Teachingnotes-Millikan'sexperiment.doc

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Topic 25.1 – Electrons & Millikan’s Experiment

In the exam you are expected to know about:

 determination of Q

 Condition for holding a charged oil droplet, of charge Q, stationary between oppositely charged parallel plates;

 QV/d = mg

 Motion of a falling oil droplet with and without an electric field;

 terminal speed, Stokes’ Law for the viscous force on an oil droplet used to calculate the droplet radius

 F = 6rv

 Quantization of electric charge.

Robert Millikan used a simple but famous experiment that served to confirm the unit electronic charge as 1.6 × 10-19_{C. He sprayed oil drops into a space between two} charged plates. Each tiny oil droplet was charged up by friction as it left the sprayer. The theory was simple; the attractive electrostatic force between the droplet and the positively charged plate would balance out the weight of the droplet. His apparatus was like this:

He would select a particular oil drop and hold it stationary by altering the voltage between the two plates.

Question 1

What are the forces acting on the plates? What is the resultant force?

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The forces on the stationary drop are like this:

We know that:

 the electric force = electric field × charge (F = Eq)  the electric field strength in a uniform field, E = V/d

It doesn't take a genius to see that:

Question 2

How might you find the weight of the drop?

Ans:

This is not a very satisfactory way because the uncertainty is too great. So another method was used. Millikan turned off the plates and watched the oil drop. Very quickly the oil drop reached terminal speed.

Question 3

What forces are acting on an object at terminal speed? What is the

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[r - radius of the sphere, v - terminal speed (m/s).]

The strange looking symbol, , is "eta", a Greek lower case letter long 'ē', the Physics Code for the coefficient of the viscosity of a fluid. The units for  are N s m-2_{. For} air,  = 1.8 × 10-5_{N s m}-2_.

Question 4

A small metal sphere of radius 0.5 mm has mass 1.0 × 10-3_{kg is dropped}

into oil of which the viscosity is 0.36 N s m-2_{. What is the terminal velocity at which it} falls?

Ans:

[Side note: this model only works for objects falling at low speed. At higher speed, turbulence has an effect.]

In question 2 we looked at the intuitive way of finding the weight by: Weight = density  volume  g

We can write this as:

So we can bring in the Stoke’s Law equation in by writing:

Rearranging and cancelling out gives us:

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So by observing the terminal speed, we can work out the radius. From that, we can work out the volume, hence the mass and weight. Although it seems long-winded, this method produces much less uncertainty than attempting a direct measurement of the radius.

Worked example

The data below is from an experiment similar to Millikan’s experiment.  Density of oil = 900 kg m-3

 Pd across the plates = 613 V  Plate separation = 0.01 m

 Viscosity of air = 1.8 × 10-5_{N s m}-2

When the voltage between the plates is turned off, the droplet falls steadily a distance of 2.50  10-3_{m in a time of 22 s. What is the charge? Take g as 9.8 m s}-2_. Work out the speed = distance ¸ time = 2.50  10-3_m_¸_{22 s = 1.14}_₁₀-4_m/s

Now work out the radius using the equation:

r = 1.02  10-6_m

Mass = volume  density =

We know that weight = force from the electric field

mg = qV/d

Rearranging gives us q = mgd V

q = 4.49  10 -15 _kg__{9.8 m s}-2__{0.01 m}₌_7.1_₁₀-19_C

613 V

This is a charge of magnitude 5 e (where e = -1.602  10-19_C)

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Question 5

In an experiment to determine the charge on a charged oil droplet, the droplet was held stationary in a vertical electric field of strength 57 kV m-1_{. After the field was} switched off, the droplet fell at a steady speed, taking 18.3 s to fall through a vertical distance of 2.0 mm

Viscosity of air = 1.8 × 10-5_{N s m}-2 Density of oil = 970 kg m-3

g = 9.8 m/s-2

(a) Calculate the speed of the droplet as it falls. (b) Show that the droplet’s radius is 9.7  10-7 m (c) Calculate the charge of the droplet.

(d) Compare this to the electronic charge. What does it suggest?

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Summary

Weight of the oil drop is balanced by force from electric field.

The radius is too small to be measured directly.

Oil drop is allowed to fall.

Speed is measured.

Stoke’s Law is used to measure radius by viscous drag.

Radius is used to work out the volume

Mass is found by volume  density

Weight is determined.

Weight = force due to electric field.

mg = Eq