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Explanations of conductors in electrostatics that I have encountered seem to describe positive charge spreading out, because you could say that lack of electrons can be thought of as abundance of protons (that in itself is not trivial to me- can any system of negative charges be replaced by another system of positive charges, creating identical field lines?).

If we theoretically removed all electrons from the conductor (this is theoretically possible, isn't it?) I would be left with a bunch of stationary positive charges (as the protons of the atoms) which are spread somewhat evenly across the volume of the conductor (and not the surface, as would have happened with mobile positive charges). As far as I know, this make electric field inside possible, which is not what textbooks and lectures indicate that happens.

What would happen in reality?

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  • $\begingroup$ To answer your parenthetical question: given a set of negative charges $-q_1, -q_2, -q_3, ...$, replace them with a set of positive charges $+q_1, +q_2, +q_3, ...$ At every point in space, the electric field due to each of the charges will flip direction. Thus, the total electric field will flip direction at every point in space. Thus, the field lines will be identical, just flipped in direction. $\endgroup$ – Michael Seifert Aug 29 '17 at 18:27
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    $\begingroup$ Obligatory what-if what-if.xkcd.com/140 $\endgroup$ – Shufflepants Aug 29 '17 at 18:42
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    $\begingroup$ Simple answer: The freely moving electrons are what let it be a conductor. If you take them all out, it's not a conductor anymore, and it's perfectly fine to have an electric field inside it. $\endgroup$ – eyeballfrog Aug 29 '17 at 19:41
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If you suddenly removed all the electrons from a piece of material, or even just the valence electrons, you would be left with a huge concentration of positive ions in a small volume, which would exert a huge electrostatic repulsion on each other. Since you no longer have the bonding influence of the electrons to counteract this repulsion, the material would blast apart, in very short order, in a process that's known as a Coulomb explosion.

To put some numbers into things, suppose that you have one cubic millimeter of iron, and you suddenly remove one electron per atom. This turns out to be about $0.00014\:\mathrm{mol}$ of iron, but because Avogadro's number is so huge, that's about $8.491\times 10^{19}$ electrons, and a corresponding charge of about $13.6\:\rm C$ in the sphere, an electrostatic charge distribution that holds about $1.6\times 10^{15}\:\rm J$ of energy, or about $385$ kilotons of TNT, i.e. about twenty times bigger than the explosion that flattened Hiroshima.

(And, obviously, that's the amount of energy that you will need to put in to be able to suddenly remove all of those electrons. In more human terms, that's a $1\:\rm GW$ power station running nonstop for 18 days. And, as mentioned in the comments, this amount of energy represents about twenty times more than the original rest mass of the iron.)

That said, if you scale things down significantly, then Coulomb explosions can become quite reasonable things and indeed important research tools. Normally you do this with small(ish) molecules and atomic clusters (so, from a few to a few hundred atoms), where you have a few hundred electrons or so (instead of tens of quintillions), and you remove them with a high-intensity, high-photon-energy beam coming from a free-electron laser (FEL). In the process you might then get single-molecule x-ray diffraction spectra, information about the initial structure from where the atoms flew off to after the explosion, or you might just learn about the physics of the ionization and explosion processes. For a nice overview, see these slides by Christoph Bostedt, or the papers in this google search.

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    $\begingroup$ Another nice illustration of Coulomb explosions: nature.com/nchem/journal/v7/n3/full/nchem.2161.html One of the author, Philip E. Mason, is thunderfoot on youtube and he has a series of videos on the subject: youtube.com/user/Thunderf00t/search?query=coulomb $\endgroup$ – user154997 Aug 29 '17 at 12:14
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    $\begingroup$ Note that a cubic millimeter of iron weigs about $7.84$ milligrams, while pumping $1.6\cdot 10^{15}$ joules of energy into it (by removing electrons and keeping the protons from blasting apart) will increase its (apparent rest) mass by $0.17$ grams, which is an increase by a factor of over $20$. $\endgroup$ – Arthur Aug 29 '17 at 12:34
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You will not be able to remove all the electrons of any conductor no matter what kind... The removal of electrons from any conductor comes from the valence electrons of the atoms in particular from the one electron most weakly connected to the nucleus.

As you keep going, removing more and more electrons from the conductor, I think you should reach a point on which, the conductor will break apart as far as the interatomic/molecular forces that keep their atoms/molecules together are overwhelmed by the coulombian force between the positive charges in this atoms/molecules.

