Principle of Millikan’s determination of the electronic charge, e (A-level only) — AQA A-Level Physics
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Millikan's oil-drop experiment determined the fundamental unit of electric charge, e, by balancing the gravitational and electric forces on charged oil droplets and observing that all measured charges were integer multiples of approximately 1.6 x 10^-19 C. This provided direct experimental evidence for the quantisation of electric charge.
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Principle of Millikan’s determination of the electronic charge, e (A-level only) explained
Principle of Millikan’s determination of the electronic charge, e (A-level only) covers 5 specification points, including Condition for holding a charged oil droplet,… and QV = mg d Motion of a falling oil droplet with….
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Open a point below for its explanation and revision guidance.
Condition for holding a charged oil droplet, of charge Q, stationary between oppositely charged parallel plates.
In Millikan’s experiment, a charged oil droplet is held stationary between two horizontal parallel plates connected to a voltage supply. For the droplet to remain at rest, the resultant force on it must be zero. The forces are the weight mg acting downwards and the electric force acting upwards. The electric force has magnitude QE, where Q is the droplet’s charge and E is the uniform electric field strength between the plates. The field is related to the potential difference V and plate separation d by E = V/d. The condition is therefore QE = mg, or QV/d = mg. This allows the charge Q to be found from measurements of V, d and m. In Millikan’s method, the droplet’s mass is determined from its terminal velocity when falling without the electric field, and the charge is found to be an integer multiple of the elementary charge e.
QV = mg d Motion of a falling oil droplet with and without an electric field; terminal speed to determine the mass and the charge of the droplet.
In Millikan’s experiment, a charged oil droplet falls under gravity and is then held by an electric field. Without the field, the droplet reaches terminal speed when its weight is balanced by viscous drag (upthrust is considered negligible). Measuring that speed gives the radius and mass using Stokes' Law. With a vertical electric field, the droplet can be held stationary. When the electric force balances the weight, $QV/d = mg$, where $Q$ is the droplet charge, $V$ is the potential difference, $m$ is the mass, $g$ is gravitational field strength and $d$ is the plate separation. Combining the measurements allows $Q$ to be determined. Repeating for many droplets shows that charge is quantised in multiples of $e$.
Stokes’ Law for the viscous force on an oil droplet used to calculate the droplet radius.
Stokes’ Law gives the viscous drag on a small sphere moving slowly through a fluid: F = 6πηrv, where η is the viscosity of the fluid, r is the sphere radius and v is its speed. In Millikan’s experiment, an oil droplet falls through air and reaches terminal speed when the resultant force is zero. At that point, weight minus upthrust equals viscous drag. Substituting Stokes’ Law into this balance gives an equation containing the unknown radius r. Solving it yields r from the measured terminal speed, oil density, air density and air viscosity. The radius then gives the droplet volume and mass, which are needed to find the charge from the electric force balance.
F = 6πηrv Significance of Millikan’s results.
Stokes’ law gives the viscous drag on a small sphere moving slowly through a fluid as F = 6πηrv, where η is the fluid viscosity, r the sphere radius and v its speed. In Millikan’s experiment, charged oil droplets fall under gravity and are held or moved by an electric field between horizontal plates. When a droplet falls at terminal velocity, the drag balances the net downward force, allowing r and then the droplet charge to be found. Repeating measurements showed every droplet charge was an integer multiple of a smallest value, the elementary charge e. This demonstrated quantisation of electric charge and gave a value for e, supporting the idea that charge exists in discrete units carried by electrons.
Quantisation of electric charge.
Quantisation of electric charge means that any electric charge is an integer multiple of a fundamental unit, the elementary charge e. In Millikan’s experiment, tiny charged oil droplets were balanced or moved by an electric field, and their charges were calculated from force balances. The measured charges were always close to whole-number multiples of about 1.60 × 10⁻¹⁹ C, never arbitrary fractions of it. This showed that charge comes in discrete packets, now understood as the charge carried by electrons and protons. The elementary charge magnitude is e = 1.60 × 10⁻¹⁹ C, and a body’s charge can be written Q = ne, where n is an integer.
