In Millikan's oil drop experiment, the mass of oi drop is obtained by observing the terminal velocity Vt of the freely falling drop in the absence of an electric field. Under these circumstances, the effective weight equals the viscous force given by stroke's law F = 6πηrVt, where 6π is a constant and η is the viscosity of air with a value(1.81*10^-5kgm/s) and r is the radius of the drop, which is r = (0.65*10^-10cm). Also, the actual weight mg = 4/3ηr^3ρg must be corrected for the buoyant force of the air. This is done by replacing ρ with ρ - ρa, where; ρ is the density of the oil with a value 9.20*10^2kg/m^2 and ρa is the density of air with value (1.29kg/m*2). 1. With all these preliminaries, show that the charge on the drop q = 18π*d/v*[√η^3*Vt^3/2(ρ - ρa)g]. Where d = 0.25cm is the separation of the plate, v = 350kv is the voltage across them that keeps the drop stationary. 2. Determine the terminal velocity with respect to the given value of the quantities.