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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.
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.
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.)
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 []
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:Quote from: Geezer on 27/06/2011 22:01:25Ah 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
Quote from: Geezer on 27/06/2011 22:57:57I don't think so Burning.You seem to be assuming the plates have some constant potential relative to the particle.QuoteNo. 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.QuoteIf 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. QuoteIf 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 []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?Quote1.) 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.Quote2.) An object with a net non-zero charge will produce an electric fieldIf it's charged, yes, and in this case it certainly is.Quote3.) 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.Quote4.) 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.Quote5.) 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.
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.
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.
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.)
Quote from: Geezer on 28/06/2011 21:28:19Now, 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?
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_equationshttp://en.wikipedia.org/wiki/Lorentz_forceThe 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.