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Calculating the time to reach a potential difference

  1. Jan 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Q: Two conducting balls of radius 0.1 m are situated 3m apart in free space. Electrons are transferred from one ball to another at a rate of 10^13 per second. How long does it take for a p.d of 100 kv to develop?
    2. Relevant equations

    where ε0 is the electric constant (permittivity of vacuum)

    3. The attempt at a solution
    First I calculated the charge needed for the particular voltage using the potential equation. Then I found C/s by multiplying the number of electrons by charge per electron. Finally I divided the charge needed by the charge per second to get the time to reach the charge. This answer does not match the book.

    I think I need to take the increase in potential of one ball and the fall in potential of the other in account. But I have no idea how to do that?

    Any help would be appreciated!
    Last edited by a moderator: Jan 25, 2016
  2. jcsd
  3. Jan 25, 2016 #2


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    Staff: Mentor

    Welcome to the PF.

    The problem implies that the electrons are transferred at a constant rate, so whatever energy is needed to overcome the growing electrastatic repulsion is apparently available, so I don't think you have to consider that.

    Can you post your detailed calculations step-by-step, including units? That will help us to look for any mistakes or oversights. Thanks. :smile:
  4. Jan 25, 2016 #3
    10^13 * 1.602 * 10^-19 = 1.602 * 10^-6 C/s
    100*10^3 V= Q/(4*pi*e0), Q = 3.34 * 10^-5 C
    t = 3.34 * 10^-5/1.602 * 10^-6
    = 20s

    But the answer should be .36s
  5. Jan 25, 2016 #4


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    2017 Award

    Staff: Mentor

    The units don't match here.
    Always work with units, it helps to find most errors. You forgot one factor.

    You also have to consider that this equation is the voltage relative to infinity, not relative to the other ball.
  6. Jan 25, 2016 #5


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    Staff: Mentor

    I think, as the OP surmised, the question is looking for the potential difference between the spheres, and as electrons are moved from one sphere they leave behind an equivalent net positive charge on the other.

    That implies the two spheres will have equal and opposite potentials near their surfaces (assuming negligible induced charge interactions with the given separation between them).
  7. Jan 25, 2016 #6
    Actually that was a typo.

    It was supposed to be 100*10^3 V= Q/(4*pi*e0 * 3). That's what I had done, but I typed it wrong.
  8. Jan 25, 2016 #7


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    Science Advisor
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    Gold Member

    Several problems here.
    You seem to have substituted the distance between the spheres, 3m, for r. r should be the radius of one sphere.
    Secondly, this only gives you the potential at the surface one one sphere due to its own charge. The potential difference depends on both potentials.
    Thirdly, the potential at one sphere is affected by the field of the other sphere.
    Fourthly, the charge distribution at one sphere will be affected by the field from the other sphere. The charge distributions on each will not be uniform. However, my understanding is that taking that into account would make it rather an advanced problem. If you pretend that each charge distribution is uniform then the potential at one sphere due its own charge is ok, but if you try to take into account the potential there due to the other sphere it will no longer be a uniform potential. Maybe with a radius one thirtieth of the distance between them it is near enough uniform.
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