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1. If "displacement current" crosses from "one plate to the other", why does the other plate have to be connected to the other side of the circuit?
You must have an electric field between the plates, therefore a potential difference. The electric field slightly shifts the electron couds from their normal position around the positive charges of atomic nuclei of the dielectric. This is a real motion of charges. The electron clouds being no more exactly centered around the positive charges, each atom constitutes an electric dipole and the displacement current represents this relocation of charges. If the dielectric is the vacuum, the electron/positron pairs that are created at random due to the quantum fluctuation, play the same role as the atomic charges of the dielectric. This explains why the permittivity ε0 is not null in vacuum, allowing also displacement currents. 2. If I measure the voltage between either plate and ground, what voltage reading do I get?
This voltage is not defined. It doesn't depend on the charge of the capacitor. It is as if you had additional real capacitors connected between each plate and the ground. You have a circuit, the voltage depends on the voltage sources that can exist along this circuit, and these additional capacities depend on the distance between each plate and the ground, on the geometry of the conductive surfaces facing one another, and on all other conductors that are near the capacitor and can be more or less connected to the ground and disturb the electric field. In practice, the magnitude is on the order of pF. It is to be noted that there is strictly no relation between the value of the capacitor which is the capacity between plates, and the capacity between each plate and the ground or the conductors in the surrounding. It is like several capacitors in series: each one must be calculated independently. In practice, all generators and electronic circuits present a capacity relative to the ground, so it is extremly difficult to modelize a real circuit when these capacities are not negligible, which is always the case at high frequencies, due to the fact that even some tenth of pF present a significant impedance at these frequencies.
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Topic: Tossing Ideas Around (Read 2876 times)
