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Charge of a sphere, thin circular hoop

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    What minimal charge should have a sphere of radius R to surpass a thin circular hoop of radius r that is uniformly distributed with the charge Q and made from resilient elastic. The elastic modulus (of resilience) is [tex]\Psi[/tex]. R>r

    2. Relevant equations
    Coulomb's law.


    3. The attempt at a solution
    IMO the electric force is [tex]F = \frac{k_e Q q}{R^2}[/tex]. What's then? I haven't got the slightest idea. Help me, please.
     
    Last edited: Apr 10, 2010
  2. jcsd
  3. Apr 10, 2010 #2

    tiny-tim

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    Hi Oloria! :smile:

    (have a psi: ψ :wink:)

    The idea is to expand the hoop …

    so what will be the "horizontal" radial force on the hoop if the charge on the sphere is -P ?
     
  4. Apr 10, 2010 #3
    Hi tiny-tim :smile:

    I think the electrici force gotta be bacbalanced by a force connected with resilience. I think F~ψ but I dunno how to write it down properly.

    I don't understand why you used "-P" and I don't know what answer you expect (I have already mentioned Coulomb's law).
     
    Last edited: Apr 10, 2010
  5. Apr 10, 2010 #4

    tiny-tim

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    Hi Oloria! :smile:

    The sphere must have negative charge, so that it can repel the hoop, and make it bigger. :wink:

    Forget the elastic force for the time being …

    just calculate the "horizontal" radial electric force from the sphere on the hoop …

    what do you get? :smile:
     
  6. Apr 10, 2010 #5
    Repel? I thought there should be Q and -q so they attract each other. I don't understand :(

    [tex]F = \frac{k_e Qq}{R^2}[/tex] I guess the distance between midpoints of these two objects is R. q is the charge on the sphere.
     
  7. Apr 10, 2010 #6

    tiny-tim

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    oops! :redface: i misread the question, yes it'll be Q and q, so the sphere makes the hoop larger :smile:
     
  8. Apr 10, 2010 #7
    Q and -q, am I right?
    And I don't know what to do next. Can you help me, please?
     
  9. Apr 10, 2010 #8

    tiny-tim

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    No, Q and q: the sphere must repel the hoop, so that the hoop is the same radius (R) as the sphere.

    What is the electric force between a sphere of charge q and a hoop of charge Q that just fits onto the equator of the sphere?
     
  10. Apr 10, 2010 #9
    [tex]F = \frac{k_e Qq}{R^2}[/tex]?
    I thought the sphere will repel a hoop so it will be impossible to start surpassing (being so close of two objects). (When the sphere is in the hoop they have to repel each other so the hoop will be larger).
     
  11. Apr 10, 2010 #10

    tiny-tim

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    It depends how fast you push the sphere, or whether the hoop is fixed somehow.

    Anyway, now match the electric force with the elastic force. :smile:
     
  12. Apr 18, 2010 #11
    But I don't know how to write down the elastic forse. Could you help me?
     
  13. Apr 18, 2010 #12

    tiny-tim

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    Consider an extremly small arc of angle 2θ radians …

    if T is the tension in the hoop, and if the radial force is F/2π per radian, find the equation relating T and F, and then let θ tend to zero. :wink:

    (alternatively, you could use the work-energy theorem)
     
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