1. The problem statement, all variables and given/known data When an oil drop falls freely, the velocity first increases, but afterwards it quickly reaches a constant speed, since the air resistance becomes equal to the weight of the oil drop. The air resistance is given by: Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v, where [itex]\eta[/itex] is the viscosity of air with a value of 1.759*10-5 N s/m2, Oil's density is 918.7 kg/m3. In Millikan's experiment, the speed of the oil drop is 5.449*10-4 m/s. - Find the radius and mass of the oil drop. Afterwards, electricity was sent through the plates, and the speed of the oil drop increases by 5.746*10-4 m/s. The upward electric force acting on the oil drop is given by: FE=qE, where E is the electric field strength: E = 3.178*105 N/C. Find the electric charge of the oil drop, q. 2. Relevant equations For no. 1: Weight = Air resistance mg = Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v, [itex]\rho[/itex]Vg=Fair=6[itex]\pi[/itex]r[itex]\eta[/itex]v, No. 2: FE=qE mg = FE + Fair 3. The attempt at a solution All right. What I did in no. 1 was to say that the weight was equal to the air resistance. Assuming the drop to be spherical, I said: [itex]\rho[/itex](4/3[itex]\pi[/itex]r3)g = 6[itex]\pi[/itex]r[itex]\eta[/itex]v. Then I solved for r: r = 2.19*10-6 m And I found m: m = [itex]\rho[/itex](4/3[itex]\pi[/itex]r3) = 4.04*10-14 kg (This seems to be too light even for a raindrop?) For no. 2: I said that the weight was equal to the sum of air resistance and the electric force: mg = 6[itex]\pi[/itex]r[itex]\eta[/itex]v + qE Solving for q gave: q = -1.31*10-18 C. However, this answer seems wrong, as Millikan tried to find the charge of the electron e from this experiment, which is 1.602*10-19 C. Have I done something wrong? In case I did, what was my mistake? Thanks a lot!