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Electrostatic pressure

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A sphere with radius R is charged with surface density σ. The charge found in a small surface Δs is repelled by the rest of the sphere, thus generating an electrostatic pressure. Find the pressure.

    2. Relevant equations

    3. The attempt at a solution
    I think I have to compute the field for the sphere, which can be done using Gauss' law. For Δs: Esph = σ/ε0
    Then I think I have to find the field for Δs and subtract it, but I don't know how to find it.
    If field is constant, ten just multiplying QΔs = Δs·σ by the field would give the force.

    Thanks for your help!
     
  2. jcsd
  3. Mar 25, 2009 #2

    rl.bhat

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    Yes. And force per unit area will be pressure.
     
  4. Mar 25, 2009 #3
    Thanks.

    Now how do I find the field for Δs?
     
  5. Mar 25, 2009 #4

    rl.bhat

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    The magnitude of the electric field at every point on the sphere is the same and is equal to Esph = σ/ε0
     
  6. Mar 26, 2009 #5
    Yes, but I what meant to is: shouldn't I find the field generated by the sphere and subtract the field generated by Δs to have the field generated by a sphere without Δs? If so, my problem is that I don't know how to find the field generated by Δs.
     
  7. Mar 28, 2009 #6
    Bump...
     
  8. Mar 29, 2009 #7

    rl.bhat

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    A body does not exert force on its own surface. So if you want to find pressure on delta s,
    assume that it is removed from the sphere and kept very close to the sphere. When you remove delta s from the sphere, charge on each surface of delta s will be 1/2*sigma*delta s. Since delta s is very small, field due sphere close to its surface does no change much. So it will be sigma/ epsilon(not). So the pressure =force /area = 1/2*sigma*delta s*sigma/epsilon(not) = (sigma)^2/2epsilon(not)
     
  9. Mar 29, 2009 #8
    OK thanks, I understand now. Just one small thing, why is the charge 1/2·Δs·σ? I thought the charge equals to surface (Δs) multiplied by the density (σ). Where does the 1/2 come from?
     
  10. Mar 29, 2009 #9

    rl.bhat

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    When the delta s is on the sphere , charge sigma delta s is only on outer surface. When you remove it from the sphere, charge distributes equally on both the surfaces. Hence 1/2*sigma*delta s.
     
  11. Mar 29, 2009 #10
    OK thanks!
     
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