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Atoms are described as having a nucleus at the center with electrons orbiting (or maybe the nucleus with a high probability of being in the center and the electrons more spread out).

If this is so, you would think that there is more negative charge concentrated at the 'perimeter' of the atom, and the negative force would dominate when interacting with something outside the atom, due to its proximity.

Thus, why do atoms have a neutral charge?

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  • $\begingroup$ "or maybe the nucleus with a high probability of being in the center" - I just love this! $\endgroup$
    – MEMark
    Commented Oct 4, 2021 at 11:13
  • $\begingroup$ See London force and van der Waals force. $\endgroup$ Commented Oct 4, 2021 at 16:33
  • $\begingroup$ physics.stackexchange.com/questions/556294/… $\endgroup$ Commented Oct 5, 2021 at 4:10
  • $\begingroup$ Even ignoring all the details about quantum physics, there is a simple geometrical mistake in your second paragraph. For any fixed point far enough outside the atom, roughly one half of the 'perimeter' is actually farther away than the nucleus and thus has less influence. With this in mind, it is not terribly surprising that the effects of electrons on the near and the far side tend to (mostly) balance out. $\endgroup$
    – mlk
    Commented Oct 5, 2021 at 11:33
  • $\begingroup$ @mlk That's a great point. I had realized that when thinking it over after posting. The "(mostly)" in your comment still remains to be quantified; and does the atom really waver between varying levels of imbalance depending on the position of all the electrons? $\endgroup$ Commented Oct 5, 2021 at 13:36

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Neutral charge just means there is no net charge when considering the entire atom/molecule. It doesn't mean there can't be a non-zero or non-symmetric electric field. This is especially true for molecules that are neutral yet still polar, such as water.

you would think that there is more negative charge concentrated at the 'perimeter' of the atom, and the negative force would dominate when interacting with something outside the atom, due to its proximity.

According to Gauss's law this isn't the case. A sphere of negative charge has the same field outside of it as a point charge at the center of the sphere.

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  • $\begingroup$ Do you mean that we can treat an orbiting electron as really smeared out into a sphere of homogenous charge, not as a charge sometimes here, sometimes there? $\endgroup$ Commented Oct 3, 2021 at 20:49
  • $\begingroup$ @ReinstateMonica In not really saying yes or no to that here. That model could be useful for some things but not for others. What I am saying is that if you assume this then Gauss's law tells us what I say in the answer. $\endgroup$ Commented Oct 4, 2021 at 0:24
  • $\begingroup$ @ReinstateMonica In reality, each interaction with the sensor will collapse the electron to an eigenstate with more discrete position, so we lose the ability to directly apply Guass's law. However, you can run the same argument from a large number of interactions with one electron, the position eigenvector of which is spherically symmetric, that you use to go from a spherically symmetric distribution of fixed charges to Gauss's law in the first place. [1/2] $\endgroup$
    – g s
    Commented Oct 4, 2021 at 5:09
  • $\begingroup$ [2/2] However, if you get very close then the pauli exclusion principle - combined with the quantized energy levels of an electron with respect to a nucleus - prevents electrons from the atom from finding themselves in your sensor's atoms, or, if you're looking at the atoms with an electron beam, makes your electron beam bounce off of the atom. I think the coulomb force interpretation of this is that, while a discrete spherical distribution of events approximates a continuous spherical distribution, it is still discrete, so a single event can still dominate if it's very, very close. $\endgroup$
    – g s
    Commented Oct 4, 2021 at 5:16
  • $\begingroup$ In [1/2] I should have said, "the position eigenvectors of which are distributed in a spherically symmetric manner" $\endgroup$
    – g s
    Commented Oct 4, 2021 at 5:18
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At long distance, $R \gg a_0$, the atom appears entirely neutral, which is good, because the energy density of Avogadro's number of electron charges is huge. A charged sphere has electrostatic energy:

$$ U = \frac 3 5 \frac{Q^2}{4\pi\epsilon_0 R} $$

With $Q=N_Ae$:

$$U= 5\times10^{19}\,{\rm J/m^{-1}}$$

That's 12,000 megatons of TNT for a 1 meter sphere and 1 gram of protons. And Earth sized sphere clocks in around 2 kT.

