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Author Topic: Why wouldn't this work? (perpetuum motion)  (Read 18746 times)

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #50 on: 28/06/2011 22:28:07 »
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.)

Potential at a point doesn't matter and can be anything you want, since you can add any global constant to the potential.  Potential variation over space does matter, since the gradient of potential tells you about the field strength, and field strength is directly proportional to force on a charged particle.  Are you saying that in your example, the potential is constant between the two plates?  If so, why?

The derivative of potential is field, so a constant potential means that the field between the plates is zero.  Why should there be zero field between the plates when they're charged?
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #51 on: 29/06/2011 00:09:56 »
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.)

Potential at a point doesn't matter and can be anything you want, since you can add any global constant to the potential.  Potential variation over space does matter, since the gradient of potential tells you about the field strength, and field strength is directly proportional to force on a charged particle.

Are you saying that in your example, the potential is constant between the two plates?  If so, why?

The derivative of potential is field, so a constant potential means that the field between the plates is zero.  Why should there be zero field between the plates when they're charged?

I'm not sure I understand your point JP. The potential between the two plates is constant because that's what capacitors do - unless they lose some charge of course.

There must be a terminology problem here :D  When I say "potential" I am referring to the potential difference measured in volts.

I am certainly not saying the electrostatic field within the capacitor is zero. As long as there is a potential difference between the plates, it can't be zero.

Try this slight variation. What do you think we will observe?

Attach the Y input of an oscilloscope, that just happens to have virtually infinite input impedance, to one terminal of the capacitor (doesn't matter which one.) Attach the ground side of the scope to the particle generation apparatus.

Now charge up the capacitor to some voltage v and remove the charging apparatus. (I'm assuming you are somewhere in deep space to minimize any effects of stray capacitance to any ground).

Observe and record the voltage shown on the scope in "free run". Connect the trigger input on the scope to start a scan when the particle generator launches a particle.

Now launch a charged particle and record the trace on the scope.
« Last Edit: 29/06/2011 03:32:14 by Geezer »
 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #52 on: 29/06/2011 04:47:21 »
Ah.  The key part is you're using potential to mean potential difference, while I'm using it to mean electrostatic potential, which are two related, but different things.  We're both right on our own descriptions, I believe.

While I think you're absolutely right about the potential difference I know you're absolutely wrong about the physics.  :p  I mentioned this before, but the potential (of either flavor) is a poisson rouge here.

Let's try this direction instead.  Everything you need to know about this problem is contained in five equations.  Four Maxwell's equations and the Lorentz force law:
http://en.wikipedia.org/wiki/Maxwell%27s_equations#Table_of_.27microscopic.27_equations
http://en.wikipedia.org/wiki/Lorentz_force

The result of all that is that Maxwell's equations tell you that the charged capacitor plates create a field that is constant in magnitude and pointing from the positive to the negative plate: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html  (I'm neglecting edge effects here, which we can do at least for the though experiment.  In reality the field "bows out" at the edges, but this won't change our results much.)

The Lorentz force law tells us that the force a particle in an electric field feels is the local field (vector) value multiplied by the charge of the particle.  (We're ignoring additional components due to the magnetic field here, since there isn't one if the plates are static.)  Since there is a non-zero field between the plates pointing from one to the other, a charge between the plates will feel a force.  QED.  No need to go to potentials.

All of this is modified slightly in the dynamics case, with a moving particle, but if the plates are big with high charge and the particle is small with small charge, electrostatics should be a pretty good model, even if it's moving.

----------------------

The potential question is actually interesting, and I think I've figured out where your argument is going wrong, but I'm headed to bed so I'll leave it at Maxwell's equations for now.  :)  Besides, potential is derived as a computational tool from Maxwell's equations and the Lorentz force, so if we don't agree on the more fundamental equations, we won't agree on potentials either.
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #53 on: 29/06/2011 05:41:40 »
Ah.  The key part is you're using potential to mean potential difference, while I'm using it to mean electrostatic potential, which are two related, but different things.  We're both right on our own descriptions, I believe.

