OCR A Level Physics A: Electric fields, Coulomb's law, potential and uniform fields (H556) – 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 section 6.2 (Electric fields) of OCR A Level Physics A (H556), Version 3.0 (March 2026), first assessment 2017. Revise with the study guide and notes. Data: g = 9.81 m s⁻²; G = 6.67 × 10⁻¹¹ N m² kg⁻²; ε₀ = 8.85 × 10⁻¹² F m⁻¹; elementary charge 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg; proton mass 1.67 × 10⁻²⁷ kg.

Questions

1. (a) Define electric field strength. [1] (b) Describe the field lines around an isolated, positively charged sphere and say what the spacing of the lines shows. [2]

2. Two tiny charged beads carry +2.0 µC and −0.80 µC, with 0.25 m between their centres. (a) Find the magnitude and direction of the force each exerts on the other. [3] (b) Find the force when the separation is doubled. [1]

3. A uniformly charged sphere of radius 5.0 cm carries +18 nC. (a) Calculate the electric field strength 0.15 m from its centre, naming the model you use. [3] (b) Find the force on a −3.0 nC charge placed there, with its direction. [2]

4. A charge of +6.0 nC is at A and a charge of −8.0 nC is at B. A point P is 0.30 m from A and 0.40 m from B, and the lines PA and PB are at right angles. Find the magnitude of the resultant field at P and its direction relative to the line from A to P. [6]

5. In a hydrogen atom an electron orbits a proton at a distance of 5.3 × 10⁻¹¹ m. (a) Calculate the electric force between them. [2] (b) Calculate the gravitational force between them. [2] (c) Find the ratio of the two forces. [1]

6. Two horizontal plates 2.5 cm apart have a p.d. of 750 V. An oil droplet carrying the charge of three excess electrons is held stationary between them. Calculate the mass of the droplet and state which plate is positive. [5]

7. A capacitor has square plates of side 8.0 cm, 0.30 mm apart. (a) Calculate its capacitance with air between the plates. [2] (b) A material of relative permittivity 3.2 fills the gap. Calculate its permittivity and the new capacitance. [2] (c) Calculate the charge stored at 12 V with the material in place. [2]

8. A proton enters midway between two horizontal plates, moving parallel to them at 1.0 × 10⁶ m s⁻¹. The plates are 5.0 cm long and 2.0 cm apart, with a p.d. of 600 V. (a) Calculate the field strength between the plates. [1] (b) Calculate the proton’s acceleration. [2] (c) Calculate the time it spends between the plates. [1] (d) Calculate its sideways displacement when it leaves. [2] (e) State whether it leaves the plates and find the angle of its path to the original direction. [2]

9. (a) Derive an expression for the capacitance of an isolated sphere of radius R in terms of R and ε₀. [3] (b) Treating the Earth as an isolated sphere of radius 6.37 × 10⁶ m, calculate its capacitance. [2]

10. A point charge of −12 nC is fixed in air. (a) Calculate the electric potential at 0.50 m and at 0.20 m from it. [2] (b) A +2.0 nC charge is moved from 0.50 m to 0.20 m. Calculate the change in its potential energy and say what the sign shows. [3] (c) State the potential energy of this charge at 0.20 m. [1]

11. A fixed sphere carries +5.0 µC. A +1.0 nC charge is pushed towards it from 0.80 m to 0.20 m (distances from the centre). (a) Calculate the force on the charge at each distance. [2] (b) Use a single trapezium to estimate the area under the force-distance graph. [2] (c) Calculate the work done exactly using potentials. [2] (d) Explain why the trapezium estimate differs from the exact value. [1]

12. An electron travelling at 4.0 × 10⁶ m s⁻¹ is fired from far away straight at a fixed sphere of radius 2.0 cm carrying −2.0 nC. (a) Calculate the electron’s initial kinetic energy. [1] (b) Calculate its distance from the sphere’s centre at closest approach. [3] (c) Calculate its speed when 1.0 m from the centre. [3] (d) Explain why the point-charge model is valid here. [2]

Answers

1. (a) Force per unit positive charge, E = F/q [1]. (b) Straight lines directed radially outwards, equally spaced [1]. Closer lines show a stronger field, so it weakens with distance [1]. Examiner insight: A definition needs “per unit positive charge”, not just “force on a charge”.

