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Charge at origin

  1. Sep 8, 2006 #1
    A charge of 2.75 μC is held fixed at the origin. A second charge of 2.75 μC is released from rest at the position (1.15 m, 0.350 m). If the mass of the second charge is 1.50 g, and its speed when it moves infinitely far from the origin is 8.6838 m/s, at what distance from the origin does the 2.75 μC charge attain half the speed it will have at infinity?

    I calculated the 8.6838 m/s by using PEi = KEf +. kq1q2/r = (mv^2)/2 => v = sqaure root of (2kq1q2/mr). I tried rearranging the equation to solve for r = 2kq1q2/mv^2, and got an answer of 4.8084 m, however, my answer is not correct. I used k = 8.99 x 10^9 Nm^2/C^2, q1 = 2.75 x 10^-6 C, q2 = 2.75 x 10^-6 C, m = 0.0015 kg, and r (for the first problem) = 1.2021 m (using Pythagorean theorem). Any ideas of where I went wrong?
     
  2. jcsd
  3. Sep 8, 2006 #2

    Andrew Mason

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    You have to solve for [itex]v = .5v_\infty[/itex]

    Since:

    [tex]v = \sqrt{2\Delta (-PE)/m} = \sqrt{\frac{2kq^2}{m}\left(\frac{1}{r_0} - \frac{1}{r}\right)}[/tex]

    and

    [tex]v_\infty = \sqrt{\frac{2kq^2}{m}\frac{1}{r_0}}[/tex]

    the condition [itex]v = .5v_\infty[/itex] is just:

    [tex]\sqrt{\frac{2kq^2}{m}\left(\frac{1}{r_0} - \frac{1}{r}\right)} = \frac{1}{2}\sqrt{\frac{2kq^2}{m}\frac{1}{r_0}}[/tex]

    solve that for r.

    AM

    PS you will note that you do not have to use k, q, m or calculate that speed at infinite separation.
     
    Last edited: Sep 8, 2006
  4. Sep 8, 2006 #3
    Okay, I figured it out, thank you!
     
    Last edited: Sep 8, 2006
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