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Homework Help: Electrostatic force between two halves of sphere

  1. Nov 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the electrostatic force of interaction between two halves of a spherical conductor of radius R carrying a charge Q.
    This thread already exists- https://www.physicsforums.com/threads/electric-force-between-two-halves-of-a-sphere.317380/ but is closed.

    2. Relevant equations
    Electrostatic pressure:
    dF/dA = σ2/2ε0

    3. The attempt at a solution
    I understand how to do it using the method outlined in the above mentioned thread, but I want to do it using the argument stated in the last post of the thread, the one that mentions this at the very end: "By the way, I think the "supposed way" to do this Griffith problem is to use the formula for surface pressure of a conductor earlier in the chapter and add up the integrate z-component of the force on the base and the side of a hemisphere. Be ware, you need to make a correct physical argument about the direction of the pressure/force."

    So, to solve it, I tried splitting up the northern hemisphere as elementary rings. Basically, here's what I'm trying to do:
    Take an elementary area dA containing a charge dq on an elementary ring of radius r. The force acting on dq is
    dF = ( σ2 dA ) / 2ε0
    Now comes my problem-
    I'm not sure what the net force on the ring should be, should it be ∫dFsinθ or ∫dFcosθ, i.e. should I take the vertical components or the horizontal components?
    So, I tried using both and got the right answer using ∫dFsinθ
    Here's what I did:
    The net force on the ring is
    F (ring) = ∫dFsinθ = ∫( σ2 2πRcosθRdθ sinθ) / 2ε0
    and F(hemisphere) = ∫F(ring)
    Integrating from 0 to π/2 I get Q2 / 32πεR , which is correct.
    But I still don't get why I'm choosing ∫dFsinθ over ∫dFcosθ.
  2. jcsd
  3. Nov 29, 2015 #2
    Because you are measuring angle θ from the vertical .(Assuming you are trying to find force on the right hemisphere)

    If you measure θ from the horizontal you need to have ∫dFcosθ . Both ways you end up with the same answer .
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