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Homework Help: Giving/Recieving Electrons

  1. Jan 4, 2005 #1
    The question goes as follows:

    The correct answers are:
    (a) +1.5q
    (b) +4q
    (c) +4q

    Although, my logic doesn't seem to give me these answers, I was wondering if anyone can explain why and how to approach this problem.

    My logic goes as follows:
    Sphere A = +5q
    Sphere B = -q
    Sphere C = 0

    "Spheres A and B are touched together and then seperated. " Resulting in Sphere A = +4q, Sphere B = 0, Sphere C = 0

    "Sphere C is then touched to sphere A and seperated from it." Resulting in Sphere A = 0, Sphere B = 0, Sphere C = +4q

    "Lastly, sphere C is touched to sphere B and seperated from it." Resulting in Sphere A = 0, Sphere B = +4q, Sphere C = 0

    Anyway, we just began this chapter and my understanding of it is extremely minimal so thats why i'm probablly going about it extremely wrong. Just wondering if anyone could help me out and explain how it works?
  2. jcsd
  3. Jan 4, 2005 #2


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    HINT: Since the spheres are conductors then two spheres that have been brought into contact with each other must be at the same potential.
  4. Jan 4, 2005 #3
    Thanks for the response.
    Ohh okay, so after

    A and B are touched, it becomes: A=+2q, B=+2q, C=0q
    then C and A touch, becoming: A=+1q, B=+2q, C=+1q
    then C and B touch, resulting: A=+1q, B=+1.5q, C=+1.5q

    Is that how it should be done?
  5. Jan 4, 2005 #4


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    The answers to question b) and c) are simple to get as the total charge must conserve.However,your reasoning for the first question is defectuos.
    A-B:A intially +4q,B initially -q.Think of it in terms of particles with positive charge and negative charge (electrons and postirons).On A there are 4 particles with positive charge and on B there is one particle with negative charge.When these 2 are put into contact,there is a flow of the particles.Let's assume the positive particles start to move.Leaves the first and annihilates with the negative particle already on B.Now on B the charge is 0.The system still isn't in equilibrium.Leaves the second.On B the charge is zero,therefore it will go there very easily.The second one goes there as well.It will equilibrium when the 2 spheres will have the same charge.Namely +2q.The charge flow will happen as long as there is nonequlilibrium in the charge quantity.So,when the 2 charged bodies enter into contact,the electric current will tend to restablish the equilibrium.

    Consider the same logics for the other 2 interactions and u'll end up with 1.5q on both C and B.


    PS.I'm sorry,i'm used to explining this in much more complicated terms.That's why i seemed to be a little sluggish. :redface:
  6. Jan 4, 2005 #5


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    Way to go!
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