While I think you're absolutely right about the potential difference I know you're absolutely wrong about the physics.  :p  I mentioned this before, but the potential (of either flavor) is a poisson rouge here.

Let's try this direction instead.  Everything you need to know about this problem is contained in five equations.  Four Maxwell's equations and the Lorentz force law:
http://en.wikipedia.org/wiki/Maxwell%27s_equations#Table_of_.27microscopic.27_equations
http://en.wikipedia.org/wiki/Lorentz_force

The result of all that is that Maxwell's equations tell you that the charged capacitor plates create a field that is constant in magnitude and pointing from the positive to the negative plate: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html  (I'm neglecting edge effects here, which we can do at least for the though experiment.  In reality the field "bows out" at the edges, but this won't change our results much.)

The Lorentz force law tells us that the force a particle in an electric field feels is the local field (vector) value multiplied by the charge of the particle.  (We're ignoring additional components due to the magnetic field here, since there isn't one if the plates are static.)  Since there is a non-zero field between the plates pointing from one to the other, a charge between the plates will feel a force.  QED.  No need to go to potentials.

All of this is modified slightly in the dynamics case, with a moving particle, but if the plates are big with high charge and the particle is small with small charge, electrostatics should be a pretty good model, even if it's moving.

----------------------

The potential question is actually interesting, and I think I've figured out where your argument is going wrong, but I'm headed to bed so I'll leave it at Maxwell's equations for now.  :)  Besides, potential is derived as a computational tool from Maxwell's equations and the Lorentz force, so if we don't agree on the more fundamental equations, we won't agree on potentials either.

JP, you are obviously delusional due to lack of sleep ;D I'm sure it will all become clear in the morning.

I'm sure the equations are quite correct, but I'm also sure that if you conduct the experiment you will discover that there was no change in the energy of the system comprising the capacitor and the particle.

Anyway, rather than throw a bunch of physics mumblespeak at me, why don't you try to tell me what you think we would see on our scope? After all, we know that we can throw as much math at it as we want, but if the experimental evidence does not support the math....

However, here's where you are obviously going wrong. You said, "Since there is a non-zero field between the plates pointing from one to the other, a charge between the plates will feel a force."

With all due respect, that is complete bollocks. A charge between the plates will be subject to a resultant force that is a function of the difference between the potential of the charge and the potential at that point in the field. If there is no difference, there will be no force.

Now, kindly explain (without a lot of hand wavy math) why there would be any difference between the potential of the charge and the potential at that point in the field.

(Somebody is terribly wrong here. I hope it is not me, but at least I've got the Laws of Thermodynamics behind me, and I don't have to do this stuff for a living.)
« Last Edit: 29/06/2011 06:04:28 by Geezer »
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #54 on: 29/06/2011 06:01:10 »
Oh, and just to "up the ante" a bit, let's assume the charge on the capacitor is not much greater than the charge on the particle. I don't think that should make much difference to the outcome.

(Anyone for Poker?)
 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #55 on: 29/06/2011 13:57:02 »
There will be very little change to the potential drop between the plates as a particle passes through them.  So what?  That has nothing to do with the force felt by the particle, which is what I've been saying.  Potential is a mathematical construct, not a physical quantity, which is why I figured it would be easiest to go to the basic equations that govern physical quantities (fields and charges).  Here's the net result of the laws of physics (encompassed by Maxwell's and Lorentz's equations):

1) There's a field between the plates.
2) Therefore a charge placed in that field feels a force.

Which of these do you disagree with?


The problem with the potential drop description is that it seems like you're assuming (though I'm a bit confused as to your argument) that if the potential difference between the plates is constant as a particle passes by, that the particle feels no force.  This isn't a law of physics.  A particle passing through the plates moves from high potential to low potential energy and gains kinetic energy in the process.  The energy of the particle is conserved.  The potential between the plates should remain constant or else energy isn't conserved in the system!
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #56 on: 29/06/2011 17:50:35 »
I can see I'm going to have to run the experiment myself.  ;D.

