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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.)

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_equations [Links inactive - To make links active and clickable, login or click here to register]http://en.wikipedia.org/wiki/Lorentz_force [Links inactive - To make links active and clickable, login or click here to register]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 [Links inactive - To make links active and clickable, login or click here to register] (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.

Ah, right, thanks! I wonder if a capacitor approximates to a whole bunch of dipoles, and would that help to model the particlar situation?

So, are you saying my original answer was correct except for entirely the wrong reasons? []

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.

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?

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.

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!