Howdy- Consider a low pass filter subjected to an AC source (i.e the "output" is the capacitor voltage). I understand mathematically how to assess the frequency reponse of such a circuit. What I am after is a conceptual description of why. Why do capacitors act as shorts to high frequency and open circuits at DC? I know that the Xc = 1/(jwC) representation shows this mathematically, but I am wondering what is really physically hapening behind the equation. Applying an AC voltage to a capacitor moves charges from one "plate" of the capacitor to the other, then back. Somehow, if I increase the rate at which I am doing this, the peak-to-peak voltage across the capacitor begins to diminish (and the balance is applied across the series resistor). Why? Thanks!