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?
Answer
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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