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Work on done on a test charge

  1. Feb 14, 2007 #1

    rgo

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    1. The problem statement, all variables and given/known data
    I am looking at quation 10 but it relates to question 9
    [​IMG]

    2. Relevant equations

    In question 9 I calculated the electric feild to be 5.4 X10^6 N/c.
    I did this by using the equation E = K(Q/r^2)


    3. The attempt at a solution

    I would assume that I would use this equation for question 10
    W = PE2 - PE1 = -qEd

    therfore

    W = -(1.5x10^-6 C)(5.4 x 10^6 N/c)(0.25m)
    W= -2.025 J

    The sign is negative because the work is in the opposite direction of the electrical field.

    This seems to be too simple. Am I missing something?

    Thanks for the help
     
  2. jcsd
  3. Feb 14, 2007 #2
    Youre moving a positive test charge from lower potential to higher potential, so the work done on the charge is potential energy at D - potential energy at B. What does that say about the sign?
     
  4. Feb 15, 2007 #3

    rgo

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    I think I got it

    Thats right, so my final fomula should be:
    W = (1.5x10^-6 c)(5.4 X10^6 N/c)(0.25m) - (1.5x10^-6 c)(0)(0.25m)
    W = 2.025 J

    Thank you for your help

    RG
     
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