Connecting three spheres with wires redistributes charge

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SUMMARY

The forum discussion centers around a problem involving three identical conducting spheres (A, B, and C) with initial charges of q_A = 0, q_B = 0, and q_C = +Q. After connecting A to B and then A to C, the final charge on sphere A is determined to be +Q/4. Participants debated the implications of wire length and charge distribution, with some arguing for a final charge of +Q/3 based on symmetric distribution. The consensus is that the problem's complexity arises from the lack of information regarding wire length and the assumption of equal potential among the spheres.

PREREQUISITES
  • Understanding of electrostatics principles, particularly charge conservation.
  • Familiarity with conducting spheres and their behavior when connected.
  • Knowledge of electric potential and its implications in charge distribution.
  • Basic concepts of capacitors and charge interactions.
NEXT STEPS
  • Study the principles of charge conservation in electrostatics.
  • Learn about the behavior of conductors in electrostatic equilibrium.
  • Explore the concept of electric potential and its calculation in systems of conductors.
  • Investigate the effects of wire length on charge distribution in connected conductors.
USEFUL FOR

Students of physics, particularly those studying electrostatics, educators teaching introductory physics concepts, and anyone interested in understanding charge distribution in conductive systems.

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.
 
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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/
 
iluvatar said:
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.
iluvatar said:
(The answer is +Q/4!)
Is this a book problem? It could be a typo.
iluvatar said:
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.
 
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.
 
scottdave said:
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.
 
rude man said:
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 .
 
iluvatar said:
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:
 
iluvatar said:

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 :

rude man said:
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
 
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
NFuller said:
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
rude man said:
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.
rude man said:
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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