Your focus
- Derive the condition QV/d = mg for a charged oil droplet held stationary between parallel plates.
- Use the condition to calculate the charge on a droplet from measurements of V, d and m.
- Explain how this condition is used in Millikan’s determination of the electronic charge.
Show all 15 objectives
- Describe how terminal speed of a falling oil droplet is used to find its mass, ignoring upthrust.
- Apply the force balance $QV/d = mg$ to a droplet held stationary in an electric field.
- Explain how repeated measurements of droplet charge provide evidence for charge quantisation.
- State Stokes’ Law and identify the meaning of each symbol.
- Derive an expression for the radius of a falling droplet from its terminal speed.
- Use the calculated radius to find the droplet volume and mass.
- State Stokes’ law and identify each quantity in F = 6πηrv.
- Use terminal velocity and Stokes’ law to determine the radius of a falling droplet.
- Explain how Millikan’s results demonstrate quantisation of electric charge and provide a value for e.
- Define quantisation of electric charge and state the value of the elementary charge.
- Apply Q = ne to find the number of elementary charges on a body.
- Explain how Millikan’s oil-drop results provide evidence for charge quantisation.
Which command word is this?
In AQA A-Level Physics, 'derive' requires a logical sequence of steps starting from fundamental principles (e.g., force balance, Stokes' law) to obtain the required expression. Marks are awarded for correct equations, substitution, and algebraic manipulation. No marks for simply quoting the final formula.
Principle of Millikan’s determination of the electronic charge, e (A-level only) exam tips
Topic Overview
Millikan's oil-drop experiment is a classic A-level physics practical that demonstrates the quantisation of electric charge. It involves balancing charged oil droplets against gravity using an electric field, allowing the charge on individual droplets to be measured. By analysing many droplets, Millikan showed that all charges were integer multiples of a fundamental value, the elementary charge e.
This topic is crucial because it provides experimental evidence for the existence of discrete charge and allows determination of e, a fundamental constant. It also reinforces understanding of forces, electric fields, Stokes' law, and terminal velocity, making it an excellent synoptic topic that connects mechanics and electricity.
Key Concepts
- →Electric charge is quantised: any charge Q is an integer multiple of the elementary charge e (Q = ne, where n is an integer).
- →The elementary charge e is the magnitude of the charge on an electron (and proton), approximately 1.6 x 10^-19 C.
- →In the experiment, oil droplets are charged by friction (or X-rays) and introduced between two horizontal plates. The electric field E = V/d can be adjusted to balance the droplet.
- →When a droplet is stationary, the electric force qE balances the effective weight (weight minus upthrust). When falling freely, terminal velocity is reached when viscous drag (Stokes' law) balances the effective weight.
- →By measuring terminal velocity, the radius and mass of the droplet can be found, allowing calculation of its charge. Repeating for many droplets reveals the common factor e.
Marking Points
- For the droplet to be stationary, the resultant force must be zero, so the upward electric force balances the downward weight.
- The electric force on a charge Q in a uniform field E has magnitude QE, and its direction depends on the sign of Q and the direction of E.
- The uniform field between parallel plates is E = V/d, where V is the potential difference and d is the plate separation.
- The weight of the droplet is mg, where m is its mass and g is the gravitational field strength.
- The condition is QE = mg, which may be written QV/d = mg.
- The charge Q can be calculated from Q = mgd/V once m, g, d and V are known.
- In Millikan’s experiment, the droplet is held stationary by adjusting V, and the measured charges are found to be multiples of e.
- Without the electric field, the droplet falls at terminal speed when the resultant force is zero: weight is balanced by viscous drag (ignoring upthrust).
- The terminal speed without the field is used with Stokes’ Law to find the droplet radius, and hence its volume and mass.
- With the electric field applied, the electric force on the droplet is $QV/d$, where $V$ is the potential difference across plates separated by $d$.
- When the droplet is held stationary, the electric force balances the weight: $QV/d = mg$, which rearranges to $QV = mgd$.
