Why wouldn't this work? (perpetuum motion)

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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.


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Offline RD

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« Reply #1 on: 20/06/2011 16:30:47 »
« Last Edit: 20/06/2011 16:32:21 by RD »

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Offline imatfaal

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Why wouldn't this work? (perpetuum motion)
« Reply #2 on: 20/06/2011 17:08:33 »
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline AlmostHuman

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« 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".
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline syhprum

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« Reply #5 on: 20/06/2011 22:14:36 »
Did not Hertz, Marconi el al look into this sort of thing ?
syhprum

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Offline AlmostHuman

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« 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.

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Offline jartza

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« 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.

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Offline Quark

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« 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.


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Offline JP

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« 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.

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Offline Quark

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« 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.

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Offline Geezer

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« 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.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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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 »

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Offline JP

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« 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.

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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.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline AlmostHuman

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« Reply #15 on: 22/06/2011 01:54:52 »
I think this diagram should help. [diagram=634_0]

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.

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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 »

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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 »

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Offline Geezer

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« 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 »
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Offline burning

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« 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.

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Offline Geezer

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« 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.
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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.
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Offline AlmostHuman

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« 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:

[diagram=635_0]

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.


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Offline JP

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« 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.

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Offline burning

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« 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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Offline AlmostHuman

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« Reply #25 on: 22/06/2011 19:03:42 »
@Geezer

Potential energy for charged particle/object in electric field is Wp=Q*E (Where Q is charge of particle/object, and E is module of vector of electric field). Now, if we do the math... we would see that this electric field isn't really as conservative as we thought it is. What do I mean by that? Let's try shooting our particle from point A between capacitor plates (like shown before) and see where particle goes and take one point in that trajectory and mark it as B. Now, if we slightly move our trajectory (try to hit B in straight line) we would see that we need to do some work that was done by capacitor to actually hit the B point again. Which shouldn't be "allowed" if this was truly conservative field. This was just to question that path independent statement.

@JP
About that gravity thing...
Are you saying that Moon/Earth are losing THEIR potential energy, since they are doing all that work?
About that particle capacitor...
The work that was done by capacitor field wouldn't be done if we didn't take a path between the plates, therefore, the work
was done by capacitors energy, not by particle's potential energy.

PS
All I want to know is where this energy comes from? And I think that our discussion is progressing quite nicely, because I think we are getting somewhere :).
I wanted this discussion to come to the point where we would try to combine two independent fields to work together, and create useful work. And since I will be occupied for next few days (I don't know if I will be able to participate here next three to five days) I'm jumping maybe too far ahead. (We didn't clear out that energy thing)

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Offline Geezer

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« Reply #26 on: 22/06/2011 20:40:46 »
Quote
Geezer said: "In running round ANY closed loop in a static electric field the field

No he didn't. That was some other geezer.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Geezer

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« Reply #27 on: 23/06/2011 00:54:18 »

All I want to know is where this energy comes from?


It doesn't. The only energy present is the energy that was initially stored in capacitor, and a lot of that was dissipated in the loop when it was introduced between the plates.

I suggest you construct a model, test it, and publish the results before we move this topic to "New Theories" [;D]
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Soul Surfer

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« Reply #28 on: 23/06/2011 08:44:50 »
A charged particle moving between one plate and the other of a charged capacitor is just the same as the electrical breakdown of that capacitor,  that is the capacitor is slightly discharged.  In the case of the non contacting loop of conductor through the capacitor as I have stated above the charge in the conducting wire just rearranges itself to match the potential of the locality clearly if it is a loop the large voltage drops quickly between the plates and builds slowly round the loop.  The energy of this rearrangement depends on the self capacitance of the wire and the rate at which it does the rearranging when the capacitor is charged depends on the self inductance of the wire loop.

I state again the total voltage drop around any closed loop in a static electric field is always zero.
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Offline JP

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« Reply #29 on: 23/06/2011 20:00:38 »
I'm  not quite sure moving a charge is the same as discharging the capacitor.  In the figure he showed above, the charge is introduced separately from the capacitor itself and doesn't touch the plates.  The charge on the plates shouldn't change and once the independent charge has been removed, the energy stored in the field between the plates should be the same as it was initially.

