$V(r) = \frac{q u(r)}{r}$$

where

$$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du \exp(-2mru)\left(1+\frac{1}{2u^2}\right)\frac{\sqrt{u^2-1}}{u^2}$$

where m is the electron mass and we work in natural units (so m is also the inverse Compton wavelength).

For distances much smaller than $\frac{1}{m}$, we have the asymptotic expansion:

$$u(r) = 1-\frac{\alpha}{3\pi}\left[2\log(m r)+2\gamma +\frac{5}{3}+\cdots\right]$$

For distances much larger than $\frac{1}{m}$, we have:

$$u(r)=1+\frac{\alpha}{4\sqrt{\pi (mr)^3}}\exp(-2m r)+\cdots$$

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I am not entirely familiar with this expansion, but I am quite sure you cannot just put there an electron mass there for a general particle deflected or bound by the potential. If anything, it should be the deflected/bound particle mass. And second, I think this might be computed either just for two electrons (e.g. identical fermions) or particles deflected/bound by a "very heavy" particle. In any case, for say two distinguishable charged particles of comparable but different mass, the effective potential would surely have different corrections. –  Void Jul 18 at 0:00
    
Simply put, this is no general correction to Coulomb's law and should not be considered a valid answer. –  Void Jul 18 at 14:01

The validity of Coulomb's Law over large distances is equivalent to bounding the mass of the photon. In quantum field theory, where one derives Coulomb's law, if the photon had a mass $m$, then the Coulomb potential gets replaced by the Yukawa potential (in natural units where $\hbar=c=1$ and Gaussian units): $$ \frac{e^{-mr}}{4\pi r}\ , $$ where $m$ is the mass of the photon. There are experimental bounds for the mass which can be viewed as an upper bound on the validity of Coulomb's law. Here is a fairly recent paper on this topic titled Upper bounds on the photon mass.

As @count-iblis (correctly) points out, you can think of the Compton wavelength of an electron (about $2\times 10^{-12}$m) as a lower bound for the validity of Coulomb's law. However, quantum field theory predicts the nature of the corrections. Given that it is the only way to "derive" Coulomb's law, I would say that it is a matter of taste as to whether Coloumb's law is actually breaking down at this length scale.

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As for large distances - it is hard to tell whether Coulomb's law applies with any correction or not. A main restriction on precision tests of Coulombs law at large distances is basically the inverse square distance fall-off of the physical effects. If we take a too small charge, it's strength falls of very quickly beyond measurability.

On the other hand, if we take a too large one, say such that we can safely detect it from a kilometer afar, we have to consider that the field intensity is a million times ($1/(10^{-3})^2 = 10^6$) stronger from a distance of one meter. Such large charges attract the ions of opposite charge e.g. of the surrounding slightly ionized air and on the other hand repulse the same-charge. Even if you are in a vacuum, your charge actually has to be somehow fixed in place by say a handle, and this handle will again provide a mechanism for charge neutralization.

This automatic charge neutralization is also the reason we do not find large macroscopic charges in nature (making it "quasi-neutral" on scales larger than say a few meters). Thus, we cannot make any conclusions from astrophysical observations since there electric fields play only a very minor role.

But if we take a look at electromagnetism as a whole, so far we have no better working theory on large scales. Radiation and magnetic fields aren't constrained by any similar cancelling effect as the electric field and Maxwell equations are tested to immense precision on terrestrial scales. This is an indirect argument showing that Coulomb's law should not get any corrections at any large length scale just for the sake of consistency of Maxwell's equations. On the other hand, on cosmic scales, we have so "flattened" data of the systems that we can consider Maxwell's equations (and thus Coulomb's law) as a "model" with which we can very slightly tamper as is done e.g. here.

On the contrary, on small distances, it is just a question of where do we draw the line. We could say Coulomb's law stops applying already at the scales of ångströms, that is atomic scales, where electrons just get frozen out instead of falling into the proton and we get the quantum energy levels. There is a natural extension of Coulomb's law through the potential acting on the quantum particle or the force acting on the mean velocity of a quantum state, so we can in principle go deeper.

