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Author Topic: Why wouldn't this work? (perpetuum motion)  (Read 18808 times)

Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« on: 20/06/2011 14:46:32 »
Hello everyone.

There's an idea that bugs me for last few years. I've tried consulting few PhDs and got no answers, and I have no materials to do this experiment on my own.
I know that this device shouldn't work, but I have no idea why.

Let's say that we have a charged capacitor. Let's use most simple version of capacitor - two parallel conductive plates that have same dimensions and we charge those plates to say... 100V (doesn't matter, really).

Place closed coil of wire around one plate. The coil should enter capacitor space near -100V charged plate, and exit capacitor space near +100V charged plate, so there should be nice difference in potential energy of free electrons in coil that are near +100V plate and those that are close to -100V plate.

In theory, capacitor energy is defined only by its voltage levels and capacitance. Since capacitor is charged, and disconnected, the voltage levels should stay the same, and since it is rigidly built, it should not change it's capacity.

Electric field should appear as soon as we charge the capacitor, and it should exist only between capacitor plates. This field should start moving free electrons in coil thus producing a current in coil circuit. Now, if we place resistor in coil circuit the current generated in coil should generate heat in resistor transforming electrical energy in heat.
The question is obvious. Only energy that we provided to this system was energy used to charge capacitor, and now we have heat constantly generated by current passing through resistor, and capacitor should maintain its voltage levels...

So, if anyone knows how capacitor will lose it's energy, please reply. Don't bother with imperfection of regular capacitor, I know it will "run dry" by itself. I want to know how it loses its energy when it transfers it to electrons in the coil.

Thank you!
PS Please excuse any grammatical and spelling mistakes that I made.



 

Offline RD

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Why wouldn't this work? (perpetuum motion)
« Reply #1 on: 20/06/2011 16:30:47 »
« Last Edit: 20/06/2011 16:32:21 by RD »
 

Offline imatfaal

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Why wouldn't this work? (perpetuum motion)
« Reply #2 on: 20/06/2011 17:08:33 »
In this house we ...

 

Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« Reply #3 on: 20/06/2011 17:15:40 »

Work is done when charging or discharging a capacitor ... 


We woild charge it only once and then disconnect it(leave it's connectors open).

Please copy this text and paste it in single-space font to see the "image".
Start of text:"

        ------------------------    Charged to +100 V (Capacitor plate)
                            _____
                      _____/     |  Coil (closed)
                 ____/           |
            ____/                |
       ____/                     |
      | ------------------------ |  Charged to -100 V (Capacitor plate)
      |__________________________|
"end of text

@imatfaal
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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #4 on: 20/06/2011 21:04:59 »
Are you saying you know it won't work but you'd like to understand why, or are you saying it does work?

If you believe it works, I'm afraid we'll have to move it to "New Theories".
 

Offline syhprum

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Why wouldn't this work? (perpetuum motion)
« Reply #5 on: 20/06/2011 22:14:36 »
Did not Hertz, Marconi el al look into this sort of thing ?
 

Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« Reply #6 on: 20/06/2011 23:45:25 »
Are you saying you know it won't work but you'd like to understand why, or are you saying it does work?

If you believe it works, I'm afraid we'll have to move it to "New Theories".

I don't know if it works. I just believe it doesn't - there's no proof either way. I'd love to know why it shouldn't work - that's all.

Perpetuum motion is, I am afraid, impossible, so this cannot work.
I say again, I've consulted few PhDs, and they didn't provide me any answers. I hope I've made myself clear about setup of this device.

Thank you for your attention.
 

Offline jartza

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Why wouldn't this work? (perpetuum motion)
« Reply #7 on: 21/06/2011 04:16:09 »
Hello everyone.

There's an idea that bugs me for last few years. I've tried consulting few PhDs and got no answers, and I have no materials to do this experiment on my own.
I know that this device shouldn't work, but I have no idea why.

Let's say that we have a charged capacitor. Let's use most simple version of capacitor - two parallel conductive plates that have same dimensions and we charge those plates to say... 100V (doesn't matter, really).

Place closed coil of wire around one plate. The coil should enter capacitor space near -100V charged plate, and exit capacitor space near +100V charged plate, so there should be nice difference in potential energy of free electrons in coil that are near +100V plate and those that are close to -100V plate.

In theory, capacitor energy is defined only by its voltage levels and capacitance. Since capacitor is charged, and disconnected, the voltage levels should stay the same, and since it is rigidly built, it should not change it's capacity.

