Question

Millikan's oil-drop experiment measures the charge on individual oil drops. Why would you try to set the electric field force on a drop equal to the gravitational force on that drop?

Answer 1

he electric field is set to produce a force that will balance the force of gravity thereby stopping the drop from falling. The gravitational force is M*g where M is the mass of the oil drop and g is the acceleration due to gravity. The electric field force is produced between two metal plate and is given by Fe = q*V/d where q is the charge , V is the voltage needed to create the electric field and d is the separation of the plates. M can be determined from the rate of fall of a drop with no electric field. Equating the forces Mg =q*V/d and solving for q we get q=M*g*d/V. Millikan found that q turned out to be an integral multiple of a particular number which was taken as the charge of an electron.