1. The Millikan Oil Drop Experiment & Elementary Charge

2. Scientists wondered … Does there exist in nature a smallest
unit of charge?
If so, what is the
magnitude of this
“elementary charge”?
The Thinker - bronze and marble sculpture by Auguste Rodin (French sculptor )
Coulomb had originally called the unit of a very small charge a Coulomb, but was it truly the smallest?

3. The Millikan Oil-Drop Experiment
FE = FG
qe = mg
since e = ∆V r
where DVb is the
balancing value
b/w the plates
q DVb = mg
q = mgr
∆Vb
( - )
negatively charged plate FE + e r Fg
Robert Millikan (American) devised and performed a series of creative experiments in He received the Nobel Prize in 1923 for his discovery.
He reasoned that the smallest charge would be for an individual e-. When tiny oil droplets are sprayed in a fine mist from an atomizer they become
electrically charged by friction (some have excess e-, others deficit of e-). Millikan hypothesized that by measuring the total charge on any oil drop, it
would be an integral multiple of the elementary charge (there was no way of knowing the excess/deficit of e- on a droplet).
He used an electrical microbalance to isolate and suspend charged oil drops and to measure the total charge on each.
A droplet was sprayed and the potential difference b/w the plates adjusted to balance the droplet w/ the same charge as the lower plate.
So, when a positively charged droplet is balanced, (CLICK)
FE = Fg  FE = qE and Fg = mg Note: Fe acts upward going from +ve to -ve
qE = mg
q = mg/E  E = V/r (b/w two parallel plates where E is constant)
q = mgr/Vb (CLICK)
where Vb = the balancing value of electrical potential difference between the plates
r = separation between the plates
(+)
positively charged plate Vb

4. Determining the Elementary Charge
q = mgr
DVb
The Elementary
Charge e = 1.602x10-19C
( - )
negatively charged plate FE + e r
q = Ne Fg
It’s possible to determine the TOTAL CHARGE on an oil drop if we know its MASS:
We can determine MASS by measuring TERMINAL SPEED when the droplet falls: FE is removed and Fg = Fdrag; m∝v2 where Fdrag is found from Stoke’s Law
Measure terminal speed under gravity alone  calculate mass
Measure balancing voltage and distance r  calculate total electric charge on the droplet from q =mgr/V
Repeated countless times until he found that the (CLICK)
smallest q = 1.602x10-19C  ELEMENTARY CHARGE!
and every droplet was a multiple of that #
e is the smallest unit of electric charge!!!
“q” was never below this value AND always an integer multiple of that value
And since Electric Charge = excess/deficit of electrons
So, q=Ne (where N = # of excess/deficit of electrons)
(+)
positively charged plate Vb
where N = # of excess/deficit of electrons

5. Example 1
Two parallel plates are placed 1.5cm apart with a potential difference of 275V between them. The upper plate is negative. An oil drop of mass 4.5x10-15kg is balanced inbetween the plates similar to Millikan’s experiment.
Is the drop charged positively or negatively?
What is the charge magnitude on the drop?
How many excess/deficit electrons does the drop possess?

6. Limitations Millikan accurately calculated “e”.
But his experiment did not adequately describe the motion of charged particles (due to the presence of air resistance).

7. Instantaneous Acceleration
a = FE m q2 q1 FE + + r2 r1
The charge q1 experiences a force FE from charge q2 pushing it away (repulsive force)
So, if charge q1 is free to move it will accelerate in the direction of FE
with an instantaneous acceleration (CLICK)
But, as “r” increases and “FE” decreases so “a” decreases (NOT constant!)
This is very difficult to analyze using Newton’s Laws  need Calculus So let’s use Energy Laws to analyze instead!
Acceleration of the charge will decrease
because as r↑, FE↓

8. Conservation of Energy
ET = EE + EK
ET’ = EE‘ + EK’ q2 q1 FE + + r2 r1
Remember that in a closed system, energy is conserved
And electric forces are conservative forces so therefore no losses to thermal etc.
We can assume frictionless, and Eg negligible
Work out the calculations on the board if not covered already.
Total Energy of the charge remains constant
because as EE decreases, EK increases
any change in a particle’s electric potential energy results in a corresponding change to its kinetic energy when moving in an electric field (and ignoring gravitational effects)

9. Special Case
Recall that parallel plates are a special case where e is uniform and constant …
Therefore Electric Force is constant since FE = qe
and so is acceleration, since a = FE m
∴ a charged particle in a uniform field moves w/ uniform acceleration

10. Special Case Therefore, between two parallel plates : - DEE = DEK
-qDV = ½ mn 2
If n ³ 10% the speed of light (v~ 3.00x107m/s) then the answer is not valid  relativistic effects increase the mass

11. Example 2
An electron is fired horizontally at 2.5 x 106 m/s between two horizontal parallel plates 7.5 cm long. The magnitude of the electric field is 130 N/C. The plate separation is great enough to allow the electron to escape. Edge effects and gravitational forces are negligible. Find the velocity of the electron as it escapes from between the plates.

12. Example 1 - EXTRA
A particle accelerator uses two parallel plates to accelerate a helium nucleus, or a particle, toward a Cesium atom in order to split it. The potential difference across the plates is 50,000V. With what speed does the a particle hit the Cs atom if it starts at rest? Is this answer valid?
ma= 6.6x10-27 kg
qa = 2e = 3.204x10-19 C
Handout Example Sheet!

13. Example 2 - EXTRA
A stray, stationary proton finds itself 450nm from a gold nucleus. How fast will it be travelling when it is 15mm away? + Au qp
450nm
15mm

14. Example 3 - EXTRA
A small sphere with a charge of 750mC and a mass of 25mg is located midway between two charged plates separated by 40cm with a voltage of 200V. The charge is pulled slowly upward by a string until it reaches the positive plate and is then released.
a) How much work does the string do on the sphere?
b) What average force does the string exert?
c) With what speed does the sphere reach the negative plate?