Let's say our particle is charged to 100 volts, the capacitor is charged to 10 volts, we connect the scope to the positive plate of the capacitor, and we aim the particle exactly half way between, and parallel to, the plates.

When the particle is a great distance from the capacitor the voltage shown on the scope is zero. (The capacitor as a whole has no charge.)

As the particle approaches the capacitor's field, the voltage on the scope ramps up to 105 volts (100 + 10/2). As the particle leaves the field of the capacitor, the voltage ramps back down to zero.

If we aim the particle so that it takes a diagonal path from the positive plate towards the negative plate, the voltage on the scope will initially ramp to 100 volts then increase rapidly towards 110 volts as the particle transits the capacitor, then it will ramp back down to zero.

In practice it would be quite difficult to set this up because any stray capacitance between the capacitor and anything else will alter the rate of change of voltage, and that will result in deflection of the particle.

   
 

Offline JP

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« Reply #57 on: 29/06/2011 18:04:16 »
But you're still claiming the particle doesn't feel a force due to the capacitor's field as it passes between the plates?  Potentials aside, how is that possible since there's a field there and charges in fields experience forces?

By the way, check out 2.13.3:
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/coursenotes/modules/guide02.pdf

That is exactly the problem we're discussing here.
« Last Edit: 29/06/2011 18:10:04 by JP »
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #58 on: 29/06/2011 18:29:47 »
Oh, I'm not saying there are no forces. There are, but they balance. It's easier for the particle to reorganize the voltages than change direction. As I said, this will only work if there is nothing that tends to "anchor" the plates to some other reference and, in reality, there are plenty of things that will tend to do that. In the artificial situation that we have here where the capacitor is totally isolated, it is quite happy to use the particle as a reference because that's the only reference it has, if you see what I mean.

I don't think I'm violating any of the math here. It's just a different way of thinking about the problem. However, I'll take a look at the link you posted.

 

Offline burning

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Why wouldn't this work? (perpetuum motion)
« Reply #59 on: 29/06/2011 19:33:37 »
I think that possibly the source of confusion is that you are confusing the potential of a charge in an external field and the potential due to the charge itself.  When we are considering the work done on the charge the former matters and the latter doesn't.  This is equivalent to the fact that when we want to find the force acting on the charge we only care about the electric field produced by external charges, not the field produced by the charge itself.

Whenever we refer to a potential as an absolute value, there is always has to be an implicit "relative to" hidden in the statement somewhere.  When we say that the capacitor is charged to 10 volts, that means that the capacitor is charged so the potential difference between the two plates is 10 volts.  Now just as there is a vector electric field due to the charges on the capacitor plates, we can also describe a scalar potential field due to the charge.  The vector electric field is the gradient of the scalar potential field.  Consequently, just as the electric field is a property of the charge on the plates alone, so is potential field.

The potential field is defined only to within an arbitrary additive constant.  Conventionally, this constant is usually assigned to make the potential at infinity equal to zero, if at all possible, but it doesn't really matter.  The positive plate will have a potential 10 volts greater than the potential at the negative plate.  The potential at the positive plate is the maximum of the potential field; the potential at the negative plate, the minimum.  Halfway between the plates, the potential is 5 volts below the positive plate and 5 volts above the negative plate.  As we go farther and farther away from the capacitor, the potential approaches this in between value.  This is field gives the potential of an external charge due to charge on the capacitor and it is completely independent of the magnitude of that external charge.

Now, when you say that your particle is charged to 100 volts, you imply that if the charge were less so would be the potential.  The potential of a charged particle due to its own charge is not something that is that commonly considered.  However, sometimes we might consider the potential difference between the surface of the particle and a point infinitely distant from the particle.  That is the best guess I can make for what we might consider to be an intrinsic potential of a charged particle.  However, I need to emphasize that if that's the case, we are talking about the potential due to the particle's own field, and this quantity is irrelevant to how the particle responds to an external electric field.

