Naked Science Forum
On the Lighter Side => New Theories => Topic started by: AlmostHuman on 20/06/2011 14:46:32
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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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if anyone knows how capacitor will lose it's energy, please reply.
Work is done when charging or discharging a capacitor ...
http://www.astarmathsandphysics.com/a_level_physics_notes/electricity/a_level_physics_notes_lost_energy_when_charging_or_discharging_capacitors.html
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In this house we ...
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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
And in what house do we praise the Lord?
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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".
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Did not Hertz, Marconi el al look into this sort of thing ?
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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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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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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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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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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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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.
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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 )
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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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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.
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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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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.
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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...
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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.
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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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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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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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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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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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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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@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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Geezer said: "In running round ANY closed loop in a static electric field the field
No he didn't. That was some other geezer.
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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]
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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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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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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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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]
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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.
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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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
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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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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.
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.
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.
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.
Completely agree.
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.
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.
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.
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.
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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?
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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.
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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.
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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.
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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.
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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.)
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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?)
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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!
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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.
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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.
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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.
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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.
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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.
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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.
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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?
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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.
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Ah, right, thanks! I wonder if a capacitor approximates to a whole bunch of dipoles, and would that help to model the particlar situation?
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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.
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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.
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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.
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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]
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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".
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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.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fi52.tinypic.com%2F2hd1rg8.png&hash=aeca455c043fbb81b1c7e7e2bfd70ea6)
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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.)
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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.
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So, are you saying my original answer was correct except for entirely the wrong reasons? [;D]
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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!
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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.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fi52.tinypic.com%2F2hd1rg8.png&hash=aeca455c043fbb81b1c7e7e2bfd70ea6)
Thank you JP, Geezer, and the rest of you.
This is the best answer I got so far, but I can't be sure about all of this. One thing isn't so obvious...
Why were we told that electric field is significantly weaker outside capacitor plates, and compared to the magnitude of the field inside capacitor, it's close to zero function?
I mean, why were all those PhDs wasting their time on that if that wasn't true? But, if that is true, how come we have that nice simulation that JP have done?
One other thing... I have seen experiment where beam of electrons were shot between capacitor plates, and that beam was deflected. Angle of deflection was linear function of voltage applied to capacitor.
And last, but not the least...
How come Moon is still orbiting the Earth? I mean, Moon's gravity pull is reason for ocean tides, and Moon is working it's 4ss off to keep that tide rolling 'round the planet.
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Thank you JP, Geezer, and the rest of you.
Why were we told that electric field is significantly weaker outside capacitor plates, and compared to the magnitude of the field inside capacitor, it's close to zero function?
I mean, why were all those PhDs wasting their time on that if that wasn't true? But, if that is true, how come we have that nice simulation that JP have done?
The region between the capacitor does have a higher field than outside. It has to: the fields from the plates add up constructively there (they point in the same direction.) Outside the plates, the fields generally don't add up fully constructively, so the total field is much less. If you assume the plates are infinitely large (a classic, but purely theoretical problem), you only get a field between the plates and it is zero outside due to them adding purely destructively outside the plates.
One other thing... I have seen experiment where beam of electrons were shot between capacitor plates, and that beam was deflected. Angle of deflection was linear function of voltage applied to capacitor.
Deflection is possible, I believe, if the particle starts far away and ends far away. Gain of energy isn't. I'm pretty certain a particle can change direction but not speed in my simulation. (We're ignoring one effect--if a particle changes direction, it radiates away energy, so it actually will lose a slight amount of energy if it changes direction.)
And last, but not the least...
How come Moon is still orbiting the Earth? I mean, Moon's gravity pull is reason for ocean tides, and Moon is working it's 4ss off to keep that tide rolling 'round the planet.
I don't see why the moon isn't losing energy due to this. I can't find a mention of it in a Google search, though. My guess is that it's not a very significant source of energy loss.
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Hm... if your simulation is correct, and I don't see why it wouldn't be, field just outside the capacitor should be extremly strong. Because electric field is negative gradient of potential.
All after all, my confusion levels have significantly dropped due to this discussion. Thank you!
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Hm... if your simulation is correct, and I don't see why it wouldn't be, field just outside the capacitor should be extremly strong. Because electric field is negative gradient of potential.
All after all, my confusion levels have significantly dropped due to this discussion. Thank you!
It should be strong, but not as strong as between the plates. The reason for this is that the electric fields tend to fall off as you move away from the charged plates. Just outside the capacitor, you're close to the plates, so you expect the field to be large. You expect it to be largest in between the plates, since the fields from the two plates point in the same direction, so they add up. Outside the plates, they may point in different directions, so one might be subtracted from the other. But if you're closer to one than to the other, the closer plate's field isn't completely canceled out, so you still get a large(ish) field.
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I was talking just about gradient of that function in your simulation. I always get same plot if I run simulation. We can see that greatest change happens at max and min of that function (at capacitor edges), and gradient is just that - the amount of change. That is if you look at this just as a math problem. But this is off-topic...
Thank you!
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Yeah, it is just a math problem, but as Geezer always points out to me, trusting the math without physical intuition can cause all sorts of problems. :) The fact that the math is backed up by physical reasoning makes me trust that simulation a lot more.