If the plates were slightly discharged, the charge on the plates would change and the field between them would be reduced, reducing the total energy stored between them.

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Offline JP

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« Reply #30 on: 23/06/2011 20:19:31 »
Maybe we can simplify the question this way:

You have a parallel plate capacitor charged up and disconnected so that the charges on the plates can't escape.  You then fire a charge between the plates on a path that's initially parallel to them.  It has high enough velocity that it makes it through the plates even though it's deflected by them. 

Coming out, it would seem that it's velocity should be higher than going in, so it gained kinetic energy.  In addition, a charge was accelerated so it should radiate away some energy.  If the charge on the plates remained the same, the field between them should be the same when the particle has gone on it's merry way.  Where did this extra kinetic and radiative energy come from, then?

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #31 on: 24/06/2011 04:49:21 »
It's really just a capacitor with a conductor looping around one of the plates. When a potential difference is applied across the plates they will charge with opposite potentials. While the plates are charging and the electric field strength is increasing, a small transient current will be induced in the loop.

When the plates are fully charged there is no potential difference between the ends of the diagonal wire because the field strength at both ends of the wire is exactly the same (although that may not be obvious).

It may help to consider what happens if the loop is connected to ground. You then have the equivalent of three capacitors. Two go between ground and each plate connection, and a third goes between the plate connections.

Here is the equivalent circuit;

[diagram=636_0]

which, of course, is equivalent to this;

[diagram=637_0]

or this;

[diagram=638_0]


 

« Last Edit: 24/06/2011 06:45:24 by Geezer »
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Offline AlmostHuman

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Why wouldn't this work? (perpetuum motion)
« Reply #32 on: 26/06/2011 21:57:46 »
Maybe we can simplify the question this way:

You have a parallel plate capacitor charged up and disconnected so that the charges on the plates can't escape.  You then fire a charge between the plates on a path that's initially parallel to them.  It has high enough velocity that it makes it through the plates even though it's deflected by them. 

Coming out, it would seem that it's velocity should be higher than going in, so it gained kinetic energy.  In addition, a charge was accelerated so it should radiate away some energy.  If the charge on the plates remained the same, the field between them should be the same when the particle has gone on it's merry way.  Where did this extra kinetic and radiative energy come from, then?

Thank you JP! That was the source of my confusion... and it still is :). I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.

PS If this particle had significant mass, and if it was deflected upwards (gravity perspective) we could get some work done, right?

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #33 on: 27/06/2011 09:55:18 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.


It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.

When you charge the capacitor with that particular capacitance to any potential, assuming the dielectric of the capacitor has no losses, the energy that went into charging that particular capacitor is still in the capacitor.

As the title of this topic implies some sort of "free energy", it has been moved to new theories. However, a new topic could be started to help understand the physical mechanisms involved within this type of capacitor.


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Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #34 on: 27/06/2011 14:30:50 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.


It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.


Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #35 on: 27/06/2011 18:58:40 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.


It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.


Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...


If I understand the setup properly, I'm not sure a charged particle would be deflected at all if the plates of the capacitor were disconnected from everything. In that situation I think the net charge on the particle is zero with respect to the capacitor.

It would be like trying to deflect electrons from the gun in a cathode ray tube with deflection plates that were not referenced to the electrons' potential. The electrons will ignore any potential on the plates and fly straight ahead.
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Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #36 on: 27/06/2011 20:12:28 »

I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.


It can't. The capacitor didn't lose any energy that wasn't there in the first place. The extra loop of wire forms part of the capacitor and, by a remarkable coincidence, it actually determines the total capacitance of the capacitor, including the loop of wire.


Yes, that's true for the wire.  But what's the explanation if you send a single charged particle through the capacitor as described above?  I can hand-wave a bit and talk about fields that aren't between the two plates and conservation of energy, but I feel like there should be a much more elegant answer that I'm missing...


If I understand the setup properly, I'm not sure a charged particle would be deflected at all if the plates of the capacitor were disconnected from everything. In that situation I think the net charge on the particle is zero with respect to the capacitor.