To look at what happens to a charged particle moving through an external electric field we are only interested in the character of the field and the value of the charge.  The electric force on the charge at any moment is equal to the electric field vector at the particle's location multiplied by the particle's charge.  The work done by the electric field if the particle moves from point A to point B is equal to the electrostatic potential difference between point A and point B multiplied by the particle's charge.  That's it.

Now, if we consider the case of a charged particle moving between the plates of a capacitor, and we give that moving charge a magnitude comparable to or greater than magnitude of the charges on the plates, we will have a scenario that will not be adequately approximated by the electrostatic description of a charged capacitor.  In this case, as the particle approaches it will exert a sufficient force on the charges on the plates to change the charge distribution significantly from what it was when the particle was distant.  This means that the electric field produced by the capacitor will be changing moment by moment and the whole thing becomes an electrodynamics problem.  People take graduate courses to learn to solve this sort of problem properly.  So under that circumstance, the charge on the incoming particle matters more than what I suggested above.  However, even in that case we would expect the particle to be attracted by the opposite sign plate and to be repelled by the same sign plate.  The particle will still be deflected in its flight.

If on the other hand we are considering a charged particle with a charge magnitude that is small compared to the charge on the capacitor, the electrostatic capacitor is an excellent approximation.  The electric field that the particle is exposed to will at all times be well approximated by the field of the capacitor with no external charges present.

All this isn't to say that there are no instances in which the combined field of the particle and capacitor (and similarly the combined potential due to those fields) is a quantity of interest.  But those circumstances would be when we are interested in the effect on a third hypothetical entity.  If we are interested in the effect of the capacitor on the particle, we need to keep the particle's field well out of the discussion.
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #60 on: 29/06/2011 21:09:09 »
Thanks Burning. I've just been going over the MIT coursework JP posted, and I'm pretty sure my idea is a complete load of bollocks! Such a nice experiment too. Oh well, JP was right - this time  ;D

I'll take some time to go through your post, but I'm sure you are quite correct.
 

Offline JP

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« Reply #61 on: 30/06/2011 17:36:19 »
So I think coming back to the question of where the charged particle is drawing it's energy from, we have:

1) The charged particle gains some kinetic energy from the plates as it passes between them due to the electric field created by the plates.

2) The total charges on the plates don't change, though they may rearrange themselves slightly as the charge passes by.  If we use a small charge on the particle and a large charge on both plates, this effect will be small.

3) If the particle starts really really far from the plates and ends really really far from the plates, you'd assume it's effect on the charge distribution of the plates to be negligible, so that the energy stored in the capacitor hasn't changed.

4) The energy in the capacitor hasn't changed, but the charge has gained kinetic energy.  How?

What I've been thinking is the answer is where the charged particle is originally, it will make the plates slightly harder to charge.  This extra energy required to charge them is gained as kinetic energy as the particle passes through the plates.  This makes a bit of sense, since it's well known that by putting a dielectric in between the plates of a capacitor, you can reduce the stored energy, thereby allowing more charge to be put on the plates with the same voltage source.  The charged particle acts in a similar way, by interacting with the charges you're putting on the plates is allows more energy to be stored in the entire particle/capacitor system with the same charge.

I'm not convinced this is the entire explanation, as I can't think of a way to really simplify this explanation and not need to go through a lot of complex calculations in order to check it. 
 

Offline Geezer

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« Reply #62 on: 30/06/2011 18:23:48 »
JP - when you say the kinetic energy has changed, am I right in thinking that's a consequence of the work done to alter the direction of the particle?
 

Offline JP

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« Reply #63 on: 30/06/2011 19:22:44 »
Yes.  The electric field does work to alter the trajectory.  The question is where it gets the energy to do this work.

