Overview
Notes summarizing Millikan’s oil-drop experiment to measure the electron charge and its link to electron mass using Thomson’s e/m ratio.
Experimental Setup and Components
- Apparatus includes x-ray source, high-voltage DC supply (up to 10,000 V), atomizer, and viewing telescope.
- Two horizontal plates inside container; top plate has a small hole for droplet entry.
- Top plate connected to positive terminal; separation between plates is D.
- X-rays ionize gas; resulting ions charge oil droplets entering the field region.
Falling Drop: Forces and Terminal Velocity
- With voltage off, droplet falls; gravity accelerates it, viscous drag opposes motion.
- Steady (terminal) downward velocity v1 measured over a known distance and time.
- Forces considered: effective gravitational force (including buoyancy) and viscous drag.
Force Equations and Formulas (Fall)
- Effective weight: FG = (4/3)πR^3(ρ − ρ_air) g.
- Stokes’ drag at terminal v1: FV = 6π η R v1.
- At terminal speed: FG = FV.
- Radius from terminal fall: R^2 = 9 η v1 / [2 g (ρ − ρ_air)].
- Once R known, compute FG using FG = (4/3)πR^3(ρ − ρ_air) g.
Rising Drop with Electric Field
- Apply voltage; uniform field E = V/D directed downward to upward depending on polarity.
- Electric force on charged drop: FE = Q E = Q V/D; direction upward for positive top plate.
- Two operating modes to find Q:
- Balance mode: FE = FG; drop suspended (v = 0), so Q = FG D / V.
- Rise with terminal velocity v2: upward motion with viscous drag downward.
Force Equations and Formulas (Rise)
- At terminal rise: FE = FG + FV (drag now 6π η R v2 downward).
- Using FG = 6π η R v1 from fall condition:
- Q V/D = 6π η R v1 + 6π η R v2.
- Charge from rise data: Q = 6π η R (v1 + v2) D / V.
Discrete Nature of Charge and Electron Charge
- Measured charges on many drops displayed as a chart of values.
- Differences between drop charges showed an integral multiple pattern.
- Smallest step (elementary charge): e = 1.602 × 10^−19 C.
- Interpretation: charges due to integer number of electrons attached to droplets.
Electron Mass via Thomson’s e/m
- Known electron charge-to-mass ratio (Thomson): (e/m) = 1.76 × 10^11 C/kg.
- Electron mass from Millikan’s e and Thomson’s e/m:
- m = e / (e/m) = (1.602 × 10^−19) / (1.76 × 10^11) ≈ 9.1 × 10^−31 kg.
Key Equations Table
| Quantity | Equation | Notes |
|---|
| Effective weight FG | FG = (4/3)πR^3(ρ − ρ_air) g | Accounts for buoyancy |
| Stokes drag | F_drag = 6π η R v | η is air viscosity |
| Terminal fall condition | FG = 6π η R v1 | Defines v1 |
| Radius from fall | R^2 = 9 η v1 / [2 g (ρ − ρ_air)] | Solve for R |
| Electric field | E = V/D | Plate voltage and spacing |
| Electric force | FE = Q E = Q V/D | On charged drop |
| Balance mode charge | Q = FG D / V | Drop suspended |
| Rise mode charge | Q = 6π η R (v1 + v2) D / V | Uses v1 and v2 |
| Elementary charge | e = 1.602 × 10^−19 C | From charge steps |
| Electron mass | m = e / (e/m) = 9.1 × 10^−31 kg | Using e/m = 1.76 × 10^11 C/kg |
Key Terms & Definitions
- Terminal velocity: Constant speed when net force is zero due to drag balancing driving force.
- Buoyant force: Upward force equal to weight of displaced air; included via (ρ − ρ_air).
- Stokes’ law: Linear drag force for small spheres at low Reynolds number, F = 6π η R v.
- Elementary charge (e): Smallest unit of charge; 1.602 × 10^−19 C.
Action Items / Next Steps
- Measure v1 by timing fall over known distance with field off.
- Compute R via terminal fall relation; then find FG.
- Apply voltage; either suspend drop or measure v2 during rise.
- Calculate Q using chosen mode; repeat for many drops to confirm quantization.
- Use e with Thomson’s e/m to confirm electron mass value.