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Homework Help: A fixed charge at origin and a moving charge repelled

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data

    A charge of +3.37µC is held fixed at the origin. A second charge of 3.37µC is released from rest at the position (1.15 m, 0.590 m) .
    (a) If the mass of the second charge is 2.90 g, what is its speed when it moves infinitely far from the origin?

    (b) At what distance from the origin does the 3.37 µC charge attain half the speed it will have at infinity?

    Reference https://www.physicsforums.com/threads/charge-at-origin.131159/ (I have the same question as someone else who already asked this question but I still couldn't get the answer)

    2. Relevant equations

    1/2mv^2 = qV
    U= kQq/r

    3. The attempt at a solution

    I already found the answer to A just using Conservation of Energy. But I can't find the answer to B.


    Someone else wrote how to find part B, but when I solved for it my answered ended solving for r as

    r= ro/3

    but when I tried this, I couldn't figure it out. My ro was Sqrt(1.6706) so I thought the answer would be Sqrt(1.6707)/3 but it wasn't right.
  2. jcsd
  3. Feb 12, 2016 #2

    Doc Al

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    You'd also use conservation of energy for part B. What would the KE be at the point in question? (Use the results from A.) What's the total energy?
  4. Feb 12, 2016 #3
    7.38087 was the velocity. KE = 1/2mv2
    So the total KE is 1/2(0.0029)(7.38087)^2

    Total Energy must be equal to KE + PE.... hmm. But to find the Potential Energy...
    (8.99 x 10^9)(3.37 x 10^-6)^2 / Sqrt(1.6707)

    So the Total Energy is both of those put together. But how is the total energy going to help me find the half the speed at infinity?
  5. Feb 12, 2016 #4


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    You know the speed and KE at infinity. At half the speed, what is the KE? What does that leave for PE?
  6. Feb 12, 2016 #5
    KE at half the speed to infinity: 1/2(0.0029)(3.690435^2 = .019748 Joules

    But we know that when the speed is at infinity, potential is equal to 0, right?

    Therefore the Potential Energy must be: KE at speed at Infinity - KE at half the speed to infinity
    so. PE = 1/2(0.0029)(7.38087)^2 - 1/2(0.0029)(3.690435^2

    But now how am I going to use this potential energy to figure out the distance? Oh wait, the equation for the potential gives me the distance between the two points??
  7. Feb 12, 2016 #6


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  8. Feb 12, 2016 #7
    So the PE is equal to... .078992001 - .019748 = .059244001 Joules.

    .059244001 = kQ0q/r I know what both of my Q0, k, q. But I have the energy in Joules and I know I need to convert it into coulombs.... I'm working with a positive charge so I need to divide my Potential Energy by the charge of a proton, which is 1.6 x 10^-19 Coulombs. Now I can find the r, but it will be in meters!
  9. Feb 12, 2016 #8


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    I hope that's not what you meant. Energy is energy, charge is charge.
    Not sure why the exclamation mark. What did you expect a distance to come out in when you use standard units? Or are you just surprised that the answer is not many metres?
  10. Feb 12, 2016 #9
    Uh oh.... I think I'm lost then lol. Well now I have the potential energy.... but I'm not sure how to use this to find the distance. I thought I could use the Formula I wrote above.. U =kq0q/r but it doesn't seem right. Am I going to use the eletric potential of a point charge at 0 and infinity? V = kq/r? I have no idea what formula i should use with the potential energy I have....
  11. Feb 12, 2016 #10


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    What makes you think it is wrong? Each charge is in Coulombs, r is in metres, and the units of k are such that the right hand side produces Joules, matching the left hand side. Just apply the formula. There is no conversion to be done.
  12. Feb 13, 2016 #11
    Thank you. End up getting the answer :)
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