I think you can check where the energy comes from by computing the energy necessary to construct a an electric dipole  both with a third charge present and without.  The difference in energies with and without should have to do with the gain of kinetic energy when you fire an external charge through the dipole.  Now, a capacitor isn't a dipole, but the same basic principles should hold in the more complex case.
 

Offline Geezer

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« Reply #64 on: 30/06/2011 19:40:33 »
Ah, right, thanks! I wonder if a capacitor approximates to a whole bunch of dipoles, and would that help to model the particlar situation?
 

Offline JP

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« Reply #65 on: 30/06/2011 19:59:55 »
Ah, right, thanks! I wonder if a capacitor approximates to a whole bunch of dipoles, and would that help to model the particlar situation?

Yeah.  The tricky part in the capacitor case is that as a conductor, it's charges are free to respond to the incoming particle as burning noted.  If we start from a single dipole, it should be quite easy to compute the potential energies of the entire system of dipole + test particle with the test particle in various positions.  The gain in kinetic energy as a particle moves through the dipole should be proportional to the change in potential energy of the initial and final configurations. 

In the capacitor case, this would manifest itself in a little extra work required to charge the capacitor with the test charge in it's initial position.  This energy is liberated as kinetic energy as the particle moves.  I'm pretty sure that's the answer, but I haven't had the time to work through it yet.
 

Offline Geezer

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« Reply #66 on: 30/06/2011 20:59:12 »
How about this?

Work was done, but the speed of the particle did not change. Only it's direction changed, so it did not take any additional energy with it when it left the capacitor.

If that's true, it means that the capacitor dissipated some energy in doing the work. We know the dielectric exerted a torque on the particle. The torque stressed the dielectric, and in so doing, produced some friction that was dissipated as heat in the capacitor.
 

Offline JP

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« Reply #67 on: 30/06/2011 21:34:04 »
That's one of the issues I was trying to resolve by thinking about external fields.  But I think that in general the particle gains kinetic energy.

I think I have an answer, though.

In a really simple example, consider taking a + charged particle and releasing it from rest near the + charged plate of the capacitor.  It will accelerate towards the - charged plate.  Clearly it gains kinetic energy, but if you compute the energy stored in the capacitor from the textbook formula that depends only on the capacitor's charge, you would think that this energy came from nowhere.

In reality, if you place a + charge near the + plate of the capacitor, it's going to take more energy to charge it, since this + charge is going to repel any charges put onto that plate.  If you place a + charge close to the - plate of the capacitor, it's going to be easier to charge it due to attraction.  So the original state of the capacitor has more stored energy than the final state even though the charge on the plates has stayed constant.  The mistake is in assuming that the capacitor's energy can be computed solely from the charges on it while neglecting any other charges in the problem.

This isn't all that surprising, actually, since if you charge up a capacitor and slide a dielectric into the gap, you decrease the stored energy (and this is felt as a force pulling the dielectric into the gap as you insert it).  Or as it's usually presented--inserting a dielectric in the gap allows more charge to be put on the plates from the same voltage source.
 

Offline Geezer

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« Reply #68 on: 30/06/2011 22:23:29 »
I think you are right. The particle does seem to be accelerated. I thought I read in here

http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/coursenotes/modules/guide02.pdf (Example 2.13.3)

that it had the same speed, but what it actually says is that the particle still has the same horizontal velocity component plus an additional orthogonal component, so its velocity did increase, and therefore it has more energy on leaving than entering.

That would suggest that if you apply a continuous stream of charged particles, you'll eventually discharge the cap, which does not seem too unreasonable.

Presumably the deflection plates in an electrostatically controlled CRT (the type used in oscilloscopes) require a continuous supply of energy as long as they are bending the beam. It's probably not too difficult to conduct an experiment to see if the energy supplied to the plates approximates to the change in energy of the beam. Mind you, it may not be that easy to find a CRT like that these days  :D
« Last Edit: 01/07/2011 05:58:15 by Geezer »
 

Offline Geezer

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« Reply #69 on: 30/06/2011 22:39:03 »
Wait a minute! It may be a lot easier than that.

