Connecting three spheres with wires redistributes charge

  • #1
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Homework Statement


There are three identical conducting spheres, A, B and C. They are initially charged as q_A = 0, q_B = 0, q_C = +Q. Initially, A and B are connected by a wire. Then the spheres are connected (by a wire) as follows:
1) A to C (while A is still connected to B)
2) Connection between A and C is dropped
3) Connection between A and B is dropped
What is the final charge on sphere A?
(The answer is +Q/4!)

Homework Equations


Conservation of charge. The topic of electric potential es still not used, this is just a beginning problem.

The Attempt at a Solution


I initially tried with symmetric distribution of charge when they are all connected but that brings me to +Q/3 as answer, not +Q/4. Trying to understand the answer led me to think that in step 1) all charge is moved to the extreme spheres (B and C, each one with +Q/2), so the charge is better separated, but what confuses me here is the electric potential is not the same (assuming that is still can be considered as Kq/R).
Any guidance to understand this simple problem is welcomed.
Thank you.
 

Answers and Replies

  • #2
scottdave
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Think of it like connecting a big uncharged capacitor to a smaller charged capacitor. So the two initially uncharged spheres act like the bigger capacitor, if that helps.
Suppose the initial charged sphere had 20 electrons removed from it to start (+q). When it hooks up to the network of bigger spheres, some electrons will come over from them and fill in the holes on the first sphere (leaving positive charges behind). Can you figure how many electrons each of the initially uncharged spheres will contribute?

You might want to read this thread. It is a different situation, but may give you some insight. https://www.physicsforums.com/threa...rged-sphere-touched-to-charged-sphere.516011/
 
  • #3
I initially tried with symmetric distribution of charge when they are all connected but that brings me to +Q/3 as answer
This is the correct answer.
(The answer is +Q/4!)
Is this a book problem? It could be a typo.
but what confuses me here is the electric potential is not the same
You're right, which is why Q/4 cannot be the answer.
 
  • #4
rude man
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Are the wires long? Makes a big difference.
If so I agree the answer is Q/3.
If not the problem is extremely challenging!

PS my reasoning is based on equal potential for all three spheres at the end.
 
  • #5
iluvatar
Think of it like connecting a big uncharged capacitor to a smaller charged capacitor. So the two initially uncharged spheres act like the bigger capacitor, if that helps.
Suppose the initial charged sphere had 20 electrons removed from it to start (+q). When it hooks up to the network of bigger spheres, some electrons will come over from them and fill in the holes on the first sphere (leaving positive charges behind). Can you figure how many electrons each of the initially uncharged spheres will contribute?

You might want to read this thread. It is a different situation, but may give you some insight. https://www.physicsforums.com/threa...rged-sphere-touched-to-charged-sphere.516011/
Hi, thanks for your answer, but as I said this is really a beginner answer, first chapter of electrostatics, without even resorting to Coulomb law and much less to capacitors ( I mean, if I have to explain this, I cannot use concepts that we have no seen yet). But, of course, I will try to think in terms of capacitors as you suggest (is not at all evident for me right now), and then try to simplify and go back to just charge conservation and interactions between touching conductors.
On the other hand, the link you sharing although useful does not help because it corresponds to two touching spheres where the result is much more easy.
 
  • #6
iluvatar
Are the wires long? Makes a big difference.
If so I agree the answer is Q/3.
If not the problem is extremely challenging!

PS my reasoning is based on equal potential for all three spheres at the end.
Hi, actually there is no info regarding the wire length ... which makes this more confusing. This question comes from the Halliday-Resnick material for interactive lectures. I will try to attach the screenshot to the post .
 
  • #7
rude man
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Hi, actually there is no info regarding the wire length ... which makes this more confusing. This question comes from the Halliday-Resnick material for interactive lectures. I will try to attach the screenshot to the post .
I would assume the wires are long :smile:
 
  • #8
iluvatar

Homework Statement


There are three identical conducting spheres, A, B and C. They are initially charged as q_A = 0, q_B = 0, q_C = +Q. Initially, A and B are connected by a wire. Then the spheres are connected (by a wire) as follows:
1) A to C (while A is still connected to B)
2) Connection between A and C is dropped
3) Connection between A and B is dropped
What is the final charge on sphere A?
(The answer is +Q/4!)

Homework Equations


Conservation of charge. The topic of electric potential es still not used, this is just a beginning problem.

The Attempt at a Solution


I initially tried with symmetric distribution of charge when they are all connected but that brings me to +Q/3 as answer, not +Q/4. Trying to understand the answer led me to think that in step 1) all charge is moved to the extreme spheres (B and C, each one with +Q/2), so the charge is better separated, but what confuses me here is the electric potential is not the same (assuming that is still can be considered as Kq/R).
Any guidance to understand this simple problem is welcomed.
Thank you.
This is an screenshot for the problem :

Are the wires long? Makes a big difference.
If so I agree the answer is Q/3.
If not the problem is extremely challenging!

PS my reasoning is based on equal potential for all three spheres at the end.
There is no actual statement regarding the wire lenght... Here is the image of the problem (from Halliday - Resnick interactive classroom questions) : http://imgur.com/LQJsxPy
 
  • #9
It's almost as if the author assumed half of the charge left sphere A and distributed itself among spheres B and C, leaving Q/4 on sphere A. But assuming the spheres are far enough apart such that they do not interact with each other, Q/3 should be the correct answer.
 
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  • #10
haruspex
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It's almost as if the author assumed half of the charge left sphere A and distributed itself among spheres B and C, leaving Q/4 on sphere A. But assuming the spheres are far enough apart such that they do not interact with each other, Q/3 should be the correct answer.
The diagram seems to show A and B rather close to each other. But even if they were touching they would get more than half the charge between them when all three are connected. Anyway, I suspect the different proximities are just to fit the diagrams to the page.
The only way I can make sense of the answer is that A and B are not connected until after A is disconnected from C.
 
  • #11
haruspex
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To get the answer of Q/4, half the charge on C would have to be transferred to A and almost immediately, by virtue of mutual repulsion of like charges, half of A is transferred to B. That would leave Q/4 on A.
But C is still connected to A, so more would be transferred then from C.
I don't see why having the spheres far enough away from each other changes anything.
It is certainly different if the spheres are close. If A is close to B but C is far from both then C would retain more than 1/3 of the total charge, but there is no way it would retain half.
 

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