OxfordAQA International AS and A-level Physics: Electric potential and capacitors (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 follow the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, exams from May/June 2020 onwards. They cover sections 3.8.3 (Electric potential) and 3.8.4 (Capacitors), which are International A-level only. Learn the content from the study guide and the revision notes. See also the course hub, the checklist and the free diagnostics.

Data: ε₀ = 8.85 × 10⁻¹² F m⁻¹, so 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²; e = 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg. Treat air as a vacuum and spheres as point charges.

Questions

1. Define absolute electric potential. State whether the potential near an isolated negative charge is positive or negative, and explain why. [3]

2. A small sphere carries a charge of +4.0 nC. (a) Calculate the potential at a point 0.025 m from its centre. [2] (b) Calculate the work done by an external agent in moving a +2.0 nC charge from infinity to that point. [2]

3. A charge of +6.0 nC and a charge of −3.0 nC are 0.20 m apart. (a) Calculate the potential at the midpoint of the line joining them. [2] (b) Find the point between the charges, measured from the +6.0 nC charge, where the potential is zero. [3]

4. An alpha particle (charge +2e, mass 6.64 × 10⁻²⁷ kg) starts from rest and is accelerated through a potential difference of 1.8 kV in a vacuum. Calculate its kinetic energy and its final speed. [4]

5. The potential V is measured at distances r from the centre of a small charged sphere.

r / cm	4.0	6.0	8.0	10.0	12.0
V / V	540	360	270	216	180

(a) Show that the data fit V ∝ 1/r. [2] (b) Determine the charge on the sphere. [2] (c) Use the slope of the line between r = 6.0 cm and r = 10.0 cm to estimate the field strength at 8.0 cm, and explain why this is larger than the value from E = Q/(4πε₀r²). [2]

6. At 0.20 m from a point charge the field strength is 9.0 × 10³ V m⁻¹. (a) Calculate the charge. [2] (b) Find the field strength at 0.50 m. [1] (c) Calculate the potential difference between 0.20 m and 0.50 m. [2] (d) A student estimates the same potential difference from the area of a trapezium under the E against r graph. Calculate the estimate and say whether it is too large or too small. [2]

7. A parallel plate capacitor has square plates of side 0.12 m, separated by 0.80 mm of air. (a) Calculate its capacitance. [2] (b) A material of relative permittivity 2.3 fills the gap. Calculate the new capacitance and the charge on one plate at 15 V. [3] (c) State what is meant by relative permittivity. [1]

8. Explain how a dielectric made of polar molecules increases the capacitance of a parallel plate capacitor. [4]

9. (a) Calculate the energy stored in a 2200 μF capacitor charged to 12 V. [2] (b) A 680 μF capacitor holds a charge of 1.5 mC. Calculate the pd across it and the energy stored. [2] (c) Describe how the energy stored is found from a graph of charge against pd. [1]

10. A 150 pF air-filled parallel plate capacitor is charged to 50 V and disconnected from the supply. A slab of relative permittivity 4.0, filling the gap, is then slid in. (a) Calculate the charge and the energy stored before the slab is inserted. [2] (b) Calculate the capacitance, pd and energy stored after insertion. [3] (c) Account for the change in stored energy. [1] (d) The experiment is repeated with the supply left connected. Calculate the charge and energy stored after insertion, and say where the extra energy comes from. [3]

11. Two parallel plates, each of area 0.015 m², are 6.0 mm apart in air with 900 V across them. (a) Calculate the capacitance. [2] (b) Calculate the charge on one plate and the energy stored. [3] (c) Calculate the field strength between the plates. [1] (d) An electron is released from rest at the negative plate. Calculate its kinetic energy and speed after it has moved 2.0 mm. [3]

Answers

1. Work done per unit positive charge [1] when a small test charge is moved in from infinity to the point [1]. Negative: the positive test charge is attracted, so the work done by the external agent bringing it from infinity (where V = 0) is negative [1]. [3] Examiner insight: Include “per unit charge”, “from infinity” and “small test charge”.

