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Plastic Balls in a Test Tube with Mass

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    One charged plastic ball is vertically above another in a test tube as shown. The balls are in equilibrium a distance "d" apart. If the mass of the upper ball is reduced by a factor of four, the distance between the balls in the test tube would become :

    1) 4d
    2) 8d
    3) 2d
    4) square root of 2d

    2. Relevant equations

    ?

    3. The attempt at a solution

    How are charge and mass related? That's what I can't understand.
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2012 #2

    rl.bhat

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    Charge decides the repulsive force and mass decides the gravitational attractive force. In equilibrium position they are equal amid opposite.
     
  4. Sep 14, 2012 #3
    Ok. I'm thinking along the lines of

    F initial = G m1m2 / r^2

    F final = G 1/4m1m2/ r^2

    But I need to put those in terms of "r" and compare, right?
     
  5. Sep 14, 2012 #4

    rl.bhat

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    By changing mass of tthe plastic ball, the weight mg changes. Now compare the two weights with the force of repulsion.
     
    Last edited: Sep 14, 2012
  6. Sep 14, 2012 #5
    Ok. What I just did was

    Gm1m2
    --------
    1/4 Gm1m2

    to get a 4:1 ratio. Yet I didn't use anything with "R."
     
  7. Sep 14, 2012 #6

    rl.bhat

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    When mass decreases, the weight decreases. To balance this the charges must move farther. What is the formula for the force of repulsion?
     
  8. Sep 14, 2012 #7
    F = kq1q2/r^2 ?
     
  9. Sep 14, 2012 #8

    rl.bhat

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    Curret. Now if the initial distance for equilibrium is d for mg , what is new distance to balance mg/4?
     
  10. Sep 14, 2012 #9
    I'm thinking:

    d initial = mg
    --------------
    d final 1/4 mg

    solving for d initial I get:

    d initial = 4d final

    But there's a square involved so shouldn't the answer be "2d?"
     
  11. Sep 15, 2012 #10

    rl.bhat

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