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A capacitor allows AC to pass through. From what I've read it's caused by the changing electric field in the space between the capacitor plates which causes a change in charge on the other plate of capacitor which drives a current and NOT by a curly changing magnetic field inducing EMF on the other plate.

This brings up a question. Since there is dielectric medium in between the plates and some of the energy will be used up in setting up a polarization of the dielectric medium it implies that not all current $i$ going one way will come out of the other plate. Is this true?

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In an ideal dielectric medium all of the energy used to set up the polarization gets returned when the polarization relaxes. In reality, the capacitor will heat up some, dissipating energy. A much smaller effect, even in the absence of a dielectric, is that all elements of AC circuits have radiative losses.

There is no lost current because the polarization charge at the surfaces of the dielectric are equal, and opposite, having no net charge. So the amount of charge attracted to the opposite plate will be equal to the charge stored on the driving plate.

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  • $\begingroup$ I am considering only the ideal case here. But during operation $i$ must be less than what would flow in a normal short circuit isn't it? $\endgroup$
    – Weezy
    Commented Oct 16, 2016 at 12:16
  • $\begingroup$ Not in the ideal case. I'm not 100% certain, but I'm pretty sure that what you're thinking about are known as transient effects that dissipate in a very short time. $\endgroup$
    – Sean E. Lake
    Commented Oct 16, 2016 at 12:20
  • $\begingroup$ Well in the ideal case when you say that energy is given up by the dielectric when the polarization relaxes it would do so when the current has been stopped and my argument is that when we stop the current, besides transient effects, we would see a short surge(maybe not noticeable) in charge flow due to the energy released by the dielectric which was not available for use during the normal operation. It is this unavailable energy that would make the capacitor a less effective short circuit even in ideal case. Practically it shouldn't matter but it's a thought. $\endgroup$
    – Weezy
    Commented Oct 16, 2016 at 12:25
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    $\begingroup$ The dielectric relaxes every cycle in an AC circuit as the charge reverses because of the reversing current. $\endgroup$
    – Sean E. Lake
    Commented Oct 16, 2016 at 12:31

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