In physics class I learned that removing the dielectric out of a charged capacitor increases the voltage and therefore the energy stored.

Could this effect be used to build a generator? The extra voltage could be extracted somehow, the dielectric put back and the capacitor recharged at the lower voltage. How efficient would this be?

  • \$\begingroup\$ Energy cannot be created nor destroyed. It can only be changed from one form to another. \$\endgroup\$ – Adam Z Nov 7 '17 at 12:40
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    \$\begingroup\$ @AdamZ energy is put in by pulling the dielectric, which requires work \$\endgroup\$ – fafl Nov 7 '17 at 12:43
  • \$\begingroup\$ Electrostatic generators are all based on capacitive voltage-multiplication. VandeGraaff and Wimshurst have moving capacitor plates (with VDG the belt is moving the charge, not changing the dielectric. Original VDG had foil patches on its belt, with contact-brushes.) Yours could have rotating flat plastic fan blades between unmoving metal plates, with rotating brushes or ion-needles to contact the plates during the voltage-peaks. Or, hold the plastic fan still, and rotate the plates and brushes. Also it should act like a motor. \$\endgroup\$ – wbeaty Nov 7 '17 at 14:51

Yes. Such a generator does exist, and has done for centuries. It is called a "Wimshurst Machine"

In a Wimshurst machine, the two insulated discs and their metal sectors rotate in opposite directions passing the crossed metal neutralizer bars and their brushes. An imbalance of charges is induced, amplified, and collected by two pairs of metal combs with points placed near the surfaces of each disk. These collectors are mounted on insulating supports and connected to the output terminals. The positive feedback increases the accumulating charges exponentially until the dielectric breakdown voltage of the air is reached and an electric spark jumps across the gap.

The machine is theoretically not self-starting, meaning that if none of the sectors on the discs has any electrical charge there is nothing to induce charges on other sectors. In practice, even a small residual charge on any sector is enough to start the process going once the discs start to rotate.

However, the energy comes from the energy put into the rotating disk(s)

enter image description here

  • \$\begingroup\$ I guess this is it. I had something in mind that works without brushes, but maybe it's just not possible. \$\endgroup\$ – fafl Nov 7 '17 at 13:29
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    \$\begingroup\$ I'm not sure a Wimhurst is exactly what the OP asked for, it reduces capacitance by separating the plates, not by withdrawing dielctric, a Wimshurst is a continuous acting electrophorous \$\endgroup\$ – Neil_UK Nov 7 '17 at 13:48
  • \$\begingroup\$ @Neil_UK Well, as Einstein might have said - it's all relative \$\endgroup\$ – Dirk Bruere Nov 7 '17 at 13:49
  • \$\begingroup\$ OK, I've sketched out what the OP's machine would look like, and it is a double-sided electrophorous. So a Wimhurst does act the same way, it just wasn't immediately apparent to me, and still isn't obvious. \$\endgroup\$ – Neil_UK Nov 7 '17 at 14:44
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    \$\begingroup\$ @fafl An AC version is possible: shove the dielectric in and out, and the voltage goes up and down. Do it with a stack of metal plates and dielectric surfaces, like a tuning capacitor in an old radio, high-speed rotating plastic blades. Add a series capacitor and a diode bridge to tap its energy during peaks. The diode Vf gives the main loss mechanism. (So perhaps use moving brushes instead.) Heh, use the output to keep the capacitor plates charged, and you have a self-exciting generator, like Wimshurst and Kelvin Thunderstorm. That, or use electrets for your dielectric plates. \$\endgroup\$ – wbeaty Nov 7 '17 at 14:44

The energy is \$E=\dfrac{C\cdot U^2}{2}\$. Removing the dielectric reduces the capacitance, that's why the voltage increases - because of the law of the energy conservation. In few words, the energy is the same.

  • \$\begingroup\$ But then why does it say the energy increases here? quora.com/… \$\endgroup\$ – fafl Nov 7 '17 at 12:44
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    \$\begingroup\$ The charge remains the same. But the work done removing the dielectric most definitely adds energy to the system. \$\endgroup\$ – Dave Tweed Nov 7 '17 at 12:50
  • \$\begingroup\$ @DaveTweed I never said that the charge changes. \$\endgroup\$ – Marko Buršič Nov 7 '17 at 13:33
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    \$\begingroup\$ But you said, "the energy is the same." Which is flat-out wrong. The point is, something remains the same, but it isn't the energy. \$\endgroup\$ – Dave Tweed Nov 7 '17 at 14:17
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    \$\begingroup\$ Electric energy is not conserved, since mechanical energy is being injected into the system. If we force the attractive capacitor-plates apart, the charge stays constant, the capacitance decreases, and the energy increases fast (going as the square of plate-spacing, for wide plates.) \$\endgroup\$ – wbeaty Nov 7 '17 at 14:35

No, I don't think you can make a generator. When you remove the dielectric, the amount of free charge (i.e. the charge actually on the metal plates) stays the same. The total energy of the plates+dielectric system also stays the same.

The plate potential goes up because the dielectric is now depolarized and has no electrostatic energy anymore. This energy has been transferred to the capacitor in the form of increased potential. You could drain off some "extra" charge to reduce the potential back to the initial value, but when you put the dielectric in again it will repolarize and reduce the potential even further.

Putting the dielectric back in is the reverse process. The total energy does not change. It requires no force to put the dielectric back in. Therefore there is no work done.

EDIT: LOL ok apparently you CAN. That Wimshurst machine thing is cool. I imagine that the asymmetry is key to producing forces.

  • \$\begingroup\$ The total energy does NOT stay the same. It requires work to remove the dielectric, which adds energy to the system. When you put the dielectric back in, you get that energy back -- there's actually an attractive force that draws it in. This is the principle by which an electrostatic motor works, and if you have a motor, you can also have a generator. \$\endgroup\$ – Dave Tweed Nov 7 '17 at 14:23
  • \$\begingroup\$ Actually try it. You'll find a large attraction-force that pulls the dielectric back into the capacitor (called "slot-effect" in electrostatics.) Pulling out the dielectric is performing work and storing energy. Also, if you separate the parallel plates, the voltage increases, and the stored energy increases faster (as the square of voltage.) It's the macro-analog of dielectric polarization (which is the EM dual of magnetization.) \$\endgroup\$ – wbeaty Nov 7 '17 at 14:30
  • \$\begingroup\$ Thx, I have to think about that one for awhile. \$\endgroup\$ – user168148 Nov 7 '17 at 14:43

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