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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.

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Numbers

What holds at terminal speed for a falling droplet, and how is Stokes' law then used?

Numbers

How is the mass of a droplet found, and what pattern did Millikan see in the charges?

Process

What happens when a field is applied to a falling droplet, and how does drag change with speed?

Key term

What does quantisation of electric charge mean?

Questions from this subtopic

Each one with the model answer underneath.

  1. 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. 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. 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. 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. 5

    Write the formula used to calculate the charge on a stationary oil droplet from the equilibrium condition.

    Model answer

    Q = mgd/V

  6. 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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