2. (a) V = Q/(4πε₀r) = 8.99 × 10⁹ × 4.0 × 10⁻⁹ / 0.025 [1] = 1.4 × 10³ V (1439 V) [1] (b) ΔW = QΔV = 2.0 × 10⁻⁹ × 1439 [1] = 2.9 × 10⁻⁶ J [1]. [4] Examiner insight: Carry the unrounded value from (a) into (b) and show the substitution.

3. (a) Each charge is 0.10 m away. V = 8.99 × 10⁹ × (6.0 − 3.0) × 10⁻⁹ / 0.10 [1] = 2.7 × 10² V [1] (b) Let x be the distance from the +6.0 nC charge: 6.0/x = 3.0/(0.20 − x) [1]. 6.0(0.20 − x) = 3.0x, so 1.2 = 9.0x [1], giving x = 0.13 m [1]. [5] Examiner insight: Show the two terms set equal, with the right distance for each charge.

4. ΔW = QΔV = 2 × 1.60 × 10⁻¹⁹ × 1.8 × 10³ [1] = 5.8 × 10⁻¹⁶ J [1]. ½mv² = 5.76 × 10⁻¹⁶ so v = √(2 × 5.76 × 10⁻¹⁶ / 6.64 × 10⁻²⁷) [1] = 4.2 × 10⁵ m s⁻¹ [1]. [4] Examiner insight: A charge of e instead of 2e halves the energy. State the charge you use.

5. (a) Product Vr: 540 × 0.040 = 21.6, 360 × 0.060 = 21.6, 270 × 0.080 = 21.6 (and the same for the others) [1]. Vr is constant, so V ∝ 1/r [1] (b) Vr = Q/(4πε₀), so Q = 21.6 / 8.99 × 10⁹ [1] = 2.4 × 10⁻⁹ C [1] (c) E ≈ ΔV/Δr = (360 − 216) / 0.040 = 3.6 × 10³ V m⁻¹ [1]. Formula: 8.99 × 10⁹ × 2.4 × 10⁻⁹ / 0.080² = 3.4 × 10³ V m⁻¹. The V–r curve is convex, so the chord slope is steeper than the slope at the midpoint [1]. [6] Examiner insight: For “show that”, work the product for several pairs of readings. In (c), say why the values differ.

6. (a) Q = E × 4πε₀r² = 9.0 × 10³ × 0.20² / 8.99 × 10⁹ [1] = 4.0 × 10⁻⁸ C [1] (b) E ∝ 1/r², so E = 9.0 × 10³ × (0.20/0.50)² = 1.44 × 10³ V m⁻¹ [1] (c) ΔV = (Q/4πε₀)(1/0.20 − 1/0.50) = 8.99 × 10⁹ × 4.0 × 10⁻⁸ × 3.0 [1] = 1.1 × 10³ V [1] (d) Area = ½(9.0 × 10³ + 1.44 × 10³) × 0.30 = 1.6 × 10³ V [1]. Too large, because the curve lies below the straight line joining its end points [1]. [7] Examiner insight: Link the curve’s shape to the direction of the error; “the graph is curved” alone is weaker.

7. (a) A = 0.12² = 0.0144 m² [1]. C = ε₀A/d = 8.85 × 10⁻¹² × 0.0144 / 0.80 × 10⁻³ = 1.6 × 10⁻¹⁰ F [1] (b) C = 2.3 × 1.593 × 10⁻¹⁰ = 3.7 × 10⁻¹⁰ F [1]. Q = CV = 3.66 × 10⁻¹⁰ × 15 [1] = 5.5 × 10⁻⁹ C [1] (c) The ratio of the material’s permittivity to ε₀, equal to the factor by which the capacitance rises when the material fills the gap [1]. [6] Examiner insight: Convert 0.80 mm to metres before substituting.

