Atomic electric field an issue

1. Oct 15, 2012

ChaseRLewis73

My question is this. I have been taught in every textbook, physics lecture, and chemistry lecture that the net charge of a stable atom is 0. The result of that should be a null electric field. But looking at any model of an atom since the electrons are spaced out significantly from the nucleus it's easy to see that the electric field is equivalent to a charge between -1 and 0 for any value outside the atom if you assume the cloud to be a sphere (such as the filled S orbital of helium).

Experiments also show that most atoms are not magnetic (at least to any consistent degree, not sure about dynamic equilibrium). This isn't possible if electrons actually move with any significance according to the maxwell equations. Since moving negative charges means a dynamic electric field. I haven't done the exact math to prove the impossibility of dynamic equilibrium but it doesn't seem very likely giving the geometric concepts.

Calculating the field potential outside the atom is fairly easy for spheical orbits like those believed to comprise the s sub-shell.

We can model the two protons as a point charge (of +2) since the distance between them is much less than the distance between the protons and the electrons (when they are squared the difference ends up less than 1/(2502)

now assume lowest energy configuration (a 1D line of a negative electron at -r a +2 positive charge at 0 and another negative value at r). Doing raw calculations the electron repulsion outweighs any attractive value. (This calculation is actually independent of r as since they all use the same value of r it cancels out into simple geometric values).

Thing is I extended this type of calculation into spherical shapes into the Pi - bonds assuming best case scenario (no additional repulsion from inner electrons). ... didn't get any better in fact it just got magnified.

My question is why do so many text books claim nuetral charge when no model of the atom geometrically predicts this if electron and proton have the same charge?

Last edited: Oct 15, 2012
2. Oct 15, 2012

Jazzdude

I read your post three times, and still cannot make much sense of it. You make a very confused impression. Atoms are neutral as an object for any sensible meaning of the word neutral. That means specifically that the electric monopole moment vanishes. What you said about magnetism also made little sense. You'll need to use quantum theory to argue about that one.

3. Oct 15, 2012

ChaseRLewis73

My question basically is how is it considered nuetral if the electric field potential outside the electron radius has points that can't be equal to zero? according to coulomb's law. Simple fact is that because charge is inverse squared to distance that being closer to the electron means you will have a non-zero electric field potential. Is the model just to simple and because of the new charge presence the electron will adjust it's radius to make U = 0? Even if so with a finite number of electrons how could it be possible to make the potential outside the entire atom negligble ... I mean most of chemistry makes the assumption of repulsion / attraction of covalent atoms (reason it attacks specific sites there is less repulsion there) so I can't see how an atom can be considered to be completely nuetral there is more to nuetrality then simply adding the total number of charges together ... geometry is very important as well.

U = kqq/r so let's assume a free electron at 1 radius past the atomic radius of helium (31 pm). Assuming point charges for proton and electron you get the equation below.

U = -k (1/4r2 - 1/r2 - 1/9r2)

It's easy to see the electric field potential in this situation is not equal to zero. So the atom will act on charge carriers that are in it's vicinity. I understand the model is a bit simple and the cloud will adjust to the new charge but none the less you get repulsion thus a proof of a non-zero electric field outside the atom.

Guess I'm nitpicking but to me proton-electron balanced doesn't seem to mean neutrally charged like it would on the macroscopic scale where something with zero net charge effectively moves through an electric field unobstructed.

4. Oct 15, 2012

Staff: Mentor

When we say "neutral" we usually mean zero net charge, no more, no less.

As you've noticed, when you have an extended charge distribution whose net charge is zero, it's possible to have a nonzero electric field outside the distribution. In some places, the field points towards the distribution (atom in your case), in other places the field points away from the distribution (atom). The "average in- or outwardness" turns out to be zero. Mathematically we describe this using Gauss's Law (one of Maxwell's famous equations of electromagnetism).

5. Oct 15, 2012

Bill_K

You seem to think that a Helium atom has one electron on one side, and one on the other! Not so, both electrons are in an S state. An S state is spherically symmetric, which means that each electron is equally likely to be on any side. This holds even taking into account the mutual repulsion between the two electrons.

A Helium atom being spherically symmetric has no electric multipole moments, and produces no external electric field.

6. Oct 15, 2012

ChaseRLewis73

Ya that was more for making a very simple calculation assuming minimal energy and everything was a point particle (very classical approach). Reason I was asking how QM handled this differently. I can understand that Gauss's law causes ∇E = ρ/ε where you have the distribution ρ = 0. Thing is spherically symmetric would seem to be kinda odd since most atoms aren't spherically symmetric only s is spherical at all according to experiments.

$2k/r^{2} - 2k/r^{2} = U = 0$ at the atomic radius but that wouldn't create a 0 net charge for things elsewhere.

Seems the charge would range from electrons occupying the same position closest to it and electrons occupying the same position farthest from it along a sphere of constant radius in a probability distribution according to the schrodinger equation. Symmetry would ruin that by dictating a constant state thus an exact positions in exchange for infinite momentum according to Heisenberg (not practical). Otherwise the charge distribution couldn't remain constant for a point particle unless wave nature some how smears the charge uniformly about the orbital. I'd think based on the energy level you'd get a distribution of some type instead since orbiting electron speed is estimated << speed of light just with average ∇E = 0.