# Physics of capacitors over frequency

1. Jul 14, 2004

### PMASwork

(not sure if this belongs here or in the Electrical Engineering section)

Howdy-

Consider a low pass RC 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!

2. Jul 16, 2004

### Goalie_Ca

Well think of it this way, in a city you might have a highway with a 2-lane bridge that everyone needs to take. Eventually the cars will pile up there... kinda like the charges. Lets say traffice is not rush hour, and its going both ways. Well, then there will always be people turning around lets say and start going the other way.

Okay, i know that's a lame analogy but that's the best i could come up with on the spot.

3. Jul 16, 2004

### Gza

To understand it physically, just look look at Ampere's law. The magnitude of the displacement current is proportional to the time derivative of the electric flux through a surface (that surface would be between the plates of the capacitor, where no "real" current passes through, yet a changing electric field sure does.) So basically increasing the frequency of your AC directly increases the displacement current between the plates of the cap. If you hooked up a DC source, the electric field wouldn't change in time, thereby zero displacement current between the caps (and zero "real" current of course.)

4. Jul 16, 2004

### wisky40

Well, also as you know capacitors are storages of energy. When D.C. is use, positive charges flow in one direction and negative charges flow in the opposite direction. As soon as the capacitor saves all the energy in the electric field between it's plates the process of the charges flowing stops.
When A.C. is use, the same phenomenon takes place, but there is one difference, the flow of charges starts to switch direction discharging the capacitor and then charging the capacitor in the other direction(changing polarity).
If the frequency of the source is change into a higher frequency, the amount of charge storage in the plates is less because the direction of the flow in going to switch faster not allowing the capacitor to get full of energy, having in this way less voltage at the plates of the capacitor. Because q>q' => V=q/C>q'/C=V'.