8. With no field the polar molecules point in random directions [1]. When the plates are charged, the field acts with opposite forces on the two charged ends, forming a couple that rotates each molecule towards alignment [1]. Layers of opposite charge appear on the dielectric surfaces, creating a field opposing the applied field [1]. The net field and so the pd fall for the same plate charge, and C = Q/V rises [1]. [4] Examiner insight: Keep the order: rotation, opposing field, smaller pd, larger C.

9. (a) E = ½CV² = ½ × 2200 × 10⁻⁶ × 12² [1] = 0.16 J (0.158 J) [1] (b) V = Q/C = 1.5 × 10⁻³ / 680 × 10⁻⁶ = 2.2 V [1]. E = ½QV = ½ × 1.5 × 10⁻³ × 2.21 = 1.7 × 10⁻³ J [1] (c) The energy is the area under the Q against V line (a triangle, ½QV) [1]. [5] Examiner insight: Choose the energy form that uses the data given.

10. (a) Q = CV = 150 × 10⁻¹² × 50 = 7.5 × 10⁻⁹ C [1]. E = ½CV² = ½ × 150 × 10⁻¹² × 50² = 1.9 × 10⁻⁷ J [1] (b) C = 4.0 × 150 = 600 pF [1]. V = Q/C = 7.5 × 10⁻⁹ / 600 × 10⁻¹² = 12.5 V [1]. E = Q²/2C = (7.5 × 10⁻⁹)² / (2 × 600 × 10⁻¹²) = 4.7 × 10⁻⁸ J [1] (c) The stored energy falls by a factor of 4.0 (1.4 × 10⁻⁷ J lost); the capacitor does work pulling the slab into the gap [1] (d) V stays 50 V, so Q = 600 × 10⁻¹² × 50 = 3.0 × 10⁻⁸ C [1]. E = ½ × 600 × 10⁻¹² × 50² = 7.5 × 10⁻⁷ J [1]. The supply moves the extra charge (2.25 × 10⁻⁸ C) onto the plates [1]. [9] Examiner insight: Say which quantity is held fixed: Q when isolated, V when connected.

11. (a) C = ε₀A/d = 8.85 × 10⁻¹² × 0.015 / 6.0 × 10⁻³ [1] = 2.2 × 10⁻¹¹ F [1] (b) Q = CV = 2.21 × 10⁻¹¹ × 900 = 2.0 × 10⁻⁸ C [1]. E = ½CV² = ½ × 2.21 × 10⁻¹¹ × 900² [1] = 9.0 × 10⁻⁶ J [1] (c) E = V/d = 900 / 6.0 × 10⁻³ = 1.5 × 10⁵ V m⁻¹ [1] (d) ΔV = Ex = 1.5 × 10⁵ × 2.0 × 10⁻³ = 300 V, so KE = eΔV = 4.8 × 10⁻¹⁷ J [1]. v = √(2 × 4.8 × 10⁻¹⁷ / 9.11 × 10⁻³¹) [1] = 1.0 × 10⁷ m s⁻¹ [1]. [9] Examiner insight: Linking (c) and (d) through ΔV = Ex avoids finding force and acceleration.

Where marks are usually lost

• Using e where the charge is 2e.
• Not converting mm, cm, pF, nC or μF to SI units.
• Adding potentials as vectors, or using the midpoint distance for both charges.
• Leaving “from infinity” or “per unit charge” out of the definition.
• Using r² in the potential formula.
• Writing capacitor energy as QV rather than ½QV.
• Not saying whether Q or V is held constant when a dielectric goes in.
• Describing dielectric action without the opposing field.

Next steps

• Re-read the revision notes and study guide.
• Check your progress on the course hub and the printable checklist.
• Take one of the free 10-minute diagnostics to find the next weak area.
• Book a free trial class if you want help with any of this.

Official syllabus

OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.8.3 and 3.8.4.