2. (a) F = Qq/(4πε₀r²) = (2.0 × 10⁻⁶ × 0.80 × 10⁻⁶) / (4π × 8.85 × 10⁻¹² × 0.25²) [1] = 0.23 N [1], attractive because the charges are unlike [1]. (b) The force falls to one quarter: 0.058 N [1]. Examiner insight: Give the direction in words.

3. (a) A uniformly charged sphere is modelled as a point charge at its centre [1]. E = Q/(4πε₀r²) = 18 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.15²) [1] = 7.2 × 10³ N C⁻¹ [1]. (b) F = qE = 3.0 × 10⁻⁹ × 7.2 × 10³ = 2.2 × 10⁻⁵ N [1], towards the sphere because unlike charges attract [1]. Examiner insight: Name the model, and measure from the centre, not the surface.

4. E from A = 6.0 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.30²) = 599 N C⁻¹, directed away from A [1]. E from B = 8.0 × 10⁻⁹ / (4π × 8.85 × 10⁻¹² × 0.40²) = 450 N C⁻¹ [1], directed towards B [1]. The fields are at right angles, so add as vectors [1]: E = √(599² + 450²) = 7.5 × 10² N C⁻¹ [1] at tan⁻¹(450/599) = 37° to the line from A to P, turned towards B [1]. Examiner insight: Show each field separately before combining; adding 599 and 450 as plain numbers is a common slip.

5. (a) F = e²/(4πε₀r²) = (1.60 × 10⁻¹⁹)² / (4π × 8.85 × 10⁻¹² × (5.3 × 10⁻¹¹)²) [1] = 8.2 × 10⁻⁸ N [1]. (b) F = Gm_em_p/r² = 6.67 × 10⁻¹¹ × 9.11 × 10⁻³¹ × 1.67 × 10⁻²⁷ / (5.3 × 10⁻¹¹)² [1] = 3.6 × 10⁻⁴⁷ N [1]. (c) Ratio = 8.2 × 10⁻⁸ / 3.6 × 10⁻⁴⁷ = 2.3 × 10³⁹ [1]. Examiner insight: Check that both forces use the same r.

6. E = V/d = 750/0.025 = 3.0 × 10⁴ V m⁻¹ [1]. Charge q = 3 × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁹ C. In equilibrium qE = mg [1]. m = qE/g [1] = 4.8 × 10⁻¹⁹ × 3.0 × 10⁴ / 9.81 = 1.5 × 10⁻¹⁵ kg [1]. The charge is negative, so the force is upwards, E points downwards and the upper plate is positive [1]. Examiner insight: Reason from the sign: the force on a negative charge is opposite to E.

7. (a) C = ε₀A/d = 8.85 × 10⁻¹² × 0.080² / 3.0 × 10⁻⁴ [1] = 1.9 × 10⁻¹⁰ F [1]. (b) ε = εᵣε₀ = 3.2 × 8.85 × 10⁻¹² = 2.8 × 10⁻¹¹ F m⁻¹ [1]; C = εA/d = 6.0 × 10⁻¹⁰ F [1]. (c) Q = VC = 12 × 6.04 × 10⁻¹⁰ [1] = 7.2 × 10⁻⁹ C [1]. Examiner insight: Carry the unrounded capacitance into part (c).

8. (a) E = 600/0.020 = 3.0 × 10⁴ V m⁻¹ [1]. (b) a = eE/m [1] = 1.60 × 10⁻¹⁹ × 3.0 × 10⁴ / 1.67 × 10⁻²⁷ = 2.9 × 10¹² m s⁻² [1]. (c) t = 0.050 / 1.0 × 10⁶ = 5.0 × 10⁻⁸ s [1]. (d) y = ½at² [1] = 0.5 × 2.87 × 10¹² × (5.0 × 10⁻⁸)² = 3.6 × 10⁻³ m [1]. (e) 3.6 mm is less than 10 mm (half the gap), so it leaves [1]. tan θ = at/v = 1.44 × 10⁵ / 1.0 × 10⁶, so θ = 8.2° [1]. Examiner insight: Find t from the horizontal motion, then compare y with half the gap.