Electric field should appear as soon as we charge the capacitor, and it should exist only between capacitor plates. This field should start moving free electrons in coil thus producing a current in coil circuit. Now, if we place resistor in coil circuit the current generated in coil should generate heat in resistor transforming electrical energy in heat.
The question is obvious. Only energy that we provided to this system was energy used to charge capacitor, and now we have heat constantly generated by current passing through resistor, and capacitor should maintain its voltage levels...

So, if anyone knows how capacitor will lose it's energy, please reply. Don't bother with imperfection of regular capacitor, I know it will "run dry" by itself. I want to know how it loses its energy when it transfers it to electrons in the coil.

Thank you!
PS Please excuse any grammatical and spelling mistakes that I made.




Think about it this way:

If lot of electrons are packed into a metal plate, then the electrons want to move away from the plate, because same charges repel.

If some positively charged object is brought near the metal plate, then electrons don't want to move away from the metal plate as much as before, because opposite charges attract.

And that's how capacitors work.

Now all we need to understand is that the aforementioned urge of the charges to leave the plate of the capacitor is what we call the voltage of the capacitor, and that the voltage diminishes whenever any opposite charges move towards the capacitor plate.
 

Offline Quark

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Why wouldn't this work? (perpetuum motion)
« Reply #8 on: 21/06/2011 12:49:24 »
From what I understand, you have this circuit which will work if the circuit is closed,"We woild charge it only once and then disconnect it(leave it's connectors open)."this is your quote that has me confused because when you disconnect your circuit from everything then you have no reference point or ground thus rendering your circuit "OPEN". Despite the voltage or amount of coloumbs in the capacitor there is no electron flow in the circuit when the circuit is not grounded or has a reference to ground.Essentially all you have is a battery that's still in it's package waiting for someone to connect it to their ipod or whatever.As far as I know the only way electrons can flow without both voltage and resistance present at the same time is by way of superconductors.

 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #9 on: 21/06/2011 13:49:49 »
It's hard to comment on your setup, since I don't understand it perfectly from your description, but I can comment on the general setup.  If you charge the capacitor then put any chunk of metal between the plates, charges on the metal will rearrange themselves such that it becomes negatively charged on the side towards the + capacitor plate and positively charged on the side towards the - capacitor plate.  The effect of this is that the total field between the plates is reduced.  Since energy is stored in the electric field, this means the total energy of the capacitor system is reduced. 

The major effect you should see here is that the coil gets "sucked in" between the plates, since the energy stored in the field gets primarily turned into kinetic energy.  If the coil isn't a good conductor, it will generate a lot of heat as the charges move, so this energy will also become heat energy.  None of this is free energy, though, since all you're doing is taking energy out of the field--and you put that energy there to begin with when you charged the capacitor.  If you try to make an engine out of this by moving the coil into and out of the capacitor, you'll find that it takes more energy to pull the coil back out as you harvested when you put it in.  So every in/out cycle loses you energy.
 

Offline Quark

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Why wouldn't this work? (perpetuum motion)
« Reply #10 on: 21/06/2011 14:01:54 »
Totally correct JP, Kirchoff's voltage law for series circuits and current law for paralell circuits conclude that power in equals power out,also the laws of thermodynamics clearly state thet you can't get more energy out of a system than what is put in.
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #11 on: 21/06/2011 19:04:18 »
I think I need a circuit diagram, but if I have the idea right, no continuous current is going to flow in the coil or the resistor.

The capacitor does not appear to be connected to anything, so there is no path for the flow of electrons. There would be a small temporary flow when the coil was introduced into the field of the capacitor, but as long as the coil is stationary in the field, no current will flow.
 

Offline RD

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« Reply #12 on: 21/06/2011 21:15:19 »
Coil + capacitor isn't that an LC oscillator circuit ? ... http://en.wikipedia.org/wiki/LC_circuit#Operation
(the oscillations die away, they don't continue perpetually )
« Last Edit: 21/06/2011 21:18:58 by RD »
 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #13 on: 21/06/2011 21:30:24 »
Coil + capacitor isn't that an LC oscillator circuit ? ... http://en.wikipedia.org/wiki/LC_circuit#Operation
(the oscillations die away, they don't continue perpetually )


It sounds like his coil isn't plugged into a circuit though.  I think we need a better description of the setup.
 

Offline Geezer

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« Reply #14 on: 21/06/2011 23:20:03 »
Looking at the description again, I suspect the coil will rapidly equalize the potentials on the plates. It's going to do pretty much the same thing as a resistor connected between the plates - i.e., discharge the cap.
 

Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« Reply #15 on: 22/06/2011 01:54:52 »
I think this diagram should help.

Clarification:
Two horizontal, unconnected lines stand for capacitor plates. They are charged with same amount of electric charges just different polarity. The triangle represents loop of insulated wire. Everything is rigid, and unmovable (except for the electrons in wire).

