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Electric Potential Energy of a sphere

  1. Aug 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A test charge of +1.0x10^-6 C is 40cm from a charged sphere of 3.2x10^-3 C. How much work was required to move it there from a point 1.0x10^2 m away from the sphere?

    2. Relevant equations

    W = [tex]\Delta[/tex]E
    W = Ee2 - Ee1

    3. The attempt at a solution

    Ee2 = (9.0x10^9)(1.0x10^-6)(3.2x10^-3)/(0.4)
    = 72

    Ee1 = (9.0x10^9)(1.0x10^-6)(3.2x10^-3)/(100)
    =0.288

    W = 72 - 0.288
    = 71.7
    = 72 (significant digits)

    I'm pretty sure my approach is correct, but I'm not sure if I should reverse the distance variables, since I would imagine it takes negative work to bring two like charges closer.
     
  2. jcsd
  3. Aug 2, 2009 #2

    rl.bhat

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    In this case work done by the electric field is negative. But the work done by the external force is positive.
     
  4. Aug 3, 2009 #3

    kuruman

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    I think it's the other way around. Since there is no sign in front of the charge on the sphere, we must assume it is positive and the electric field generated by it must be radially out. This is also the direction of the (repulsive) electric force on the positive test charge. If the charge moves out from 40 cm to 100 m, the work done by the electric force is positive because the force doing the work and the displacement are in the same direction.
     
  5. Aug 3, 2009 #4

    rl.bhat

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    In this problem charge is moved from 100 cm to 40 cm.
     
  6. Aug 3, 2009 #5

    kuruman

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    Yes, of course. I misread the problem.
     
  7. Aug 20, 2009 #6
    I got the same answer that was posted.

    To clarify the distance, the test charged is moved from a position 100 m away from the charged sphere to a position 40 cm away from the charged sphere.
     
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