It would be like trying to deflect electrons from the gun in a cathode ray tube with deflection plates that were not referenced to the electrons' potential. The electrons will ignore any potential on the plates and fly straight ahead.

Hmm... I suspect we don't understand the setup in the same way, then.  The way I read it is that you've basically charged the two plates up then disconnected them while still charged.  With nowhere else to go, the charge remains on the plates, so you have two parallel plates with equal but opposite charges.  If you introduce a charged particle between them, it will feel a force.  If it's an electron, the force will point from the negative to the positive plate.

My question is whether there's an easy way to explain the conservation of energy in the case where you fire a particle through the two plate system so that it escapes out the other side.  Simplistically you'd think that the particle would pick up extra velocity as a result of this force it feels between the plates, so it has more energy coming out than going in.  The charge on the plates doesn't change.  Therefore, how is energy conserved?  The coil is a poisson rouge so far as this question goes.

It is, of course, conserved, and I suspect it has to do with taking into account motion as well as magnetic fields and also realizing the fields aren't confined entirely to the region between the plates.  It feels like there should be a more straightforward explanation, but it's escaping me...

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #37 on: 27/06/2011 20:35:10 »
Hmm... I suspect we don't understand the setup in the same way, then.  The way I read it is that you've basically charged the two plates up then disconnected them while still charged.  With nowhere else to go, the charge remains on the plates, so you have two parallel plates with equal but opposite charges.  If you introduce a charged particle between them, it will feel a force. 

I don't think it will feel any force at all  [;D]

As the plates are not referenced to anything, the plates will reference themselves to the charged particle. The potential on the particle (from the plates) will be zero.

In reality there would be some capacitance between the plates and ground, so they would have some common reference, but as described, there is no common reference. The plates are truly "floating".

Think of it this way - if you measure the voltage between either of the plates and ground, it will always be zero. The electron does the same thing.
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Offline JP

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« Reply #38 on: 27/06/2011 20:51:15 »
The plates will be at different potentials with respect to each other.  Ground is irrelevant here.  But you don't even need potential at all.  Two parallel plates of opposite charges create a field in the gap.  A charged particle in a field experiences a force. 

Of course, the plates need to be mechanically held in place (but not electrically grounded) and the charge free to move, but I believe that was stated in the original description of the system.

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #39 on: 27/06/2011 21:31:21 »
B-b-b-but, if a particle carries charge, it has to be charged with respect to something. So, you can only measure the charge on a particle with respect to some reference. There's no "absolute" charge - at least I don't think there is.

In this case, as the plates and the particle share no common reference, it's the particle that's the "immovable" object. The capacitor adjusts it's potentials to be relative to the particle rather than the other way around.

If the particle was charged to a million volts (relative to some reference), when the particle shot between the plates, the plates would be elevated to a million volts relative to the particle's reference, plus and minus their relative potentials.
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Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #40 on: 27/06/2011 21:43:54 »
B-b-b-but, if a particle carries charge, it has to be charged with respect to something. So, you can only measure the charge on a particle with respect to some reference. There's no "absolute" charge - at least I don't think there is.

There is an absolute charge, though.  Charge isn't like voltage, where everything is defined with respect to a reference.  Charge is a fundamental physical quantity that comes in discrete units.  If I have a ball with 1 Coulomb of charge and one of 2 Coulombs of charge repelling each other, the resulting physics is quite different than if I had 0 Coulombs and 1 Coulomb or +1/2 Coulomb and -1/2 Coulomb, so you can't just pick an arbitrary reference.  (Zero charge physically means that there are as many negative as positive charges in something and negative and positive charges are fundamental properties of subatomic particles.)

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #41 on: 27/06/2011 22:01:25 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.
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Offline Geezer

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« Reply #42 on: 27/06/2011 22:27:27 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

I should be a bit more precise. The point where the particle is in the capacitor's field will have a potential equal to the potential of the particle when it was charged relative to some reference. The plates will be at potentials relative to that reference because that's the easiest thing for them to do, and there is nothing to prevent them assuming those potentials.

