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It is said that atoms with the same number of electrons as protons are electrically neutral, so they have no net charge or net electric field.

A particle with charge cannot exist at the same position and time as another; an electron cannot be positioned at the location of a proton, at any single point in time, without displacing the proton.

Assuming the above is correct, how can a single electron cancel out the entire electric field of a proton? I don't think there is any position a single electron can take, that would result in the entire electric field of the proton being cancelled out - it seems like it will always be only partially cancelled out.

For simplicity, let's look at a single hydrogen atom that we consider to be electrically neutral. It has one proton and one electron, so at any single point in time, there will be a partial net electric field (because the electron will never be in a position where its field can completely cancel out the proton's field), and the electric field from the electron will only cancel out part of the field from the proton. So at this single point in time, there will be a net electric field from the proton. So how can this atom be considered to be electrically neutral, with no net charge or field?

Here is a graphical representation of two sources of electric fields interacting:

enter image description here

As you can see from the image, only part of the (equal but opposite) electric fields produced by both sources are affected by each other. To have the field from one source cancel out the other, completely, we would need to position the sources in the same location, at the same time, which is not possible.

I know that I'm wrong, so please correct me.


marked as duplicate by Jon Custer, John Rennie, Kyle Kanos, Martin, ZeroTheHero Mar 9 at 1:13

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    $\begingroup$ For a more precise description of the intended meaning of the phrase "no net electric field" in this context, search for Gauss's Law. $\endgroup$ – Harry Johnston Mar 4 at 23:31
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    $\begingroup$ This question might qualify as a duplicate of physics.stackexchange.com/q/267371 $\endgroup$ – user2647513 Mar 5 at 5:10
  • $\begingroup$ If the two fields did cancel each other out compeletely, even close to the atom or molecule, there would be no polar molecules, no hydrogen bonds, no DNA, no us. So you are wrong knowing you are wrong, because you are right. $\endgroup$ – Peter A. Schneider Mar 7 at 13:12
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    $\begingroup$ Possible duplicate of Is there an electric field around neutral atoms? $\endgroup$ – Keith Mar 7 at 22:19

It is said that atoms with the same number of electrons as protons are electrically neutral, so they have no net charge or net electric field.

This is a great over-simplification, which I am sure you have already determined (based on why you are asking this question). You can have objects that are polarized where, overall, they have no "net charge", yet the distribution of charge is very important. The example you give is an excellent one of a dipole, where the net charge is $0$, yet $\mathbf E\neq 0$ at distances away from the dipole.

Really, the idea of electrically neutral is a macroscopic description meaning that if we look in this general area we will see that the number of positive charges exactly balances our the number of negative charges. However, as we "zoom in" we will find this to not be the case for an "electrically neutral" body, since (neglecting QM) we will have point charges at specific locations, and most of the "charge density" will be $0$ due to no charges being present at all, and then "infinite" (or at least really large) at the locations of the charges.

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    $\begingroup$ 'Neglecting QM' seems a strange idea when modelling an atom. The 'electrons' are not at specific locations. $\endgroup$ – Keith Mar 7 at 5:26
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    $\begingroup$ Rather like how newspaper images are printed using half-tones. Viewed from a distance, they appear to be printed in shades of grey or smooth colours, but viewed close up, the individual dots and white spaces can be seen. $\endgroup$ – CJ Dennis Mar 7 at 7:37
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    $\begingroup$ @Kieth It looks like this question is in the framework of classic electrostatics, where point particles have definite locations. Models don't have to be 100% correct to use them and learn things from them. $\endgroup$ – Aaron Stevens Mar 7 at 11:40
  • $\begingroup$ The trouble is that the intuition in the question that charges don't balance out exactly is correct both in a classical sense and in proper quantum mechanical sense. Moreover it is physically important in explaining the observed interactions between neutral molecules and even forces on a macro scale. The answer doesn't explain the important and significant forces that result such as London forces. $\endgroup$ – matt_black Mar 7 at 17:09
  • $\begingroup$ @matt_black Yes there is important application of these things, but the question isn't asking about forces that arise from things like this. I think your answer does a good job doing this, so if someone is interested in this they can go to your answer. $\endgroup$ – Aaron Stevens Mar 7 at 17:51

If something is 'Electrically neutral' this means that the algebraic sum of its electric charges, however distributed, is zero.

