Millikan Oil Drop

In our experiments, a modified version of Millikan’s apparatus is used. An
atomizer produces a large number of oil droplets between the plates. By selective
application of an electric field, one droplet is singled out for observation. The droplets
are contained in a glass-walled compartment, with horizontal plates only about 3 mm
apart and applied voltages in the 100 to 350-volt range. Fields lower than Millikan’s are
used since the droplets are also smaller than his. The droplets are observed with a highpower
microscope and a high-sensitivity closed-circuit television system.
Theory:
(I) Measurement of the mass and radius of the droplet.
Consider a spherical drop of density ρ and radius a falling under gravity in air of
density σ. Stokes’ law states that for a spherical body falling with velocity v in a
medium of viscosity η, the viscous resistive force is given by 6πηav. The
Archimedean upward force on the body will be 4πa 3 σg/3. If the falling body has
reached its terminal velocity v and is therefore no longer accelerating, then the
weight of the body must equal the upward force plus the viscous force.
Therefore,
3 3
4πa ρg 4πa σg
= 6πηav
+
3 3
weight viscous upward force
of drop force due to air
4 3
and ( ) 6
3 π a ρ − σ g = πηav
.
Typically, σ is close to ρ/1000 and therefore, within the accuracy of this
4 3
experiment, can be ignored. Then 6
3 π a ρ g = πηav
and 9η
v
a = .
2ρ
g
(II) Static Experiment
If an electric field E is now applied so that the drop, if charged, remains
suspended between the two plates, then the weight of the drop will be balanced by
the upward electric force. Thus if e’ is the total charge on the drop, then
eV '
mg = e'
E = s
where V s is the voltage across the plates and d is the plate
d
mgd ⎛4
3 ⎞ d
separation, and e'
= , or e'
= ⎜ πa ρ ⎟ g . Substituting in our equation above
V
⎝ ⎠ V
to eliminate a we get
s
3
s
1/2
⎧⎪
⎛ 9η
⎞ ⎫⎪v
Kv
e' = ⎨6πµ
d⎜
⎟ ⎬ =
⎪⎩
⎝ ρ g ⎠ ⎪⎭
V V
3/2 3/2
2
s s
where K is the constant within the brackets.
(III) Dynamic Experiment
If the applied voltage is increased above V s , the droplet will rise in this applied
field. If V is now the applied voltage, and the droplet has reached its terminal
velocity, then
3 3
4 π a ρ g eV ' 4 a g
+ 6πηav2
= +
π σ
3 d 3
weight viscous electric upward force
of drop force field force due to air
Solving this for e’, as for the static case, gives the equation:
,