Specification: AQA A-level Physics (7408) 3.7.3.1 | Paper: 2 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Coulomb’s law: the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation. Like charges repel; unlike charges attract.
Point charge: a uniformly charged sphere behaves, outside the sphere, as if all its charge were concentrated at its centre, so r is always measured centre to centre.
Permittivity of free space, ε₀: the constant in Coulomb’s law for charges in a vacuum, ε₀ = 8.85 × 10−12 F m⁻¹. The permittivity of air is so close to ε₀ that charges in air are treated as if they were in a vacuum.
The equation
force = (charge 1 × charge 2) ÷ (4π × permittivity of free space × separation²) F = Q₁Q₂ ÷ (4πε₀r²)
| Quantity | Symbol | Unit |
|---|---|---|
| electrostatic force (on each charge) | F | newtons (N) |
| charges | Q₁, Q₂ | coulombs (C) |
| separation (centre to centre) | r | metres (m) |
| permittivity of free space | ε₀ | farads per metre (F m⁻¹) |
Rearranged: r = √(Q₁Q₂ ÷ 4πε₀F) | Q₁ = 4πε₀r²F ÷ Q₂ | Q = √(4πε₀r²F) for two equal charges | Convert µC (× 10−6), nC (× 10−9) and cm to m before you substitute, and square r.
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) Two point charges of +3.0 µC and +5.0 µC are 0.20 m apart in air. Calculate the electrostatic force between them, to 2 significant figures, and state whether it is attractive or repulsive. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 3.0 µC = 3.0 × 10−6 C; 5.0 µC = 5.0 × 10−6 C
F: F = Q₁Q₂ ÷ (4πε₀r²)
I: F = (3.0 × 10−6 × 5.0 × 10−6) ÷ (4π × 8.85 × 10−12 × 0.20²)
F: F is already the subject
A: F = 1.5 × 10−11 ÷ 4.448… × 10−12 = 3.37… N = 3.4 N (2 s.f.), repulsive because both charges are positive
Q2 (Core) Two small charged spheres carry charges of +4.0 nC and −6.0 nC. Their centres are 3.0 cm apart in air. Calculate the force between them, to 2 significant figures, and state its type. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 4.0 nC = 4.0 × 10−9 C; 6.0 nC = 6.0 × 10−9 C; 3.0 cm = 0.030 m
F: F = Q₁Q₂ ÷ (4πε₀r²)
I: F = (4.0 × 10−9 × 6.0 × 10−9) ÷ (4π × 8.85 × 10−12 × 0.030²)
F: F is already the subject
A: F = 2.4 × 10−17 ÷ 1.000… × 10−13 = 2.4 × 10−4 N (2 s.f.), attractive because the charges are opposite
Watch out: the minus sign only tells you the force is attractive. Use the size of each charge and state the direction in words.
Q3 (Core) Two point charges, each +2.0 µC, repel each other with a force of 0.90 N in air. Calculate their separation, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 2.0 µC = 2.0 × 10−6 C
F: F = Q₁Q₂ ÷ (4πε₀r²)
I: 0.90 = (2.0 × 10−6 × 2.0 × 10−6) ÷ (4π × 8.85 × 10−12 × r²)
F: r² = 4.0 × 10−12 ÷ (4π × 8.85 × 10−12 × 0.90) = 0.03996… m², so r = √0.03996…
A: r = 0.1999… m = 0.20 m (2 s.f.)
Watch out: after rearranging you have r², not r. Remember the final square root.
Q4 (Core) Two identical small spheres carry equal charges. When their centres are 15 cm apart in air, the force between them is 20 mN. Calculate the charge on each sphere, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 15 cm = 0.15 m; 20 mN = 0.020 N
F: F = Q₁Q₂ ÷ (4πε₀r²), with Q₁ = Q₂ = Q, so F = Q² ÷ (4πε₀r²)
I: 0.020 = Q² ÷ (4π × 8.85 × 10−12 × 0.15²)
F: Q² = 0.020 × 4π × 8.85 × 10−12 × 0.15² = 5.004… × 10−14 C², so Q = √(5.004… × 10−14)
A: Q = 2.237… × 10−7 C = 2.2 × 10−7 C (0.22 µC, 2 s.f.)
Q5 (Exam) In a hydrogen atom the electron and proton are separated by 5.3 × 10−11 m. Calculate (a) the electrostatic force between them and (b) the acceleration of the electron caused by this force, both to 2 significant figures. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – (a) force
F: F = Q₁Q₂ ÷ (4πε₀r²)
I: F = (1.60 × 10−19)² ÷ (4π × 8.85 × 10−12 × (5.3 × 10−11)²)
F: F = 2.56 × 10−38 ÷ 3.123… × 10−31
A: F = 8.19… × 10−8 N = 8.2 × 10−8 N (2 s.f.), attractive
Step 2 – (b) acceleration
a = F ÷ mₑ = 8.19… × 10−8 ÷ 9.11 × 10−31 = 8.99… × 1022 m s⁻²
A: 9.0 × 1022 m s⁻² (2 s.f.), towards the proton
Watch out: square the whole of (5.3 × 10−11), including the power of ten. Many calculator slips come from squaring only 5.3.
Q6 (Exam) Two point charges exert a force of 12 mN on each other. One charge is doubled and the separation is tripled. Calculate the new force, to 2 significant figures.
Show solution
F: F = Q₁Q₂ ÷ (4πε₀r²), so F ∝ Q₁Q₂ ÷ r²
I: new force = 12 × 2 ÷ 3² = 12 × 2 ÷ 9
F: new force = 24 ÷ 9 = 2.66… mN
A: 2.7 mN (2 s.f.)
