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Homework Help: Find the electric potential energy of this system of charges

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    A point charge q_1 = 86.5 uC is held fixed at the origin. A second point charge, with mass m = 5.90×10−2 kg and charge q2 = -2.95 uC, is placed at the location (0.323 m, 0)
    1) Find the electric potential energy of this system of charges.
    2)If the second charge is released from rest, what is its speed when it reaches the point (0.121 m, 0)?

    2. Relevant equations

    W = EPEa - EPEb

    3. The attempt at a solutionI think the above equation is used in 1 but I'm not sure how to start it. For 2, I'm not sure how to figure out the problem using the x,y coordinates given. Any help would be appreciated
  2. jcsd
  3. Feb 3, 2010 #2
    alright so what you are basically doing is bringing a charge from infinity to a point (0.323,0) The reason you have to think about it this way is that the only way that charge is in existance, but not part of the initial system's energy is if it is an infinite distance away. so what you want to do is take EPEi(i denoting infinity), and subtract from it EPEb.

    For your second question think about it like gravity. You have the earth at the origin. you want to find the speed when something falls from 323 meters in the air to 121 meters in the air. So you are going to find the difference in potential energy. So the equation is the same. So you are going to have kq/0.121 - kq/0.323. that will give the kinetic energy gained by the charge. Throw that energy in the E = 0.5mv² and there you go. solve for v
  4. Feb 3, 2010 #3
    oh and by the way EPEi is always 0 since r is infinity
  5. Feb 4, 2010 #4
    ok so I'm still not exactly comprehending number 1. so basically we "start" at infinity and move to a point (.323), which in this case we find the difference (kq/.323- kq/0). so is this the EPE of the charges? there seems like there would be more to this problem.
    for 2, i am getting a negative number when i subtract kq/.121 - kq/.323. thus i cant take a sq root. am i doing something wrong?
    thanks for all your help
  6. Feb 4, 2010 #5
    the negative is not an issue in this case. The negative is a seperate issue a lot of times in EM. It has something to do with who the charge is being moved by and so on. But in this case you are only looking for a difference in energy. so just make the negative number a positive one.

    for the first question, the 0 you put on the denominator is actually an infinity. which puts kq/infinity which is equal to 0. so you are basically finding the potential of kq/0.323
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