So, consider an average shorted capacitor connected to an alternating current power source. This capacitor has a certain capacitance, voltage, displacement current, etc. associated with it. Assume the dielectric of this capacitor is air. Let us increase the frequency of this source. Recall that this would cause more current to leak through the dielectric in the capacitor, since the dielectric is not perfect. I believe the current leakage is in the form of displacement current. At a certain point, the frequency is so high that the capacitor is leaking current completely, i.e. the dielectric is no longer effective.
If we take this capacitor and then increase the distance, the capacitance will go down, whereas the voltage goes up, for a given charge. Eventually we increase the distance so far that one plate is very close to the source, while the other is very close to the load.
The voltage and frequency are both very high, and so, the circuit is complete by means of displacement current from one capacitor to another.
So does this work, or is there something I am overlooking? If anyone needs me to explain it better, just ask and I'll try to help you understand what I am saying.
If we take this capacitor and then increase the distance, the capacitance will go down, whereas the voltage goes up, for a given charge. Eventually we increase the distance so far that one plate is very close to the source, while the other is very close to the load.
The voltage and frequency are both very high, and so, the circuit is complete by means of displacement current from one capacitor to another.
So does this work, or is there something I am overlooking? If anyone needs me to explain it better, just ask and I'll try to help you understand what I am saying.