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Homework Help: Charge on two suspended metal balls

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Two very small metal coated foam spheres, each of mass 2.80x10^-6kg, are attached to nylon threads 45cm long and hung from a common point. When the spheres are given equal quantities of negative charge, each supporting thread makes an angle of 15 degrees with the vertical. Find the charge on each sphere.

    2. Relevant equations
    F=(kQ1Q2)/r^2 in the form of Q=√((Fr^2)/k) as Q1 and Q2 will be the same.
    probably Pythagoras and sine/tangent/cosine ratios

    3. The attempt at a solution
    ok, so r or the distance between the charges was pretty obvious. If the strings are 0.45m long and the angle at the point where the strings combine is 30° (2x15°), then its going to be an equilateral triangle so r=0.45m (I think). I suppose we are assuming k=9X10^9. Noe for the force. The force of gravity will be 2.80x10^-6x9.8 (F=ma), which equals approximately 2.7x10^-7. I then did (2.7x10^-7)/tan(90-15), which equals about 7.4x10^-8. I thought that equaled half the force of repulsion of the sphere, but I'm not sure. If thats right then the total force of repulsion is 1.5x10^-7. So then
    Q=√((1.5X10^-7x0.45^2)/9x10^9). That equaled 1.8X10^-9. But the actual answer is 6.66X10^-9. I know my rounding is having quite and effect on it, so i did it again, using as many decimals as I could but I was still quite a bit off. What am I doing wrong?
  2. jcsd
  3. Feb 11, 2013 #2
    An equilateral triangle has angles of 60° - this one has a top angle of 30°.
  4. Feb 11, 2013 #3
    oh wow, oopsies!
  5. Feb 11, 2013 #4
    I don't really understand your solution attempt here.
    Start by drawing a free-body diagram for one of the charges and resolve the forces along the x and y directions.
    You should be able to eliminate the tension leaving only 1 unknown ( the charge ).
  6. Feb 11, 2013 #5
    First and foremost, when did you learn that an equilateral triangle has an angle 30°? It's not an equilateral triangle!
    This one will be isosceles!
    Next why are you taking the weight as it is? Take its horizontal components!
    Then I don't get why you are dividing by tan75°. Could you explain that?
  7. Feb 11, 2013 #6
    Rookie mistake. You are unworthy of this forum. Next time go to yahoo answers...
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