What is the Charge on a Suspended Oil Drop in Millikan's Experiment?

[washablemarker]

Homework Statement

Robert Millikan received a Nobel Prize for determining the charge on the electron. To do this he set up a potential difference between two horizontal parallel metal plates. He then sprayed a drops of oil between the plates and adjusted the potential difference until drops of a certain size remained suspended at rest between the plates. Suppose that when the potential difference between the plates is adjusted until the electric field is 10,000 N/C downward, a certain drop with a mass of 3.27 x 10^-16 kg remains suspended. what is the magnitude of the charge on this drop?

Homework Equations

umm... i suppose E = Kc x (q/r^2) and f = Kc x (q1)(q2)/r^2 would apply. I'm not sure if i need anything else though.

The Attempt at a Solution

i have got the right answer i think, but i have no idea how what is going on.

(3.27 x 10^-16)(9.8) = 3.2046 x 10^-15 = qE
E = 3.2 x 10^-19

i understand that here, mass is being multiplied by free fall acceleration, and then set equal to charge x electric field, but i don't know where this came from. does it have anything to do with the unit of electric force being Newtons, or kg x m/s^2. some explanation would be greatly appreciated.

 

[Vagrant]

Draw a free body diagram of the oil drop and work out the conditions for it to remain suspended.

 

[washablemarker]

yeah thanks. i figured it out minutes after. it's basically just electric field against gravity. thank you.