Milikan's oil drop experiment: polarity of oil drops, why there is no vacuum and the oil drops got suspended in the Electric field From what I understand, I'm not sure if this is simplified but Milikan used a spray bottle as an atomiser in which small oil droplets would get charged negatively and fall through to a chamber that has NO vacuum such that they can reach terminal velocity. Then they fall through a hole and fall into an Electric field acting upwards. Since the oil drops are negative they will be attracted up and have a weight down. These forces are balanced until the oil drop is suspended so $F_{\text{net}}=0$ and through fluid mechanics, he deduced the oil drop's mass, then equated forces and then hence found charge.
However I have 3 questions:

*

*Does the atomiser always strip the oil droplets of electrons, thereby ALWAYS giving it a negative charge? And if so why is that the case?

*Why is it necessary that there is no vacuum. When I was taught it, they said it was essential to the experiment but why does the oil drop need to reach a terminal velocity?

*Even if the forces are balanced such that $F_{\text{net}}=0$, how did Milikan ensure the oil droplets were suspended? Couldn't the oil drops just keep going downward at a constant velocity especially since they reached a terminal velocity in the chamber?

 A: To answer your questions in order:
Question #1:
The drops can acquire either a positive or negative charge,
consisting of one or more electron charges.  I have seen apparatus
that included a small radioactive source to reliably supply free
charges to passing oil drops.
But any positively charged oil drops would be swept downward by the applied electric field even faster than gravity and disappear from the microscope view.
Question #2
The presence of an atmosphere, and the need for the drop to reach terminal velocity, have to do with the second part of the experiment. The terminal velocity of the drop in the absence of a field, combined with the density of the oil and some fluid flow dynamics gives the mass of the drop. This lets the experimenter solve the equation:$$mg=QV$$where $m$ is the mass of the drop, $g$ is the acceleration of gravity, $Q$ is the charge on that drop, and $V$ is the voltage necessary to hold the drop suspended
Question #3
To answer this part, this how the experiment is carried out:

*

*Spray the oil and look through the microscope with the electric field off;


*You see many drops of oil as points of light.  Some are dropping quickly (heavy ones) others are dropping more slowly.  All are falling at their particular terminal velocity;


*Turn on the electric field at $0$ volts, and slowly increase it.  Some drops pick up speed and fall out of view;  they're the positive ones;


*Others reverse their course and shoot up; they're light, highly negatively charged drops;


*The interesting ones slow down, some more than others.  They are small, negatively charged;  exactly the ones you want.


*As you increase the voltage, these drops will begin to slow even more, and even reverse course.  Pick one drop, and adjust the voltage until that drop remains stationary.  Other drops will still be falling or rising but your chosen drop is doing neither.


*Watch the drop. Correct the voltage for any slight drift. Record the voltage.


*Now increase the voltage to lift the drop up, turn off the field, and time the drop as it falls. Catch it with the field, lift it back up, and repeat the previous steps.  Play PONG with the drop.
So Millikan did not just apply one voltage to match gravity.  He increased and decreased the voltage to slow, stop, and lift different drops.
Some complications:
Sometimes the chosen drop changes its behavior in mid-experiment, when it picks up/loses some charge.
Often the microscope part of the system reverses  directions.  Drops fall "up" in the field of view, or rise "down"
Thanks to @rob and his comment for this reference to Millikan's original paper https://history.aip.org/exhibits/gap/Millikan/Millikan.html
