7.6 The Millikan Oil Drop Experiment: Determining the
Elementary Charge
Physicist Robert Millikan determined experimentally the charge
on a single electron:

He started by creating a uniform electric field
between two parallel plates. By using a power
supply with a variable voltage he could control
the magnitude of the electric field.
 
V
d

He then sprayed a fine mist of oil from an atomizer. The tiny oil
droplets become negatively charged by friction with the nozzle as
they are sprayed.

Since a charged particle experiences a force,
, when in
an electric field, Millikan was able to adjust the voltage precisely so
that the force on the particle by the electrical field balanced the
force on the particle by the gravitational field:
F  q
FE  Fg
q  mg

We can rearrange this equation to express the charge q as a
function of the mass of the particle (m) and the balancing voltage
(ΔVb)
q
mg

mgd
q
Vb
Where ∆d is the plate separation distance.

Millikan determined the mass by turning off the electrical field and
measuring the final speed of the droplets as they fell under gravity
onto the bottom plate.

Since it was impossible for Millikan to determine the number of
extra electrons in any particular droplet, he repeated the
experiment many times and looked for a pattern. The charge on
any droplet turned out to be a multiple of some small value. There
were no trials that had results smaller than this elementary charge.

The elementary charge on one single electron was determined to be
e= -1.602 x 10-19C

To determine the total charge q on an object with an excess or
deficit of N number of electrons use the equation.
q=Ne
EXAMPLES
Example 1: Find the number of electrons needed to give an oil drop a charge of 4*10-8 C.
Example 2: A 5*10-15 kg oil drop is balanced between two plates, separated by 2 cm, with
a voltage of 600 Volts. Find the charge on the drop.
SOLUTION
Example 1: Find the number of electrons needed to give an oil drop a charge of 4*10-8 C.
q=Ne
N = q/e
Solution 1: N = 4*10-8/1.6*10-19 = 2.5*1011 electrons.
Example 2: A 5*10-15 kg oil drop is balanced between two plates, separated by 2 cm, with
a voltage of 600 Volts. Find the charge on the drop.
Solution 2: V = Ɛ * r
600 = Ɛ (0.02)
Ɛ = 30,000 N/C
mg = qƐ
(5*10 )(9.8) = q (30,000)
-15
Therefore q=1.6*10-18 Coulombs, or 10 electrons.
Textbook Work
Section 7.6: The Millikan Oil Drop Experiment
Chapter 7 Review
Self-Quiz - Page 369 # 18
Review - Pages 370- 375 # 117, 5565, 6790
- Practice p. 364# 13
- Page 365 # 16