OxfordAQA International AS and A-level Physics: Coulomb's law and electric field strength (9630) – Revision Notes

These notes follow the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. They cover sections 3.8.1 (Coulomb’s law) and 3.8.2 (Electric field strength), which are marked International A-level only. For full explanations, read the study guide, then test yourself with the practice questions. The course hub and printable checklist show where this unit sits. To find your weak spots first, take one of the free diagnostics.

Data used below: ε₀ = 8.85 × 10⁻¹² F m⁻¹; e = 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg; g = 9.81 N kg⁻¹.

Definitions and statements to learn

• Coulomb’s law (3.8.1): the force varies with the product of the two charges and falls with the square of the distance between them. It acts along the line joining them: like charges repel, unlike charges attract.
• Air as a vacuum: air changes the force by a negligible amount, so use ε₀ for charges in air.
• Charged sphere: the charge may be considered to be at the centre, so measure r from the centre.
• Electric field: a region in which a charge experiences a force.
• Electric field strength E: the force per unit positive charge at a point. E = F/Q. It is a vector, pointing the way a positive charge would be pushed.

Equations and units

Quantity	Equation	Unit
Force between point charges	F = Q₁Q₂ / (4πε₀r²)	N
Field strength (definition)	E = F/Q	N C⁻¹ (same as V m⁻¹)
Uniform field	E = V/d	V m⁻¹
Work done in a uniform field	Fd = QΔV	J
Radial field	E = Q / (4πε₀r²)	N C⁻¹
Force on a charge in a field	F = QE	N
Acceleration of a charged particle	a = QE/m	m s⁻²

The factor 1/(4πε₀) is about 8.99 × 10⁹ N m² C⁻². Convert nC (10⁻⁹), μC (10⁻⁶) and cm (10⁻²) before substituting.

Must-know distinctions

Compare	Key difference
Uniform and radial field	Uniform: E = V/d, parallel equally spaced lines, same everywhere. Radial: E = Q/(4πε₀r²), lines spread out, E falls as 1/r²
Force F and field strength E	F is on a particular charge, E is a property of the point. F = QE
Electric and gravitational forces	Electric can attract or repel, gravity only attracts. Both obey an inverse-square law
Direction for positive and negative charges	A positive charge is pushed along E. A negative charge is pushed against E
Source charge and test charge	In E = Q/(4πε₀r²), Q is the charge that makes the field, not the charge feeling the force

Method boxes

Coulomb’s law calculation

1. Convert all charges to coulombs and r to metres.
2. Use r from centre to centre (for spheres).
3. Substitute into F = Q₁Q₂/(4πε₀r²) using magnitudes.
4. Give direction from the signs: unlike attract, like repel.

Scaling (ratio) problems

1. Write F ∝ Q₁Q₂/r².
2. Multiply by the change in each charge.
3. Divide by the square of the change in r.

Charged particle entering a uniform field at right angles

1. E = V/d.
2. F = QE and a = F/m.
3. t = L/v (the speed along the plates is constant).
4. y = ½at², v_y = at.
5. Angle of leaving = tan⁻¹(v_y/v). Check y is less than half the gap.

Field line sketches

• Positive point charge: radial lines outwards.
• Negative point charge: radial lines inwards.
• Parallel plates: equally spaced straight lines at right angles to the plates, from positive to negative, with slight outward bending only at the edges.
• Lines never cross. Closer lines mean a stronger field. The arrow shows how a positive test charge would be pushed.

Derivation to reproduce

1. Work done moving charge Q across a uniform field: W = Fd, with F = QE.
2. Work done is also W = QΔV.
3. So Fd = QΔV.
4. Substituting F = QE: QEd = QΔV, and Q cancels.
5. E = ΔV/d.

Gravitational and electrostatic forces compared

For an electron and a proton, the ratio of electrostatic to gravitational force is

Fₑ / F_g = e² / (4πε₀ G mₑ mₚ) = 2.3 × 10³⁹

The separation cancels, so the ratio is the same at every distance. Gravity is negligible between subatomic particles.

