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Pith ball Charges

  1. Jun 26, 2008 #1
    Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine strings and are not moving. If the angle that each string makes with the vertical is q = 42.6 , and the charges on the two balls are equal, what is that charge (in microC)?

    [​IMG]

    I equated the forces in x and y to zero:

    Tcos42.6-117.6=0
    T=159.76

    Tsin42.6+Fe=0
    plugging in 159.76
    Fe=108

    then i used the equation Fe=kQ^2/r^2

    i solved for r by doing this:
    r=2(1.2sin42.6)
    r=1.6 m

    I'm not sure if my steps are right. Can someone help me? Thanks in advance!
     
  2. jcsd
  3. Jun 26, 2008 #2
    What's tan theta equal to?

    Yep, it seems right what you have right now.




    tan 42.6 = Electric force / gravitational force


    Select to get more hint
     
  4. Jun 26, 2008 #3
    let's a better way to find Fe, which is 108 N. Then plugging it into the equation then solving for Q, the answer I came up with is 178 microC, but the answer is wrong =(
    I don't know where I've made a mistake.
     
  5. Jun 26, 2008 #4

    alphysicist

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    Homework Helper

    Hi chukie,

    The weight of 117.6 N you find here does not look correct to me. (You don't have units here, but your last post indicates that you meant it to be Newtons.) Do you see what went wrong?
     
  6. Jun 26, 2008 #5
    ahhh!! i see now! i forgot to convert grams into kg! thanks so much!
     
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