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Charge for a Grain of Dust on the Moon? 
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#1
Feb1913, 09:28 PM

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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! 


#2
Feb1913, 10:32 PM

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Draw a freebody diagram for the grain of dust  what are the forces acting on it?



#3
Feb1913, 11:18 PM

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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?



#4
Feb1913, 11:30 PM

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Charge for a Grain of Dust on the Moon?
What about if I used the equation Fnet = m * a, which would be 1.6*10^9?



#5
Feb2013, 12:02 AM

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However  if the dust thingy is levitating, doesn't that mean the net force on it is zero? 


#6
Feb2013, 12:07 AM

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Sure, so if the net force is zero, then would I need an equation using the acceleration? Thanks for your help.



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