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Module 8: From the Universe to the Atom
Quick questions on Millikan's oil drop experiment: HSC Physics Module 8
12short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.
What is why the experiment was needed?Show answer
Thomson's 1897 measurement of $e/m$ for the electron was the charge-to-mass ratio, not the charge itself. To separate the two and find both the mass and charge of the electron, an independent measurement of $e$ alone was required.
What is the apparatus?Show answer
Robert Millikan's 1909 experiment (refined through about 1913) used:
What is two methods?Show answer
Stationary method (the simplest to describe). Adjust the voltage until a chosen droplet hangs motionless. The electric force on the charge balances gravity:
What is results?Show answer
Millikan measured thousands of drops over many years. Every measured charge was a positive integer multiple of a single value:
What is worked example?Show answer
A drop of mass $5.0 \times 10^{-15}$ kg is held stationary between plates 5.0 mm apart with potential difference 460 V. Find the charge on the drop.
What is modern view?Show answer
Charge quantisation in units of $e$ is observed in every macroscopic system. Quarks have charges of $\pm e/3$ and $\pm 2e/3$, but they are confined inside hadrons and cannot be isolated as free particles. The smallest free charge is the electron's $-e$ (or its antiparticle's $+e$), exactly the unit Millikan measured.
What is stationary method?Show answer
Adjust the voltage until a chosen droplet hangs motionless. The electric force on the charge balances gravity:
What is falling-and-rising method?Show answer
With the field off, the droplet falls at terminal velocity $v_g$ set by gravity vs viscous drag. With the field switched on (in the direction that drives the negative droplet upward), it rises at terminal velocity $v_E$ set by net electric force vs drag. Combining $v_g$ and $v_E$ eliminates the radius-dependent constants and gives the charge $q$ directly.
What is confusing $E$ and $V$?Show answer
Between parallel plates, $E = V/d$. The field is in V/m and is uniform between the plates; the voltage is the work per unit charge to move from one plate to the other.
What is setting up the force balance with the wrong sign?Show answer
The electric force must be opposite to gravity (upward, for a negative droplet) to balance it. Always check by drawing the free-body diagram.
What is reporting a non-integer multiple of $e$ without comment?Show answer
If your calculation gives $q/e = 2.4$, you should say either that experimental error puts the actual value at 2 or 3, or that the problem is testing your ability to round. Real charges are integer multiples of $e$.
What is saying Millikan measured the mass of the electron directly?Show answer
He measured the charge $e$. The mass of the electron then follows from Thomson's $e/m$.