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?
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.
Infinities can be tricky.
The force between two charged particles varies inversely with the square of the distance between them. The energy required to increase the distance between two oppositely-charged particles from d1 to d2 is the integral of the force over that path. Even if d2 is infinite, this integral has a finite value.
This result generalizes to large collections of charges on, say, the plates of a capacitor. What this means in terms of your question is that the capacitance of the two plates does not actually tend toward zero as they are moved apart, and the voltage does not go to infinity. One way to interpret this result is to say that each plate individually has some minimum value of capacitance to the universe "at large".
It may help to visualize this not as two parallel plates, but rather as two concentric spheres, and allow the outer sphere to grow to infinite radius.
It may also help to draw the analogy with gravity, which is another inverse-squared force. An object falling to the surface of the Earth, even from infinitely far away, has a finite amount of energy (and a finite velocity) when it arrives.
Charge will stay on a capacitor's plates unless that charge can be carried elsewhere. If the charged plates are isolated, then pulled apart in a vacuum, they'd keep their charge indefinitely. Dust, humidity, air itself, can all carry off that nonzero charge.
Like charges repel, so they spread out over the surface of a conductor. The plate or plate assembly wouldn't really be pushed apart until we're talking incredibly dense charges. Even then, I'd expect to capture dust, ionize the air, or shed "conductor" atoms one at a time rather than cause the plate to fall apart on any more macroscopic scale.
Keep in mind that the charge doesn't necessarily remain on the plates. Especially charging a Leyden jar to high voltage, the charges will jump the air gap between the foil and jar to sit directly on the dielectric surface. You can remove the plates, but the charge remains with the jar. Here's a good demonstration https://www.youtube.com/watch?v=9ckpQW9sdUg
But you could accomplish the original thought experiment by using a double layer dielectric. Construct a capacitor out of two pieces of foil, with two sheets of plastic between them. Then charge it up to high voltage, remove the foil, and the plastic sheets will remain stuck together. Then peel the plastic sheets apart, and in a perfect insulating medium they should remain fully charged. In air, their surface charge density would be too high and bleed some off as soon as they're separated, down to the 26.55 microcoulombs per square meter limit http://www.coe.ufrj.br/~acmq/efield.html ... though that's still enough to stick a balloon to the ceiling :)
Reasoning just mathematically and assuming you have your plates perfectly insulated. As long as the plates are large compared to the plate size A (A is the area, not a length, but this is an approximation anyway), the system will behave as a plate capacitor : U =Q d / ( A epsilon). So voltage will go up linear with the distance between the plates.
Note that the energy gain comes from the force needed to separate them.
But when the distance between the plates gets larger than A the two plates will start behaving more and more like charged points with U= Q Q /(d 4 pi epsilon). This goes to 0 in the limit of d going to infinity.
Now, the force needed to separate them further drops very quickly, never becoming exactly 0 but much smaller than the increase in separation distance.
You can't just look at this mathematically. In that the ability of the capacitor to hold a charge depends on the electrostatic field between the plates and that field will weaken with distance, the voltage will not indefinitely increase. It should increase to a point as it can be likened to a rubber band being stretched and storing energy, but eventually that rubber band will break and the energy will redistribute. Imagine plates being seperated until all you had was a straight wire with a plate on either end. Do you think the charges wouldn't redistribute and electrons on one end wouldn't move toward the other in which a deficiency exists? We're told to not rely on intuition, but this is where visualization is more powerful than a formula.
If you want to watch high voltage capacitor plates being moved apart, try a Wimshurst Machine
Theory is here
