What Charge Must an Oil Droplet Have to Balance Gravity in an Electric Field?

[Magna1]

Don't know where to go Get stuck at every corner.
Here it is
"A small droplet of oil of 50^10*-6m radius is sitting inside of an electric field 500^10*3 N/C which is pointing straight up. What must be the charge on the oil droplet so that the electrostatic force cancels out due to gravity? The density of oil is 800kg kg/m^3


I don't know where I am going with this. I used E=kQ/r^2 knowing E as 500^10*3 then igured Q as 138.86^10*-15C. I s this even close.

Thanks
"Learning Never Stops"

 

[FZ+]

Did you just use the radius they supplied? If so, you are overestimating the problem...

Things to note:

(a) They have given you the electric field strength, from which you can directly relate charge to force.

(b) They have given you radius, and density, from which you can calculate the force from gravity.

Equate the two and solve.

 

[Magna1]

and then?

I figured out the F due to Gravity (Fg. I used the E and found the charge on Q. but how do I tie the 2 numbers together? I figured out the Fg by using the density=m/v but once again How do I put the two numbers together?


"Learning Never Stops"

 

[Doc Al]

Originally posted by Magna1
I figured out the F due to Gravity (Fg. I used the E and found the charge on Q. but how do I tie the 2 numbers together? I figured out the Fg by using the density=m/v but once again How do I put the two numbers together?

The force due to gravity equals the force due to the E field. The force due to the E field depends on Q, which is what you are trying to find. Set the forces equal and solve for Q.

 

[Abelard]

So can you give concrete equations to equate with?