Watch out: tripling r divides the force by 9, not 3. Inverse square means square the factor.
Q7 (Exam) A +25 nC point charge and a −40 nC point charge are moved from 10 cm apart to 6.0 cm apart in air. Calculate the force between them at each separation and the increase in the force, to 2 significant figures. (ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 25 nC = 2.5 × 10−8 C; 40 nC = 4.0 × 10−8 C; 10 cm = 0.10 m; 6.0 cm = 0.060 m
F: F = Q₁Q₂ ÷ (4πε₀r²), with Q₁Q₂ = 2.5 × 10−8 × 4.0 × 10−8 = 1.0 × 10−15 C²
I: at 0.10 m: F = 1.0 × 10−15 ÷ (4π × 8.85 × 10−12 × 0.10²) = 8.99… × 10−4 N
I: at 0.060 m: F = 1.0 × 10−15 ÷ (4π × 8.85 × 10−12 × 0.060²) = 2.49… × 10−3 N
F: increase = 2.49… × 10−3 − 8.99… × 10−4
A: forces 9.0 × 10−4 N and 2.5 × 10−3 N (both attractive); increase = 1.59… × 10−3 N = 1.6 × 10−3 N (2 s.f.)
Check with the ratio: (0.10 ÷ 0.060)² = 2.78, and 8.99… × 10−4 × 2.78 = 2.5 × 10−3 N
Q8 (Exam) A small sphere carrying +20 nC is fixed on an insulating stand. A second small sphere of mass 0.50 g carrying +50 nC is free to move vertically directly above it. Calculate the separation of their centres when the second sphere is held at rest by the electrostatic force alone, to 2 significant figures. Treat the air as a vacuum. (ε₀ = 8.85 × 10−12 F m⁻¹, g = 9.81 N kg⁻¹)
Show solution
Convert first: 0.50 g = 5.0 × 10−4 kg; 20 nC = 2.0 × 10−8 C; 50 nC = 5.0 × 10−8 C
Step 1 – weight to balance
W = mg = 5.0 × 10−4 × 9.81 = 4.905 × 10−3 N, so the upward electrostatic force must also be 4.905 × 10−3 N
Step 2 – separation
F: F = Q₁Q₂ ÷ (4πε₀r²)
I: 4.905 × 10−3 = (2.0 × 10−8 × 5.0 × 10−8) ÷ (4π × 8.85 × 10−12 × r²)
F: r² = 1.0 × 10−15 ÷ (4π × 8.85 × 10−12 × 4.905 × 10−3) = 1.833… × 10−3 m²
A: r = 0.0428… m = 0.043 m (4.3 cm, 2 s.f.)
Watch out: grams must become kilograms before you find the weight; 0.50 g is 5.0 × 10−4 kg, not 0.50 kg.
Q9 (Stretch) For the electron and proton in a hydrogen atom, separated by 5.3 × 10−11 m, calculate the ratio of the electrostatic force to the gravitational force between them, to 2 significant figures. Explain why the ratio does not depend on the separation. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg, mₚ = 1.67 × 10−27 kg, G = 6.67 × 10−11 N m² kg⁻², ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – electrostatic force
FE = e² ÷ (4πε₀r²) = 8.19… × 10−8 N (as in Q5)
Step 2 – gravitational force
F: FG = Gmₑmₚ ÷ r²
I: FG = 6.67 × 10−11 × 9.11 × 10−31 × 1.67 × 10−27 ÷ (5.3 × 10−11)²
A: FG = 3.61… × 10−47 N
Step 3 – ratio
FE ÷ FG = 8.19… × 10−8 ÷ 3.61… × 10−47 = 2.26… × 1039
A: 2.3 × 1039 (2 s.f., no unit). Both forces follow an inverse square law, so the ratio is e² ÷ (4πε₀Gmₑmₚ) and the r² cancels.
Watch out: this is why gravity is ignored when dealing with forces between charged particles; the electrostatic force is about 1039 times bigger.
Q10 (Stretch) Two identical small conducting spheres, each of mass 0.20 g, hang from the same point on insulating threads 50 cm long. They are given equal charges and come to rest with each thread at 10° to the vertical. Calculate the charge on each sphere, to 2 significant figures. Treat the air as a vacuum. (ε₀ = 8.85 × 10−12 F m⁻¹, g = 9.81 N kg⁻¹)
Show solution
Convert first: 0.20 g = 2.0 × 10−4 kg; 50 cm = 0.50 m
Step 1 – separation of the centres
r = 2 × 0.50 × sin 10° = 0.1736… m
Step 2 – electrostatic force from equilibrium
vertically T cos θ = mg; horizontally T sin θ = F, so F = mg tan θ
F = 2.0 × 10−4 × 9.81 × tan 10° = 1.962 × 10−3 × 0.1763… = 3.459… × 10−4 N
Step 3 – charge
F: F = Q² ÷ (4πε₀r²)
I: 3.459… × 10−4 = Q² ÷ (4π × 8.85 × 10−12 × 0.1736…²)
F: Q² = 3.459… × 10−4 × 4π × 8.85 × 10−12 × 0.1736…² = 1.160… × 10−15 C²
A: Q = 3.40… × 10−8 C = 3.4 × 10−8 C (34 nC, 2 s.f.)
Watch out: r is the distance between the two spheres (2L sin θ), not the thread length and not L sin θ.
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