When we get beyond quantum mechanics and pass to quantum field theory, we find that the term "potential" or "force" doesn't really have a meaning anymore. We get corrections to the energy levels in atoms which can be in principle assigned to a different potential as described by Count Iblis, but this is a conventional step. An example of such a convention would be the Lamb shift, which is either "classically explained" by an extra Dirac delta infinity in the Coulomb potential or by a fluctuation of the electron position (as explained in the link). Nevertheless, the effects are purely of a quantum nature and it is always just a formal handle to retrofit the quantum results to a "corrected" potential or force.

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From Classical Electrodynamics by JD Jackson Chapter 1 Section I.2

The inverse square law is known to hold over at least 25 order of magnitude in length!

Earlier:

The laboratory and geophysical tests show that on length scales of order $10^{-2}$ to $10^7$ m, the inverse square law holds with extreme precision. At smaller distances we must turn to less direct evidence often involving additional assumptions. For example, Rutherford's historical analysis of the scattering of alpha particles by thin foils substantiates the Coulomb law of force down to distances of the order of $10^{-13}$ m, provided the alpha particle and the nucleus can be treated as classical point charges interacting statically and the charge cloud of the electrons can be ignored. All of these assumptions can be, and have been, tested, of course, but only within the framework of the validity of quantum mechanics, linear superposition (see below), and other (very reasonable) assumptions. At still smaller distances, relativistic quantum mechanics is necessary [...] quantum electrodynamics (the relativistic theory of point electrons interacting with massless photons) holds to distances of the order of $10^{-18}$ m.

The earlier parts of the section give an overview of some of the historically important experiments and bounds.

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Well... you don't really measure electric/magnetic forces at distances much larger than several meters, but that's because electric potentials are difficult to build up. I guess on the small end, it's a little more difficult, but the strong force is essentially the only force that matters inside of nuclei.

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One does not always do direct tests for the validity of Coulomb's law which anyway is a consequence of Maxwell's equations. Indirect tests are the way to go. –  suresh Jul 18 at 0:11

Coulomb's Law describes the force between two electrically charged particles.

$$|F|=k_e{|q_1q_2|\over r^2}\qquad $$

This equation is valid for ANY distance and the force goes to zero at infinity. This means that theoretically, the Coulomb force exists between all charged particles.

Note that two conditions must be satisfied for this equation to hold true.

  1. The particles must be point charges
  2. The particles must be stationary with respect to one another.

Because this force obeys the inverse square law, like jhobbie mentioned, the force really isn't applicable at distances greater than a few meters.

At smaller distances, as the wikipedia page notes

Coulomb's law holds even within atoms, correctly describing the force between the positively charged atomic nucleus and each of the negatively charged electrons. This simple law also correctly accounts for the forces that bind atoms together to form molecules and for the forces that bind atoms and molecules together to form solids and liquids. Generally, as the distance between ions increases, the energy of attraction approaches zero and ionic bonding is less favorable. As the magnitude of opposing charges increases, energy increases and ionic bonding is more favorable.

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Classically, you are right. Quantumly, Count Iblis is on a better track. Coulomb's law is, from a QFT perspective, a 1-loop non-relativistic limit of $e^- + e^- \rightarrow e^- + e^-$ scattering, and thus not exact at all scales. –  ACuriousMind Jul 17 at 22:18
    
@ACuriousMind Just out of curiosity... was my answer downvoted simply because it was off on the quantum aspect? Still new and trying to determine best way to answer questions. –  jkeuhlen Jul 18 at 13:50
    
I've not downvoted it, since it is classically correct, so I cannot speak for the actual reason, but I cannot see any other reason, so that's probably it. –  ACuriousMind Jul 18 at 13:56
    
Okay thanks for the input. –  jkeuhlen Jul 18 at 13:59

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