A tiny negatively charged oil drop is held stationary in an electric field between two horizontal parallel plated as shown.  The mass is 2.5 ´ 10^-10 kg.

(a)    What are the two forces acting on the drop and in which direction do they act?

(b)   The drop is stationary.  What can be said about the two forces?

(c)    What is the charge on the oil drop?                                                                   

(a) The force acting downwards is the weight (P)

The upwards force is the force of electrostatic attraction (P)

(b) The two forces balance out (P)

(c) E = V/d = 3000 V ¸ 0.025 m = 120 000 V/m (or N/C) (P)

E = F/Q Þ Q = F/E = (2.5 ´ 10^-10 kg ´ 10 N/kg) ¸ 120 000 N/C = 2.08 ´ 10^-14C (P)