 depleting the source. Consequently, the ordinary capacitor will simply release as much energy as work (to move the plates and dielectric) as it stored. Hence, it will also produce dissipation of the source via the amount of energy stored in the capacitor. You still get "free energy" stored in the capacitor, but also dissipate the source by an equal amount.
Rigidized Capacitors Must Be Used
Only rigidized capacitive collectors are useful in free energy devices. Such capacitors are in fact actually available, e.g., as calibration standards, but they are extremely expensive ($400 to $2,000 or so each, for a capacitance reaching about 1 uf).
So, capacitive type collectors must be rigidized, if used in
overunity circuitry. Even so, in a single integrated circuit,
although one collects free energy, one will use half of what was
collected to dissipate the source. Not all the remaining half
will be discharged through the load; some will be discharged in
other circuit and component losses. Hence, there will always be
less work done in the load than is done in the source to kill it,
by a conventional two wire single closed circuit. In my second
referenced paper (Feb.94), I included precise proof that this is
true. One must use energy collection and shuttling between two
isolated circuits, and the load discharge current must not pass
back through the primary source of potential.
We have previously provided precisely how to utilize capacitive collectors in our two referenced papers. We point out here that the capacitors must be calibration standard capacitors, or specially made rigidized capacitors.
It Does Not Require Electron Current to Charge An Ideal Capacitor
For the benefit of the skeptic, this is already proven. We simply list references (2) and point out the equation that represents the energy K in a charged capacitor. Here we have K = ½(CV)2. It is totally the displacement current
dØ/dt flowing (from a higher potential) onto the charging plate that produces the higher potential Ø on that charging plate, and hence a V between the two plates, one of them (the "ground" side) being held at a constant potential. The mass displacement current component dm/dt of the electron current dq/dt has nothing whatsoever to do with energy accumulation; it has only to do with the dissipation of energy that is happening simultaneously in all losses and loads in the circuit loop.
We reiterate that most ordinary capacitors have terrible internal movement, and accomplish as much energy dissipation as they do energy collection by permitting dq/dt and work performed upon the plates and dielectric to move them. The standard two-wire circuit also guarantees that all such dq/dt current "through" the capacitor is passed back through the source against its back EMF, doing an equal amount of work in the source to dissipate its separation of charges and "destroy" the source.
An ideal capacitor does not pass dq/dt, but only massless displacement current as theorized by Maxwell to save current continuity in a circuit containing a capacitor, and hence to save Ampere's current law. That is, an ideal capacitor is a dm/dt blocking device. However, the capacitors utilized in normal circuits are not ideal capacitors at all. By allowing the plates to move, electron current dq/dt is created on both sides of the capacitor. Otherwise there would not be a ground return dq/dt, but only a ground return dØ/dt. This dØ/dt would not and does not push electrons back up through the source against its back EMF; else the ground side of the source, which is engaged in continuous dØ/dt exchanges with the vacuum, would produce destructive amperage dØ/dt in the battery or potentialized source, against its back EMF, while it was simply sitting on the shelf. In fact, a flow of dØ/dt continually runs from the vacuum to the positive terminal, then through the inside of the battery to the negative terminal, and thence back to the surrounding vacuum. Also, the incoming flow from the vacuum "splits" at the positive terminal, where one branch flows inside the source to the negative term