I had always thought that a capacitor with a dielectric transmits AC precisely because it polarizes in response to the applied field. To tell you the truth, I realize now that I had two different views on how this process works: One: the AC is oscillating on both sides of the dielectric and the dipoles of the latter are turning round in harmony with those oscillations. The other: an oscillation of the AC happens in a conductor in one side of the dielectric; the dipoles of the dielectric oscillate confronting that oscillation, so that the + charge faces the electrons when they approach and the – charge faces them when they fly away and this oscillation of the dipoles sort of “emits” a new AC through the conductor of the other side (something akin to the electron transitions in an atom emitting an EM wave). As to why the transmission is better (reactance is lower) for higher frequencies, I was thinking that it is so because the lower frequency did not have enough energy to make the dipoles turn round. But reading this text in Wikipedia (entry about Electrical Reactance), “Driven by an AC supply, a capacitor will only accumulate a limited amount of charge before the potential difference changes polarity and the charge dissipates. The higher the frequency, the less charge will accumulate and the smaller the opposition to the current.” it seems that it is the opposite, that is to say, the AC goes through the dielectric in spite of the polarization, because it is not affected by the latter to the extent that the frequency is high enough… What is closer to the truth? Can anyone provide some explanation helping me to visualize the process?