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Help! Electricity and work

  1. Sep 14, 2005 #1
    Two identical point charges of q = +1.25 x 10-8 C are separated by a distance of 1.10 m. What is the minimum amount of work required by an external force to move them closer together so that they are only 0.60 m apart?


    I have no idea how to do this. I tried using -q(Vb-Va), but that got me 1x10^-6, which is wrong.


    The figure below shows three charges at the corners of a rectangle of length x = 0.45 m and height y = 0.29 m. What is the minimum amount of work needed to move the +2.7 µC charge to infinity?

    [​IMG]


    Again, same thing, I tried using that formula, and Vb would be 0 since it is to infinity, but I got -.4779, which is wrong.
     
  2. jcsd
  3. Sep 15, 2005 #2

    mukundpa

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    Homework Helper

    The total amount of electrostatic potential energy stored in a system of n point charges is
    U = 0.5(q1V1 + q2V2 + q3V3 + ... qnVn)
    Where q is a charge and V is the potential at that point (q) due to all other charges.
     
  4. Sep 15, 2005 #3
    I think the idea here is that you just flip the sign around from positive to negative depending on the context. If you are bringing the charges closer together you are adding work to the system so it should be a positive value of work. If you are bringing the charges from close back to infinity then I think you just switch to a minus sign and report a negative amount of work (system is losing energy). Either way, the signs are the only thing that really differ (the answer is the same, you just change the sign to plus or minus depending on the context). I always get confused about which is which on these types of problems as well...
     
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