Expert: Richard J. Raridon - 3/11/2006

Question
A negatively charged oil drop in a milikan oil drop apparatus is observed to fall 1.0mm in a time of 24.7s in the absence of the electric field . the same drop can be held be stationary in a constant electic feild of 2.37*10^4N/C. The viscoustiy of air is 1.80*10^ -5 Ns/m^2 and the density of the oil is 824kg/m^3. the viscous drag f on a sphere of radius r in a medium of viscousity n is given as 6(pi)nrv. neglecting the buoyancy force of the air calculate

the radius of the oil drop
the charge on the oil drop
the number if excess electrons on the oil drop  

Answer
Well, you obviously have a book containing the equation for the viscous drag.  Doesn't it also have an equation for the radius?   It's r = 3[nv/(2pg)]^1/2 where v is the terminal velocity (which you can calculate from the information given, in m/s), p (really rho) is the density, and g is the acceleration due to gravity.  
The charge q = [18(pi)/E][n^3v^3/(2pg)]^1/2
and the divide the charge by 1.6x10^-19 to get the number of excess electrons on the drop.