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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.
Quote from: burning on 22/06/2011 06:19:33Considering 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.
Geezer said: "In running round ANY closed loop in a static electric field the field
All I want to know is where this energy comes from?
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?
I need someone really good with this stuff to explain to me how the capacitor lose it's energy by that process.
Quote from: AlmostHuman on 26/06/2011 21:57:46I 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.
Quote from: Geezer on 27/06/2011 09:55:18Quote from: AlmostHuman on 26/06/2011 21:57:46I 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...
Quote from: JP on 27/06/2011 14:30:50Quote from: Geezer on 27/06/2011 09:55:18Quote from: AlmostHuman on 26/06/2011 21:57:46I 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.