Question 1
This question is about the experiment to measure the charge on an electron performed by American physicists Robert Millikan and Harvey Fletcher in the early years of the twentieth century.
Consider the forces on an oil drop falling at terminal velocity through air, as shown in Fig. 1.1.
Ignoring any upthrust forces, state why the drag force D must be equal and opposite to the weight W of the drop when it is falling at terminal velocity.
[1]
Answer
The drag force is given by the equation D = 6airrv where r is the radius of the oil drop and v is the terminal velocity. The symbol
air represents the viscosity of air; this is a measure of the resistance that air has to the motion of objects in it.
Show that the terminal velocity of a drop of mass 1.8 10–15 kg is about 7
10–5 m s–1. The buoyancy of air may be ignored.
density of oil = 8.7 102 kg m–3
viscosity of air = 1.8 10–5 Pa s
[2]
Answer
The oil drop is given a negative charge and falls into a region of uniform electric field, as shown in the diagram.
Draw six lines on Fig. 1.2 to represent the uniform field between the plates. Assume
that the oil drop does not distort the field.
[2]
Answer
[2]
Answer
When Millikan and Fletcher performed this experiment with many oil drops, they found that the charge on the drop was always a multiple of about 1.6 10–19 C (to two significant figures). Why does this suggest that there is a fundamental quantity of electric charge and how does the modern model of the structure of hadrons suggest that 1.6
10–19 C may not be the smallest quantity of charge that a particle can possess?
[2]
Answer