Charge for a Grain of Dust on the Moon?


by m00nbeam360
Tags: charge, dust, grain, moon
m00nbeam360
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#1
Feb19-13, 09:28 PM
P: 20
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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Simon Bridge
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#2
Feb19-13, 10:32 PM
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Draw a free-body diagram for the grain of dust - what are the forces acting on it?
m00nbeam360
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#3
Feb19-13, 11:18 PM
P: 20
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?

m00nbeam360
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#4
Feb19-13, 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?
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#5
Feb20-13, 12:02 AM
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What about if I used the equation Fnet = m * a, which would be 1.6*10^-9?
That statement is meaningless.
However - if the dust thingy is levitating, doesn't that mean the net force on it is zero?
m00nbeam360
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#6
Feb20-13, 12:07 AM
P: 20
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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#7
Feb20-13, 12:33 AM
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Have you drawn the free body diagram?
Have you identified the different forces on the dust?


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