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Question: A metal sphere of radius R carries a total charge Q

  1. Jan 31, 2006 #1

    siddharth

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    Question: A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the "northen" hemisphere and the "southern" hemisphere?

    The force for a small area 'da' on the outer surface should be

    [tex] df = \sigma da E [/tex]

    What E to use? My guess is that it should be averaged over both sides of the surface.
    So,

    [tex]E = \frac{1}{2} (\frac{Q}{4 \pi \epsilon_0 R^2} + 0)[/tex]
    [tex] \vec{E} = \frac{Q}{8\pi \epsilon_0 R^2} \vec{e_r} [/tex]

    and

    [tex] \sigma = \frac{Q}{4 \pi R^2} [/tex]

    And the answer should be

    [tex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \frac{Q^2 \sin\theta}{32 \pi^2 \epsilon_0 R^2}(\sin\theta \cos\phi \vec{e_x} + \sin\theta \sin\phi \vec{e_y} + \cos\theta \vec{e_z}) d\theta d\phi [/tex]

    Is my approach correct?
     
    Last edited: Jan 31, 2006
  2. jcsd
  3. Jan 31, 2006 #2

    Meir Achuz

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    Yes, but you should follow Tesla's advice and recognize from the beginning that the x and y components of F vanish due to the axial symmetry.
     
  4. Jan 31, 2006 #3

    siddharth

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    Thanks Meir Achuz. That would have been my next step. Stating that F_x and F_y are zero by symmetry.
     
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