In a measurement of the electron charge by Millikan's method, a potential difference of 1.5 kV can be applied between horizontal parallel metal plates 12 mm apart. With the field switched off, a drop of oil of mass 10-14 kg is observed to fall with constant velocity 400 μm s-1. When the field is switched on, the drop rises with constant velocity 80 μm s-1. How many electron charges are there on the drop? (You may assume that the air resistance is proportional to the velocity of the drop, and that air buoyancy may be neglected.)
(The electronic charge = 1.6 * 10-19 C, the acceleration due to gravity = 10 m s-2.)
2. The attempt at a solution
As I understand I need to find Q?
In my book I have this formula: Q = 6 π r η v / E, where η is the coefficient of viscosity of air. The formula is applicable "when an electric field has been applied such that the drop is stationary." There are also formulas when there is no electric field, but I think in this case we have electric field.
But we don't have the radius and have the distance between plates instead.
I am also not sure whether I need to find Q, or some other value. Also how to find η?