@Phoneboy As I understood it, the field generated by the displacement current is exactly the same as if a wire were connecting the plates, i.e. the field rotates round the wire in a circle, is this not correct?
Anyway appreciate your info this is the kind of stuff I need to know, and I am showing my own lack of knowledge in this field.
So its helpful when others can prevent me (and others) from wasting time!
I am pleased though at the amount of people on the case and looking for free energy nowadays as compared to say 5 years ago. The brainwashing is falling away and people now accept that overunity is possible and are actively looking for it, and hopefully soon we can achieve a pracical way to do it.
But anyway., if parallel plates don't produce curl then at a quick guess I am going to say a coaxial type arrangement will be better?? But i don't know.!
So yes please let us know your ideas and thoughts on this subject!
best regards
Make an RF transmitter and put a coil in the radiated field to pick up the energy from the EM waves.
But don't place the coil in the nearfield i.e. where its back emf flux will influence the transmitter coil and cause real power to be taken at the input, but place it just outside the nearfield so that the signal is still relatively high but will have no effects on the transmitter and therefore can't take more energy from the source.
As the field is radiated in all (or many) directions you can place multiple coils in the field, like sails in the wind so to speak, each deriving power to a real load.
Although the total transmitted power will not be more than that input to the system, the output can be more because the output power is not limited and is due to the magnetic field. Its only the back emf flux that normally kills the source flux and demands more real power to be input
The output current x number of turns is limited because that is what generates back emf flux and kills the source flux, but the higher in frequency you go the more volts you get per turn of the coil, hence higher power for the same level of flux. Thus output power is not dependent on input power.
(12-06-2025, 10:15 AM)dgreen264 Wrote: @Phoneboy As I understood it, the field generated by the displacement current is exactly the same as if a wire were connecting the plates, i.e. the field rotates round the wire in a circle, is this not correct?
Anyway appreciate your info this is the kind of stuff I need to know, and I am showing my own lack of knowledge in this field.
So its helpful when others can prevent me (and others) from wasting time!
I am pleased though at the amount of people on the case and looking for free energy nowadays as compared to say 5 years ago. The brainwashing is falling away and people now accept that overunity is possible and are actively looking for it, and hopefully soon we can achieve a pracical way to do it.
But anyway., if parallel plates don't produce curl then at a quick guess I am going to say a coaxial type arrangement will be better?? But i don't know.!
So yes please let us know your ideas and thoughts on this subject!
best regards
You're correct that the displacement current is essentially the same as if there was a wire connecting the plates, but the source of the magnetic field is the A vector and that has a 1/r gradient perpendicular to the vector outside of the wire. That gradient causes vorticity/eddies like a river where the current flow is faster in the center. The center of the eddies are the magnetic field. That's why I said that if the field is uniform between the plates of the capacitor you wouldn't see a magnetic field. This dude explains it pretty well. https://montalk.net/notes/327/transverse...inal-waves
(12-08-2025, 05:55 PM)phoneboy Wrote: You're correct that the displacement current is essentially the same as if there was a wire connecting the plates, but the source of the magnetic field is the A vector and that has a 1/r gradient perpendicular to the vector outside of the wire. That gradient causes vorticity/eddies like a river where the current flow is faster in the center. The center of the eddies are the magnetic field. That's why I said that if the field is uniform between the plates of the capacitor you wouldn't see a magnetic field. This dude explains it pretty well. https://montalk.net/notes/327/transverse...inal-waves
So how should be the correct arrangement of capacitor (plates or other shapes) to achieve this?
A.
(12-08-2025, 05:55 PM)phoneboy Wrote: You're correct that the displacement current is essentially the same as if there was a wire connecting the plates, but the source of the magnetic field is the A vector and that has a 1/r gradient perpendicular to the vector outside of the wire. That gradient causes vorticity/eddies like a river where the current flow is faster in the center. The center of the eddies are the magnetic field. That's why I said that if the field is uniform between the plates of the capacitor you wouldn't see a magnetic field. This dude explains it pretty well. https://montalk.net/notes/327/transverse...inal-waves
So how should be the correct arrangement of capacitor (plates or other shapes) to achieve this?
A.
