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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.8.1 (Coulomb’s law) and 3.8.2 (Electric field strength) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. Section 3.8 is International A-level only. Use the study guide and the revision notes. The course hub and printable checklist show where this unit sits. Try a free diagnostic first if you are unsure.

Data: G = 6.67 × 10⁻¹¹ N m² kg⁻²; g = 9.81 m s⁻²; ε₀ = 8.85 × 10⁻¹² F m⁻¹; e = 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg; proton mass 1.67 × 10⁻²⁷ kg. Give two or three significant figures.

Questions

1. State Coulomb’s law for point charges, and explain why the same equation can be used for charges in air. [3]

2. Two small spheres carry charges of +2.4 μC and −0.90 μC. Their centres are 0.35 m apart in air.

(a) Calculate the magnitude of the force on each sphere and state whether it is attractive or repulsive. [3] (b) The separation is doubled. State the new force. [1]

3. A conducting sphere of radius 3.0 cm carries a charge of +40 nC. A small sphere carrying +1.5 nC is held 5.0 cm from the surface of the large sphere. Calculate the force on the small sphere and state its direction. [4]

4. An electron and a second electron are 1.0 nm apart.

(a) Calculate the electrostatic force between them. [2] (b) Calculate the gravitational attraction between the two electrons. [2] (c) Calculate the ratio of the electrostatic force to the gravitational force, and state how this ratio changes if the separation is doubled. [2]

5. (a) Describe the electric field lines around an isolated negative point charge. [2] (b) Describe the field lines between two oppositely charged parallel plates, and say what they show about field strength. [2]

6. (a) Define electric field strength. [1] (b) A test charge of +3.5 nC feels a force of 2.8 × 10⁻⁵ N at a point in a field. Calculate the field strength at that point. [2] (c) State the direction of the force on a −3.5 nC charge placed at the same point, relative to the field. [1]

7. A small sphere carries +12 nC.

(a) Calculate the electric field strength at 0.25 m from its centre. [2] (b) Calculate the field strength at 0.50 m from its centre. [1] (c) Calculate the charge on a sphere that would produce a field strength of 2.0 × 10⁴ N C⁻¹ at 0.50 m from its centre. [2]

8. Two parallel plates are separated by a distance d and have a potential difference ΔV. Starting from the work done in moving a charge Q from one plate to the other, show that the field strength between the plates is E = ΔV/d. [3]

9. Parallel plates 1.8 cm apart have a p.d. of 540 V across them. An electron starts from rest next to the negative plate and crosses to the positive plate.

(a) Calculate the field strength in the gap. [1] (b) Calculate the force on the electron. [1] (c) Calculate the acceleration of the electron. [1] (d) Calculate the speed of the electron when it reaches the positive plate. [3]

10. A helium nucleus (mass 6.64 × 10⁻²⁷ kg, charge +2e) enters the region between two horizontal parallel plates at 4.0 × 10⁵ m s⁻¹, parallel to the plates. The plates are 6.0 cm long and 2.0 cm apart, with a p.d. of 150 V.

(a) Calculate the electric field strength in the gap. [1] (b) Calculate the force on the nucleus. [1] (c) Calculate the acceleration of the nucleus across the plates. [1] (d) Calculate the time the nucleus spends between the plates. [1] (e) Calculate the sideways displacement of the nucleus when it leaves the plates. [2] (f) Describe the shape of the path between the plates and explain why it has this shape. [2] (g) A proton enters in the same way at the same speed. Calculate its sideways displacement on leaving. [1]

11. A tiny oil droplet of mass 2.4 × 10⁻¹⁵ kg is held stationary between horizontal plates 1.2 cm apart with a p.d. of 350 V. The droplet is negatively charged.

(a) Calculate the field strength in the gap. [1] (b) Calculate the charge on the droplet. [2] (c) Calculate how many electrons’ worth of charge this is. [1] (d) State which plate is positive. [1] (e) The p.d. is halved. Explain why the droplet now accelerates, and calculate the size and direction of its acceleration. [3]

Answers

1. The force varies directly with the product of the two charges [1] and inversely with the square of the distance between them [1]. Air changes the force by a negligible amount, so it can be treated as a vacuum and ε₀ used [1]. [3] Examiner insight: “Proportional to the charges and the distance” misses both the product and the inverse square.

2. (a) F = Q₁Q₂/(4πε₀r²) = 2.4 × 10⁻⁶ × 0.90 × 10⁻⁶ / (4π × 8.85 × 10⁻¹² × 0.35²) [1] = 0.16 N [1]. The charges are unlike, so the force is attractive [1]. [3] (b) r doubles, so F falls to a quarter: 0.1585 / 4 = 0.040 N [1] [1] Examiner insight: Convert μC to C before substituting. Say “attractive” or “repulsive”, not just a number.

3. The charge acts as if at the centre, so r = 3.0 + 5.0 = 8.0 cm = 0.080 m [1]. F = 40 × 10⁻⁹ × 1.5 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.080²) [1] = 8.4 × 10⁻⁵ N [1], repulsive, so directed away from the large sphere (like charges) [1]. [4] Examiner insight: Using 0.050 m gives 2.2 × 10⁻⁴ N. Show the line r = 0.080 m so your reasoning is visible.