I was talking about the coil, just to amp up the effect, but i think this is enough to demonstrate what I had in mind.

Thank you all, and I am very sorry for missing "Create a new diagram" link when I started the topic.
 

Offline JP

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« Reply #16 on: 22/06/2011 02:02:24 »
So the coil makes contact with neither capacitor plate?

According to that diagram, when you charge the capacitor, the charges in the coil will rearrange a bit, then stop moving.  Some heat will be given off while the charges rearrange.  Then everything will be nice and stationary.  There won't be perpetual motion since there isn't motion.
« Last Edit: 22/06/2011 02:18:31 by JP »
 

Offline AlmostHuman

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« Reply #17 on: 22/06/2011 02:20:24 »
So the coil makes contact with neither capacitor plate?
True. Coil doesn't make any contact with capacitor.
Hm... There is a field inside capacitor. In my diagram that field is going to try to move electrons vertically - up or down.
Let's say it wants them moving upwards. Only way electrons can move upwards in wire between capacitor plates is if they go diagonally from lower left corner to upper right corner. So, field is pushing them that way, and when they are out of the capacitor, there is no more field to push them further, and more importantly, there is no field to stop them from moving along the wire. So, those electrons were accelerated by capacitor field, and only resistance of that wire is stopping them from moving indefinitely, but we need them only to reach capacitor entrance to repeat the process.
Edit: Why do you need motion in this setup to have moving electrons?
That was my idea of that... thing, whatever it is.

PS After all, I don't believe it would work, but yet I haven't figured out why and how...
« Last Edit: 22/06/2011 02:42:04 by AlmostHuman »
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #18 on: 22/06/2011 04:47:10 »
The diagram helps!

The diagonal line is an insulated wire, but it is a good conductor. While everything is stationary there will be no potential difference along its length when it is a static field, and therefore no current will flow through it. However, the wire does have inductance, so there would be a transient EMF and current flow when the loop was introduced into the field of the capacitor.

The other issue is that the wire is going to act to reduce the energy that can be stored in the capacitor. The field in the vicinity of the diagonal wire will collapse.

BTW, if this did produce continuous current flow you would also be able to get a continuous current flow from a plate elevated above the surface of the Earth. You can get a very short duration flow from such a device, but only while the plate equalizes the potential in its vicinity.

The fundamental problem with your device is that because the charge on the plates is static, the field between them is also static. If you were to make the field alternate by continuously reversing the polarity of the plates you could extract energy from the loop, but it would always be a bit less than the energy you were putting into the plates.
« Last Edit: 22/06/2011 04:48:45 by Geezer »
 

Offline burning

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Why wouldn't this work? (perpetuum motion)
« Reply #19 on: 22/06/2011 06:19:33 »
An important fact to keep in mind is that the field is not zero outside the plates of the capacitor.  For a parallel plate capacitor, where the plate separation is small compared to the plate dimensions, it is an excellent approximation that the field between the plates is uniform and that outside of the plates it is zero everywhere.  However, it is only an approximation, and your thought experiment is an area where applying the approximation is inappropriate.

Considering your diagram, the "base" of your triangular loop is all going to be close to a single electrostatic potential.  The potential drop along the side passing between the plates will essentially the same as the potential drop as the straight up and down side.  As you've drawn your capacitor, the "fringe field" along the straight up and down side will be just about as strong as the field in between the plates.  Now you could stretch out that side so that it passed through space mostly far away from the capacitor where the fringe field is weak.  However, in doing so you also make the length of the side much longer.  Getting into mathematics speak, the potential difference is the path integral of the electric field, and it will always be the case that the weakness of the field will always be offset by the fact that you are integrating it over a longer path.
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #20 on: 22/06/2011 06:38:57 »

Considering your diagram, the "base" of your triangular loop is all going to be close to a single electrostatic potential.  The potential drop along the side passing between the plates will essentially the same as the potential drop as the straight up and down side.


Jolly good, but unless there is a very large current flowing through the triangular loop, there won't be any potential drops, anywhere.

There will only be a current in the loop while it is being inserted into the field. When it stops moving in a static field, no current will flow.
 

Offline Soul Surfer

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Why wouldn't this work? (perpetuum motion)
« Reply #21 on: 22/06/2011 09:29:54 »
The simple answer to this question is that the electric field is NOT confined to the capacitor plate.  In running round ANY closed loop in a static electric field the field force always integrates to zero. It is part if the fundamental equations of electromagnetism so although there is an accelerating field between the plates there is a precisely opposing field in travelling round the rest of the loop so the "pressure" (voltage) on the electrons to move is exactly equal in both directions so the electrons do not move round the wire and no current flows.

