Posted by
**Frank** on
.

In Millikan's experiment, an oil drop of radius 1.516 µm and density 0.860 g/cm3 is suspended in chamber C (Figure 22-14) when a downward-pointing electric field of 1.92 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.)

Calulate the mass of the drop as (4/3)*pi r^3*(density), and convert it to kg. Multiply by g = 9.8 m/s^2 to get the gravitational force (weight) in N.

Set the sum of the weight W and the electrostatic force

F = q E,

equal to zero for a non-accelerating (suspended) oil drop. Forces are considered positive downward. E is the electrical field, 1.92*10^5 N/C. Solve for the charge q in Coulombs. It should have a negative sign.

I got it, thanks!