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Around the edges of a metal, the density of electrons is higher than the rest of the metal. I think this is because every electron is surrounded by neighbours pushing them away but once you get to the edge, there's nothing pushing you back away from the edge towards the middle (because the metal ends there so there are no more electrons).

My question is, would the same thing not happen in the very centre of the metal? So, if we had aluminium circle and represented the electron density as a heat map, would it have a dot in the middle?

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  • $\begingroup$ Could you please point to the source stating that the density of electrons around the edges is higher than in the rest of the metal? I assume you are talking about a neutral piece of metal, i.e., the net charge in a piece of metal is zero. $\endgroup$ – V.F. Oct 17 '18 at 19:39
  • $\begingroup$ physics.stackexchange.com/questions/43068/… i assume it because they don't have neighbours pushing them away. $\endgroup$ – tgmjack Oct 17 '18 at 19:40
  • $\begingroup$ in a sharp edge theres more surface area.. $\endgroup$ – tgmjack Oct 17 '18 at 19:42
  • $\begingroup$ In this post, the metal is charged, so the electrons gathering at the surface or edges are excessive electrons. If the metal is not charged, there won't be any accumulation of electrons at the edges or in the middle. So, the answer to your question is negative - there won't be any dot in the middle of an aluminum circle. $\endgroup$ – V.F. Oct 17 '18 at 19:47
  • $\begingroup$ You are welcome. $\endgroup$ – V.F. Oct 17 '18 at 19:53
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I'm sure this will be superseded by a more expert answer, but here's my simple conceptual answer.

I'm going to use a very simple 1D model in order to explain my reasoning.

Imagine a 1D interval, of length $L$, with 5 point charges of the same charge, all of which are free to move along the interval.

This is similar to the situation where free electrons are free to move within a metal.

The question is: What is the equilibrium configuration? Will the points charges bunch up?

By symmetry, one point charge will be in the middle at $L/2$, with 2 charges either side, and the configuration is symmetric about the centre.

Due to mutual repulsion, there will be one charge at each end of the interval i.e. one at $x=0$ and one at $x=L$.

So, our configuration so far has charges at $x=0$,$x=L/2$, and $x=L$ which I'll call charges 1, 3, and 5 respectively.

The positions of the two remaining charges will be $x=\alpha L/2$ and $x=L(1-\alpha/2$, by symmetry, and these charges I'll call 2 and 4, respectively.

In the absence of any boundary effects at $x=0$ and $x=L$, the remaining two charges will be closer the ends than the mid-point due to contributions from the opposite side of the interval. That is, $\alpha=1/2$ if we consider only the repulsion from charge 1 and 3, but when we include charges 4 and 5 there must be a displacement towards $x=0$, so $\alpha<1/2$.

If you were to extend this argument for $N$ charges and estimate a line density, you would see the line density increases towards the boundary of the interval.

I am assuming this reasoning will extend to 3D in a similar "simple" volume (one with no holes).

Now, I have completely neglected any surface effects. Firstly, the electrons are only "free" within the bulk of the metal. Secondly, electrons can be induced to leave the surface of a metal via the photoelectric effect. Third, the constant interactions on the interface with the atmosphere (chemistry leading to, e.g., the formation of oxides) must greatly affect this reasoning. Lastly, this reasoning does not even include the effects of a background lattice.

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