This does not imply that there is no electric field in its vicinity. Plenty of neutral bodies – even, it is believed, the neutron – have electric fields, for just the reason you have pointed out.

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    $\begingroup$ +1 for mentioning the neutron so I didn't have to. $\endgroup$ – J.G. Mar 5 at 9:59
  • $\begingroup$ True, but actually explaining some of those forces (especially, for example, London forces) would have made a better answer. $\endgroup$ – matt_black Mar 7 at 17:10
  • $\begingroup$ A longer answer, certainly. $\endgroup$ – Philip Wood Mar 8 at 18:16

Building upon other answers, we must first differentiate between net charge and electric field - an atom with an equal number of equally charged positive and negative particles will have no net charge, but may still have an electric field, depending on the arrangement of the charge, as in a dipole.

Now, your intuition is correct, it doesn't seem valid that a hydrogen atom is a dipole simply because of the spatial distribution of the subatomic particles. Based on the answer to this question:

Is there an electric field around neutral atoms?

Electrons are not actually "orbiting" the nucleus, and in fact they are not spatially localized, their probability distributions are spread over the nucleus, which leads to a symmetric distribution and neutralization of the overall electric charge in space.

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    $\begingroup$ +1 for "not orbiting" $\endgroup$ – Jasper Mar 5 at 15:06
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    $\begingroup$ But -1 because even a full QM treatment doesn't lead to the perfect neutralisation of the nuclear charge. Otherwise there would be no dispersion forces between molecules which are a product of quantum fluctuations in the electron distributions. $\endgroup$ – matt_black Mar 7 at 17:14

It is said that atoms with the same number of electrons as protons are electrically neutral, so they have no net charge or net electric field.

Certainly they have no net charge, but it is not true that they have no net electric field. However, you may have heard this because at a long enough distance, the electric field of objects with no net charge can be negligible compared to the electric field of a charged object.

At distances much larger than the separation between the electrons and nucleus, the magnitude of the electric field of a net neutral object decays more rapidly than that of a charged object. Typically, at such distances, the electric field is dominated by a dipole term. While the electric field of object with a net charge scales $1/r^2$, where $r$ is the distance from the charged object, the electric field of a dipole scales as $1/r^3$. So if you double the distance, the electric field of a charged object is a quarter of what it was, while the electric field of a dipole is an eighth of what it was. Hence, at large distances, the electric field of net neutral objects can sometimes be considered negligible compared to the electric field of charged objects.

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    $\begingroup$ I believe that in the original statement, the phrase "net electric field" was intended to refer to Gauss's Law. So it isn't exactly untrue, just way too ambiguous. $\endgroup$ – Harry Johnston Mar 6 at 21:20

As it has been noted in Wood's answer, electrical neutrality just means that the algebraic sum of the electric charges is zero. It does not imply anything about the presence of fields. Notice, that the same situation holds also for electrolytic solutions, so no special role is played by the quantum nature of the charges.

About the fields, I would like to add that it is true that there is nothing forbidding to have non-zero fields in a globally neutral system. However, we should also take into account the observation time. Measurements of electric fields correspond to time average of the fields. Therefore, if a short time measurement on microscopic scale could measure a non-zero field, a time average over macroscopic times could give average macroscopic fields close to zero.

  • $\begingroup$ For an electron in a 2P state - is this what gives the hydrogen atom its tiny magnetic field? $\endgroup$ – Rick Mar 4 at 23:24
  • $\begingroup$ In the case of the hydrogen atom in a 2p state it is not neutrality responsible of the magnetic field but the fact that 2p states carry a current. One could have magnetic field also in the case of non-neutral ions. $\endgroup$ – GiorgioP Mar 5 at 6:06

Your intuition is correct and physically important but a little more complicated than you describe

When we say atoms are electrically neutral we mean that objects far away from the atom see no net electrical field (atoms are very small, so even if the effect you envisage exists it will only be apparent for objects close to the atom).

