Specification: AQA A-level Physics (7408) 3.12.1.4 Turning points option | Paper: 3 Section B | Data and formulae booklet: the balance condition QV/d = mg is given in the exam; Q = ne and the rearranged forms are not
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Stationary (held) drop: with the pd adjusted so the drop is stationary, the upward electric force equals the weight, QV/d = mg. There is no viscous force because the drop is not moving.
Quantisation of charge: Millikan found that the charge on every drop was a whole-number multiple of 1.6 × 10−19 C. Charge comes in discrete packets of size e, the charge of the electron.
Direction of the force: the electric force on a negatively charged drop is towards the positive plate, so to hold it up the top plate must be positive. The field strength between the plates is E = V/d.
The equation
electric force = weight QV/d = mg and Q = ne, where n is a whole number
| Quantity | Symbol | Unit |
|---|---|---|
| charge on the drop | Q | coulombs (C) |
| pd between the plates | V | volts (V) |
| plate separation | d | metres (m) |
| mass of the drop | m | kilograms (kg) |
| gravitational field strength | g | metres per second squared (m s⁻²), equivalent to N kg⁻¹ |
| number of electron charges | n | no unit (a whole number) |
Rearranged: Q = mgd ÷ V | V = mgd ÷ Q | m = QV ÷ gd | n = Q ÷ e | Convert mm to m (× 10−3) before you substitute. If n is not close to a whole number, check your working.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An oil drop carries a charge of 3.2 × 10−19 C. It is between two horizontal parallel plates 0.0050 m apart with a pd of 500 V between them. Calculate the electric force on the drop, to 2 significant figures.
Show solution
F: F = QV/d
I: F = 3.2 × 10−19 × 500 ÷ 0.0050
F: F is already the subject
A: 3.2 × 10−14 N (2 s.f.)
Q2 (Core) An oil drop of mass 4.9 × 10−15 kg is held stationary between horizontal plates 0.0080 m apart when the pd between them is 600 V. Calculate the charge on the drop, to 2 significant figures. (g = 9.81 m s⁻²)
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F: QV/d = mg
I: Q × 600 ÷ 0.0080 = 4.9 × 10−15 × 9.81
F: Q = mgd ÷ V = (4.9 × 10−15 × 9.81 × 0.0080) ÷ 600
A: 6.409… × 10−19 C = 6.4 × 10−19 C (2 s.f.), which is 4e
Q3 (Core) An oil drop of mass 3.3 × 10−15 kg has 3 excess electrons. Calculate the pd needed to hold it stationary between horizontal plates 6.0 mm apart, to 3 significant figures. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Convert first: 6.0 mm = 6.0 × 10−3 m; Q = 3e = 3 × 1.60 × 10−19 = 4.8 × 10−19 C
F: QV/d = mg
I: 4.8 × 10−19 × V ÷ 6.0 × 10−3 = 3.3 × 10−15 × 9.81
F: V = mgd ÷ Q = (3.3 × 10−15 × 9.81 × 6.0 × 10−3) ÷ 4.8 × 10−19
A: 404.6… V = 405 V (3 s.f.)
Q4 (Core) A drop carrying a charge of 2e is held stationary between horizontal plates 4.0 mm apart when the pd is 250 V. Calculate the mass of the drop, to 2 significant figures. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Convert first: 4.0 mm = 4.0 × 10−3 m; Q = 2 × 1.60 × 10−19 = 3.2 × 10−19 C
F: QV/d = mg
I: 3.2 × 10−19 × 250 ÷ 4.0 × 10−3 = m × 9.81
F: m = QV ÷ gd = (3.2 × 10−19 × 250) ÷ (9.81 × 4.0 × 10−3)
A: 2.03… × 10−15 kg = 2.0 × 10−15 kg (2 s.f.)
Watch out: leaving d as 4.0 instead of 4.0 × 10−3 m gives a mass 1000 times too small.
Q5 (Exam) A negatively charged oil drop of mass 6.2 × 10−15 kg is held stationary between horizontal plates 5.0 mm apart when the pd is 380 V. (a) State which plate is positive. (b) Calculate the charge on the drop, to 3 significant figures. (c) Deduce the number of excess electrons on the drop. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) polarity
the electric force must be upwards; a negative charge is attracted to the positive plate, so the top plate is positive
Convert first: 5.0 mm = 5.0 × 10−3 m
Step 2 – (b) charge
F: QV/d = mg
I: Q × 380 ÷ 5.0 × 10−3 = 6.2 × 10−15 × 9.81
F: Q = mgd ÷ V = (6.2 × 10−15 × 9.81 × 5.0 × 10−3) ÷ 380
A: 8.00 × 10−19 C (3 s.f.)
Step 3 – (c) number of electrons
n = Q ÷ e = 8.00… × 10−19 ÷ 1.60 × 10−19 = 5.00…
A: 5 excess electrons
Q6 (Exam) In a repeat of Millikan’s experiment, a student obtains these charges for five drops: 3.18, 4.83, 6.35, 9.62 and 11.2 (all × 10−19 C). (a) Show that the results support the idea that charge is quantised in units of 1.60 × 10−19 C. (b) Use all five results to find a value for e, to 3 significant figures.
Show solution
Step 1 – (a) divide each charge by 1.60 × 10−19 C
3.18 ÷ 1.60 = 1.99; 4.83 ÷ 1.60 = 3.02; 6.35 ÷ 1.60 = 3.97; 9.62 ÷ 1.60 = 6.01; 11.2 ÷ 1.60 = 7.00
every value is very close to a whole number (2, 3, 4, 6 and 7), so each charge is an integer multiple of 1.60 × 10−19 C
Step 2 – (b) best value of e
total charge = 35.18 × 10−19 C; total number of charges = 2 + 3 + 4 + 6 + 7 = 22
e = 35.18 × 10−19 ÷ 22 = 1.599… × 10−19 C
A: 1.60 × 10−19 C (3 s.f.)
