More on Millikan's Oil Drop Experiment
(©2004, François G. Amar, All rights reserved)

As the book states, Millikan designed an apparatus that allows charged oil droplets to be held stationary in
the field of view of a magnifying eyepiece. The balance condition can be depicted as follows:

                                F(electric)
                                     L
                                       |
                                       |
                     droplet    O
                                       |
                                       |
                                      V
                               F (gravitational)

Now recall that the electric force is given by

        F(electric) = qE

where q is the charge in Coulombs and E is the known electric field between
the charged plates (see figure 2.5 in BLB) and is in units of Nt/Coul.

The gravitational force is

        F(gravitational) = - mg

where m is the mass of the droplet and g is the acceleration due to gravity at
the surface of the earth (g = 9.8 m/s²)

The mass of the droplet is obtained by measuring the radius, R, of the stationary
drop in the telescope and then using the formula for the volume of a sphere.

    mass droplet = volume sphere x density of oil = (4pR³/3) x r

To get the charge on the droplet, we write the force balance equation:

        F(electric) + F(gravitational) = 0
            qE     +      (-mg)      = 0
or
         qE = mg
so
        q = mg/E = 4pR³rg/3E
 
 

Problem to try-- Determine the charge on a droplet given the following data:

R= 1.64x10^-4 cm ; E=1.92x10⁵ Nt/coulomb ; r = 0.851 g/cm³ (density of oil)

Be careful converting units! Answer in a few days. You will be responsible for this

material on the TEST!

Here now is the ANSWER TO THIS PROBLEM