4. (a) Fₑ = e²/(4πε₀r²) = (1.60 × 10⁻¹⁹)² / (4π × 8.85 × 10⁻¹² × (1.0 × 10⁻⁹)²) [1] = 2.3 × 10⁻¹⁰ N [1] [2] (b) F_g = Gm²/r² = 6.67 × 10⁻¹¹ × (9.11 × 10⁻³¹)² / (1.0 × 10⁻⁹)² [1] = 5.5 × 10⁻⁵³ N [1] [2] (c) Ratio = 2.30 × 10⁻¹⁰ / 5.54 × 10⁻⁵³ = 4.2 × 10⁴² [1]. It is unchanged, because both forces are proportional to 1/r² [1]. [2] Examiner insight: A comparison needs the ratio calculated, not just two numbers. Saying that r cancels shows why the ratio is fixed.

5. (a) Straight lines spreading out from the charge [1], with the arrows pointing in towards it [1]. [2] (b) The lines are parallel, equally spaced and at right angles to the plates, pointing from the positive plate to the negative plate [1]. Equal spacing shows the field strength is the same everywhere between the plates (uniform) [1]. [2] Examiner insight: Include arrowheads, and link spacing to field strength in words.

6. (a) The force per unit positive charge [1] at a point. [1] (b) E = F/Q = 2.8 × 10⁻⁵ / 3.5 × 10⁻⁹ [1] = 8.0 × 10³ N C⁻¹ [1] [2] (c) Opposite to the field direction [1]. [1] Examiner insight: “Per unit charge” alone is incomplete: say positive charge. The unit N C⁻¹ should be stated with the answer.

7. (a) E = Q/(4πε₀r²) = 12 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.25²) [1] = 1.7 × 10³ N C⁻¹ [1] [2] (b) r doubles, so E falls to a quarter: 4.3 × 10² N C⁻¹ [1] [1] (c) Q = E × 4πε₀r² = 2.0 × 10⁴ × 4π × 8.85 × 10⁻¹² × 0.50² [1] = 5.6 × 10⁻⁷ C [1] [2] Examiner insight: Rearrange the equation before substituting. In part (b), a clearly stated ratio argument is quicker than recalculating.

8. Work done moving charge Q across the gap against force F is W = Fd [1]. The work done per unit charge is the p.d., so W = QΔV [1]. Equating and using F = QE: QEd = QΔV, so cancelling Q gives E = ΔV/d [1]. [3] Examiner insight: A “show that” needs every step. Starting from E = V/d proves nothing.

9. (a) E = V/d = 540 / 0.018 = 3.0 × 10⁴ V m⁻¹ [1] [1] (b) F = eE = 1.60 × 10⁻¹⁹ × 3.0 × 10⁴ = 4.8 × 10⁻¹⁵ N [1] [1] (c) a = F/m = 4.8 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 5.3 × 10¹⁵ m s⁻² [1] [1] (d) v² = 2ad [1] = 2 × 5.27 × 10¹⁵ × 0.018 [1], so v = 1.4 × 10⁷ m s⁻¹ [1] [3] Examiner insight: Carry unrounded values forward. The energy route, eV = ½mv², gives the same speed.

10. (a) E = 150 / 0.020 = 7.5 × 10³ V m⁻¹ [1] [1] (b) F = QE = 2 × 1.60 × 10⁻¹⁹ × 7.5 × 10³ = 2.4 × 10⁻¹⁵ N [1] [1] (c) a = F/m = 2.4 × 10⁻¹⁵ / 6.64 × 10⁻²⁷ = 3.6 × 10¹¹ m s⁻² [1] [1] (d) t = L/v = 0.060 / 4.0 × 10⁵ = 1.5 × 10⁻⁷ s [1] [1] (e) y = ½at² [1] = ½ × 3.61 × 10¹¹ × (1.5 × 10⁻⁷)² = 4.1 mm [1] [2] (f) A parabola [1]. The speed along the plates is constant, and there is a constant acceleration at right angles to it [1]. [2] (g) The proton has half the charge and a quarter of the mass, so a is twice as large. The time is the same, so y = 8.1 mm [1]. [1] Examiner insight: Use the charge 2e in part (b). In part (f), “constant force at right angles to a constant velocity” explains the parabola; “it curves” does not.

11. (a) E = 350 / 0.012 = 2.9 × 10⁴ V m⁻¹ [1] [1] (b) QE = mg [1], so Q = 2.4 × 10⁻¹⁵ × 9.81 / 2.92 × 10⁴ = 8.1 × 10⁻¹⁹ C [1] [2] (c) 8.07 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 5.04, so 5 electrons [1] [1] (d) The force on the droplet is upwards, so the field points downwards and the upper plate is positive [1] [1] (e) Halving the p.d. halves E, so the upward force falls to mg/2 [1]. The net force is mg/2 downwards [1], so a = g/2 = 4.9 m s⁻², downwards [1]. [3] Examiner insight: State the equilibrium condition (electric force equals weight) before substituting. In part (e), give a direction.

Where marks are usually lost

• Taking r to the surface of a sphere rather than to its centre.
• Forgetting to convert nC, μC and cm to C and m.
• Giving no direction when the question asks for one.
• Using the charge e instead of 2e for a helium nucleus.
• Writing “it curves” instead of “parabola”, with a reason.
• Starting a “show that” from the result.
• Defining E as “force per charge” without saying positive charge.

Next steps

• Re-read the revision notes and study guide.
• Check the course hub and tick off the printable checklist.
• Take one of the free 10-minute diagnostics to see what to revise next.
• Book a free trial class if you want help with any of these questions.

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).