I think you are forgetting that the "spring" is much weaker on the pullback stroke because you have removed charge at the top of the stroke. Anyway, the answer is no, I don't think it will "work" if by "work" you mean the ability to take out more energy than you put in all told. But that's just my old-fashioned opinion.
Humbugger
Yes, I now see that the Coulomb force will be 1/2 on the return stroke so the result force will be 1/2 of the Coulomb force in total. There will be small losses in the capacitor charge so we will also need to recharge the capacitor for losses. But all in all, all we need to input is enough energy to keep the capacitor plate in weight resonance. Will this give us more energy out than we did put into the system? You say no, I say maybe because we only need to pay for the first initial charge of the capacitor and can repeat the cycle a lot of times before we need to top up the capacitor charge again. So we end up with the question, do we need to use more energy to fight the 1/2 Coulomb force up stroke than we get back from the double capacitor energy at the double distance? Anybody up for doing math on a real life capacitor? Plate=round radius 50mm, plate distance 0,1mm at closest and 0,2mm distance at top stroke. Voltage on plates 12 volt. Dielectric= Air. GL.
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Topic: Konstantin Meyl -- Tesla Scalar Wave Theory (Read 51469 times)
