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The usual explanation goes as follows. Charge flows until the applied electric field is canceled by the induced field, giving a total field $\vec{E} = \vec{0}$. This is reasonable if one gives as an example a rectangular slab of conductor with a constant applied field $\vec{E}_0$ throughout. However this is just one simple conductor with a simple applied field.

It's not immediately clear to me that this explanation works for all shapes and all applied fields $\vec{E}$. Why should all conductors always have an electrostatic configuration? (Maybe I should ask, why should all closed conductors always have a static configuration? I don't want any batteries and wires and whatnot. Just some arbitrarily shaped, on it's own, conductor).

Ohm's law reads

$$ \vec{J} = \sigma \vec{E}$$ I've been told that this is an experimental result and I'm happy to except $\vec{E} = \vec{0}$ given the static condition $\vec{J}_{\_} = \rho_{\_}\vec{v}_{\_} = \vec{0}$ or $\vec{v}_{\_} = 0$ (I use minus signs as subscripts just to emphasis that there are also plus charges - I want the electrostatic condition to clearly state that there is no velocity in the free charges. The plus charges, nuclei, are presumably static). However, all textbooks should then say that $\vec{E} = \vec{0}$ is an experimental observation which is to be accepted. [As pointed out in the comments, assuming that electrostatics exists for conductors is equivalent to assuming $\vec{E} = \vec{0}$ within a conductor - so to rephrase my title: why is electrostatics possible? Or equivalently, why is $\vec{E} = \vec{0}$ possible for conductors? This paragraph is me saying "I'm okay if the answer is to except as an experimental fact that conductors can reach statics". And actually, I think this is the answer to this question. Yet, textbooks don't quite say it this way. At least not explicitly. The following paragraphs ask if a theoretical reasoning exists]

Textbooks try to give a proof/argument that $\vec{E} = \vec{0}$ (electrostatics) is always possible. They reason that it's always possible with induced charges (mainly using the simple geometry of a rectangular slab with a simple electric field as visual proof). However, this shouldn't be convincing. For any surface and any applied field? If such an argument exists, given the definition that a conductor has an infinite amount of charge to ensure that any $\sigma$ can be produced, we should be able to prove:

For a conductor with an arbitrary surface S in an external electric field $\vec{E}$, you can always find a $\sigma(S)$ such that

$$\int_{S'} \frac{1}{4\pi\epsilon_0}\frac{\sigma (\vec{r}')dS'}{\|\vec{r} - \vec{r}'\|^3}(\vec{r} - \vec{r}') = -\vec{E}\; \; \text{for points $\vec{r}$ in the conductor} \\ \text{and} \\ \vec{K} = \sigma\vec{v} = \vec{0} $$

It seems kind of magically that conductors can do this. Is such a proof possible? The $\sigma$ always exists?

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    $\begingroup$ The answer to the title question is yes, by definition. If there's an electric field within a region where charge is free to move, e.g., within a conductor, charge will accelerate contradicting the electrostatic assumption. But you ask a subtly different question in the 2nd paragraph of the text. $\endgroup$ – Alfred Centauri Jun 21 '18 at 0:08
  • $\begingroup$ @AlfredCentauri Oh yeah I see how my title gets at that, and I understand what you are saying (I think). But also, just knowing what a conductor is, you are led to the possible conclusion that statics is impossible with conductors present. In order to prove that statics is possible, I would like to know why a surface charge configuration $\sigma$ is always possible, for any shape and any applied bizarre field $\endgroup$ – DWade64 Jun 21 '18 at 0:12
  • $\begingroup$ I think if we start out with "electrostatics is an experimental fact" from this it has to follow, $\vec{E} = \vec{0}$ in the conductor, which implies $\rho = 0$ which implies charge can only pile up on the surface (you have a $\sigma$) which means the integral I show has to exist as well as the second condition. So from the experimental fact that statics exists, everything else follows and my question disappears. This is of course assuming that we did the experiment for every possible conductor shape with every possible applied field. Other than this, I don't think a theoretical proof exists $\endgroup$ – DWade64 Jun 21 '18 at 0:32
  • $\begingroup$ Also, I have intuitive experience with balls rolling down into the bottom of a valley. If it can (no road blocks to impede its path), it will. I don't have intuitive experience with charges and electricity. I should accept as a definition that conductors are objects which can always reach a minimum. There are no stumbling blocks. It's still weird to me how you can reach a minimum with any surface though. It's something I don't have experience with so I was looking for a theoretical "proof/explanation" but I think I comes down to accepting statics as an experimental fact for conductors $\endgroup$ – DWade64 Jun 21 '18 at 0:45
  • $\begingroup$ It's quite astounding that any surface can produce literally any equal and opposite applied field. Any field. It seems unrealistic and plainly incorrect. If there is a theoretical "proof/explanation" for such an amazing result, it probably comes from Laplace's equation/uniqueness theorem's, and it probably relies heavily on the idealized (but technically incorrect) view that when charge flows to the surface, it doesn't leave behind equal and opposite charge in the volume of the conductor. Charge flowing to the surface doesn't alter the idealization that $\rho$ remains $0$ within the conductor $\endgroup$ – DWade64 Jun 21 '18 at 2:08

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