Stokes' Law (AQA A Level Physics)

Revision Note

Dan Mitchell-Garnett

Last updated

Motion of a Falling Oil Droplet

Motion of the Oil Droplet without the Electric Field

  • Millikan's experiment was designed to determine the charge of oil droplets, and therefore the charge on electrons

    • To calculate charge using the condition for stationary oil droplets, however, the mass of each oil droplet was required

  • With the electric field turned off, the oil droplets fell at terminal velocity

    • This means their weight, mg, was balanced by drag forces as they passed through the air

    • Millikan timed the oil drops as they passed the window of his apparatus to determine their speeds

A diagram showing weight and drag forces on an oil droplet

12-1-6-drag-forces-on-a-oil-droplet

At terminal velocity, speed is constant and resultant force is zero, so weight is balanced by drag forces

Stokes' Law

Viscous Drag Force using Stokes' Law

  • In fluid dynamics, Stokes' law allows drag force on an object through a fluid (such as air) to be calculated, provided:

    • The object is small

    • The object is spherical

    • The speed is low

  • Stokes' law states that, for an object with radius moving at velocity v  through a fluid with viscosity η, the viscous drag force, F, is:

F space equals space 6 straight pi eta r v

Calculating Mass from Stokes' Law

  • Recall that the drag force and weight are equal at terminal velocity, so:

6 straight pi eta r v space equals space m g

  • Both mass and radius are unknown, but we can write mass in terms of radius for a sphere of known density, ρ :

m space equals space rho V space equals space fraction numerator rho 4 straight pi r cubed over denominator 3 end fraction

  • Combining these equations allowed Millikan to determine the radius, and therefore mass, of an oil drop whose velocity was known

    • Recall the condition for a stationary oil drop:

fraction numerator Q V over denominator d end fraction space equals space m g

  • This now gave Millikan enough information to calculate the charge on an oil drop if he knew its terminal velocity and the potential difference  required to make it stationary

Significance of Millikan's Results

  • Millikan repeated his experiment for many different oil droplets

    • They all had different charges, but he noticed an interesting pattern in his results

    • Every value of charge was an integer multiple of 1.60 × 10−19 C

Table of results from Millikan's Experiment

Oil Droplet Number

Charge on Droplet (× 10−19 C)

Charge divided by 1.60 × 10−19 C

1

1.59

0.994

2

11.1

6.94

3

9.54

5.96

4

15.9

9.94

5

6.36

3.98

Millikan's results showed that the charge on each oil drop was an integer multiple of 1.60 × 10−19 C

Quantisation of Electric Charge

  • Each negative oil droplet had been ionised, so carried an integer number of electrons

    • The fact that each droplet's charge was quantised into integer multiples of 1.60 × 10−19 C suggested that this was the charge of each electron

Worked Example

Using Stokes' Law, show that the droplet's mass can be expressed in terms of its velocity v, the viscosity of air η, the density of the oil ρ  and acceleration due to gravity g, as:

fraction numerator 6 straight pi eta v over denominator g end fraction square root of fraction numerator 9 eta v over denominator 2 rho g end fraction end root

Answer:

Step 1: Write down the relevant equation from the data and formulae sheet

  • Stokes' law:

F space equals space 6 straight pi eta r v

Step 2: Apply Stokes' law to the droplet at terminal velocity

  • At terminal velocity (in the absence of an electric field), weight is equal to viscous drag force:

m g space equals space 6 straight pi eta r v

Step 3: Write mass in terms of radius and density

  • We need to get rid of the radius term as it will not be in our final expression

    • We can do this by writing radius in terms of the required quantities

  • Recall the equation relating mass, density and volume of a sphere, Vs :

m space equals space V subscript s space rho

m space equals space fraction numerator 4 straight pi r cubed rho over denominator 3 end fraction

Step 4: Write an expression for the droplet's radius

  • Substitute the expression for mass in terms of radius and density and rearrange for radius:

fraction numerator 4 up diagonal strike straight pi r to the power of up diagonal strike 3 space 2 end exponent rho g over denominator 3 end fraction space equals space 6 up diagonal strike straight pi eta up diagonal strike r v

r squared space equals space fraction numerator 9 eta v over denominator 2 rho g end fraction

r space equals space square root of fraction numerator 9 eta v over denominator 2 rho g end fraction end root

Step 5: Substitute this into the expression for mass

  • We can finally eliminate radius from the equation in step 2:

m space equals space fraction numerator 6 straight pi eta r v over denominator g end fraction

m space equals space fraction numerator 6 straight pi eta v over denominator g end fraction square root of fraction numerator 9 eta v over denominator 2 rho g end fraction end root

Examiner Tips and Tricks

"Show that" questions are one of the rare moments in an exam you can be totally confident you have the correct answer, but don't get complacent. The marks are awarded for clear and logical working out. Don't try to skip a step and hope you can convince your examiner, be thorough.

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Dan Mitchell-Garnett

Author: Dan Mitchell-Garnett

Expertise: Physics Content Creator

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.