But your simple intuition is complicated because electrons in atoms are not point particles existing at a single point in the atom. They are more like standing waves spatially distributed around the nucleus (because of Heisenberg and quantum mechanics). This means things are a lot more complex than they would be if electrons were simple point particles revolving around the nucleus generating a moving electrical dipole.

But, the basic intuition that the overall field generated by the electron clouds surrounding atoms and molecules is not zero is correct. And this is very important in chemistry and for the physical properties of molecules.

The electrons in asymmetric molecules are often distributed in ways that generate electrical dipoles. The simple hydrogen chloride molecule (despite being neutral overall), for example, has a strong electrical dipole because chlorine pulls electrons away from hydrogen.

But this probably isn't the key thing you were asking about. Even neutral molecules without built-in dipoles have some features where the distribution of electrons can generate momentary electrical fields. While the simple picture of an electron as a point particle revolving around a nucleus isn't a good way to picture this, even quantum mechanics has to admit that there is uncertainty about the electron's position in its orbital cloud. This uncertainty generates temporary fluctuating electrical fields because the distribution doesn't precisely offset the electrical field of the nucleus. The detailed quantum explanation of this is pretty complicated but the picture of fluctuating electrical fields captures some of the idea intuitively.

More importantly these fluctuating fields are very important in the real world. They are the basic explanation of van Der Waals (or, more correctly London) forces between molecules and between atoms. When one molecule has a fluctuating field, it can generate a field in a neighbouring molecule that is close to it. This generates a weak force that declines rapidly with the distance between the molecules (it is proportional to r-6 where r is the intermolecular distance). For neutral molecules with no built-in dipole, this is a major factor in their physical properties. Gases like N2 would never become liquids if the force did not exist. Simple hydrocarbons and many other non-polar molecules would not liquefy at room temperature without these forces. Even in many polar molecules the London force accounts for the majority of the molecule-molecule interaction (see wikipedia).

London forces can even be important at the macroscopic level: they are the underlying force that allows geckos to walk up walls.

So, yes, even neutral atoms or molecules have fluctuating electrical fields because of the mismatch between the "position" of the electrons and the nucleus. And these fluctuations are important because they generate forces between atoms and between molecules. These are very important in determining the physical properties of many substances and can even be seen at macroscopic scales.


While many answers address the question by the OP perfectly, I am just adding what I think might be added to these answers as a separate answer.

The electric potential of a generic charge-distribution is exactly described via the multipole expansion which is a power series in the inverse powers of $r$. The zeroth order term in this series (which goes as $\frac{1}{r}$) is a monopole term whose strength is proportional to the total charge (or, if you wish, the total monopole moment) of the configuration. The first order term in this series is a dipole term whose strength is proportional to the total dipole moment of the configuration. The second order term would be proportional to the total quadruple moment of the configuration, and so on.

So, unless the charge distribution is simply consisting of a point-charge, the dipole picture is also an approximation (that is to say that the exact electric potential has contributions from the quadrupole moment, and so on) and is only one order in $\frac{1}{r}$ better than the monopole approximation. So, in this sense, at sufficiently large distances from the atom, the approximation of an atom as a neutral monopole is not far worse than its approximation as a dipole with a non-zero dipole moment.


This is really an extended comment.

The question as stated makes classical assumptions around orbiting charged point particles.

This is not the experimental fact.

One of the major drivers for the invention of QM was precisely that such a classical model does not work very well. Apart from the electron in fact being de-localised, classical EM requires that an accelerating particle radiate and hence an orbit would gave to decay.

So, please understand that we learn the classical model partly to understand why it does not work very well and consequently need another approach.

The experimental fact is that a neutral atom does have a symmetric charge distribution. See John Rennie's answer

  • $\begingroup$ That's not right -- there is an external electromagnetic field even in quantum mechanics. The naive reasoning ("the electron wavefunction is spherically symmetric, so the electromagnetic field is exactly zero by the shell theorem") is incorrect because the electromagnetic field is also quantum. The true quantum state has entanglement between the electron and the field; the field is not just in its vacuum state. $\endgroup$ – knzhou Mar 8 at 12:54

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