Worked reminders

Coulomb force. Charges +1.5 μC and +2.5 μC are 0.20 m apart in air. F = (1.5 × 10⁻⁶ × 2.5 × 10⁻⁶) / (4π × 8.85 × 10⁻¹² × 0.20²) = 3.75 × 10⁻¹² / 4.45 × 10⁻¹² = 0.84 N, repulsive.

Radial field and force. A sphere carries +24 nC. At 0.40 m from its centre, E = 24 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.40²) = 1.3 × 10³ N C⁻¹, directed away from the sphere. A −2.0 nC charge there feels F = QE = 2.7 × 10⁻⁶ N, towards the sphere. At 0.10 m the field would be (0.40/0.10)² = 16 times greater.

Deflection. A singly charged ion (mass 3.2 × 10⁻²⁶ kg) enters plates 4.0 cm long and 1.0 cm apart at 2.5 × 10⁵ m s⁻¹, at right angles to the field. The p.d. is 200 V.

E = 200/0.010 = 2.0 × 10⁴ V m⁻¹. F = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N. a = F/m = 1.0 × 10¹¹ m s⁻². t = 0.040 / 2.5 × 10⁵ = 1.6 × 10⁻⁷ s. y = ½at² = 1.3 mm. v_y = at = 1.6 × 10⁴ m s⁻¹, so the angle is tan⁻¹(1.6 × 10⁴ / 2.5 × 10⁵) = 3.7°.

Wording that earns credit

• “The force on a positive charge is along the field direction, so the electron’s force is opposite.”
• “The horizontal speed is constant because there is no force in that direction; the vertical acceleration is constant, so the path is a parabola.”
• “The lines are equally spaced and parallel, so the field strength is the same everywhere between the plates.”
• “The ratio does not depend on r because both forces are inversely proportional to r².”

Quick self-test

1. State Coulomb’s law in words.
2. Find the force between charges of +1.5 μC and +2.5 μC placed 0.10 m apart in air.
3. The separation of two charges is halved. By what factor does the force change?
4. Define electric field strength and give two equivalent units.
5. Find E at a distance of 0.15 m from the centre of a sphere carrying +9.0 nC.
6. Two parallel plates 1.2 cm apart have a p.d. of 900 V. Find E between them.
7. An electron is placed in a uniform field of 1.5 × 10⁴ V m⁻¹. Find its acceleration.
8. A point is moved from r to 4r from a point charge. By what factor does E change?
9. Why is the weight of the electron in question 7 ignored?
10. What is the shape of the path of a charged particle between the plates when it enters at right angles to the field?

Answers

1. The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation. It acts along the line joining them.
2. F = 3.75 × 10⁻¹² / (4π × 8.85 × 10⁻¹² × 0.10²) = 3.75 × 10⁻¹² / 1.112 × 10⁻¹² = 3.4 N.
3. F ∝ 1/r², so halving r multiplies F by 4.
4. Force per unit positive charge, E = F/Q. Units: N C⁻¹ and V m⁻¹.
5. E = 9.0 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.15²) = 3.6 × 10³ N C⁻¹.
6. E = 900/0.012 = 7.5 × 10⁴ V m⁻¹.
7. F = eE = 1.60 × 10⁻¹⁹ × 1.5 × 10⁴ = 2.4 × 10⁻¹⁵ N. a = F/m = 2.4 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 2.6 × 10¹⁵ m s⁻², opposite to the field.
8. E ∝ 1/r², so E falls to 1/16.
9. Its weight is mg = 8.9 × 10⁻³⁰ N, which is far smaller than the electric force of 2.4 × 10⁻¹⁵ N.
10. A parabola, then a straight line after it leaves the field.

Where marks are usually lost

• Putting r to a sphere’s surface rather than its centre.
• Forgetting to square r, or squaring it twice.
• Not converting nC, μC and cm.
• Giving a direction that ignores the sign of the charge.
• Using E = V/d for a radial field.
• Writing the derivation without stating the two expressions for work done.
• Describing the path as a “curve” without saying it is parabolic and why.
• Leaving out units, or quoting a force in N C⁻¹.
• Not checking whether the particle hits a plate.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.8.1 Coulomb’s law and 3.8.2 Electric field strength (International A-level only).