# Generating current using capacitor

If you get a capacitor that is two plates and a dielectric and next to it you put two plates close but not in contact at opposite sides and the capacitor in the middle is charged, each of its plates will act as a Leyden jar to induce charges in the outside plates.

When the outside plates are connected they will induce current because a huge potential difference is created between the outside plates.

The potential difference is created the same way the outside aluminum foil on a Leyden jar is charged, when the water inside the bottle is charged.

If I fixed the two outside plates (white plates) in one place and rotated the capacitor (red plates) at 60Hz so that the electric field is charging fast just like a magnetic generator, will it produce constant current if the capacitor itself is not discharged?  b. If the capacitance between the plates of the capacitor in the middle is 1µF and that capacitance between the plates of the capacitor and the outside plates is 5µF. If the capacitor is rotated at 60Hz will the current be generated be

Q= 5µF x 1200v =6

current per second 6 x 60Hz x 2 (2 is number of plate discharges per cycle) =720 amperes

What do you think?

• "... when the outside plates are connected they will induce current because a huge potential difference is created between the outside plates." (1) You are implying that there is an unmentioned huge voltage source somewhere in this setup. (2) You have a direct wired connection between the two outer plates. As a result you won't get any voltage difference. If this is a free-energy device I can tell you now that there will be no success ahead. Can you edit to clarify? Feb 13, 2021 at 10:18
• I strongly recommend that you ask this question on a physics site. Feb 13, 2021 at 11:39
• "Current per second 6 ...". Current per second isn't a thing. Current is the rate of change of charge, $Q = \frac {di}{dt}$ and $1 \ \text A = 1 \text {C/s}$. Where did '6' come from? Feb 13, 2021 at 11:44
• @successahead: I think: "Build it and see if it works." You keep making suggestions for various influence machines and asking if they will work. Try building one.
– JRE
Feb 13, 2021 at 15:27

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 1200 V, 5 µF capacitor1 whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

The source impedance of your generator is two of these in series, and this puts a limit on how much power you can get out of it. If you match the source impedance with a load of 1060 Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 400 mA through the combination, for a total power delivered of about 170 W.

1 Assuming air is the dielectric, you'll need a spacing on the order of 1 mm in order to withstand the voltage, which means that the area of each plate needs to be about 566 m2, roughly the same order of magnitude as a baseball diamond.

The basic question seems to be "is it possible to make a generator based on electric charge and capacitance?" The answer to that question is yes. Electrostatic machines that are the "duals" of electromagnetic machines have been constructed and are currently being manufactured. There is currently ongoing research and development of such machines for use as motors and generators. The word "dual" indicates that, in electrostatic machines, electric fields, voltage and capacitance assume the roles of magnetic fields, current and inductance electromagnetic machines.

https://player.pbs.org/viralplayer/3036744232/

https://www.c-motive.com/tech

https://en.wikipedia.org/wiki/Electrostatic_motor

How about we try some numbers?

• You want one microfarad of capacitance between your moving plate and a fixed plate.
• You'll need some space between the fixed and the moving plate - if they are too close together they'll drag and you don't want that.
• Capacitance of a parallel plate capacitor is given by $$\C = \epsilon_0\frac{A}{d} \$$ where $$\C \$$ is in farads, A is in the area of the plates square meters, $$\d\$$ is the separation distance between the plates in meters, and $$\\epsilon_0\$$ is $$\8.8541878128(13)×10^{−12}\$$.

As you can see, it takes a very large area and a very small separation to get large capacitances.

Assuming a separation of 0.5 mm, and a capacitance of 1 microfarad, you'll need an area of about 60 square meters. Spinning. At 60 rotations per second - that's 3600 rotations per minute. That's a thing nearly eight meters across rotating at a very high rate - with a clearance of 0.5 mm between the surfaces.

Do you see a problem here?

Your idea is impractical, even if by some chance you got the physics and mechanics correct.