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Imagine the following induction scenario: a super large charge +Q is brought close to a real conductor with limited supply of charges; that is, the total charge of all electrons inside this conductor is way less than Q.

What will happen to the conductor and the charges? Will the charges inside the conductor flow outside?

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What happens when a super large charge is brought close to a conductor with limited supply of charges?

The charges inside the conductor will move in response to the external field, with all available negative charge accumulating on the surface facing the external charge $+Q$ and all available positive charge accumulating on the opposite side the conductor.

So, how is this situation different from a regular induction in a good conductor?

In a good conductor, the charges would move until the field inside the conductor became zero. In a poor conductor, like the one you've described, even after maximum possible separation of charges, the field inside would still not be zero.

Will the charges inside the conductor flow outside?

If the applied external field is very strong, some charges could escape the conductor due to field emission or air ionization, but this would not be unique for a poor conductor: whenever a conductor is placed in electrical field, some charges end up on its surface and try to escape - and they would if the field was strong enough to overcome work function.

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This is a very deep question. As with many deep questions, the answer is "it depends on what level of idealization you want to work at, or how many nitty-gritty physically realistic effects you want to incorporate."

Practically speaking, the answer is that your thought experiment, in which all the electrons get stripped away, can't really happen in practice with a realistic macroscopic object that we would usually think of as a "conductor". It's very difficult to charge anything to a net charge above around $Q = 10^{-6}$ Coulombs without it ionizing the air around it and spontaneously discharging - and the smaller the charged "particle", the harder it is to charge it. (I'm talking about a completely isolated object with a net charge, not just charge separation like in a battery or a capacitor.) That's because Coulomb repulsion makes it really difficult to put more charge onto something that already has a net charge.

$10^{-6}$ Coulombs is only $10^{-11}$ moles, so in order for your nearby conductor to be completely depleted, it would need to be super-microscopic, so the classical picture would pretty much break down - you'd really need to consider its atomic structure. Moreover, the electrons are crucial for physically holding the conductor's atoms together, so even if you could strip away all of its electrons, it would physically far apart long before all the charge succeeded in separating.

(Further complicating matters, at the incredibly strong field strengths that you're considering, you would probably need to consider the effect of vacuum polarization in Quantum Electrodynamics - spontaneous production of electron-positron pairs out of the vacuum which would tend to generate their own internal electric fields that partially cancel the "external" field. The point is that this situation is incredibly complicated and requires a fully quantum-mechanical rather than classical treatment.)

But we can slightly modify your thought experiment to make it much more realistic. Most of the electrons in a solid aren't actually mobile "free charges" - only those near the Fermi surface are. Stripping away the "bound charges" far below the Fermi level requires much stronger, "ionizing" electric fields, as described above. But in semiconductors, you can realistically deplete all the mobile or "free" charge carriers near the Fermi level, which is conceptually similar to what you're describing. This creates what's called a "depletion region", whose physics has been well studied - see the linked Wikipedia page.

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Assuming the conductor is not grounded, when you approach the massive charge $+Q$, the opposite charges on the surface on the conductor will accomodate on the side facing the charge $+Q$, while the same sign charges will move to the opposite side of the conductor, thus creating an induced dipole.

However, the charges will not necessarily flow "outside", as remember that they're still bound to the surface of the conductor. However, you will effectively have a potential difference between the surface of the conductor facing the external charge $+Q$ and the charge itself. If this potential difference grows enough as to surpass the ionization potential of the medium between the conductor and the charge (i.e. air), then the medium will be ionized and the charge could "jump" (that is, transfer itself) towards the external charge $+Q$.

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