Hi Andy. You can use trifilar coreless coil with 2 of the 3 wires as capacitor plates and third wire as pickup. Not sure if it's OU. In my experiments the ESR of cap seems to change when the 3rd winding is induced and current taken out. V across third wire will get almost as high as V across cap. I did not discharge cap into load in my experiments. I will try it again some time. This was all using roll of 2.5mm twin and earth house wire and spark gaps and I didn't measure much. Was just playing.
Kind regards, Sandy.
(12-04-2025, 06:58 AM)dgreen264 Wrote: Yes, the wound capacitor is right and tested
The opossite is right too is say a capacitor inside of a coil get the electrical field of the magnetic field
at 90 degrees
Regards
Hi all
Just wanted to put this idea out there... its regarding the displacement current of a capacitor.
A capacitor stores energy in the electrostatic field between the plates, but little or no attention is normally given to the fact that a magnetic field is created between the plates when current flows in the wires.
This is called displacement current and produces a (magnetic) field within the capacitor that is equivalent to that which would be created if a wire were to connect the two plates together.
A coil placed between the two plates of the capacitor can receive this oscillating magnetic energy and create an ac voltage on its terminals. A load can be placed on the coil's output and real power can be drawn from it. When current (and hence power) is taken from this coil, the coil generates its own magnetic field which opposes the original field generated by the displacement current (Lenz's law).
The net field will be reduced due to this back flux from the coil (superposition theorem applies). But as there is no 'primary' coil generating the magnetic flux, there is no mechanism for the back flux to reflect electrically back onto the input circuit thus demanding real power input as in a normal transformer.
In the capacitor-generated flux case, all that happens is that the flux is diminished due to the back flux, and this limits the power out because the net flux reduces and hence the output voltage - which is proportional to net flux - reduces.
But the way I see it, there is now way for this to change the input impedance. The input still looks totally reactive i.e.no real power, and can be resonated away to practically zero, allowing for losses etc.
There is no way for this reduction in magnetic field (due to load current) to electrically get back onto the input circuit.
So this would be free energy then?
If anyone can make suggestions or comments as to why this wouldn't work and where I am missing something in the theory, please do tell! Guests cannot see images in the messages. Please register at the forum by clicking here to see images. Guests cannot see images in the messages. Please register at the forum by clicking here to see images.
Let's do some calculations for you know how much power you can get
E = B * c
Let's think in a parallel plates capacitor, square plates of L length, d as the distance between the plates, k the relative dielectric permitivity
Let's connect the capacitor to an AC voltage source
v(t) = V* Sin( w*t )
So
E = V/d
B= E/c
As capacitor is connected to the AC power source there is reactive power and not real power extracted from the source, so total energy extracted from the source is zero
When the capacitor is wound the magnetic induction inside the coil is:
So power you can get increase with surface of the capacitor, voltage and frequency and the less separation of the plates the more power you can get, you can design for the kw level you want
Thanks for the analysis @Sparky2026. I would have to plug in some real numbers to see how feasible it is with regards to capacitor physical dimensions vs power output.
One problem you would have to deal with is capacitive coupling between the hv ac end of the capacitor and the coil wiring. You would probably have to put an earthed plate between the high end capcitor plate and the coil to prevent current drain from cap to coil, which obviously would degrade performance as it wouldn't be working as intended (i.e. taking energy from the magnetic field)
Do you think there is any mileage in the RF coil idea? It would be like a radio receiver where it receives power from the field but doesn't affect the transmitted signal power. The only difference being that you are extremely close to the transmitter, but still not close enough to have Lenz's law effects on the transmit cooil
Here is my idea for a circuit relating to this topic.
HV supply charges the inter winding capacitance of L1 and L3. S1 closes and the stored charge is discharged through L4, a torroidal inductor, almost saturating L4's core. At the same time, a voltage is induced in L2 and is stepped down to charge the output cap. After S1 opens, L4 discharges back through D1 into the "capacitor" of L1 and L3, partially refilling it and the cycle repeats.
As a place to start, if L1,L2 and L3 are comprised of a trifilar winding made from 30m of 2.5mm twin and earth wire on retail spool, we might get about 2nF of capacitance. So L4 would be a very small inductance to be saturated by this small discharge.
I haven't built this circuit. I'm just sharing my ideas. Kind regards, Sandy.