Watch out: no drop has a charge of 5e here, but that doesn’t matter. Quantisation means every charge is a whole multiple of e, not that every multiple must appear.
Q7 (Exam) A negatively charged oil drop is held stationary between horizontal plates 5.0 mm apart when the pd is 420 V. X-rays are used to ionise the air, and the drop picks up one extra electron. It is now held stationary at 315 V. (a) Deduce the number of electron charges on the drop before and after. (b) Calculate the mass of the drop, to 2 significant figures. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) compare the two charges
mg and d are unchanged, so QV = mgd is constant: Q₁ × 420 = Q₂ × 315, giving Q₂ ÷ Q₁ = 420 ÷ 315 = 4/3
Q₂ = Q₁ + e, so Q₁ + e = (4/3)Q₁, giving Q₁ = 3e and Q₂ = 4e
A: 3e before, 4e after
Step 2 – (b) mass
F: QV/d = mg
I: (3 × 1.60 × 10−19) × 420 ÷ 5.0 × 10−3 = m × 9.81
F: m = QV ÷ gd = (4.8 × 10−19 × 420) ÷ (9.81 × 5.0 × 10−3)
A: 4.11… × 10−15 kg = 4.1 × 10−15 kg (2 s.f.)
Watch out: a larger charge needs a smaller pd to balance the same weight. Q and V are inversely proportional here.
Q8 (Exam) An oil drop is held stationary between horizontal plates when the pd is 400 V. The pd is suddenly increased to 500 V. Calculate the initial acceleration of the drop, to 3 significant figures, and state its direction. (g = 9.81 m s⁻²)
Show solution
Step 1 – forces at 400 V
QV/d = mg, so the electric force at 400 V equals mg
Step 2 – forces at 500 V
F ∝ V, so the electric force is now (500 ÷ 400) × mg = 1.25mg upwards; resultant force = 1.25mg − mg = 0.25mg upwards
Step 3 – acceleration
F: F = ma
I: 0.25mg = ma
F: a = 0.25g = 0.25 × 9.81
A: 2.45 m s⁻² upwards (3 s.f.). The drop starts from rest, so there is no viscous force at that instant.
Watch out: the acceleration does not last. As the drop speeds up, the viscous drag grows until the drop rises at a constant terminal speed.
Q9 (Stretch) An oil drop of radius 1.2 μm is held stationary between horizontal plates 6.0 mm apart when the pd is 590 V. The density of the oil is 880 kg m⁻³. Ignore the upthrust of the air. (a) Calculate the charge on the drop, to 2 significant figures, and the number of electron charges it carries. (b) The radius is measured with an uncertainty of 5%. Estimate the resulting percentage uncertainty in the charge. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Convert first: 1.2 μm = 1.2 × 10−6 m; 6.0 mm = 6.0 × 10−3 m
Step 1 – mass of the drop
volume = (4/3)πr³ = (4/3)π × (1.2 × 10−6)³ = 7.238… × 10−18 m³
m = ρ × volume = 880 × 7.238… × 10−18 = 6.369… × 10−15 kg
Step 2 – (a) charge
F: QV/d = mg
I: Q × 590 ÷ 6.0 × 10−3 = 6.369… × 10−15 × 9.81
F: Q = mgd ÷ V = (6.369… × 10−15 × 9.81 × 6.0 × 10−3) ÷ 590
A: 6.35… × 10−19 C = 6.4 × 10−19 C (2 s.f.); n = 6.35… ÷ 1.60 = 3.97…, so 4 electron charges
Step 3 – (b) uncertainty
Q ∝ m ∝ r³, so the percentage uncertainty in Q is 3 × 5% = 15% (working with the extremes: 1.05³ = 1.157…, about 16%)
Watch out: the radius is cubed, so it dominates the uncertainty. This is why Millikan measured r so carefully (from the terminal speed, using Stokes’ law).
Q10 (Stretch) An oil drop of radius 0.85 μm is held stationary between horizontal plates 4.0 mm apart when the pd is 278 V. The density of the oil is 880 kg m⁻³. (a) Calculate the field strength between the plates. (b) Ignoring upthrust, calculate the charge on the drop, to 3 significant figures, and deduce the number of electron charges. (c) The density of air is 1.29 kg m⁻³. Calculate the percentage change, to 2 significant figures, in your value of Q if the upthrust of the air is included, and comment on whether ignoring upthrust is justified. (g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Convert first: 0.85 μm = 8.5 × 10−7 m; 4.0 mm = 4.0 × 10−3 m
Step 1 – (a) field strength
E = V/d = 278 ÷ 4.0 × 10−3 = 69 500 V m⁻¹
Step 2 – (b) mass and weight
m = (4/3)πr³ρ = (4/3)π × (8.5 × 10−7)³ × 880 = 2.263… × 10−15 kg; mg = 2.220… × 10−14 N
F: QV/d = mg, so Q = mg ÷ E
I: Q = 2.220… × 10−14 ÷ 69 500
A: 3.195… × 10−19 C = 3.20 × 10−19 C (3 s.f.); n = 3.195… ÷ 1.60 = 1.99…, so 2 electron charges
Step 3 – (c) including upthrust
upthrust = weight of air displaced, so the effective weight is proportional to (ρoil − ρair) = 880 − 1.29 = 878.71 kg m⁻³
Q falls by a fraction 1.29 ÷ 880 = 0.00146…
A: a change of 0.15% (2 s.f.). This is far smaller than the uncertainty in r³, so ignoring upthrust is justified.
Watch out: only include upthrust when the question gives you the density of air. Otherwise assume QV/d = mg exactly.
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