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Electric Forces Lab

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data

    When the pith balls are suspended, all forces are balanced. Assuming a mass of 0.060g for each pith ball, calculate the electric force which is repelling the pith balls. Hint: draw a free body diagram and apply Newton's 2nd Law seperately in the x and y directions. Assuming an equal charge on each pith ball, calculate the charge in Coulombs of each pith ball.

    The pith balls form an isosceles triangle with the stand.

    2. Relevant equations

    Fg = mg

    Fe = [tex] \frac{kq1q2}{r^{2}} [\tex]

    Fe = mg

    3. The attempt at a solution

    Angle of pith balls = [tex] tan^{-1} (0.10/0.0125) [/tex] = 82.87

    Because gravity is acting at an angle, so the acceleration due to gravity wouldn't be 9.81m/s^2 right?

    Fg = (6.0*10^-5) * (Tan 82.87) * (9.81) = 4.705 * 10^-3

    Fg = Fe

    4.705 * 10^-3 = k q^2 / r^2

    r^2 = 6.25*10^-4

    q = sqrt[ (Fg)(r)^2) / k ]

    q = sqrt[ (4.705 * 10^-3) (6.25*10^-4) / (8.99*10^9) ]

    q = 7.2*10^-7 C

    I'm not sure if my use of 'angles' was correct. If anyone can double check my work I would really appreciate it. Thanks in advance.
     
  2. jcsd
  3. May 28, 2009 #2

    Redbelly98

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    Gravity will act vertically downward, not at an angle.

    You are forgetting about 1 of the 3 forces acting on each ball. What prevents the balls from falling due to gravity?
     
  4. May 28, 2009 #3
    I'm not sure I follow. The best I can come up with is this: Fnet = Fe - Fg

    Isn't Fnet equal to zero, because the pith balls are suspended?

    2nd attempt:

    Fg = Fe(tan 82.87)

    q = sqrt[ ( (m)*(g)*(r)^2) / ( (k)*(tan82.87) ) ]

    q = 1.3*10^-16C


    The charge seems to be reasonable, especially when considering my original answer.

    I've drawn a right angle triangle to represent the forces, I have Fe as the adjacent side, Fg as the opposite, and Fnet as the hypotenuse.
     
  5. May 28, 2009 #4

    Redbelly98

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    Yes, Fnet is zero since the balls are not moving.

    But if Fnet were the hypotenuse, then it couldn't equal zero ... so it is not the hypotenuse.

    Note that you have (hopefully) Fe and Fg acting at right angles to each other. There is no way they can combine to get Fnet=0 unless there is another force present.

    Let me ask it this way: when there is no charge present, what keeps the balls from falling?
     
  6. May 28, 2009 #5
    The string attached to the stand keeps the pith balls from falling. . . I really hope that didn't read as incredibly foolish.

    So far I've established (with your help of course) that Fe is operating horizontally and Fg is operating veritcally.

    A third resultant force created by these two forces is also present.

    Fnet is zero.

    Fnet = F.unknown - Fe - Fg

    0 = F.unkown - Fe - Fg


    Am I on the right track?
     
  7. May 28, 2009 #6
    Could I write the equation like this:

    tan 82.87 = Fg / Fe

    ?
     
  8. May 28, 2009 #7

    Redbelly98

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    Yes, the string exerts a force on the balls to keep them from falling.

    Yes.

    Looks good so far.

    Yes. (I'll take your word for it that the 82.87 degrees is correct)
     
  9. May 28, 2009 #8
    Thanks a lot for your help. I really do appreciate you taking the time out of your day for me. I hope you have a pleasant night/evening.
     
  10. May 28, 2009 #9

    Redbelly98

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    You're welcome!
     
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