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  • $\begingroup$ So in a conductor that did not break apart, there will always be electrons balancing some of the positive charge, but the claim is that most of the missing electrons (in the case of net positive charge) will be those of the atoms on the surface? $\endgroup$ – Yiftach Aug 29 '17 at 10:06
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    $\begingroup$ Yes, the missing electrons will be from the atoms in the surface, even if you initially take electrons from the inside*, the electrons of the surface atoms migrate to the inner regions in order to establish equilibrium (in this case*). The calculation made above shows an interesting property in my view: if you try to rip out 1e/atom in the mentioned iron cube, the amount of spent energy will be twice the energy of the mass of this cube (via E=mc^2). $\endgroup$ – Rodrigo Fontana Aug 30 '17 at 2:54
  • $\begingroup$ This says to me that you will not be able to take that amount of electrons in that case. For instance if you think as an example in the bond between quarks in the nucleus, less then 1% of their masses are put into "bond" energy... (And thinking that the strong forces there are in some way orders of magnitude higher than the coulombian forces, I think this much of "eletric energy" will collapse the structure long before you achieve the situation... Just guessing, though). $\endgroup$ – Rodrigo Fontana Aug 30 '17 at 2:54
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The key to the answer is where you equate a lack of electrons with an abundance of protons. That is a very misleading analogy. The correct analogy is actually to equate electrons (carriers of a negative charge) with holes (which is the absence of an electron where one should be. Holes are positively charged).

Protons are fixed in place (at least in a solid, and if you ignore Brownian motion and the like). They are "frozen" into the nucleus of each atom.

One gram of copper, or one centimeter of copper wire, contains a specific number of atoms, and therefore a specific number of protons, and that doesn't change no matter what.

Now in a neutral substance, there is one electron for each proton on average. Individual atoms can lose one electron, or even two, and that makes them positive ions. However, when that happens, these ions have a strong attractive force on electrons. Removing the first electron from a neutral atom tends to be easy (in a conductor). Removing a second one becomes more difficult, and removing a third electron, or even more, becomes incrementally more difficult, and eventually impossible simply because the atom becomes more and more positively charged, and therefore attractive to electrons.

But hypothetically assume that you could remove all electrons from your conductor, and you could prevent electrons from the surrounding air to get back in. The first thing that would happen is that all the remaining atomic nuclei would repel each other. Your conductor would disintegrate (probably with a huge explosion).

Now if you could prevent that, too, you still couldn't have an electric field. An electric field exists between two charges. An electron does not "have" an electric field. A proton does not have an electric field.

So if you could remove all electrons from a conductor (which you can't), and could prevent it from flying apart (which you can't), there would be nothing left that could generate an electric field.

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  • $\begingroup$ As for the scenario of the second to last sentence- why is it incorrect to say the remaining protons would create an electric field between them? $\endgroup$ – Yiftach Aug 30 '17 at 18:54
  • $\begingroup$ There is no charge differential between them. It would be kind of like talking about the wind direction between two places that have the same barometric pressure - there is no wind, therefore it is meaningless to talk about the direction of the non-existent wind. Fields are all about directions (or more accurately, gradients). $\endgroup$ – Kevin Keane Sep 1 '17 at 3:04
  • $\begingroup$ Why, though? When all electrons are removed, the remaining protons leave us with a solid, non-conducting charged sphere. As far as I know, at each point with radius r there is electric field pointing outwards (all charge between 0 and r is treated like a single charge at the center point , while the charge between r and R cancels out at r similar to what we see in a conducting sphere). That would imply there is electric field (which is what makes the sphere explode)- is that incorrect? $\endgroup$ – Yiftach Sep 2 '17 at 8:52
  • $\begingroup$ Yes, that's pretty incorrect. The field lines aren't really pointing "outward" but rather towards the opposite electric charge. In everyday life, that may not matter much because there are opposite electric charges all over the place. The point charge in your example is simply far enough way from the opposite charge so that it looks as if the field lines extend straight out. Oh, and your solid sphere made up of protons would be highly conductive. Holes (lack of electrons) are charge carriers, just as electrons are. $\endgroup$ – Kevin Keane Oct 18 '17 at 4:18

protected by Qmechanic Aug 30 '17 at 10:57

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