Millikan's Oil-Drop Experiment: Is my reasoning correct?

[Mason Smith]

Homework Statement

Show that the electric field needed to make the rise time of the oil drop equal to the its field free time is
ε = (2mg)/q

Homework Equations

Newton's second law F = ma
The force of gravity F_G = mg, where g = 9.8 ms^-2
The force of buoyancy F_b = bv
The force of an electric field on a charge ε = F/q, where ε = the strength of the electric field and q = the charge of the particle

The Attempt at a Solution

First, I added the forces for the oil drop falling when the electric field is "off." I added the buoyant force that results from terminal velocity and the force of gravity.
F = ma = 0 = bv - mg = 0 [1]
This can be rearranged to
bv = mg [2]

Second, I reasoned that when the electric field is "on," the sum of the forces must be
F = ma = εq - mg = mg [3]
I rewrote this as
ε = (2mg)/q

Was my approach correct for solving this problem?

 

[mfb]

The "force from the velocity" is not related to buoyancy, it is just drag. If you take it into account for the downwards motion you should also do it for the upwards motion to keep the symmetry.