# Charge for a Grain of Dust on the Moon?

by m00nbeam360
Tags: charge, dust, grain, moon
 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!
 Homework Sci Advisor HW Helper Thanks P: 13,110 Draw a free-body diagram for the grain of dust - what are the forces acting on it?
 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?
 P: 20 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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