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Work Done on a particle

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A -2.0*10^-3 C charge is 0.2 m away from a -6.0*10^-3 C charge. How much work is must be done on the first charge to move it to a distance of 0.9m?

    2. Relevant equations
    F = qs*qt*k/r^2

    W = F * d* cos theta

    3. The attempt at a solution
    (-2.0*10^-3)(-6.0*10^-3)(9*10^9)/0.2^2
    =2700000 N
    (-2.0*10^-3)(-6.0*10^-3)(9*10^9)/0.7^2
    =220408.1633 N

    W = 2700000N * .2 = 540000
    W = 220408.1633 * .7 = 154285.7143

    The sum of those is 694285.7143


    This is obviously not right because the answer is supposed to be -4.2*10^5
     
  2. jcsd
  3. Apr 17, 2016 #2
    I think It is easy to use potential method, first calculate the potential difference 0.2m to 0.9m due to the second charge,
    the use the eqn, W=qV
     
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