# Charge for a Grain of Dust on the Moon?

1. Feb 19, 2013

### m00nbeam360

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. Feb 19, 2013

### Simon Bridge

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

3. Feb 19, 2013

### m00nbeam360

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. Feb 19, 2013

### m00nbeam360

What about if I used the equation Fnet = m * a, which would be 1.6*10^-9?

5. Feb 20, 2013

### Simon Bridge

That statement is meaningless.
However - if the dust thingy is levitating, doesn't that mean the net force on it is zero?

6. Feb 20, 2013

### m00nbeam360

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

7. Feb 20, 2013

### Simon Bridge

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