If the particle is "half way" between the plates, the plates will have equal positive and negative potentials relative to the potential of the particle. If the particle is closer to the positive plate, the negative plate will have a greater absolute potential (more negative) and so on. Whatever happens, any forces experienced by the particle because of the electric field will be equal and opposite because the potentials of the plates adjust to the potential of the particle.

(Phew! - Wipes sweat from forehead.)
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Offline burning

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« Reply #43 on: 27/06/2011 22:28:57 »
Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Um, no.  Stop thinking about potential for a moment and think about charges and electric fields.  As JP pointed out, charge is absolute, and we really can charge a capacitor so that one plate has an excess of positive charge and the other plate has an equal magnitude excess of negative charge.  These physically separated, non-zero charge distributions will produce an electric field.  This electric field will be strongest between the plates (although it will be present elsewhere).  If we fire a charged particle in between the plates it will experience an electric force (it experiences it before it gets between the plates, but the force is strongest between the plates).  That force will produce a change in kinetic energy of the particle.  Now, since the field is non-zero outside, the net work done on the particle very much depends on where it started and where it ends up, but it will definitely be non-zero for a large set of finite trajectories.

OK, now what about the electrostatic potential?  It is true that absolute electrostatic potential has no meaning.  It is arbitrary where we define 0 potential.  However, electrostatic potential difference is another matter.  The potential difference between two points is equivalent to the work done by the field per unit charge on a particle travelling between those points.  To assert that this difference is nonexistent is tantamount to saying that there is no electric field.  

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Offline burning

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« Reply #44 on: 27/06/2011 22:32:45 »
If the particle is "half way" between the plates, the plates will have equal positive and negative potentials relative to the potential of the particle. If the particle is closer to the positive plate, the negative plate will have a greater absolute potential (more negative) and so on. Whatever happens, any forces experienced by the particle because of the electric field will be equal and opposite because the potentials of the plates adjust to the potential of the particle.

(Phew! - Wipes sweat from forehead.)

Except the plates have opposite charges.  Let's say we have the negatively charged plate on the left and the positively charged plate on the right.  Now let's fire a negatively charged particle between the plates.  It is repelled by the negatively charged plate, so it is pushed to the right.  It is attracted by the positively charged plate so it is pulled to the right.  These forces do not cancel, they work together.

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Offline Geezer

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« Reply #45 on: 27/06/2011 22:57:57 »
I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle. If they did, the particle would certainly be deflected. However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum  [:D]
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Offline burning

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« Reply #46 on: 28/06/2011 02:20:26 »
I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle.

No. I am stating that if the capacitor has been charged and then disconnected from the circuit, as specified by the OP, then there is a constant potential difference between the two plates.  If we fire a charged particle through the electric field it will not maintain a constant potential difference with either plate.

Quote
If they did, the particle would certainly be deflected.  However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum  [:D]

Sorry, I can't understand what you mean by the rest of this.  I ask again that we step away from the electrostatic potential and look at the charges and the electric field.  Which if any of the following assertions do you consider to be false and why?

1.) It is possible to charge a capacitor, by which we mean that we create a net negative charge on one plate and a net positive charge of equal magnitude on the other plate.

2.) An object with a net non-zero charge will produce an electric field

3.) In the case of the capacitor, the electric field is strongest between the plates.  This is because the direction of the field from the positive plate matches that of the field from the negative plate in this region.

4.) If a charged particle passes through a region with an electric field, it will experience an electric force.

5.) In the case of a charged particle between the plates of a capacitor, it will be attracted toward the plate with the opposite sign and be repelled by the plate with the same sign.

