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As we know, the Coulomb potential is in the form \begin{equation} V(\mathbf{r}) = Q/\mathbf{r} \end{equation} Mathematically, as $\mathbf{r} \rightarrow 0$, $V(\mathbf{r}) \rightarrow \infty$. But is this [as $\mathbf{r} \rightarrow 0$, $V(\mathbf{r}) \rightarrow \infty$] physically true? I can’t imagine at an extremely short distance, say, on a sub-nuclear scale, we would encounter an infinite electric potential (indeed, quarks carry electric charges and produce electromagnetic fields). If this is not the case, how should we modify the Coulomb potential and describe electromagnetic interactions at extremely short distances? (Sure, I know that there are other interactions such as strong and weak interactions at this scale, but they are not what I am concerned here.)

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The Coulomb potential stays the same at shorter distances but what happens is that the value of the electron charge gets renormalized. The standard story behind this is that at normal energies, there is an "electron cloud" which screens the "true" charge of the electron, which leads to an observable effective value that we are familiar with from every day life. The more technical story is that you need to compute vacuum polarization diagrams, which contribute to the potential.

It turns out though that the perturbative expansion in QED is not always valid, since the value of the charge seems to grow at shorter length scales, so beyond a certain energy scale (or equivalently under a certain length scale), the theory breaks down. This signals the need for a more fundamental theory that can explain what exactly happens at these short scales.

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The Coulomb potential formula with fixed values of electric charges isn't verified to be valid "all the way down" and it seems unlikely, because at some small distance electric potential energy of two point charges becomes so large a negative number, it was never observed. For example, if a negative electron got too close to a positive electron (positron), the formula would predict very large negative energy of bound state. But the largest energy we know we can extract is $2m_ec^2$ where the two particles disappear and energy turns into radiation. This excludes validity of the formula for distances smaller than $\frac{Ke^2}{2m_ec^2}\approx $ 1e-15 m (around the size of proton).

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  • $\begingroup$ Please don't mix engineering notation and $\LaTeX$/MathJax: it looks ugly and confusing, especially when followed by a unit symbol. The proper way to format such a quantity is $10^{-15}\,\mathrm{m}$, or, even better, $1\,\mathrm{fm}$ (use the right-click menu of these examples to see the code). $\endgroup$
    – Ruslan
    Jun 8 '20 at 15:58
  • $\begingroup$ That's a matter of taste, I prefer 1e-15 m because it is much less typing and does not require typesetting engine to be readable. Although you're right inside dollars it renders ugly - I'll try to fix it. $\endgroup$ Jun 8 '20 at 21:50
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$V(r) = Q/r$ is the Coulomb potential for a pointlike charge. Yes, it is valid for any distance $r$ and it goes to infinity as the distance from the charge goes to zero. If this is physically true depends on what you mean with that, which is not trivial.

When you consider subatomic particles, however, you have to take into account the nuclear forces, and in particular the strong nuclear force, which would prevail.

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  • $\begingroup$ If electric potential really (physically) goes to infinity at very short distances (say, on a subatomic or even nuclear scale), then electromagnetic forces, rather than strong nuclear forces, would prevail. Conversely, if the strong nuclear forces prevail, then the electric potential can't be infinite, and this implies the Coulumb potential needs to be modified, doesn't it? $\endgroup$
    – Shen
    Jun 8 '20 at 13:10
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Coulomb potential is accurate even at small distances. In fact it is using this very form that the Schrödinger equation for a hydrogen atom is solved to surprising accuracy. This is enough to model the electron in the atom to good accuracy.

Further corrections and for nuclear behaviour we need to incorporate nuclear forces.

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  • $\begingroup$ Nuclear forces are an unnecessary complication. Just consider positronium, where both electron and positron are point-like. The relevant subtlety is rather the radiative corrections calculated in QED. $\endgroup$
    – Ruslan
    Jun 8 '20 at 11:41
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For electrons no derivation from 1/r behaviour has been found and the electron radius is below the detection limit. Protons experimentally have a radius of about 0.8 fm below which the 1/r potential no longer holds.

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  • $\begingroup$ This is the point: At what distance does the 1/r potential no longer hold? and how do you know it? Further, below this distance, how should we describe electric potential and electromagnetic interactions? $\endgroup$
    – Shen
    Jun 8 '20 at 15:05

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