- The mass $m$ is obtained from the radius and the density of the oil, so $Q$ can be calculated from $Q = mgd/V$.
- Repeating the measurement for many droplets gives values of $Q$ that are integer multiples of the elementary charge $e$, demonstrating charge quantisation.
- Stokes’ Law states that the viscous force on a small sphere is F = 6πηrv, where η is the fluid viscosity, r is the sphere radius and v is the speed.
- At terminal speed, the forces on the falling droplet balance: weight minus upthrust equals viscous drag.
- Weight is mg, and upthrust is the weight of displaced air, which can be expressed using the air density and the droplet volume.
- Substituting F = 6πηrv into the terminal-speed force balance gives an equation that can be solved for the droplet radius r.
- The radius is used to calculate the droplet volume and hence its mass from the oil density.
- Accurate values of viscosity, densities and terminal speed are needed because the radius depends on these quantities.
- Stokes’ law F = 6πηrv applies to a small sphere moving slowly through a viscous fluid, where η is the viscosity of the fluid, r is the sphere radius and v is the sphere speed.
- At terminal velocity the resultant force on a falling droplet is zero, so viscous drag balances the net downward force, which is weight minus upthrust.
- Measuring terminal velocity and using F = 6πηrv allows the droplet radius r to be determined, because η is known for the fluid and v is measured.
- With the radius known, the weight and upthrust can be calculated, and applying an electric field lets the charge on the droplet be found from the balance of electric force and other forces.
- Millikan’s repeated results showed droplet charges were whole-number multiples of a smallest charge, demonstrating quantisation of electric charge.
- The smallest common charge gave a value for the elementary charge e, now known as the magnitude of the charge on an electron.
- The experiment provided evidence that electric charge is carried in discrete units rather than being continuous.
- Quantisation means electric charge exists only in discrete amounts that are integer multiples of the elementary charge e.
- The elementary charge magnitude is e = 1.60 × 10⁻¹⁹ C, equal to the magnitude of the charge on an electron or proton.
- For a charged body, Q = ne where n is an integer, so possible charges include 0, ±e, ±2e and so on.
- Millikan’s oil-drop results supported quantisation because measured droplet charges clustered around integer multiples of e.
- The value of e obtained from the smallest common charge is consistent with the charge on a single electron.
- Quantisation explains why charge is not transferred in arbitrarily small continuous amounts in everyday electrostatic processes.
Examiner Tips
- 💡Begin by stating that the droplet is in equilibrium, so the resultant force is zero.
- 💡Write the force balance as QE = mg and then substitute E = V/d to obtain QV/d = mg.
- 💡When calculating, keep consistent SI units and rearrange carefully before substituting numbers.
- 💡Sketch the forces on the droplet for both cases: falling without the field and held stationary with the field. Label weight, viscous drag, and electric force.
- 💡Write the force balance equation before substituting numbers, and check that each symbol has its correct SI unit.
- 💡When using $QV/d = mg$, rearrange algebraically for the required quantity before inserting values to reduce arithmetic slips.
- 💡State Stokes’ Law in symbols and define each symbol, including the SI unit of viscosity.
- 💡Show the terminal-speed force balance clearly before substituting Stokes’ Law, so the examiner can follow the derivation of r.
- 💡Keep density terms distinct: use the oil density for the droplet mass and the air density for upthrust.
- 💡When deriving or using F = 6πηrv, state clearly that it applies to a small sphere moving slowly through a viscous fluid and define each symbol.
- 💡In calculations, first find terminal velocity from distance and time, then use F = 6πηrv to find r before attempting to find charge.
- 💡When explaining significance, link the numerical results to quantisation and to a value for e rather than only describing the apparatus.
- 💡Use Q = ne and show that n is an integer when interpreting Millikan-type data.
- 💡Quote e = 1.60 × 10⁻¹⁹ C with the correct unit and sign convention when explaining charge quantisation.
- 💡Link the experimental observation of whole-number multiples directly to the word quantisation in your answer.
