field is created and exists as E-[( Ø)(q)]/|q|.
An Ideal Capacitor Is an Electron Current Blocker
We point out that, if the capacitor's components are ideal, completely rigid, and do not physically move, then the capacitor is a "dm/dt blocker." If the charges really were frozen in place, then the potential would flow across the plates at the speed of light, via the flow of excess massless displacement current dØ/dt. In that case, an ammeter would not show the classical "exponential fall-off" of the current with time; the electron current dq/dt would occur as a single-point Dirac delta function at t=0, and would be zero thereafter. And no electrons would be able to move in zero time. The voltage would show an instantaneous adjustment to the charged value with a single step-function, and the capacitor would charge up fully, instantly, with no work (energy loss) whatsoever being done. And this charge-up of the capacitor would not dissipate in the slightest the source furnishing the voltage; there would be no electron current dq/dt through the back EMF of the source, hence no work inside it to deplete its separation of charges.
Problems With Ordinary Capacitors
However, most ordinary capacitors are much more than just an ideal capacitor. The plates move, the dielectric moves, etc. due to the forces created upon them by the E-fields created upon the trapped charges in them. The spatial translation of the resulting force moving the plates constitutes work; i.e., it dissipates some of the flowing dØ/dt energy. Each movement of the plates and/or dielectric carries with it all its internally trapped charges. The movement of those charges constitutes a substantial longitudinal electron current dq/dt, when compared to the longitudinal "drift" electron current in normal circuits. [Electrons spend most of their time moving radially in a wire, not down it.] This "moving plate and its transported charges" make an electron current, which pumps the inert electrons in the ground return line back through the back EMF of the source, 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