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This question already has an answer here:

I cannot understand how capacitors work.

  1. 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?

  2. 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

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marked as duplicate by Alfred Centauri, user36790, CuriousOne, John Rennie, AccidentalFourierTransform May 2 '16 at 9:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 1. en.wikipedia.org/wiki/Capacitor#Parallel-plate_model $\endgroup$ – user83548 May 1 '16 at 21:58
  • $\begingroup$ 2. circuits function because charge accumulates in the plates, it does not jump across the plates $\endgroup$ – user83548 May 1 '16 at 22:00
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  1. You reference the equation giving the electric field near a finite point charge. There is no finite point charge in a capacitor (unless we count a single electron, but I think you'll find a single electron won't produce a very large field measurement on a human scale...). The charge is distributed uniformly, and as you (you're a test charge) get closer to one plate, most of the charge is trying to push you sideways in various directions, not toward or away from the plate. The amount of charge which is relevant decreases as you get very close to a plate. If you know calculus you can write an integral to find that the field is exactly uniform.

  2. EDIT: I misunderstood your second question, see Ilja's answer and Bruce's comment.

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  1. As another answer pointed out, your formula is for electric field around an isolated point charge. It doesn't apply to the case of parallel plate capacitor. Normally we use Gauss's Law to find the electric field between the plates of the capacitor. We know that the field between the plates will be uniform from the differential form of Gauss's Law

$$\nabla\cdot{}E=\frac{\rho}{\varepsilon_0}$$

  1. You are correct that charge carriers don't flow through the dielectric of the capacitor. However, they are close enough together that a carrier introduced on one plate will repel a carrier away from the other plate. In order to keep things consistent, we say there is a displacement current flowing in this region. Displacement current is not caused by moving charge, but by changes of the electric field.
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  1. there are lots of questions and explanations, why the field of an infinite plain is homogeneous and does not depend on distance. That is the approximation involved: that the plates are big. So outside the plates the fields add to zero, and inside it's double.
  2. No, charge is not brought from one plate to the other. If you have an alternating current, it will indeed flow - but not through the capacitor - the charges will flow into the capacitor and back again. If you have direct current, it will flow into the capacitor and stop after a very short time, when the capacitor is "full".
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