Introduction: The Quest for the Smallest Charge
In the previous chapter, we saw how J.J. Thomson discovered the specific charge (the ratio of charge to mass, \(e/m\)) of the electron. However, he couldn't determine the actual charge of a single electron. This is where Robert Millikan stepped in with his famous oil-drop experiment in 1909. This experiment is a "turning point" because it proved that electric charge isn't a continuous fluid, but comes in tiny, identical "packets."
By the end of these notes, you will understand how Millikan used tiny droplets of oil, some clever math involving Stokes's Law, and a bit of patience to measure the fundamental unit of charge.
The Experimental Setup
Millikan used an atomiser to spray a fine mist of oil droplets into a chamber. Some of these droplets became charged due to friction as they left the nozzle, or by using X-rays to ionise the air inside. These droplets then fell through a small hole into a region between two horizontal metal plates.
Why oil? Millikan used oil because, unlike water, it doesn't evaporate quickly. This allowed him to observe a single droplet for a long time.
The plates were connected to a high-voltage supply, creating a uniform electric field (\(E\)) between them. Millikan could view the droplets through a microscope, which was fitted with a scale (a graticule) to measure how fast the drops were moving.
Phase 1: The Falling Droplet (No Electric Field)
When the electric field is switched off, a droplet falls through the air under the influence of gravity. Don't worry if this seems like a lot of forces at once; we can break it down easily!
As the droplet falls, three forces act on it:
1. Weight (\(mg\)), acting downwards.
2. Upthrust (\(U\)), acting upwards (due to the air the droplet displaces).
3. Viscous Drag (\(F\)), acting upwards (friction from the air).
The droplet quickly reaches a terminal velocity (\(v_1\)). At this point, the forces are balanced:
\(Weight = Upthrust + Viscous Drag\)
\(mg = U + F\)
Using Stokes's Law
For very small spheres moving slowly through a fluid (like air), the viscous drag is given by Stokes's Law:
\(F = 6\pi \eta r v_1\)
Where:
• \(\eta\) (the Greek letter eta) is the viscosity of the air.
• \(r\) is the radius of the droplet.
• \(v_1\) is the terminal velocity.
Why do we do this? We can't put a single microscopic oil drop on a weighing scale! By measuring the terminal velocity (\(v_1\)), Millikan could use the density of the oil (\(\rho_{oil}\)) and air (\(\rho_{air}\)) to calculate the radius (\(r\)) and therefore the mass (\(m\)) of the droplet. This is the "secret weapon" of the experiment.
Phase 2: The Stationary Droplet (With Electric Field)
Once Millikan knew the mass of a specific droplet, he turned on the electric field. By carefully adjusting the potential difference (\(V\)) between the plates, he could make the droplet hover perfectly still.
When the droplet is stationary, the electric force (\(F_E\)) acting upwards exactly balances the "effective weight" (Weight - Upthrust) acting downwards.
The condition for a stationary droplet is:
\(Electric Force = Weight - Upthrust\)
\(EQ = mg - U\)
Since we know that the electric field strength \(E = V/d\) (where \(V\) is potential difference and \(d\) is the plate separation), we can write:
\(\frac{QV}{d} = mg - U\)
By rearranging this, Millikan could calculate the charge (\(Q\)) on that specific droplet:
\(Q = \frac{(mg - U)d}{V}\)
Quick Tip: In many exam questions, upthrust (\(U\)) is considered negligible compared to weight because air is so much less dense than oil. If the question tells you to ignore upthrust, the equation simplifies to \(QV/d = mg\).
Charge Quantisation: The Big Discovery
Millikan repeated this thousands of times for different droplets. He found that the charge on a droplet was never just any random number. Instead, every single charge he measured was an integer multiple of a specific base value.
This is known as charge quantisation. It can be expressed as:
\(Q = ne\)
Where:
• \(Q\) is the total charge on the droplet.
• \(n\) is an integer (\(1, 2, 3, ...\)).
• \(e\) is the elementary charge (the charge of a single electron).
Millikan determined the value of \(e\) to be approximately \(1.6 \times 10^{-19}\) C. This proved that charge is not continuous; you can have 10 electrons or 11 electrons, but you can never have 10.5 electrons!
Summary of the Steps
1. Measure terminal velocity (\(v_1\)) with no field to find the droplet's radius and mass using Stokes's Law.
2. Apply electric field and adjust voltage until the droplet is stationary.
3. Equate forces (\(EQ = mg - U\)) to calculate the charge \(Q\) on the droplet.
4. Compare results from many drops to find the smallest common factor, which is the electronic charge \(e\).
Common Mistakes to Avoid
1. Forgetting Upthrust: In the "A-level only" section, always check if the density of air is given. If it is, you must include upthrust in your calculation.
2. Confusion between \(E\) and \(e\): Remember that capital \(E\) is the Electric Field Strength (in \(Vm^{-1}\)), while lowercase \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) C).
3. Units: Make sure the plate separation \(d\) is in metres and the mass is in kilograms!
Key Takeaways
• Stokes's Law (\(F = 6\pi \eta r v\)) allows us to determine the size of the droplet without weighing it.
• The Stationary Condition (\(QV/d = mg - U\)) is where we solve for the charge \(Q\).
• Quantisation means charge exists in discrete packets; the smallest packet is the electronic charge \(e\).
• Millikan's result combined with Thomson's \(e/m\) allowed scientists to finally calculate the mass of the electron.
Did you know? Millikan's experiment was so precise that his value for the electronic charge was within 1% of the value we use today!