AQA A-Level Physics 7408
Measurements and Their Errors
Particles and Radiation
Waves
Mechanics and Materials
Electricity
Further Mechanics and Thermal Physics
Fields and Their Consequences
Nuclear Physics
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Medical PhysicsOptional topic
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Turning Points in PhysicsOptional topic
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Turning Points in Physics
Principle of Millikan's determination of the electronic charge, e
- 4 things to know
- Paper 3, section B
Overview
The oil drop experiment, Stokes' law, terminal speed, and the quantisation of charge.
What you need to know
Answer each one before you reveal it.
How is the mass of a droplet found, and what pattern did Millikan see in the charges?
What happens when a field is applied to a falling droplet, and how does drag change with speed?
What does quantisation of electric charge mean?
Questions from this subtopic
Each one with the model answer underneath.
- 1
In Millikan's experiment, in which direction does the electric force act on a negatively charged oil droplet held stationary between the plates?
Model answer
The electric force acts upward on the negatively charged droplet.
- 2
When an oil droplet is held stationary in Millikan's apparatus, in which direction does the gravitational force act?
Model answer
The gravitational force acts downward.
- 3
What is the condition for the forces acting on a charged oil droplet when it is held stationary between the parallel plates?
Model answer
The electric force and gravitational force are balanced (equal in magnitude and opposite in direction).
- 4
State the equation for the equilibrium condition of a stationary charged oil droplet between parallel plates, and define each symbol.
Model answer
QV/d = mg, where Q is the droplet charge, V is the potential difference between plates, d is the plate separation, m is the droplet mass, and g is gravitational field strength.
- 5
Write the formula used to calculate the charge on a stationary oil droplet from the equilibrium condition.
Model answer
Q = mgd/V
- 6
Calculate the charge on the droplet when mass of droplet = 5.44 × 10⁻¹⁵ kg, gravitational field strength = 9.81 m s⁻², plate separation = 0.006 m, potential difference = 1000 V.
Model answer
Charge on the droplet = (5.44 × 10⁻¹⁵ × 9.81 × 0.006) ÷ 1000 = 3.20 × 10⁻¹⁹ C.
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