# What is the maximum surface charge density of aluminum?

I understand that the maximum free charge carrier density for aluminum has been measured using the Hall effect (in the case of electric current). However, I'm not clear how to determine the maximum surface charge density to which aluminum (or any conductor) can be charged, assuming the neighboring medium does not breakdown.

Say for instance we had a parallel plate capacitor with an idealized dielectric that could withstand infinite potential across it. What is the max surface charge density that the plates could be charged to? I assume that at some point all of the surface atoms are ionized.

Is this simply the volumetric free carrier density multiplied by the atomic diameter?

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Do you meant the maximum surface charge density one can charge aluminium to? That is you imply aluminium specimen is charged, not neutral, right? – Yrogirg Aug 13 '12 at 8:52

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Right, under idealized conditions, how much surface charge is aluminum capable of being charged to? – thrusty Aug 13 '12 at 16:01
Anybody, how does Hall effect helps to measure maximum free charge carrier density? I though one measures just free charge carrier density. By the way, I'm starting a bounty, to answer the question "What is the maximum potential one can charge a conductor to?" – Yrogirg Aug 15 '12 at 7:40

## 3 Answers

I think you can estimate the maximal surface charge density as follows. The energy needed to remove an electron from a solid to a point immediately outside the solid is called work function $W$. For aluminum $W$ is about $4.06-4.26$ eV. The thickness of the charged layer on the surface of a conductor is about several Fermi lengths $$\lambda_{F}=\left( 3\pi^{2}n_{e}\right) ^{-1/3},$$ where $n_{e}=N/V$ is the total electron number density for the conductor. I think that the charge starts to drain from the surface of a conductor when $E\lambda_{F}$ is of the order of the work function, where $E=4\pi\sigma$ is the electric field near the surface: $$eE\lambda_{F}=4\pi\sigma\left( 3\pi^{2}n_{e}\right) ^{-1/3}\sim W,$$ hence $$\sigma\sim\frac{W}{4e}\left( \frac{3n_{e}}{\pi}\right) ^{1/3}.$$

About the question Yrogirg. I think that the question is not quite correct. The charge are distributed on the surface of a conductor in such a way that the electric potential is a constant in the body of the conductor. The «stability» of charge on the surface is greatly dependent on the geometry of the object in question. Sharp points require lower voltage levels to produce effect of charge «draining» from the surface, because electric fields are more concentrated in areas of high curvature, see, e.g. St. Elmo’s fire.

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Oh, how could I forget about the work function. And yes you are right, it is the potential difference between the surface and the point just right outside it that matters (electric field outside the surface), not the potential difference between the surface and infinity as I though first. If the electron is just out it will be repelled away. – Yrogirg Aug 15 '12 at 14:23
what's $n_e$? "total electron number density for the conductor" --- is that the free charge density? I want to find value for $\sigma$. Might there be some kind of spin repulsion at such densities in addition to the electric one? – Yrogirg Aug 15 '12 at 18:04
No it is not free charge density. It is what I wrote $N/V$, where $N$ is the total number of electrons in the metal. According to Landau theorem the spectrum of low energy excitations of strongly interacting Fermi liquid coincides with the free fermion gas (which is strongly degenerate for low temperatures) that is how you can estimate the Fermi momenta and thus Fermi length. In fact, for this estimation you can use the experimental Fermi energy for aluminum. – Grisha Kirilin Aug 15 '12 at 18:17
However, the question "what is $n_e$?" is not important. The only important point is the thickness of charged layer. For plasma, the thickness of the charged layer on the surface is several Debye radii, for metal — several Fermi lengths. – Grisha Kirilin Aug 15 '12 at 18:25
@Yrogirg See e.g. A. Abrikosov "Fundamentals of the theory of metals" § 2.2. Quasiparticles in an isotropic Fermi liquid – Grisha Kirilin Aug 15 '12 at 18:37
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There is no sensible answer to this question.

You can put any amount of charge on a blob of aluminum sitting in a vacuum, or surrounded by an ideal insulator. Why not?

If you put an awful lot of electrons on a blob of aluminum sitting in a vacuum, the electrons will eventually start shooting off by thermionic emission, and most of the excess charge will be gone after, let's say, 1 day. If you put even more electrons, most of them will be gone after 1 millisecond. But there is no "maximum" really, just a gradual speed-up of the discharging. Even 1 excess electron will not be stable for eternity.

If you subtract electrons instead of adding them, certainly nothing will happen. Well, I guess positively-charged atomic nuclei could fly off if the charge was significant enough. Again, this process does not let you say that a certain amount of charge is "the maximum possible", it's just a process that happens more and more frequently as the charge increases.

If you add or subtract an awful lot of electrons from a blob of aluminum surrounded by insulator, the insulator will eventually break down. If you have an ideal insulator that cannot break down, then nothing will happen no matter how many electrons are there.

You seem to have the idea that all electrons must come from surface atoms, so if you take away every electron from every surface atom, then it will be impossible to take away any more charge. But if you think about it, that's sort of a weird idea, when there's still all those electrons inside the metal! In fact, the idea is not correct. The "surface" where an insulator can store charge is not infinitesimal, nor necessarily exactly one atom thick. It's actually a depth equal to the so-called Debye length. If you subtract loads of electrons -- every electron from every "surface" atom -- the Debye length will just increase allowing you to scrape electrons out of atoms residing farther and farther from the surface.

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Wouldn't it be difficult to scrape off anything but valence electrons from the surface atoms? This would leave inner-shell electrons on the surface atoms to continue screening the external field, limiting the penetration depth. – thrusty Aug 17 '12 at 6:26
Yes ... you need an extra 1500 volts to pull off a 1s electron from aluminum compared to a valence electron at the same spot. (I got that number from x-ray data.) So, as voltage increases, you will pull valence electrons out of a thicker and thicker depth from the surface, until the only remaining valence electrons are so far inside that there is a 1500V drop between the remaining valence electrons and the surface atoms. Then as the voltage increases even more, you will simultaneously pull off valence electrons from the inside and core electrons from the outside. – Steve B Aug 17 '12 at 13:29
"If you put even more electrons, most of them will be gone after 1 millisecond." So you admit that with more electrons it would be easier for an electron to emit? At some point (charge) it would be so easy that no additional energy, say from heat, would be required, that's what is meant by the maximum charge. Though your remark about thermoemission is important. – Yrogirg Aug 18 '12 at 5:30

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@Yrogirg, you're saying "Maximum charge is the point where the potential energy of an electron monotonically decreases as it goes from inside the metal to the vacuum". OK, with this definition, I think the maximum charge is some large and finite value. But I don't think this criterion corresponds to what thrusty was asking about. It is, after all, pretty irrelevant in the real world. A bit below this threshold, the charge spits out almost just as fast as it does a bit above the threshold, due to thermal fluctuations and quantum tunneling. – Steve B Aug 18 '12 at 12:32

Given the work function from the answer above, no electron can get as close as

$\frac{1}{4\pi\epsilon_0} e^2/r = 4.06eV$, at the range it would rather leave the aluminium than get closer to another electron.

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it gives the distance of $3.5 \times 10^{-10}$ meters, the same estimation as one derived from the Grisha's model. – Yrogirg Aug 21 '12 at 8:20
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