Editing Oil drop experiment (section)

===Method===
{{Refimprove section|date=December 2010}}

[[Image:Scheme of Millikan’s oil-drop apparatus.jpg|thumb|372x372px|Diagram in Millikan's original 1913 paper]]
Initially the oil drops are allowed to fall between the plates with the electric field turned off. They very quickly reach a [[terminal velocity]] because of friction with the air in the chamber.  The field is then turned on and, if it is large enough, some of the drops (the charged ones) will start to rise. (This is because the upwards electric force ''F''<sub>E</sub> is greater for them than the downwards gravitational force ''F''<sub>g</sub>, in the same way bits of paper can be picked by a charged rubber rod). A likely looking drop is selected and kept in the middle of the field of view by alternately switching off the voltage until all the other drops have fallen. The experiment is then continued with this one drop.

The drop is allowed to fall and its terminal velocity ''v''<sub>1</sub> in the absence of an electric field is calculated. The [[drag (physics)|drag]] force acting on the drop can then be worked out using [[Stokes' law]]:

:<math>F_{u} = 6\pi r \eta v_1, \,</math>
where ''v''<sub>1</sub> is the terminal velocity (i.e. velocity in the absence of an electric field) of the falling drop, ''η'' is the [[viscosity]] of the air, and ''r'' is the [[radius]] of the drop.

The weight '''''w''''' is the volume ''D'' multiplied by the density ''ρ'' and the acceleration due to gravity '''''g'''''. However, what is needed is the apparent weight. The apparent weight in air is the true weight minus the [[upthrust]] (which equals the weight of air displaced by the oil drop). For a perfectly spherical droplet the apparent weight can be written as:

:<math>\boldsymbol{w}=\frac{4\pi}{3}r^3(\rho-\rho_\textrm{air})\boldsymbol{g}.</math>

At terminal velocity the oil drop is not [[acceleration|accelerating]]. Therefore, the total force acting on it must be zero and the two forces ''F'' and <math>{w}</math> must cancel one another out (that is, <math>F = {w}</math>). This implies
:<math>r^2 = \frac{9 \eta v_1}{2 g (\rho - \rho_\textrm{air})}. \,</math>

Once ''r'' is calculated, <math>{w}</math> can easily be worked out.

Now the field is turned back on, and the electric force on the drop is
:<math>F_E = q E, \,</math>
where ''q'' is the charge on the oil drop and ''E'' is the electric field between the plates. For parallel plates
:<math>E = \frac{V}{d}, \,</math>
where ''V'' is the potential difference and ''d'' is the distance between the plates.

One conceivable way to work out ''q'' would be to adjust ''V'' until the oil drop remained steady.  Then we could equate ''F''<sub>''E''</sub> with <math>{w}</math>. Also, determining ''F''<sub>''E''</sub> proves difficult because the mass of the oil drop is difficult to determine without reverting to the use of Stokes' Law.  A more practical approach is to turn ''V'' up slightly so that the oil drop rises with a new terminal velocity ''v''<sub>2</sub>.  Then
:<math>q\mathbf{E}-\mathbf{w}=6\pi\eta(\mathbf{r}\cdot \mathbf v _2)=\left|\frac{\mathbf v_2}{\mathbf v_1}\right| \mathbf{w}. </math>