capacitor
I'm a mechanical engineering student and I'm working on a project that involves a high voltage capacitor.
I understand that when the separation between the plates of a charged capacitor is increased, the voltage increases. But I'd really like to know what happens to the plates if the capacitor is fully charged , disconnected from the charging circuit and then the plates are moved apart from each other by an infinite distance. Will each plate remain charged?
Your capacitor misconceptions aside, it is possible to deliver power by effectively one wire. Driving an antenna is an example of one wire power delivery. Power is delivered to the antenna via the transmitter and the antenna radiates the energy as an electromagnetic wave.
Your gap is too wide. Make it very, very very narrow. And rolled into a cylinder. Like a real capacitor.
Yes, +q could be less than -q, but only if the attraction/repulsion effects of electrons in the connecting wires were nearly as large as the attraction/repulsion down between the capacitor plates. (In that case the plates wouldn't be a near-perfect electrical shield for the fields produced by the wires.) But with real-world capacitors, this doesn't happen, and instead the field between the plate is totally enormous compared to the tiny fields produced by electrons in the wires. If +q only differs from -q by a millionth of a percent, we ignore it. See Engineer's capacitor vs. Physicist's capacitor, a split metal ball, versus two separate balls.
For capacitors used in circuitry, if we dump some charge on one capacitor terminal, exactly half of it will seemingly migrate to the other terminal. Weird. But "physicist-style capacitors" with small, wide-spaced plates are different, and an extra electron on the wire will make +q not equal to -q.
In detail: if the capacitance across the plates is 10,000pF, and the capacitance to Earth of each wire and plate is 0.01 pF, then the opposite plate's charges will ignore any small +q and/or -q on the connecting wires. The attraction/repulsion of electrons in the wires doesn't significantly alter the enormous +q and -q on the inner side of the capacitor plates.
Engineers use real-world components: wide capacitor plates with very narrow gaps; gaps the thickness of insulating film. But if you were a physicist, your capacitors might be metal spheres with large gaps between, or metal disks where the space between the plates was large when compared to their diameter. (Or you'd draw a capacitor symbol where the gap between plates was enormous and easy to see.) In this case the attraction/repulsion of electrons on the connecting wires would have an effect on the balance of +q -q between capacitor plates.
PS
Another weird concept: make a solid stack of thousands of disc capacitors: foil disk, dielectric disk, foil disk, etc. Use half-inch wide disks, and stack them up into a narrow foot-long rod. Now connect one end to 1,000 volts. The same kilovolt will appear on the other end! The rod is acting like a conductor. Yet its DC resistance is just about infinite. Series capacitors! Each little capacitor induces charge on the next and the next, all the way to the end.
Best Answer
Charge = capacitance x voltage (\$Q=C\cdot V\$)
If the capacitor has a voltage across its plates and the supply is disconnected, the charge remains irrespective of the distance so, if distance increases (and capacitance falls) then voltage increases proportionally. If the plates are taken to an infinite distance, the voltage becomes infinite.
It should be noted that the energy "held" in the capacitor increases as the plates are pulled apart i.e.
Energy = \$\dfrac{CV^2}{2}\$
The increase in energy comes about because work (joules) has to be done to move the plates physically apart i.e. there is a force needed to open up the gap. This, I believe keeps all the conservation of energy and charge equations happy and smiling. Remember, that on a regular capacitor, there is an attractive force between the two oppositely charged plates and it is this force that is trying to stop the plates from being pulled-apart.
If the capacitor plates remain connected to the supply, as the distance increases the voltage must stay the same so therefore charge is reduced (because C reduces) and this pushes current back into the power source.