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Homework Help: Work to move a charge

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A +35[tex]\mu[/tex]C charge is placed 32 cm from an identicle +35[tex]\mu[/tex]C charge. How much work would be required to move a +.5[tex]\mu[/tex]C test charge midway between them to a point 12cm closer to either of the charges?


    2. Relevant equations
    Wext = -qV
    V = kQ/r


    3. The attempt at a solution
    Q= +35[tex]\mu[/tex]C, q = +.5[tex]\mu[/tex]C
    I found the initial V, Vi = 2KQ/.16 and then the final V, Vf = KQ[1/.28 +1/.04]
    Then I did W = -q(Vf-Vi) and got -2.5J
    However, the answer is +2.5 J. This makes sense that the answer is positive- you're moving a + charge from an area of lower to higher potential. Why doesn't this agree with my formula though?
     
  2. jcsd
  3. Jan 27, 2009 #2

    Doc Al

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    Staff: Mentor

    Because you are finding the work done by you to move the charge, which is qΔV, not the work done by the field, which is -qΔV.
     
  4. Jan 27, 2009 #3
    Oh, so Wext = qΔV. So for energy considerations, is it W(by E) + PEi + KEi = PEf + KEf + Wext?
     
  5. Jan 27, 2009 #4

    Doc Al

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    In this particular case, in which you are moving the charge with the least amount of energy (no excess kinetic energy).
    In general, I would say: PEi + KEi + Wext = PEf + KEf
     
  6. Jan 27, 2009 #5
    Ok. If another problem asks about W from E, would Wext just be negative in the equation you provided?
     
  7. Jan 28, 2009 #6

    Doc Al

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    Staff: Mentor

    If you want the work done by the electric field, use W = -qΔV.
     
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