So what happens if you charge up an air gap capacitor, and remove and short the plates ?
The main idea is that energy is in the fields, while the wires guide the location in space for those fields. If you don't have guiding wires, then the field will spread all over and propagate just like the light or the radio waves propagates all over.
About the air (or vacuum) capacitor, if you dismantle/separate the plates, then the field will follow the new position of the plates. The energy is not "anchored" to empty space. The energy is in the fields, the fields are "anchored" to the charges, and the charges are "anchored" with the metal (as in, the charges prefer to stay with the metal plates of the capacitor). Overall, the field appear "anchored" between the two plates of the capacitor. So, in this case, the field will follow the plate movement, and if you touch the plates, the capacitor discharges.
Then, why did the energy in that video remained in the glass, and didn't remain "anchored" to the plates? The field was trapped in the glass, because when the glass dielectric was removed, charges from the plates were transferred (by induction) to the glass, and the induced charges in the glass remained attached to the glass (glass is a very good insulator), until the capacitor was re-assembled. Vacuum space does not have charges, so the dissectable capacitor trick doesn't work in vacuum.
Air do have charges, so the trick should work with air, just that the air is a gas, the molecules and the charges attached to the air will mix very fast. Any induced charges during removing an air dielectric, will move with the air and neutralize each other very fast. So it won't work with gas dielectric, and most probably won't work with a liquid dielectric either.
The glass was able to keep the induced charges separated because it is a very good insulator
and a solid. Any charge placed on glass remains "sticked" in its place. Can not move.
However, there's another very funny experiment one can do while trying to dismantle an air capacitor:
- Imagine you have two horizontal metal plates, one above the other, so a capacitor
- you charge your capacitor from a 9V battery, then remove the battery
- the capacitor is now charged, if you connect a very high impedance voltmeter, it will measure 9V (the voltmeter has to be very high impedance, so it won't discharge the capacitor much during measurement)
- now, leave the voltmeter connected (still showing 9V), and try to disassemble the capacitor by lifting the top plate away
- keep an eye on the voltmeter while lifting the top plate
To everybody's surprise (well, at least to my surprise when I've seen this for the first time) when you lift the top plate, the voltage increases like crazy!
![Shocked :o](https://www.eevblog.com/forum/Smileys/default/shocked.gif)
And if you bring back the top plate in the initial position, the voltage is 9V again. And if you lower the top plate even more (under the initial position, but without touching the bottom plate), the voltage goes less than 9V. In fact, you can generate any voltage (of the same polarity) by simply changing the distance between the plates.
![Grin ;D](https://www.eevblog.com/forum/Smileys/default/grin.gif)
So, the voltage changes with distance, in theory between zero and infinity. But why always the same sign of voltage. Why if you go to "negative distance" the other side of the plate, the sign of voltage doesn't change? That's another can of worms question.