There will be a brief rearrangement of electrons if the coil is in place when the capacitor is charged so that the voltage on the coil matches that on the capacitor and that is all that can happen.  If the field is reversed the rearrangement will take place again and an AC field will transfer energy but it requires input energy to do this and neglecting losses they always balance out.
 

Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« Reply #22 on: 22/06/2011 11:05:09 »
Geezer said: "In running round ANY closed loop in a static electric field the field force always integrates to zero". Hmm... not so sure about that in this case. Even if we do the math without any approximation, there should be some difference in this case. The fact that approximation is possible means that field is close to zero function (compared to field between the plates) outside the plates. No matter how strong the field is outside the plates it is significantly stronger between the plates (or so I was told ;) ). Having all this in mind, I really cant integrate field force to zero around this loop.
But, even if this is totally bad setup, we're missing the point.
To find a way to ask a question, I'll leave just charged and disconnected capacitor.
The diagram:



Let's say that this system is zero friction (you can add friction later)

If a charged particle with initial velocity v0 and kinetic energy Ek1=m*v02/2(going just horizontally) and charge -Q0 enters our capacitor like illustrated in diagram, the force of the capacitor electric field should add vertical component to that velocity Δv during its motion between the plates. So, when our charged particle exits capacitor, it's kinetic energy should be Ek2=Ek1+m*Δv2/2.
Since capacitor wasn't discharged by this (can't see where would charges +Q and -Q really go), it's energy should stay the same.
To summarize all this:
We had capacitor with it's energy Ec which didn't change, we had a particle with its energy that did change (at cost of what?).

I hope that this really explains my confusion about energy of a field.

PS We have similar experiment repeating itself for millions of years, only this time, it's not an electric, but gravity field.
Moon is orbiting the Earth and it's gravity is crating ocean tides (a loooooooot of energy), and is probably the reason of Earth's core being molten and hot(even more energy), and yet, Moon's gravitational field is powerful enough to keep astronauts from flying out of it's orbit.

 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #23 on: 22/06/2011 14:50:16 »
In the case you drew here, the potential energy of the moving charges is converted to kinetic energy.  The capacitor's energy doesn't change.  Total energy of capacitor + charge is constant.  Even though the charge is far from the capacitor to begin with, it still has potential energy.

The simplest case of this gravity.  If an astronaut from a distant planet came to visit the earth, he would still fall under the earth's gravity, even though he didn't really feel that gravity on his homeworld.  This can be explained because he has potential energy with respect to the earth on his homeworld even though he's far away from the earth, and this potential energy turns into kinetic energy as he falls.
 

Offline burning

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Why wouldn't this work? (perpetuum motion)
« Reply #24 on: 22/06/2011 15:59:25 »

Considering your diagram, the "base" of your triangular loop is all going to be close to a single electrostatic potential.  The potential drop along the side passing between the plates will essentially the same as the potential drop as the straight up and down side.


Jolly good, but unless there is a very large current flowing through the triangular loop, there won't be any potential drops, anywhere.

There will only be a current in the loop while it is being inserted into the field. When it stops moving in a static field, no current will flow.

I wasn't trying to say there would be current in the loop.  You are correct that the field inside the conductor will be zero and therefore the potential within the conductor will be constant.  However, I was trying to simplify matters by considering only the electrostatic potential of the capacitor electric field.

Consider the loop not as a conductor but just as a mathematical entity for the moment.  Pick any two points, A and B, on the loop.  Now the conservative nature of the electric field dictates that there is, in the static case, a fixed potential difference between A and B.  Specifically, if we travel clockwise around the loop from A to B we won't see any difference in the net change of potential than if we travel counterclockwise from A to B.

Now if the OP were correct that the field is non-zero and uniform between the plates and zero everywhere else, it would not be true that the potential difference was path independent.  If we selected both A and B to be in the zero field region, we would get a non-zero change in potential if we took the part of the loop that went between the plates as our path, but we would get zero change in potential if we took the part that is always outside the plates.

So the point is that the electric field as described by the OP is non-conservative and therefore not possible.  If somehow we were able to create a static non-conservative force field, then yes we could use it to generate a perpetual current.  However, we can't create such a force field.  The field outside the plates is non-zero, and if we were to slog through the detailed math, we would find that the electrostatic potential due to the field of the capacitor is in fact a path independent quantity.  It doesn't depend on which way we go around our loop as a mathematical entity.  Therefore, if we put a physical conductor in place of the mathematical entity, the free charges within the conductor will rearrange themselves slightly to ensure that the field within the conductor is zero, but there will not be a persistent current.
 

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Why wouldn't this work? (perpetuum motion)
« Reply #24 on: 22/06/2011 15:59:25 »

 

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