## Charge for a Grain of Dust on the Moon?

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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 Recognitions: Homework Help Draw a free-body diagram for the grain of dust - what are the forces acting on it?
 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?

## 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?

Recognitions:
Homework Help
 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?

 Sure, so if the net force is zero, then would I need an equation using the acceleration? Thanks for your help.
 Recognitions: Homework Help Have you drawn the free body diagram? Have you identified the different forces on the dust?