At closer range, the electron cloud is extremely important. Its ability to move from atom to atom drives most, if not all, of chemistry. Distortions (e.g., polarizability) are behind the behavior of dielectrics, indices of refraction, and non-linear optics. And then there is magnetism and molecular biology and protein folding and all that.

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TL; DR They don't have to - ions do exist, but relatively rare compared to the number of neutral atoms and molecules encountered in everyday life, and relatively short-living.

There are different ways to look at this problem:

Electron cloud
If we consider a single atom, then its electron has a probability distribution around a nucleus (one can always choose our system of reference to be centered at the nucleus). Since the electron cloud stretches infinitely far away from the nucleus, the atom appears approximately as neutral only when we look at a Gaussian surface of a very large radius (as compared to the atomic radius, i.e., the average thickness of the probability cloud).

Discreteness of charge
Another possible starting point is the charge quantization: since charge comes in units of charge equal to that of proton and electron, the non-neutral atom must have an excess of at least one proton or one electron. For an atom with an excess electron one can then calculate its stability as compared to that with no excess electrons. Of course, negatively and positively charged ions exist and often stable, but they easily lose or acquire the excess/missing electrons when interacting with other atoms. The parameters of the interaction are such that the neutral configuration is simply more stable.

Macroscopic charge neutrality
An object containing many atoms will attract excess charge till it becomes neutral, which is why collections of many ionized atoms are rarely observed. Also, a collection of ionized atoms would not be stable due to repulsive Coulomb interactions, which means that neutrality of charge is a condition of stability of macroscopic objects.

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    $\begingroup$ "ions do exist, but rarely encountered in everyday life and relatively short-living." Can you expand on this? A lot of biological events involve ions. $\endgroup$ Commented Oct 4, 2021 at 11:00
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    $\begingroup$ @BioPhysicist Sure, ions are everywhere, but in much smaller concentrations than neutral atoms/molecules. If you take $NaCl$ in whater, the number of $Na^+$ and $Cl^-$ aions is much smaller than the number of neutral water molecules. $\endgroup$
    – Roger V.
    Commented Oct 4, 2021 at 11:06
  • $\begingroup$ True, they are "small" concentrations, but they still matter. Cells work to maintain a potential difference across their membranes, and action potentials rely on movement of ions. I guess it depends on what "everyday life" means :) $\endgroup$ Commented Oct 4, 2021 at 11:12
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    $\begingroup$ @BioPhysicist it needs to be interpreted in the context of this question, which I interpret as: Why do most atoms have equal amounts of positive and negative charge? If one looks around, most of the matter around is not ionized plasma, so one could claim that the ions are "rare". Personally, I am not a fan of arguments about semantics or precise phrasing, as long as there is no risk to be misunderstood. $\endgroup$
    – Roger V.
    Commented Oct 4, 2021 at 11:21
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    $\begingroup$ @BioPhysicist I adjusted the phrasing. $\endgroup$
    – Roger V.
    Commented Oct 4, 2021 at 11:46
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If the question is why do we observe more neutral atoms than ions, then we can look to statistical mechanics and argue that it is energetically favorable. The boltzmann distribution tells us that the probability of higher energy states is lower than the probability of lower energy states$^*$. There is an energy associated to assembling charged objects,

$$U = \frac{\epsilon_0}{2} \int \mathbf{E}^2 {\rm{d}}^3\mathbf r$$

integrated over all space - since Coulomb is long range, this aspect can be very significant. So randomly allowing matter to be governed by the electrostatic force (and no other interaction) would tell us that matter should not be charged at the large scale, otherwise $\mathbf E$ will not vanish at long range and the long range part of the integral is important.

Clearly there are other effects, but we can actually see these effects when we look at ions in solution. Water separates salts into ions, and this is from minimizing a balance between the energy associated to entropy (from temperature) and the $U$ mentioned beforehand.

$^*$ ignoring effects from degeneracy

P.S. At first, I started to think that the water-ion effect is from the water moving around the ion to make the charges smaller and reduce the energy $U$ further, but I didn't think water's weak electrical bonding will be shorter than the ionic bond. It appears that it is marginally smaller according to this study, but I imagine that the rearrangement of charge has a smaller effect than the reduction of energy from increasing entropy (all pairings between $N=[0,\infty]$ water molecules and the $n\ll N$ ions vs all pairings of an equal number of sodium and chlorine ions)

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