How does capacitance work? I have a circuit whit a AC source a capacitor and a resistance all in series. I find that the difference of potential between the capacitor leads begin to change after some instants as it should. But my question is, if the resistance is very high, say infinite, the capacitor have no tension per se, but it experiences high voltage as a whole in respect to the ground. What i don’t understand is why there is not electric field between the plates.
I’m sorry if the question isn’t clear but I really cannot understand the implication of the definition of “potential”.
 A: I'd like to expand on leftaroundabout's answer as the question actually was phrased "what is capacitance" and this seems to be the core of the OP's problem (your feeling of the meaning of potential is probably correct though).
The antenna-analogy is a bit vague. You could see the capacitor as two reservoirs of electrons, linked so that when one fills up, the other is emptied and vice versa. In reality for example you have two plates near each other, each one can store electrons and the other feels the field from the other plate.
Imagine you start with one reservoir empty and the other full.
You can then send in a current into the first end of the capacitor for a short time, filling up the empty reservoir, and during this time the full reservoir in the other end will empty out into the other wire. Thus, during this short time, it looks as if a current is passing straight through although no electrons are passing the di-electric!
When the first reservoir is full though, the current stops. You then need to reverse the current, emptying it while the other reservoir pulls back its released electrons to refill, and now it looks as if our reversed current is passing straight through the capacitor again.
This is why AC (an alternating current) seemingly pass through a capacitor - an AC source simply "wiggles" the current back and forth so you continously fill and empty the reservoirs at both sides.
So your AC will pass through the cap and you get the same measured potential on both ends of the cap, in the case of infinite load (infinite R).
Important things I've left out:
The cap doesn't fill up the "reservoir" at a constant rate of course. As it is filled, the field from the stored electrons oppose the current (their potential is increased) so they fill up slower and slower, the reverse when it's emptied.
Also the frequency of the AC has to be compared to the size of the reservoirs (the capacitance!) when you consider how easy the current will cross (the impedance!).
A: If you choose $R\gg\omega/C$ , then most the voltage will drop off in the resistor. You may now argue that, for any voltage to drop off in a resistor, there needs to be some current in it. And there is, in fact, such a current.
Imagine the condensator plates as two antennas. Yes, pretty strange antennas, but antennas they are. What you essentially do is: you send an AC signal from one of these antennas over to the other one. If the receiver antenna has no significant load to feed (that is, if $R$ is big) then its potential will pretty much exactly follow the potential of the sender antenna. In other words,


*

*All the voltage you send into the circuitry will end up at the resistor.

*The voltage along the capacitor will be very small.
A: I think I got this. The capacitor it’s not even involved in the process! It’s like a dangling  wire. The real thing is that the AC source drain electrons from the ground, lowering the ground potential, but the ground is conventionally 0, so the potential of whatever is on the other side must go up!
