# Is charge point-like or a smear?

Coulomb gave the law for the force between two static charges while considering them to be points in space. But the differential form of Gauss' Law talks about charge densities, a thing possible only if charges are smeared out in space.

Even Feynman addresses to the problem in his lectures when he says that on solving for the electrostatic energy in the field of a point charge we get infinity as the limit.

So do we know now that whether charges are point-like or smeared out?

• – Qmechanic Aug 6 '19 at 18:24
• @Qmechanic, you edited the tags on the assumption that OP wants an answer from the POV of particle physics. I think it's more likely they want an answer about electromagnetics (i.e. how is the differential form of Gauss' Law justified if electrons are point particles). – The Photon Aug 6 '19 at 20:32
• @ThePhoton: To me the question reads as obviously being a question about particle physics, especially given the second paragraph. – user4552 Aug 6 '19 at 21:04
• @BenCrowell, the 2nd paragraph is probably asking about section 8-6 in Feynman Vol 2, available online here. This is part of a discussion of electrostatics, not particle physics. – The Photon Aug 6 '19 at 21:19
• Indeed the reference in my question pertains to the same section in Feynman Vol 2 – Gariman Singh Aug 7 '19 at 2:35

It's not a trivial matter to define this question in such a way that it has a definite answer, and you certainly can't get a good answer within classical physics.

Even Feynman addresses to the problem in his lectures when he says that on solving for the electrostatic energy in the field of a point charge we get infinity as the limit.

Yes, this is a nice way approaching the issue. Now consider that classical electromagnetism is inherently a relativistic theory, so $$E=mc^2$$ applies. For a particle with mass $$m$$, charge $$q$$, and radius $$r$$, we would expect that the inertia $$m$$ of the particle can't be greater than $$\sim E/c^2$$, where $$E$$ is the energy in the electric field. This results in $$r\gtrsim r_0=ke^2/mc^2$$, where $$r_0$$ is called the classical electron radius, although it doesn't just apply to electrons.

For an electron, $$r_0$$ is on the order of $$10^{-15}$$ meters. Particle physics experiments became good enough decades ago to search for internal structure in the electron at this scale, and it doesn't exist, in the sense that the electron cannot be a composite particle such as a proton at this scale. This would suggest that an electron is a point particle. However, classical electromagnetism becomes an inconsistent theory if you consider point particles with $$r\lesssim r_0$$.

You can try to get around this by modeling an electron as a rigid sphere or something, with some charge density, say a constant one. This was explored extensively ca. 1900, and it didn't work. When Einstein published the theory of special relativity, he clarified why this idea had been failing. It was failing because relativity doesn't allow rigid objects. (In such an object, the speed of sound would be infinite, but relativity doesn't allow signaling faster than $$c$$.)

What this proves is that if we want to describe the charge and electric field of an electron at scales below $$r_0$$, we need some other theory of nature than classical E&M. That theory is quantum mechanics. In nonrigorous language, quantum mechanics describes the scene at this scale in terms of rapid, random quantum fluctuations, with particle-antiparticle pairs springing into existence and then reannihilating.

• “[...]we need some other theory of nature than classical E&M. That theory is quantum mechanics.” (Emphasis mine.) Would it not be more correct to say QED/QFT rather than QM? Especially given that virtual particles are mentioned in the next sentence. – kkm Aug 8 '19 at 6:45
• @kkm QFT is "a kind" of quantum mechanics. It does not modify the basic postulates at all. States are still vectors in a Hilbert space, observables are still hermitian operators on said space, the dynamics is still governed by a Hamiltonian. It just so happens that you have infinitely many degrees of freedom giving rise to a few complications - like the non-aplicability of the Stone Von-Neumann theorem giving rise to inequivalent Hilbert space constructions, Haag's theorem, and so on. Even so, it is still QM. – user1620696 Aug 8 '19 at 11:36

But the differential form of Gauss' Law talks about charge densities, a thing possible only if charges are smeared out in space.

Actually differential Gauss' law is valid even for point charges. For point charge $$q$$ at point $$\mathbf x_0$$, instead of charge density, we use charge distribution $$\rho(\mathbf x) = q\delta(\mathbf x-\mathbf x_0)$$.

Even Feynman addresses to the problem in his lectures when he says that on solving for the electrostatic energy in the field of a point charge we get infinity as the limit.

That problem is a separate question. There are consistent theories both for point and extended charges with finite energy in both cases. Neither kind of theory can supply us with hint on whether real particles are points or extended bodies. This must be investigated by experiments.

So do we know now that whether charges are point-like or smeared out?

For electrons, we don't know; all experiments are consistent with point particle, but it can be extended body of small enough size. The current decades old limit on electron size is somewhere near 1e-18 m.

For protons, based on scattering experiments and their understanding in terms of quantum theory of scattering, these are believed to have non-zero size (of charge distribution) around 1e-15 m.

• Lalinsky you said “There are consistent theories both for point and extended charges with finite energy in both cases” could you provide a link to descriptions of such theories? – Dale Aug 7 '19 at 11:09
• Since protons contain three quarks, the question is whether the quarks themselves are point charges. The quarks as a group would have non-zero size. – Monty Harder Aug 7 '19 at 18:44
• – Ján Lalinský Aug 8 '19 at 19:07

It depends on scale.

Electrons can usually be viewed as point-like when viewed on a scale much larger than an individual atom.

But semiconductors often have on the order of $$10^{12}- 10^{23}$$ free electrons per $${\rm cm^{3}}$$, depending on temperature and doping. Copper, as an example of a metal, has about $$10^{23}$$ free electrons per $$\rm cm^3$$.

In these materials, if the volume you're considering is even a few $$\rm \mu m^3$$, the error produced by assuming the charge is smeared out instead of localized in thousands or trillions of points is very small.

If you're studying some system with only a few charge carriers present, then you might need to consider the charge to be localized to make accurate predictions about it.

• But what is charge in reality? – Gariman Singh Aug 6 '19 at 18:23
• @GarimanSingh, charge is a property of certain subatomic particles that causes them to attract and repel each other, and to produce electromagnetic waves. It doesn't exist on its own without those particles. – The Photon Aug 6 '19 at 18:25
• For an EE answer, see What is a Charge?. For a more physics-y answer, search "what is charge" on this site. – The Photon Aug 6 '19 at 18:33
• Nevertheless, electrons are considered point particles hence point charges. – my2cts Aug 6 '19 at 19:23
• This answer seems a little off to me. It treats the question as if it were a question about condensed matter physics, whereas it seems clear to me that an appropriate answer is one involving the relevant particle physics. The OP is asking about properties of the electron itself, not anything specific to some form of matter such as semiconductors or copper. – user4552 Aug 6 '19 at 19:55

Electrons are considered point charges. Protons have a radius somewhat smaller than a femtometer. There is a controversy over the precise value of the proton radius. So a proton can be considered a smear, albeit a very tiny one.

• Viewed on the scale of an individual atom, or in an electron gas, they're not well modeled as point particles. – The Photon Aug 6 '19 at 20:41
• @ThePhoton The charge distribution given by the wave function certainly is not a point charge. However, the wave function is NOT the particle. – my2cts Aug 7 '19 at 12:25