Millikan's oil drop experiment precisely measured the charge of individual oil droplets falling through an electric field. By measuring the droplets' terminal velocities when falling and rising, Millikan calculated their electrical charges, which were always integer multiples of a single fundamental unit of charge - the electron's charge. Millikan's experiment was the first to directly measure the discrete, quantized nature of electric charge and determine the electron's charge as -1.6022 x 10-19 Coulombs.

3. The Discovery of the Electron's Charge
In 1897 J. J. Thomson demonstrated that
cathode rays, a new phenomenon, were made
up of small negatively charged particles, which
were soon named electrons. The electron was
the first subatomic particle ever discovered.
Through his cathode ray experiments,
Thomson also determined the electrical charge-
to-mass ratio for the electron.

4. Millikan's oil-drop experiment was performed by
Robert Millikan and Harvey Fletcher in 1909. It
determined a precise value for the electric charge of
the electron, e. The electron's charge is the
fundamental unit of electric charge, because all
electric charges are made up of groups (or the absence
of groups) of electrons. This discretization of charge is
also elegantly demonstrated by Millikan's experiment.

5. The unit of electric charge is a fundamental physical
constant and crucial to calculations within
electromagnetism. Hence, an accurate determination
of its value was a big achievement, recognized by the
1923 Nobel prize for physics.

6. Millikan's Apparatus

9. Millikan's experiment is based around observing charged oil
droplets in free fall and in the presence of an electric field. A
fine mist of oil is sprayed across the top of a Perspex
cylinder with a small 'chimney' that leads down to the cell (if
the cell valve is open). The act of spraying will charge some
of the released oil droplets through friction with the nozzle
of the sprayer. The cell is the area enclosed between two
metal plates that are connected to a power supply. Hence an
electric field can be generated within the cell and its strength
varied by adjusting the power supply. A light is used to
illuminate the cell and the experimenter can observe within

10. Terminal velocity
As an object falls through a fluid, such as air or water, the
force of gravity will accelerate the object and speed it up.
As a consequence of this increasing speed, the drag force
acting on the object, that resists the falling, also
increases. Eventually these forces will balance (along
with a buoyancy force) and therefore the object no longer
accelerates. At this point the object is falling at a constant
speed, which is called the terminal velocity. The terminal
velocity is the maximum speed the object will obtain

11. Theory
Millikan's experiment revolves around the motion of
individual charged oil droplets within the cell. To
understand this motion the forces acting on an individual
oil droplet need to be considered. As the droplets are
very small, the droplets are reasonably assumed to be
spherical in shape. The diagram next shows the forces
and their directions that act on a droplet in two scenarios:
when the droplet free falls and when an electric field
causes the droplet to rise.

13. The most obvious force is the gravitational pull of the Earth on the
droplet, also known as the weight of the droplet. Weight is given by
the droplet volume multiplied by the density of the oil (ρoil) multiplied
by the gravitational acceleration (g). Earth's gravitational acceleration
is known to be 9.81 m/s2 and the density of the oil is usually also
known (or could be determined in another experiment). However, the
radius of the droplet (r) is unknown and extremely hard to measure.

14. As the droplet is immersed in air (a fluid) it will experience an upward
buoyancy force. Archimedes' principle states that this buoyancy force is
equal to the weight of fluid displaced by the submerged object.
Therefore, the buoyancy force acting on the droplet is an identical
expression to the weight except the density of air is used (ρair). The
density of air is a known value.

15. The droplet also experiences a drag force that opposes its motion. This
is also called air resistance and occurs as a consequence of friction
between the droplet and the surrounding air molecules. Drag is
described by Stoke's law, which says that the force depends on the
droplet radius, viscosity of air (η) and the velocity of the droplet (v).
The viscosity of air is known and the droplet velocity is unknown but
can be measured.

16. When the droplet reaches its terminal velocity for falling (v1), the
weight is equal to the buoyancy force plus the drag force.
Substituting the previous equations for the forces and then
rearranging gives an expression for the droplet radius. This allows
the radius to be calculated if v1 is measured.

17. If a droplet is charged it will now experience an electrical force in
addition to the three previously discussed forces. Negatively charged
droplets will experience an upwards force. This electrical force is
proportional to both the electric field strength and the droplet's
electrical charge (q).

18. When a voltage is applied to the brass plates an electric field
is generated within the cell. The strength of this electric field
(E) is simply the voltage (V) divided by the distance
separating the two plates (d).

19. If the electric field is strong enough, from a high enough voltage, the negatively
charged droplets will start to rise. When the droplet reaches its terminal velocity
for rising (v2), the sum of the weight and drag is equal to the sum of the electrical
force and the buoyancy force. Equating the formulae for these forces, substituting
in the previously obtained radius (from the fall of the same droplet) and
rearranging gives an equation for the droplet's electrical charge. This means that
the charge of a droplet can be determined through measurement of the falling and
rising terminal velocities, as the rest of the equation's terms are known constants.

21. Experimental Method
Firstly, calibration is performed such as focusing the microscope
and ensuring the cell is level. The cell valve is opened, oil sprayed
across the top of the cell and the valve is then closed. Multiple
droplets of oil will now be falling through the cell. The power
supply is then turned on (to a sufficiently high voltage). This causes
negatively charged droplets to rise but also makes positively
charged droplets fall quicker, clearing them from the cell. After a
very short time this only leaves negatively charged droplets
remaining in the cell.

22. The power supply is then turned off and the drops begin to fall. A
droplet is selected by the observer, who is watching through the
microscope. Within the cell, a set distance has been marked and the
time for the selected droplet to fall through this distance is measured.
These two values are used to calculate the falling terminal velocity.
The power supply is then turned back on and the droplet begins to rise.
The time to rise through the selected distance is measured and allows
the rising terminal velocity to be calculated. This process could be
repeated multiple times and allow average fall and rise times, and
hence velocities, to be calculated. With the two terminal velocities
obtained, the droplet's charge is calculated from the previous formula.

23. Results
This method for calculating a droplet's charge was repeated for a large number
of observed droplets. The charges were found to all be integer multiples (n) of
a single number, a fundamental electric charge (e). Therefore, the experiment
confirmed that charge is quantized.
A value for e was calculated for each droplet by dividing the calculated droplet
charge by an assigned value for n. These values were then averaged to give a
final measurement of e. Millikan obtained a value of -1.5924 x 10-19 C, which is
an excellent first measurement considering that the currently accepted
measurement is -1.6022 x 10-19 C.

24. Questions
• Why do we use oil and not water when determining the charge of
an electron?
• How was the value of 'n' calculated for the problem described in
this article?
• What is the acceleration of the droplet if the electric force is
equal but opposite to that of gravity?
• How do the oil droplets acquire either the negative or the positive
charge?