Gravity and its effect on electrons

Question

If gravity pulls everything downward why do electrons in a vertically held wire with a voltage across it not experience a downward pull?

Answer 1

An electron does feel a downward pull from the Earth. It's worth while getting a feel for how this pull compares with the electromagnetic forces it feels. To do this I'll compare two forces:

• the force the electron feels at the Earth's surface because the Earth is pulling on it gravitationally (this is just the weight of the electron, of course);
• the electrostatic force it would feel if the Earth was replaced by a proton at the centre of the Earth.

OK, so for the first of these we can work it out from first principles (I could cheat and just say that I know that $g\approx 10\,\mathrm{ms^{-2}}$ but I want to actually give the equations so you can see it).

What I will do, in both cases is assume Newton's shell theorem, which says that the effect of a spherically-symmetric distribution of mass, outside the distribution, is equivalent to a point mass at its centre.

So, for the gravitational force we can use Newton's law of gravity and the shell theorem to say that

$$F_G = -\frac{Gm_eM}{R^2}$$

Where $G$ is the gravitational constant, $m_e$ is the mass of an electron, $M$ is the mass of the Earth, $R$ is the radius of the Earth. Well, we can look up values for these things and we get

$$F_G \approx -9\times 10^{-30}\,\mathrm{N}$$

(And if you do the simpler calculation $m_e g$ you'll see that this is indeed right.)

So that's the force the electron feels due to the Earth's gravity. What's the electrostatic force it would feel due to a proton at the centre of the Earth? Well,

$$F_E = -\frac{1}{4\pi\epsilon_0}\frac{q_e q_e}{R^2}$$

Where $\epsilon_0$ is the permittivity of free space, $q_e$ is the charge on the electron (which is the same as the charge on the proton). And we can look up the numbers again, and we get

$$F_E \approx -5.7\times 10^{-42}\,\mathrm{N}$$

Which is comfortably less: the gravitational pull on an electron from the entire Earth is indeed more than the electrostatic pull from a single proton at the centre of the Earth. It's more by about $1.6\times 10^{12}$.

Well, $1.6\times 10^{12}$ protons have a mass of about $2.6\times 10^{-15}\,\mathrm{kg}$: about 2.6 picograms.

So this means that, in order to exert an electrostatic force on the electron equal to the force exerted on it by the entire Earth, you would need about 2.6 picograms of protons at the centre of the Earth: a ratio of about $2.3\times 10^{39}$ (or 1.4 femtograms of electrons, a ratio of about $4.2\times 10^{42}$).

Gravity is really, really, weak.

Answer 2

First of all: the force exerted on something with the mass of an electron by gravity is really small.

Removing an electron from an electrically neutral conductor would mean there's suddenly energy there¹, and now another electron would try to absorb that, because it would be forced to by the electrical field of the local positive charge.

Furthermore, electrons are never at rest, but always are subject to extensive random motion and typically, oscillations; they thus sometimes approach each other, then get slowed down by the opposing electrical field of the other electron, and thus tend to fill the whole conductor. You can think of that much like the osmosis happening with solubles in water. (or of an electron gas)

Then, maybe you'd want to abandon the idea of the electron as something being a "small ball of little mass with a negative elementary charge". That's a useful understanding when you need to explain quantized current flows, and for easy understanding of basic chemistry, but it's an insufficient model for what happens with electrons in metals.

In metals, we consider electrons to be unlocated in the usual sense of location; they don't have a certain place at any time, but just a function that we can use to calculate a probability density of their place and/or energy. So, you never "pull" on a single electron with your gravity, but on all the electrons in a certain energy band; that's basically just a term in the potential in Schroedinger's equation, but compared to usual other effects (periodic structure of atoms in crystals, external electrical fields,…), gravity is so weak, that you can safely ignore it for most purposes.

Answer 3

I just wanted to add to Dampmaskin. The electric force between two electrons is $$K\frac{e^2}{d^2}$$ where $$K=9. 10^9$$. The gravitational force is $$\frac{Gm^2}{d^2}$$. The ratio of both gives $$Fg/Fe = G/K (m/e)^2 = \frac{6.67 10^-11}{9. 10^9}\frac{10^{-60}}{10^{-38}} = 10^{-40}$$ That is how small the gravitational force is compared with the electrical force.