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How much work is done when moving a charge?

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    In Figure 24-60, we move a particle of charge +2e in from infinity to the x axis. How much work do we do? Distance D is 6.30 m.

    http://img216.imageshack.us/img216/6987/fig2460.th.gif [Broken]


    2. Relevant equations
    U = kQq/r
    U=-W


    3. The attempt at a solution
    Ok since there's two stationary charges (2e and e), I figured you could use the superpositioning principle and claculate the work done to move the 2e from infinity to the x-axis against the stationary 2e plus the work done to move it against the stationary e charge. Summing them would give the answer.

    U1= (k)(2e)(2e)/(2d) = 7.233E-29 J

    U2 = (k)(e)(2e)/d = 7.23E-29 J

    U(total) = 7.23E-29 + 7.23E-29 = 1.4466E-28 J

    U = -Work
    1.4466E-28 = -Work
    Work = -1.4466E-28 J

    Is my assumption that the superpositioning principle can be used false? Thank you in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 4, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your use of superposition is perfectly correct (although I didn't check your arithmetic). One error is in the sign of the work done: The work done equals the change in energy, in this case W = ΔU (not -ΔU).
     
  4. Apr 4, 2009 #3
    Thank you very much Doc, you've saved the day once more! I just have one more question left that I'm struggling with but I'm gonna try it some more. If I need help, I'll start a new thread to make everything more organized.

    Thank you once again Doc!
     
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