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What is the force of electric interaction between a copper ball
of radius R carrying net charge +Q, and a point charge +3Q
located at distance 2R from the center of the ball?

Please, help me people. My girlfriend will dump me, unless
I score at least B on this assignment... she's got all A's.

2007-03-09 03:15:45 · 2 answers · asked by Alexander 6 in Science & Mathematics Physics

My girlfriend uttered something about Apollonius circles, whatever that might be.
Then, again, she's got all A's.

2007-03-09 04:55:27 · update #1

copper ball cannot be an equipotential surface

Well, it *has* to be.

2007-03-09 09:48:43 · update #2

2 answers

F = (1/4πε) (3Q²/4R²)

Seems too simple? Check this link on "Field outside a shell". Newton first realized that in determining the gravitational force of a shell, it's equivalent to a point with the same mass at center. Same thing works for electrostatics.

The answer gets more complicated if the charges on the copper ball is free to move around in response to the presence of another point charge not on it. I'll have to think about that one.

All right, we probably can look at this as setting up an electrostatic field where the sphere represents a equipotential surface. More work. The circle of Apollonius would be an equipotential surface between 2 charged particles, so, hmm?

Still does not make sense, because the copper ball cannot be an equipotential surface if the charges on it are non-uniformly distributed because of the presence of the +3Q outside point charge. This is quite a difficult problem. Either it's uniformly charged, in which case the first result is correct, or it's not, in which case no circle of Apollonius is going to help you. The study of non-uniform distribution of charges on irregular surfaces for minimum electrostatic potential is definitely a non-trivial one!

Okay, here comes the "I'm gonna kill you for this!" answer. If Q represents a SINGLE charge, then obviously the answer is

F = (1/4πε) (3Q²/9R²)

because the single charge on the ball has gone as far as possible from the point charge, or 3R.

2007-03-09 03:28:00 · answer #1 · answered by Scythian1950 7 · 1 1

Subatomic debris do no longer appear like something. The suggestions of "appear like" and "stable" shop on with to a greater international, no longer that of such tiny debris. those brightly coloured balls you have been shown in common college that represented nucleons and electrons are almost as deceptive as they're clarifying. Even touch belongs to the macroscopic international. in case you walk right into a stable wall, what's particularly occurring is the amassed effect of the electron fields of the molecules repelling one yet another, as mediated by skill of the electromagnetic rigidity.

2016-11-23 17:19:29 · answer #2 · answered by demeritte 4 · 0 0

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