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Charge for a Grain of Dust on the Moon?

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    How much charge in Coulombs is required to levitate a motionless grain of dust 10 cm above the surface of the moon? Assume the dust grain is a point mass with mg = 1*10^-9g. The gravitational acceleration at the surface of the moon is 1.6m/s^2. Assume charge on the surface of the moon acts as a point source beneath the grain equal in charge to the grain itself.


    2. Relevant equations
    E = F/q0, a = (q0/m)*E, e = 1.60*10^-19 C, E = 8.988*10^9 N*m^2/C^2


    3. The attempt at a solution
    Any help would be greatly appreciated! I figured that the first step is to find the variables known, so the acceleration is 1.6m/s^2 = (q0/(1*10^-9g))*(8.988*10^9), so would I just need to solve for q0? Thanks so much!
     
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  3. Feb 19, 2013 #2

    Simon Bridge

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    Draw a free-body diagram for the grain of dust - what are the forces acting on it?
     
  4. Feb 19, 2013 #3
    Would there only be one force against it towards the surface of the moon? Or is there also one from the moon pointing towards the grain?
     
  5. Feb 19, 2013 #4
    What about if I used the equation Fnet = m * a, which would be 1.6*10^-9?
     
  6. Feb 20, 2013 #5

    Simon Bridge

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    That statement is meaningless.
    However - if the dust thingy is levitating, doesn't that mean the net force on it is zero?
     
  7. Feb 20, 2013 #6
    Sure, so if the net force is zero, then would I need an equation using the acceleration? Thanks for your help.
     
  8. Feb 20, 2013 #7

    Simon Bridge

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    Have you drawn the free body diagram?
    Have you identified the different forces on the dust?
     
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