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Repulsion forces

  1. Oct 8, 2014 #1

    Lea

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    1. The problem statement, all variables and given/known data
    A metal sphere of radius R carries a total charge Q. What is the force
    of repulsion between the "northern" hemisphere and the "southern" hemisphere?

    2. Relevant equations


    3. The attempt at a solution
    Since the sphere is metallic, its conducting and so the electric field inside the sphere is 0, and outside the sphere it is perpendicular to the surface and is radially outwards, and so, the force exerted by the northern hemisphere on the southern hemisphere should be 0. However, the answer appears to be other than zero. why?
     
  2. jcsd
  3. Oct 8, 2014 #2

    gneill

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    Staff: Mentor

    The net field inside is zero because the all the individual fields from the individual charges distributed around the sphere cancel each other out in the interior. That does not say that those same fields don't serve to push all the charges to the skin of the sphere, keeping them spread out evenly over the surface.

    As far as the mutual repulsion of hemispheres goes, you've still got two charge distributions pushing on each other.
     
  4. Oct 8, 2014 #3

    Lea

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    But those charge distributions have an electric field that is radially outwards, how can they push on each other?
     
  5. Oct 8, 2014 #4

    gneill

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    Staff: Mentor

    Each charge has a spherically symmetric field radiating outward around itself. Exterior to the sphere the net effect of summing all the fields leaves just the radially outward part. Within the sphere they all cancel. The overall effect is to be able to perceive no field in the interior of the sphere and a radially outward field exterior to the sphere.

    At the surface where the charges reside, they are feeling the effects of the net radial field. This is what keeps those charges confined to the outer surface. The fields aren't mutually cancelled from there outward, so each of the individual charges is "feeling" a net field that looks as if all the charge Q were placed as a point charge at the center of the sphere (and the conducting sphere is ignored).
     
  6. Oct 9, 2014 #5

    Lea

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    Thank you! I understand now.
     
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