The energy removed from the cap should equal the energy consumed in accelerating the mass of the particle from zero to whatever the orthogonal component of the velocity is when it "leaves" the field.

Wait another minute! Isn't there something called an "electrostatic voltmeter" that operates on that principle? (It's been a long time, so I could be wrong.)

EDIT: No, it does not work that way. http://en.wikipedia.org/wiki/Electrostatic_voltmeter 

I think I was confusing it with something that was sometimes referred to as a "magic eye".
« Last Edit: 01/07/2011 03:25:58 by Geezer »
 

Offline JP

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« Reply #70 on: 01/07/2011 14:46:04 »
Ok, I ran a quick numerical simulation.  This plot is of the potential (vertical axis) over space for two plates that extend from -20 to 20 on the x-axis (the one pointing towards you) and places at -10 and 10 on the y-axis.  The plates themselves correspond to the peak and trough.  Since the potential energy of a particle placed near the plates is it's charge times this potential, if left to move freely with the plates held fixed, the particle will tend to roll either down or up this potential.  The kinetic energy gained is the difference between the particle's starting and ending positions.

You can see that as the particle escapes from the region between the plates, it will eventually reach the same potential energy with which it entered the plate region.  This means the field extending beyond the edges of the plates slows it down by pulling on it.

 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #71 on: 01/07/2011 17:34:56 »
Nice sim JP! I'm not quite sure what to conclude from it though. Does it say that the kinetic energy on entry is the same as the kinetic energy on exit, or should we conclude it is different?

If the field extending beyond the plates decelerates the particle on exit, won't that be cancelled out by the a similar acceleration on entry, and, if that's true, don't we end up with the change in the trajectory simply increasing the kinetic energy?

I was thinking it's quite simple to measure the energy removed by the particle, but maybe it's too simple! All you need to know is the particle's mass, its initial velocity and the deflection angle (although the angle might be a bit of an approximation as the particle never really really leaves the field.)
 

Offline JP

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« Reply #72 on: 01/07/2011 18:25:17 »
If you start the particle really far away, fire it between the plates, and catch it really far away, it will essentially give up all the kinetic energy it gained between the plates.  You can tell because as you move really far away, everything is equipotential. 

If you start or stop it at other points which aren't equipotential to each other, it's gained or given up some of it's initial energy.

For example, if the particle is positively charged, it will be attracted towards the negatively charged plate if I fire it straight down the center, so it will roll downhill in that figure.  But as it tries to leave the region between the plates, it's got to climb back out of that low potential region, so it loses kinetic energy. 

The field-based description of this is that as it moves through this region, it gets pulled by the negatively charged plate and pushed by the positively charged plate, moving it closer to the negatively charged plate by the time it exits the region between the plates.  But as it tries to escape, it's now nearer the negative plate than the positive plate, so it feels a stronger force towards the negative plate than it feels a push from the positive plate, which tends to pull it back towards the capacitor.  This slows it down.  When you collect it far away, it's given up most of the energy it started with.
 

Offline Geezer

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Why wouldn't this work? (perpetuum motion)
« Reply #73 on: 01/07/2011 19:17:37 »
So, are you saying my original answer was correct except for entirely the wrong reasons?  ;D
 

Offline JP

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Why wouldn't this work? (perpetuum motion)
« Reply #74 on: 01/07/2011 22:34:46 »
So, are you saying my original answer was correct except for entirely the wrong reasons?  ;D

If only this was a true/false quiz! :)

Yeah, your potential way of looking at it was the easiest in the end, if only you'd been right about the details!
 

The Naked Scientists Forum

Why wouldn't this work? (perpetuum motion)
« Reply #74 on: 01/07/2011 22:34:46 »

 

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