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Electroscope forces

  1. Feb 8, 2005 #1
    2. A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 24-g spheres at the ends. When charged, nearly all the charge resides on the spheres. If the wires each make a 30 degree angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

    OK, firstly, do I consider the spheres as really small point particles and ignore their mass? I want to, but I bet that's wrong. Second, do I have to set up some kind of force diagram, with tension and such? I think that is where I am stuck. Actually, I just find this problem to be extremely obnoxious. My sense is that the net force is equal to zero ... I have to admit, I have been kind of staring idly at this problem so if you have at least a hint, I would appreciate it. I understand generally how electroscopes work, but not with forces, etc. thrown in.
  2. jcsd
  3. Feb 8, 2005 #2
    Consider just one sphere. If there is no coloumb force, what is the force which brings it to it's original vertical position? The tangential component of the gravitational force. Equate this force to the tangential component of the coloumb force.

    [tex]F_c \cos(30) = mgcos (60) [/tex] where

    [tex]F_c = \frac{KQ^2}{r^2} [/tex]
  4. Feb 8, 2005 #3
    OK, yeah, I think I did basically what you said.

    I found mg cos (60) = .024 x 9.8 x cos (60) and then multiplied that by cos (30) to get F = .136 N.

    Then, I multiplied that by r^2 and divided by k, took the square root and got Q = 3.03 x 10^-6. Does that make sense? I guess it could be that small, but really I have no idea.
  5. Feb 8, 2005 #4

    devide by cos (30).

    Also, Your problem asks you to find the charge applied to the electroscope. So the answer would be 2Q and yes the charge is going to be in micro coloumb range.
  6. Feb 8, 2005 #5
    K, yes, that makes sense. Thanks for your help.
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