Charge for a Grain of Dust on the Moon?

[m00nbeam360]

Homework Statement

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.

Homework Equations

E = F/q0, a = (q0/m)*E, e = 1.60*10^-19 C, E = 8.988*10^9 N*m^2/C^2

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!

 

[Simon Bridge]

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

 

[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?

 

[m00nbeam360]

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

 

[Simon Bridge]

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]

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

 

[Simon Bridge]

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