I cannot understand how capacitors work.   Electric field is supposedly uniform between the parallel plates of a capacitor.  But how?  Since $E=\frac{kq}{d^2}$, shouldn't the electric field approach infinity as one approaches either of the two plates ($d=0$), while it would be smaller in the middle? There is air between the capacitors, right?  Air isn't a conductor, is it?  Circuits in which the wires are interrupted by air aren't closed, are they?  Yet circuits still function with capacitors in them and charge is brought from one plate to the other while there is air between the plates.  What is going on there?   Thank you
When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is $${\bf E}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$ The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates. This result can be obtained easily for each plate. Therefore when we put them together the net field between the plates is $${\bf E}=\frac{\sigma}{\epsilon_0}\hat{n}$$ and zero everywhere else. Here, $\sigma$ is the surface charge density on a single side of the plate, or $Q/2A$, since half the charge will be on each side.   But in a real capacitor the plates are conducting, and the surface charge density will change on each plate when the other plate is brought closer to it. That is, in the limit that the two plates get brought closer together, all of the charge of each plate must be on a single side. If we let $d$ denote the distance between the plates, then we must have $$\lim_{d \rightarrow 0}{\bf E}=\frac{2\sigma}{\epsilon_0}\hat{n}$$ which disagrees with the above equation. Where is the mistake in this reasoning?  Or more likely, do our textbook authors commonly assume that we are in this limit, and that this is why the conductor behaves like a perfectly thin charged sheet?
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