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Feb 28th 2015, 12:07 AM   #1
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charge of particles (an easy problem)

Hi,
This problem is a-little basic for this forum, but can you assist with the following (refer to attachment). I've sketched out how I think the particles will behave, given a -2Q particle interacts with the +ve particles.

But my problem is, if the particles of opposite charge are touching. The attractive force become infinite.

F = k (q(1). q(2) )/ r^2

r -> 0, F -> inf.

Do the particles move? the problem states that they're fixed(??)

Your thoughts.
Any assistance will me much appreciated.
Attached Files
 hw-problem.pdf (443.3 KB, 2 views)

Feb 28th 2015, 12:17 AM   #2
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 Originally Posted by psyclone Hi, This problem is a-little basic for this forum, but can you assist with the following (refer to attachment). I've sketched out how I think the particles will behave, given a -2Q particle interacts with the +ve particles. But my problem is, if the particles of opposite charge are touching. The attractive force become infinite. F = k (q(1). q(2) )/ r^2 r -> 0, F -> inf. Do the particles move? the problem states that they're fixed(??) Your thoughts. Any assistance will me much appreciated.
Just add to that.
To calculate the applied charges, I'd have to use a field equ, F = E/q, where E is the electric field as a result of the other particles acting on +3Q,

But how do I (or would you) formulate a field equation for all particles?

 Feb 28th 2015, 05:29 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 This is way simpler than you're making it. The problem asks about charges, not forces. So think what happens when a -2 charged sphere touchs a +1 sphere - the total charge is aharde by both spheres and they end up equal charge, with each carrying the average charge. Thus they both end up with charge -1/2. It's that easy. Now continue on - what happens when the sphere which now has charge -1/2 touches the second sphere with charge +2? And so on. Last edited by ChipB; Feb 28th 2015 at 07:32 AM.
Feb 28th 2015, 06:19 PM   #4
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 Originally Posted by ChipB This is way simpler than you're making it. The problem asks about charges, not forces. So think what happens when a -2 charged sphere touchs a +1 sphere - the total charge is aharde by both spheres and they end up equal charge, with each carrying the average charge. Thus they both end up with charge -1/2. It's that easy. Now continue on - what happens when the sphere which now has charge -1/2 touches the second sphere with charge +2? And so on.
That's It! Thanks.

Particle R + -2Q = -1/2Q (averaged)
Particle S - 1/2Q = +3/4Q (averaged)
Particle T + 3/4Q = +15/8Q (averaged)

Ans: +15/8Q

and,
I thought the smallest unit of charge transferable is 1 electron?
Can I ask though, what's aharde means?

 Feb 28th 2015, 08:16 PM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 "Aharde" is a typo; should be "shared."
Feb 28th 2015, 10:00 PM   #6
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 Originally Posted by ChipB "Aharde" is a typo; should be "shared."
ok, (my apologies) I read that again, and I should have worked out you meant shared.

But getting back to the other question in my last post.

I can see how if there are two particles and one with a negative charge, the Net Charge is -1/2Q, (when in contact). But how does this net charge carry over to the next pair? Or take the pair (+Q, -2Q) the residual charge is -Q, but we don't know which particle has the extra charge, so we say both have the same Net Charge?

So the answer of +15Q/8, is actually just an average, but in reality (if measured) it could have a higher or lower charge. So we're calculating the 'probable mean' or 'expected value' of the charge?

I'm just trying to gain a better understanding of the problem.

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