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Net Charge

  1. Dec 18, 2008 #1

    Air

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    The problem statement, all variables and given/known data
    Initially, sphere A has a charge of –10e and sphere B has a charge of +20e. The spheres are made of conducting material and are identical in size. If the spheres then touch, what is the resulting charge on sphere A? a) +10e, b) -10e, c) -5e, d) +5e

    The attempt at a solution
    Sphere B -> [itex]20e-\left(\frac{+20e-10e}{2}\right) = 20e - 5e = 15e[/itex]. So there is a loss of 5e in Sphere B. This would mean that the only -5e remains on Sphere B?

    The Problem
    However, the answer is +5e. Why is this? There is a loss of 5e in Sphere B. This would mean that the only -5e remains on Sphere B?
     
  2. jcsd
  3. Dec 18, 2008 #2

    Doc Al

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    Staff: Mentor

    No. For some reason, you took the average charge but then subtracted it from Sphere B. Why?

    Instead, ask yourself: What's the total charge on both spheres together? Once they touch, how does that total charge divide itself between the two spheres?
     
  4. Dec 18, 2008 #3

    Air

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    Only negative charge moves so only negative changed one should change? After they touch, Sphere B would remain as 20e?

    Together, the net change is +10e, but Sphere B contributes more than Sphere A?
     
  5. Dec 18, 2008 #4

    Doc Al

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    Staff: Mentor

    True.
    No. If charge moves between spheres, they both must change.
    No.

    Right. (That's the equivalent of 10 missing electrons.)
    Why?

    The two spheres are identical. When they touch, the charge redistributes evenly between the two spheres.
     
  6. Dec 18, 2008 #5

    Air

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    So there is a distribution of +5e on both spheres. If we add this to the previous values (original charge value) then wouldn't we get:

    Sphere A: -10e+5e = -5e and not +5e.

    :blushing: (Sorry, I'm still confused.)
     
  7. Dec 18, 2008 #6

    Air

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    I think I've understood. I've read this thread: https://www.physicsforums.com/showthread.php?t=58634

    So, the net change from Sphere 1 and 2 is +10e. But, for it to be stabl;e and equlibrium, they must seperate equally hence Sphere 1 and 2 will both have +5e charge? Am I correct?
     
  8. Dec 18, 2008 #7

    Doc Al

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    Yes, you are correct.

    What actually happens is that 15 electrons will move from A to B in order to equalize the charge distribution, thus:
    A goes from -10e - (-15e) = +5e;
    B goes from +20e + (-15e) = +5e.
     
  9. Dec 18, 2008 #8

    Air

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    Oh I see. Thanks for the help. You've been very patient and helpful. It was encouraging.

    The reason that I was getting confused is because in Fundamentals of Physics, Chapter 21, Question 4, you have to use average to find the charge left over in the positive charge. Why is that different?
     
  10. Dec 18, 2008 #9

    Doc Al

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    Staff: Mentor

    If you post that question, I can see what's different.
     
  11. Dec 18, 2008 #10

    Air

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    The figure below shows three pairs of identical spheres that are to be touched together and then seperated. The initial charge on them are indicated. Rank the pair according to (a) the magnitude of the charge transferred during touching and (b) the charge left on the positively charged sphere, greatest first.
    [​IMG]
     
  12. Dec 18, 2008 #11

    Air

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    I understand it now because your post number 7 explain why that works but why does for positive you have to add the average and for negative you have to take away the average?
     
  13. Dec 18, 2008 #12

    Doc Al

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    Staff: Mentor

    In all cases the average charge tells you what the final charge on each sphere will be. To find the change in charge, just subtract final - initial.

    For case (1) in your last post, you start with +6e and -4e. So the final charge on each will be (+6e -4e)/2 = +1e. Thus the change in charge on left sphere will be: +1e - 6e = -5e (it gained 5 electrons); on the right sphere, it will be: +1e - (-4e) = 5e (it lost 5 electrons).

    Make sense? (Since the question asks for the magnitude of the charge transferred, the answer would be 5e.)
     
  14. Dec 18, 2008 #13

    Air

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    Yes, that makes sense. Thanks a million. :smile:
     
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