9. (a) Potential at the surface (point-charge model): V = Q/(4πε₀R) [1]. C = Q/V [1] = Q ÷ Q/(4πε₀R) = 4πε₀R [1]. (b) C = 4π × 8.85 × 10⁻¹² × 6.37 × 10⁶ [1] = 7.1 × 10⁻⁴ F [1]. Examiner insight: A derivation shows the starting equations, then the cancellation.

10. (a) V = Q/(4πε₀r): at 0.50 m, V = −216 V [1]; at 0.20 m, V = −540 V [1]. (b) ΔV = −540 − (−216) = −324 V [1]. ΔE_p = qΔV = 2.0 × 10⁻⁹ × (−324) = −6.5 × 10⁻⁷ J [1]. The negative sign is a loss of potential energy: the field does the work as unlike charges move closer [1]. (c) E_p = Vq = −540 × 2.0 × 10⁻⁹ = −1.1 × 10⁻⁶ J [1]. Examiner insight: Keep the negative sign on Q in V.

11. (a) F = Qq/(4πε₀r²): at 0.80 m, 7.0 × 10⁻⁵ N [1]; at 0.20 m, 1.1 × 10⁻³ N [1]. (b) Area ≈ ½(7.0 × 10⁻⁵ + 1.12 × 10⁻³) × 0.60 [1] = 3.6 × 10⁻⁴ J [1]. (c) W = q(V₂ − V₁) with V = Q/(4πε₀r): V₁ = 56.2 kV, V₂ = 225 kV [1]; W = 1.0 × 10⁻⁹ × 168.7 × 10³ = 1.7 × 10⁻⁴ J [1]. (d) The curve is concave up, so the straight top of the trapezium lies above it, giving an over-estimate [1]. Examiner insight: Name the curve’s shape and the direction of the error, not just “it is an estimate”.

12. (a) Ek = ½mv² = 0.5 × 9.11 × 10⁻³¹ × (4.0 × 10⁶)² = 7.3 × 10⁻¹⁸ J [1]. (b) At closest approach all kinetic energy is potential energy [1]: 7.29 × 10⁻¹⁸ = Qe/(4πε₀r) [1]. r = 2.0 × 10⁻⁹ × 1.60 × 10⁻¹⁹ / (4π × 8.85 × 10⁻¹² × 7.29 × 10⁻¹⁸) = 0.39 m [1]. (c) At 1.0 m, E_p = 2.88 × 10⁻¹⁸ J [1]. Ek = 7.29 × 10⁻¹⁸ − 2.88 × 10⁻¹⁸ = 4.41 × 10⁻¹⁸ J [1]. v = √(2Ek/m) = 3.1 × 10⁶ m s⁻¹ [1]. (d) The sphere is uniformly charged, so outside it the field is that of a point charge at its centre [1]. The electron never gets closer than 0.39 m, well outside the 2.0 cm radius [1]. Examiner insight: Energy conservation is the efficient route here.

Where marks are usually lost

• Measuring r from the surface of a sphere instead of from its centre.
• Adding field magnitudes from two charges when they are not along the same line.
• Dropping the sign of Q in V = Q/(4πε₀r) and getting the wrong sign for the energy change.
• Using E = V/d for a point charge, or the inverse square law between plates.
• Using ε₀ where the material’s ε is needed.

Next steps

• Revision notes
• Study guide
• OCR A Level Physics hub and checklist
• All free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

OCR Level 3 Advanced GCE in Physics A (H556), Version 3.0 (March 2026), published by OCR: section 6.2 (Electric fields), learning outcomes 6.2.1(a)-(d), 6.2.2(a)-(d), 6.2.3(a)-(c) and 6.2.4(a)-(e).