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Three magnetism problems

  1. Mar 2, 2005 #1

    Finished! Thanks to all that helped out.
    Last edited: Mar 2, 2005
  2. jcsd
  3. Mar 2, 2005 #2
    first one [tex] q_{1} + q_{2} = 9 * 10^{-6}C [/tex]
    thus [tex] q_{1} = -q_{2} + 9 * 10^{-6}C [/tex]
    [tex] F = k \frac{q_{1} q_{2}}{r^2} [/tex]
    in th thirdequation substitute a relation from the second equation to find just one charge. Once you have that charge sub back into 2 and solve for hte other charge.

    Second one is correct F = ma = qE = eE thus a = eE/m nothing complicated about it

    Third one (since you dont know E explicitly treat it like an unknown)
    fir the electron
    a = eE / m , v1 = 0, t = ?, d = 0.04m you know a kinematic equation to solve for this Find the expression for the time. You dont need a speicific numberic answer variables are inevitable here
    now for the proton v1=0, a = eE/m, t = what you found, d = ?
    once again you know the relation beween these variables. Sub in the t that you got and solve.
    For simplicity do not evaluate the eE term in teh first case just keep it like it is, it cancels out in teh second part
  4. Mar 2, 2005 #3

    Andrew Mason

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    When you work out the expression for the charge, as Stunner has shown, you get a quadratic equation. So it is a little trickier than he suggests.

    For 3) use [itex]d = \frac{1}{2}at^2[/itex] so [itex]t = \sqrt{2d/a} = \sqrt{2dm/f}[/itex] where f = qE

  5. Mar 2, 2005 #4
    Many thanks. I'm glad that I seemed to be on the right track with the two I was having trouble with, there were just little things I didn't keep track of that I missed. I even had the q1=9-q2 thing, I just forgot to implement it.
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