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Offline JP

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« Reply #47 on: 28/06/2011 04:37:39 »
Geezer, I think burning is absolutely right and if you can point out which of his points is wrong, maybe we can understand why we disagree.  In the meantime, I also have a specific disagreement with one thing you said:

Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Force is what produces the deflection.  Potential is not force.  Potential is a useful computational tool for getting the electric field and force may be computed from the electric field (in the statics case).  It's very close to potential energy, since it's the integral of the field over distance.  That's why you can arbitrarily re-zero potential energy by picking a reference point.  Just like potential energy, it's differences that matter, not magnitude.  You can't do the same with field or with electric forces.

http://en.wikipedia.org/wiki/Electric_potential#In_electrostatics
« Last Edit: 28/06/2011 05:51:46 by JP »

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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #48 on: 28/06/2011 07:36:01 »
Geezer, I think burning is absolutely right and if you can point out which of his points is wrong, maybe we can understand why we disagree.  In the meantime, I also have a specific disagreement with one thing you said:

Ah yes, but charge and potential are not the same things, and it's the potential that produces the deflection. In this case, there is no potential between the particle and the plates, so there is no deflection.

Force is what produces the deflection.  Potential is not force.  Potential is a useful computational tool for getting the electric field and force may be computed from the electric field (in the statics case).  It's very close to potential energy, since it's the integral of the field over distance.  That's why you can arbitrarily re-zero potential energy by picking a reference point.  Just like potential energy, it's differences that matter, not magnitude.  You can't do the same with field or with electric forces.

http://en.wikipedia.org/wiki/Electric_potential#In_electrostatics

Yes - I agree. That's why I corrected my post.

It will take me some time to respond to Burnings points. Stay tuned  [;D]
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Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #49 on: 28/06/2011 21:28:19 »
OK - Here we go again.

I don't think so Burning.

You seem to be assuming the plates have some constant potential relative to the particle.

Quote
No. I am stating that if the capacitor has been charged and then disconnected from the circuit, as specified by the OP, then there is a constant potential difference between the two plates.


Yes - I fully agree with you there.

Quote
If we fire a charged particle through the electric field it will not maintain a constant potential difference with either plate.

That's where it gets tricky. I'll come back to that.
 
Quote
If they did, the particle would certainly be deflected.  However, while the relative potentials of the plates remains constant, their mean potential can have any arbitrary value at any instant, and the particle can change their mean potential because there is absolutely nothing to prevent it from doing so.

If the particle did deflect in this situation, it would be analogous to operating a lever without a fulcrum  [:D]

Sorry, I can't understand what you mean by the rest of this.  I ask again that we step away from the electrostatic potential and look at the charges and the electric field.  Which if any of the following assertions do you consider to be false and why?

Quote
1.) It is possible to charge a capacitor, by which we mean that we create a net negative charge on one plate and a net positive charge of equal magnitude on the other plate.

Completely agree.

Quote
2.) An object with a net non-zero charge will produce an electric field

If it's charged, yes, and in this case it certainly is.

Quote
3.) In the case of the capacitor, the electric field is strongest between the plates.  This is because the direction of the field from the positive plate matches that of the field from the negative plate in this region.

Completely agree.

Quote
4.) If a charged particle passes through a region with an electric field, it will experience an electric force.

I'd say it slightly differently. I'd say that it will definitely interact with the electric field and behave according to a very specific set of rules.

Quote
5.) In the case of a charged particle between the plates of a capacitor, it will be attracted toward the plate with the opposite sign and be repelled by the plate with the same sign.

Disagree. That does not seem to take into account the potential at the point in the field and the potential of the particle. It's entirely possible for the particle to travel through the capacitor without experiencing any deflection.


I've thought of several ways to explain what I think is happening, but this might be the most straightforward and hopefully it will either make sense, or fall apart quite rapidly.

As Burning points out above, the capacitor as a whole is "neutral" - it has no net positive or negative charge. Consider it to be, as the diagram implies, totally isolated from everything after it has been charged up.

Now, along comes a charged particle with a potential of +x volts that has travelled from a great distance. The charged particle enters the electric field of the capacitor and, if I have it right, the potential of the electric field at that point will now be +x volts (I think that's the most crucial part of my argument.)

Also, I think the capacitor as a whole is no longer "neutral". It now has a charge equal to the charge of the particle. The potential difference between the plates does not change, but the potentials on the plates are now relative to the potential of the particle.

Let's see if we can blow up that description before we go any further.
« Last Edit: 28/06/2011 21:39:12 by Geezer »
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