4.2.2Tools and Inquiry — IB Physics skills
Millikan’s Oil Drop Experiment — IB Physics SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 31 of 33 Unit quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5The Apparatus
Tiny charged oil drops are suspended between two horizontal metal plates connected to a variable power supply.
Oil drops
- Oil drops A spray produces tiny drops that pick up a few extra or missing electrons through friction.
Plates
- Plates A potential difference across horizontal plates creates a nearly uniform electric field E = V/d between them.
Microscope
- Microscope The drops are viewed through a microscope, so one drop can be tracked and its motion observed.
Balance
- Balance The voltage is adjusted until a drop is held stationary, so the electric force balances its weight.
Part 2 of 5Finding the Charge
1
Adjust the voltage until a selected drop hovers, so that the upward electric force equals the downward weight.
2
At balance the forces are equal: qE = mg, and with E = V/d this gives qV/d = mg.
3
Rearrange to q = mgd/V, where d is the plate separation and V is the balancing voltage.
4
The mass of the drop is found by turning off the field and measuring its terminal velocity, using Stokes' law.
5
Repeat for many drops and compare the charges, which fall into whole-number multiples of a smallest value.
Part 3 of 5Conclusions
Conclusions
Quantisation
- Quantisation Every measured charge is a whole-number multiple of 1.6 × 10⁻¹⁹ C, so charge exists only in discrete amounts.
Elementary charge
- Elementary charge The smallest unit of free charge, e, is the charge on a proton and the magnitude of the charge on an electron.
Sources of error
- Sources of error Evaporation of the drop and air movement change its mass and motion, so careful measurements are repeated.
Significance
- Significance The experiment provided the first accurate value of e, which helped to find the mass of the electron.
| Measured charge (C) | Multiple of 1.6 × 10⁻¹⁹ C |
|---|---|
| 3.2 × 10⁻¹⁹ | 2 |
| 4.8 × 10⁻¹⁹ | 3 |
| 8.0 × 10⁻¹⁹ | 5 |
Part 4 of 5Worked Example
Question
- An oil drop of mass 1.63 × 10⁻¹⁵ kg is held stationary between plates 8.0 mm apart when the voltage is 266 V.
- Find its charge in terms of e = 1.6 × 10⁻¹⁹ C.
- (g = 9.81 m s⁻²)
- At balance q = mgd/V = (1.63 × 10⁻¹⁵ × 9.81 × 8.0 × 10⁻³) / 266.
- Numerator = 1.28 × 10⁻¹⁶; q = 1.28 × 10⁻¹⁶ / 266 = 4.8 × 10⁻¹⁹ C.
- Number of elementary charges = 4.8 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3.
Answer q = 4.8 × 10⁻¹⁹ C, which is 3e.
Part 5 of 5Exam Checklist & Tips
4.2.2 – Key Points for the Exam
- Quantisation of charge The fact that charge exists only as whole-number multiples of the elementary charge.
- Elementary charge The magnitude of the charge on an electron or proton, 1.6 × 10⁻¹⁹ C.
- Uniform electric field A field of constant strength and direction, as between parallel plates.
- Terminal velocity The steady speed reached when the drag balances the other forces.
Common Mistakes
- Stating that the drop is stationary because there is no force on it, when the forces balance.
- Forgetting to convert the plate separation from millimetres to metres.
- Saying the measured charges are random values rather than multiples of one smallest charge.
- Mixing up the direction of the electric force; for a negatively charged drop it points opposite to the field.
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