Dr_Nate
Science Advisor
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Yes, this is true.tech99 said:If the capacitor contains a dielectric, though not with a vacuum, I would have expected electrons to be moving back and forth in response to the applied electric field, constituting an electric current.
The numbers are comparable. The electrons don't need to move far in an oscillation. Dielectric breakdown occurs because the electrons are ripped from their ions; this usually occurs due to a large voltage.phinds said:If so, the number is trivial compared to the actual current in the circuit. If there are very many, you would be having dielectric breakdown.
In a sense, you are right. The free and bound charges contribute to the AC conductivity ##\sigma(\omega)## in the generalized form of Ohm's rule: ##\vec J(\omega)=\hat \sigma(\omega)\vec E(\omega)##. If I measured the conductivity at a single frequency, I wouldn't know how much was from free charge and how much was from bound charge. But, if I knew the conductivity over a large range of frequencies, I can see characteristic shapes in the graph due to bound and free charges.sophiecentaur said:Just to be difficult, I could point out that any net movement of charge through a metal conductor with a high frequency AC would be more or less the same as the movement of charges when the molecules in a capacitor's dielectric are polarised cyclically.