- 💡Always define your symbols clearly and state the equations you use. In derivations, show each step logically, including the force balance equations for both stationary and falling states.
- 💡When calculating the charge, ensure you use consistent SI units and carry through calculations carefully. Pay attention to powers of ten, especially when dealing with small radii and charges.
- 💡In explanations, emphasise that the experiment provides evidence for quantisation by showing that charges are integer multiples of e. Use the phrase 'elementary charge' and state its value with units.
Common Mistakes
- Omitting the electric field relation: the error is using QV = mg without including plate separation; the correction is QV/d = mg.
- Forgetting that forces must balance: the error is equating electric force to mass rather than to weight; the correction is QE = mg.
- Assuming the droplet is uncharged: the error is ignoring that Q must be non-zero for an electric force to act; the correction is that the droplet carries a charge Q.
- Including upthrust when balancing forces on the falling droplet; the correction is to ignore upthrust as it is negligible for oil droplets in air, so weight equals viscous drag.
- Using the droplet diameter instead of its radius in Stokes’ Law; the correction is to calculate the radius from the terminal speed and then use that radius consistently.
- Confusing the distance between the plates $d$ with the distance the droplet falls; the correction is to ensure $d$ in $V/d$ is the plate separation.
- Using the droplet diameter in place of the radius in Stokes’ Law; the correction is to use r, the radius, and to convert any diameter measurement by halving it.
- Omitting upthrust from the force balance; the correction is to include the buoyancy force so that weight minus upthrust equals viscous drag at terminal speed.
- Treating Stokes’ Law as valid for any speed or size; the correction is to note that it applies to small spheres moving slowly with laminar flow around them.
- Treating F = 6πηrv as the weight of the droplet: correct this by stating that it is the viscous drag force, while weight is mg and upthrust is the weight of displaced fluid.
- Using the droplet’s instantaneous velocity instead of its terminal velocity: correct this by measuring the steady speed reached when the resultant force is zero.
- Assuming every droplet carries exactly one elementary charge: correct this by explaining that droplets carry integer multiples of e, and the smallest common value is identified from many measurements.
- Writing that charge can take any value: correct this by stating that charge is quantised as integer multiples of e.
- Confusing e with the base of natural logarithms: correct this by using e specifically for the elementary charge magnitude, 1.60 × 10⁻¹⁹ C.
- Forgetting that n must be an integer in Q = ne: correct this by checking that calculated values of n are whole numbers within experimental uncertainty.
- Students often think Millikan measured the charge of a single electron directly. In fact, he measured the charge on oil droplets, which could carry multiple excess electrons, and inferred e as the greatest common divisor of the measured charges.
- Students may forget to include upthrust and viscous drag in force balances. In air, upthrust is small but not negligible for precise work, and viscous drag is essential for determining the radius from terminal velocity.
- Some students believe that the droplet must be stationary for the charge to be measured. While holding it stationary simplifies the force balance, the charge can also be found from the terminal velocity when the electric field is on, but the stationary method is more common in exams.
Revision Plan
- 1Day 1-2: Review the theory. Watch a video of the experiment, read the derivation of the charge formula, and make a flowchart of the steps.
- 2Day 3-4: Practice calculations. Work through problems finding charge from given radius, voltage, etc., and vice versa. Focus on unit conversions and significant figures.
- 3Day 5-6: Learn the experimental details. Memorise how droplets are charged, how terminal velocity is measured, and why multiple measurements are needed.
- 4Day 7-8: Attempt exam-style questions. Focus on 6-mark explanations and derivations. Mark your answers against mark schemes and note key phrases.
- 5Day 9-10: Active recall and summary. Create flashcards for key equations and concepts. Teach the experiment to someone else or write a concise summary from memory.
Exam Question Types
- 📋Calculation of charge on a droplet given radius, voltage, plate separation, and densities. Advice: Write down the force balance equation, calculate effective weight, then q = F_eff / E. Watch units.
- 📋Derivation of the expression for charge in terms of terminal velocity and other quantities. Advice: Start from Stokes' law and the force balance at terminal velocity, then substitute into the stationary force balance. Show all steps clearly.
