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Electric Force on a charged conductive body |
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| Dec16-12, 08:01 AM | #18 |
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Electric Force on a charged conductive body
If I understand your scheme, V0(T) is the potential due to disks at the points on the surface of the ball,
But how do you know this? It is a function of position, so it varies point to point. I think this is unknown, too. Unless the charge distribution on the disks is rigid? |
| Dec16-12, 08:15 AM | #19 |
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Yes for the sake of calculation you assume it to be constant. And then you go on to calculate the sigmas on the disks (because the ones on the ball just changed), and then you do it again for the charges on the ball, and so on you iterate it (since it converges to a value).
To be frank, I don't know this to be right for a fact, but some of my notes from the lectures had a similar case, and I sort of extrapolated it from there, and my numeric results seem to be correct (the pictures of equipotential lines seem very nice). |
| Dec16-12, 08:54 AM | #20 |
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Aha, you iterate! Great, I think then it is OK. What I was trying to explain above was a method where you do not iterate, but you would have to introduce more points and include the charges on the disk into the unknowns. Then you would find everything from one set of equations. But your method is seems equally good.
So, in the end, is the ball repelled from the plate it was touching originally? (I saw some paper with some strange cases in electrostatics when the force can point in counter-intuitive direction.) |
| Dec20-12, 04:18 AM | #21 |
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Actually it is much simpler, I kind of complicated things a bit with the V0 part (it is actually V0=0). It's actually like you said at the beginning, and there is no need to iterate anything. Although the result I get is the same either way.
Yes of course the ball is repelled from the one disk, and attracted by the other one. Now I have the job of calculating the time it would take the ball to cross from one side to the other, after I have calculated the force at a bunch of points, this, I think will be the source of another thread. |
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