Even if the atom is neutral (equal numbers of protons and electrons), the electrons and nucleus form an electrical dipole, so there is still an electric field around them, even though the total charge is zero. Is this reasoning correct? So, every atom has an electric field, even if it is too weak? If not, why?

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    $\begingroup$ Consider the hydrogen atom in the spherically symmetric 1s state. $\endgroup$ Jul 10, 2016 at 18:29

3 Answers 3


In an atom the electrons do not have a position. They are delocalised over the whole atom and in the absence of any external field this results in a symmetric charge distribution. So an isolated neutral atom is spherical and has no external field.

However if you bring two atoms near each other then their charge densities will develop fluctuations that are correlated. That is, both atoms will develop small but non-zero fluctuating dipoles and indeed higher multipoles. The interaction between these is the origin of the London dispersion force. So in these circumstances the atoms do have a fluctuating external field, though the long time average still remains zero.

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    $\begingroup$ So it seems like the "neutral" atom is really neutral due to Quantum Mechanics and the classical electrostatic doesn't work so well (I haven't studied much about Quantum Mechanics). However, why would a symmetrical charge distribution make the atom neutral? Does it has to do with Quantum Mechanics too? $\endgroup$
    – Jp_
    Jul 10, 2016 at 19:36
  • $\begingroup$ @JoãoPaulo: An atom is neutral because it has equal amounts of positive and negative charge i.e. equal numbers of protons and electrons. If you ionise the atom by removing an electron then the resulting ion is still spherically symmetrical but now has a net charge (of $-e$). The spherical symmetry just means it it has no electric dipole or higher moments. $\endgroup$ Jul 10, 2016 at 19:41
  • $\begingroup$ Sorry, I wrote it wrong. What I meant by neutral, in yhe comment, is that it has no electric field around (caused by the dipole electron-proton). So, there is no such electric field because of Quantum Mechanics, right? What about the symmetrical charge distribution, why would it make there's no electric-field around the atom? $\endgroup$
    – Jp_
    Jul 10, 2016 at 19:50
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    $\begingroup$ @JoãoPaulo: Look at the definition of the dipole moment. A spherically symmetric charge distribution necessarily has a zero dipole moment. $\endgroup$ Jul 10, 2016 at 19:53
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    $\begingroup$ @JohnRennie In your comment you say "...but now has a net charge (of $-e$)". I assume you are saying the resulting ion charge is positive $e$. By $-e$, do you mean minus the negative electron charge resulting in plus? I assume that is what you mean. But is that the right way this to say it is a positive charge, why not say $+e$ which I have seen before. $\endgroup$
    – K7PEH
    Jul 11, 2016 at 3:47

Strictly speaking, yes, I believe neutral atoms have an electric field (though very weak). For example, if you have a hydrogen atom in the ground state, the absolute magnitude of the wave function of the electron decreases exponentially with radius (for a sufficiently large radius), so there is a (very small) positive total charge within a sphere with an arbitrarily large radius.


Atoms are in the quantum mechanical regime. The electrons around the nucleus occupy orbitals ( not orbits) i.e. probability loci where a measurement will find an electron.

These orbitals have quantum numbers that give them shape. When the angular momentum quantum number l is equal to zero, the orbital is isotropic and neutrality is maintained. But for higher l values there are shapes, as seen in this table. Thus the negative charge has a shape, that leaves space for the positive charge of the nucleus to come out, creating positive and negative fields. These are the fields that generate the bonds for molecules, solids and liquids.

Edit after comments discussion

It is not simple, because this is a quantum mechanical frame.

Here is a measured orbital distribution for hydrogen , and it is symmetric.

h orbitals

The difference with the images in the table linked above is due to the fact that the experiment shows a time averaged distribution of the probable locations of the single electron around the proton for many randomly oriented hydrogen atoms. At any given time t, the electron will be at a single point with respect to the proton, and a field shape will appear. When two hydrogen atoms approach each other, the proton is not completely shielded by the electron, but an effective potential between the two atoms can be established leading to the bond that makes shared orbitals for the two electrons and creating the stable H2.

So yes, there is an instantaneous spill over field, whose probable strength that depends on the probabilities given by the orbitals .

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    $\begingroup$ Though in the absence of an external field the atom will be in a superposition of all the $p$, $d$, $f$ etc orbitals and will be spherically symmetric. So even with incompletely populated orbitals it still won't any shape other than a sphere. $\endgroup$ Jul 10, 2016 at 18:46
  • $\begingroup$ @JohnRennie There are spill over fields . Did you look at the table? $\endgroup$
    – anna v
    Jul 10, 2016 at 18:51
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    $\begingroup$ If you take a isolated boron atom, $1s^22s^22p^1$ then it does not have a single electron in e.g. the $2p_x$ orbital. It has, in effect, a third of an electron in each of the $2p_x$, $2p_y$ and $2p_z$ orbitals and the resulting charge distribution is spherical. The charge distribution only develops a dipole if you apply an external field. $\endgroup$ Jul 10, 2016 at 18:55
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    $\begingroup$ $B_2$ is a stable molecule and exists in the gas phase. However it's too reactive to hang around under normal lab conditions. $\endgroup$ Jul 10, 2016 at 19:16
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    $\begingroup$ "At any given time t, the electron will be at a single point with respect to the proton" This phrase bothers me to no end and I strongly suspect that it's just plain wrong. $\endgroup$
    – LLlAMnYP
    Jul 11, 2016 at 9:08

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