Is the Electrostatic Motor Affected by Changes in Earth's Ionization?

AI Thread Summary
The discussion centers on the impact of Earth's changing electrostatic field on the operation of an electrostatic motor, particularly its functionality during nighttime when solar ionization decreases. Participants debate the nature of the Earth's electrostatic field, questioning whether it is static or dynamic, and emphasize that energy extraction requires a changing field. The conversation also touches on the potential for creating a capacitor using a metal aerial and ground, suggesting methods to harness energy from atmospheric signals. Concerns are raised about the need for credible sources to support claims regarding the electrostatic field's behavior. The thread concludes with a call for evidence to substantiate the discussion points.
mrchiller
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Hello I wanted some feedback on the idea I have about this electrostatic motor. One thing not mentioned is that the electrostatic field of the Earth is changing from day to night because of the ionization from solar rays is stronger during the day. Yet I wonder if the motor would stilll run for most of the night.

Any thoughts about the other article I wrote "An Electromagnetic Generator System."?

The ideas are posted here:
http://mrchillersalternativeenergy.blogspot.com/
 
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mrchiller said:
Hello I wanted some feedback on the idea I have about this electrostatic motor. One thing not mentioned is that the electrostatic field of the Earth is changing from day to night because of the ionization from solar rays is stronger during the day. Yet I wonder if the motor would stilll run for most of the night.

Any thoughts about the other article I wrote "An Electromagnetic Generator System."?

The ideas are posted here:
http://mrchillersalternativeenergy.blogspot.com/

What do you mean by "electrostatic field of the Earth"? Is it a static field at any given time? You can only get cyclic energy out of a changing field, not a static field. Otherwise we could extract energy from gravity in a cyclic way...
 
If we create a potential difference between a metal aerial and ground this creates a capacitor out of the sky. Then by putting a diode bridge cross the aerial to ground which outputs to the commutator, the AC signals of the sky in resonance with the aeirial to ground system would output DC. We can create electric repulsion of the electrets by switching this potential via a commutator to keep the torque.

This is why I simplified the forces as force of electret <--- qE = (B^2)/mu_not --->magnetic force. The magnetic force comes from magnetic pressure*A ~ F where A = area. Magnetic pressure = [(B^2)/2mu_not]. This might be more accurate if: F=qE then the magnetic equivalent must be (Phi)*H or (Volts-second)*(Amps/meter).

Look up:
http://en.wikipedia.org/wiki/Magnetic_pressure
This is used when describing induction effects so it represents in this case the displacement current that creates the opposing electric field between the electret and the metal plates.
http://en.wikipedia.org/wiki/Electric_force
Also
http://en.wikipedia.org/wiki/Faradays_law
where it talks about curlE = -dB/dt
then reference to this page and look at the radio wave diagram.
http://en.wikipedia.org/wiki/Radio_waves
As you can see there is a Curl E per unit of time moving radially outward which creates -dB/dt in an aerial if in resonance. Because the electret is really a type of capacitor in itself so charge is transferred from the aerial via ~RF? AC energy to the electret motor.
 
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mrchiller said:
If we create a potential difference between a metal aerial and ground this creates a capacitor out of the sky. Then by putting a diode bridge cross the aerial to ground which outputs to the commutator, the AC signals of the sky in resonance with the aeirial to ground system would output DC. We can create electric repulsion of the electrets by switching this potential via a commutator to keep the torque.

This is why I simplified the forces as force of electret <--- qE = (B^2)/mu_not --->magnetic force. The magnetic force comes from magnetic pressure ~ F/A where A = area. Magnetic pressure = [(B^2)/mu_not](A). This might be more accurate if: F=qE then the magnetic equivalent must be (Phi)*H or (Volts-second)*(Amps/meter).

From my quick Google search and reading, the E field looks to be largely static. Nothing to rectify.

You need to provide some links to credible articles that indicate otherwise, or this thread will be closed.
 
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