- 📋Explanation of how the experiment demonstrates quantisation of charge. Advice: Describe the procedure of measuring many droplets, plotting or analysing charges, and finding a common factor. Mention that charges are integer multiples of e.
- 📋Practical-based question on sources of error and improvements. Advice: Discuss the effects of Brownian motion, evaporation, and the need for repeated measurements. Suggest using smaller droplets or waiting for terminal velocity.
Command Word Expectations (AQA)
In AQA A-Level Physics, 'derive' requires a logical sequence of steps starting from fundamental principles (e.g., force balance, Stokes' law) to obtain the required expression. Marks are awarded for correct equations, substitution, and algebraic manipulation. No marks for simply quoting the final formula.
For 'explain', you must give reasons or mechanisms. In this context, explaining how the experiment shows quantisation requires stating that measured charges are integer multiples of a common value, and that this value is e. A mere description of the procedure without linking to quantisation will not gain full marks.
For 'calculate', you must show your working, substitute correct values, and give the answer with appropriate units and significant figures. Marks are typically awarded for the correct formula, correct substitution, and correct final answer. An answer without working may not receive full marks if an error is made.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: In a Millikan oil-drop experiment, a droplet of radius 1.2 x 10^-6 m is held stationary between two horizontal plates separated by 5.0 mm with a potential difference of 420 V. The density of the oil is 850 kg m^-3 and the density of air is 1.2 kg m^-3. Calculate the charge on the droplet. (Viscosity of air = 1.8 x 10^-5 Pa s, g = 9.81 m s^-2)
- 1.Step 1: Identify given facts: r = 1.2 x 10^-6 m, d = 5.0 x 10^-3 m, V = 420 V, ρ_oil = 850 kg m^-3, ρ_air = 1.2 kg m^-3, η = 1.8 x 10^-5 Pa s, g = 9.81 m s^-2.
- 2.Step 2: Calculate the effective weight (weight minus upthrust): F_eff = (4/3)πr^3(ρ_oil - ρ_air)g = (4/3)π(1.2 x 10^-6)^3(850 - 1.2)(9.81) = 6.02 x 10^-14 N.
- 3.Step 3: The electric force qE balances the effective weight when stationary: qE = F_eff. The electric field E = V/d = 420 / (5.0 x 10^-3) = 8.4 x 10^4 V m^-1.
- 4.Step 4: Therefore q = F_eff / E = (6.02 x 10^-14) / (8.4 x 10^4) = 7.17 x 10^-19 C.
- 5.Step 5: Compare with elementary charge: q/e = (7.17 x 10^-19) / (1.6 x 10^-19) ≈ 4.48, which is not an integer. This suggests either experimental error or that the droplet carries a charge that is not an integer multiple of e, which is impossible; likely a calculation or measurement error. In practice, such a result would be discarded or re-measured.
Question: Describe how the terminal velocity of an oil droplet falling under gravity is used to determine its radius in Millikan's experiment, and hence explain how the charge on the droplet is found when it is held stationary in an electric field.
- 1.Step 1: With the electric field switched off, the droplet falls and quickly reaches terminal velocity v when the viscous drag (Stokes' law: F_d = 6πηrv) balances the effective weight (weight minus upthrust): 6πηrv = (4/3)πr^3(ρ_oil - ρ_air)g.
- 2.Step 2: Rearrange to find the radius: r = sqrt(9ηv / (2(ρ_oil - ρ_air)g)).
- 3.Step 3: With the electric field switched on and adjusted so the droplet is stationary, the electric force balances the effective weight: qE = (4/3)πr^3(ρ_oil - ρ_air)g. Since E = V/d, q = (4/3)πr^3(ρ_oil - ρ_air)gd / V.
- 4.Step 4: Substitute the expression for r from step 2 into the equation for q to obtain q in terms of measured quantities v, V, d, η, ρ_oil, ρ_air, and g.
- 5.Step 5: Repeat for many droplets and find the greatest common divisor of the charges, which is the elementary charge e.