AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Stokes’ Law for the viscous force on an oil droplet used to calculate the…
Full specification wording
Principle of Millikan’s determination of the electronic charge, e (A-level only): Stokes’ Law for the viscous force on an oil droplet used to calculate the droplet radius.
Stokes’ Law gives the viscous drag on a small sphere moving slowly through a fluid: F = 6πηrv, where η is the viscosity of the fluid, r is the sphere radius and v is its speed.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 3 of 5
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Stokes’ Law gives the viscous drag on a small sphere moving slowly through a fluid: F = 6πηrv, where η is the viscosity of the fluid, r is the sphere radius and v is its speed. In Millikan’s experiment, an oil droplet falls through air and reaches terminal speed when the resultant force is zero. At that point, weight minus upthrust equals viscous drag. Substituting Stokes’ Law into this balance gives an equation containing the unknown radius r. Solving it yields r from the measured terminal speed, oil density, air density and air viscosity. The radius then gives the droplet volume and mass, which are needed to find the charge from the electric force balance.
What earns the marks
- Stokes’ Law states that the viscous force on a small sphere is F = 6πηrv, where η is the fluid viscosity, r is the sphere radius and v is the speed.
- At terminal speed, the forces on the falling droplet balance: weight minus upthrust equals viscous drag.
- Weight is mg, and upthrust is the weight of displaced air, which can be expressed using the air density and the droplet volume.
- Substituting F = 6πηrv into the terminal-speed force balance gives an equation that can be solved for the droplet radius r.
- The radius is used to calculate the droplet volume and hence its mass from the oil density.
- Accurate values of viscosity, densities and terminal speed are needed because the radius depends on these quantities.
Common mistakes and how to put them right
Using the droplet diameter in place of the radius in Stokes’ Law.
Put it right: Use r, the radius, and to convert any diameter measurement by halving it.
Omitting upthrust from the force balance.
Put it right: Include the buoyancy force so that weight minus upthrust equals viscous drag at terminal speed.
Treating Stokes’ Law as valid for any speed or size.
Put it right: Note that it applies to small spheres moving slowly with laminar flow around them.
Exam tips
- State Stokes’ Law in symbols and define each symbol, including the SI unit of viscosity.
- Show the terminal-speed force balance clearly before substituting Stokes’ Law, so the examiner can follow the derivation of r.
- Keep density terms distinct: use the oil density for the droplet mass and the air density for upthrust.
Try the lesson
The start of the Principle of Millikan’s determination of the electronic charge, e (A-level only): Stokes’ Law for the viscous force on an oil droplet used to calculate the droplet radius. lesson, made from checked teaching notes for this specification point.
Have a go first
Stokes' law gives the viscous drag on a small sphere of radius r moving at speed v through a fluid of viscosity η. Which expression is correct?
Listen
Stokes’ Law gives the viscous drag on a small sphere moving slowly through a fluid: F = 6πηrv, where η is the viscosity of the fluid, r is the sphere radius and v is its speed. In Millikan’s experiment, an oil droplet falls through air and reaches terminal speed when the resultant force is zero. At that point, weight minus upthrust equals viscous drag. Substituting Stokes’ Law into this balance gives an equation containing the unknown radius r. Solving it yields r from the measured terminal speed, oil density, air density and air viscosity. The radius then gives the droplet volume and mass, which are needed to find the charge from the electric force balance. A strong answer earns marks for these points. Stokes’ Law states that the viscous force on a small sphere is F = 6πηrv, where η is the fluid viscosity, r is the sphere radius and v is the speed. At terminal speed, the forces on the falling droplet balance: weight minus upthrust equals viscous drag. Weight is mg, and upthrust is the weight of displaced air, which can be expressed using the air density and the droplet volume. Substituting F = 6πηrv into the terminal-speed force balance gives an equation that can be solved for the droplet radius r. A common slip: Using the droplet diameter in place of the radius in Stokes’ Law. Instead, use r, the radius, and to convert any diameter measurement by halving it. Exam tip: State Stokes’ Law in symbols and define each symbol, including the SI unit of viscosity.
Key words
Say what each word means before you turn the card.
Stokes’ Law
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Terminal speed
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Upthrust
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Viscosity
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See every definition
- Stokes’ Law
- The viscous force on a small sphere is F = 6πηrv, where η is the fluid viscosity, r is the sphere radius and v is the speed.
- Terminal speed
- The speed reached when the resultant force is zero, so that weight minus upthrust equals viscous drag.
- Upthrust
- The weight of displaced air, which can be expressed using the air density and the droplet volume.
- Viscosity
- The quantity η in Stokes’ Law, whose accurate value is needed because the radius depends on it.
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