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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?

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    \$\begingroup\$ define 'fully charged. The amount of charge you can place onto a capacitor/two-plates is limited by the dielectric withstand. Too much and it will break down. If you are talking about "fully charged" being at the corona inception voltage AND then moving the plates apart & assuming in a perfect vacuum then they would remain charged \$\endgroup\$
    – user16222
    Mar 9, 2014 at 10:35
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    \$\begingroup\$ The main point of my question is the retention/losing of charge from the plates. Whether or not the capacitor is fully charged is not particularly important. \$\endgroup\$ Mar 9, 2014 at 10:42
  • \$\begingroup\$ Thanks, guys. You're all awesome! You're the reason why I love StackExchange. \$\endgroup\$ Mar 9, 2014 at 16:47

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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.

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  • \$\begingroup\$ Thanks a lot, Andy. Just one more thing please. If there is an attractive force between the plates, will there also be repulsive force between the pieces of a plate that is made of many small pieces packed together? I assume that since the pieces of one plate all have the same positive or negative charge, they will repel each other. Am I right? \$\endgroup\$ Mar 9, 2014 at 15:27
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    \$\begingroup\$ @DavidHeisnam This is correct - look up gold-leaf electroscope - a very,very thin plate of gold is galvanically attached at one end to a vertical fixed plate that can receive charge. As voltage (aka charge) increases, the gold plate moves away from the fixed plate. \$\endgroup\$
    – Andy aka
    Mar 9, 2014 at 16:16
  • \$\begingroup\$ Now I remember having read about gold-leaf electroscope in high school. Thanks again, Andy. You have been really helpful. \$\endgroup\$ Mar 9, 2014 at 16:45
  • \$\begingroup\$ Oops! I think you commented on the wrong post! \$\endgroup\$ Mar 9, 2014 at 17:00
  • \$\begingroup\$ @DavidHeisnam - a comment was made by HL-SDK but he's removed it so now my comment (above) doesn't now make much sense!!! \$\endgroup\$
    – Andy aka
    Mar 9, 2014 at 17:11
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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.

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    \$\begingroup\$ To expand on this a little: it is often convenient to imagine that the plates of a capacitor are infinite planes, ignoring the edge effects. In this simplified case, the voltage is linear with the distance between the plates, and so it would indeed tend to infinity as the distance increased. Meanwhile, in the real world, once the distance between the plates becomes comparable with (or larger than) the size of the plates themselves, the linear relationship no longer holds and the voltage remains finite. \$\endgroup\$ Mar 10, 2014 at 2:05
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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.

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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 :)

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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.

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  • \$\begingroup\$ OP specified disconnecting the capacitor first, so there would be no straight wire. \$\endgroup\$ Feb 15 at 10:19
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If you want to watch high voltage capacitor plates being moved apart, try a Wimshurst Machine

Theory is here

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