Edexcel IAL Physics: Electric and Magnetic Fields — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Electric and Magnetic Fields revision notes

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Section A

1. Distinguish between a radial and a uniform electric field. [2]

2. State the equation linking the charge stored on a capacitor to its capacitance and potential difference, and state how the charge remaining on a capacitor changes as it discharges through a fixed resistor. [2]

3. State Faraday’s law and Lenz’s law. [2]

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Section B

4. Two parallel plates 5.0 mm apart have a potential difference of 240 V across them.

(a) Calculate the electric field strength between them. [2]

(b) An electron enters the field perpendicular to the field lines. Describe and explain the shape of its path. [3]

(c) A charged particle enters a magnetic field perpendicular to the field lines instead. State the shape of the path and explain why it differs from (b). [3]

5. A 470 μF capacitor is charged to 12 V and then discharged through a 15 kΩ resistor.

(a) Calculate the charge stored initially. [2]

(b) Calculate the energy stored initially. [2]

(c) Calculate the time constant. [2]

(d) Calculate the charge remaining after 10 s. [3]

(e) Explain why the energy stored is ½QV rather than QV. [2]

6. A coil of 250 turns and area 1.8 × 10⁻³ m² is in a magnetic field of flux density 0.42 T. The field is reduced to zero in 0.15 s.

(a) Calculate the initial flux linkage. [2]

(b) Calculate the average induced e.m.f. [2]

(c) Explain, using Lenz’s law and conservation of energy, why the induced current opposes the change. [3]

7. A point charge of +3.0 μC and a point charge of −5.0 μC are placed 0.20 m apart in a vacuum.

(a) Calculate the magnitude of the electrostatic force between the two charges, and state whether it is attractive or repulsive. [2]

(b) Calculate the electric potential at the midpoint between the two charges. [3]

8. A capacitor discharges through a fixed resistor. The charge Q remaining on the capacitor is recorded at regular time intervals, and a graph of ln Q (y-axis) against t (x-axis) is plotted.

(a) Explain how the value of RC can be found from the gradient of this graph. [3]

(b) State what is meant by the time constant of a capacitor discharge circuit. [2]

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Answers

1. A radial field has field lines directed towards or away from a point charge, and the field strength obeys an inverse square law [1]. A uniform field has parallel, equally spaced field lines and constant field strength, as between charged parallel plates [1].

2. Q = CV [1]. As the capacitor discharges, the charge decreases exponentially with time, falling by the same fraction in equal time intervals [1].

3. Faraday: the induced e.m.f. is proportional to the rate of change of flux linkage [1]. Lenz: the induced current acts in the direction that opposes the change producing it [1].

4. (a) E = V ÷ d = 240 ÷ 5.0 × 10⁻³ [1] = 4.8 × 10⁴ V m⁻¹ [1].

(b) A parabola [1]. The electric force acts in a constant direction (perpendicular to the initial velocity) [1], giving constant acceleration in that direction while the perpendicular velocity is unchanged — exactly like projectile motion [1].

(c) A circle [1]. The magnetic force is always perpendicular to the velocity [1], so its direction changes continuously as the particle turns, acting as a centripetal force rather than a constant-direction force [1].

5. (a) Q = CV = 470 × 10⁻⁶ × 12 [1] = 5.64 × 10⁻³ C [1].

(b) E = ½CV² = 0.5 × 470 × 10⁻⁶ × 144 [1] = 0.0338 J [1].

(c) τ = RC = 15 × 10³ × 470 × 10⁻⁶ [1] = 7.05 s [1].

(d) Q = Q₀e^(−t/RC) = 5.64 × 10⁻³ × e^(−10/7.05) [1] = 5.64 × 10⁻³ × e^(−1.418) = 5.64 × 10⁻³ × 0.242 [1] = 1.37 × 10⁻³ C [1].

(e) The p.d. across the capacitor rises from zero to V as charge accumulates [1], so the average p.d. during charging is V/2, and the energy is the area under the Q–V graph [1].

6. (a) Flux linkage = NBA = 250 × 0.42 × 1.8 × 10⁻³ [1] = 0.189 Wb turns [1].

(b) e.m.f. = Δ(NΦ) ÷ Δt = 0.189 ÷ 0.15 [1] = 1.26 V [1].

(c) If the induced current assisted the change, it would increase the flux, which would induce a larger current, which would increase the flux further [1] — creating energy from nothing [1]. Opposing the change means work must be done against the induced effect, and it is that work which supplies the electrical energy [1].

7. (a) F = kQ₁Q₂ ÷ r² = 8.99 × 10⁹ × (3.0 × 10⁻⁶ × 5.0 × 10⁻⁶) ÷ 0.20² [1] ≈ 3.4 N [1], attractive, since the two charges have opposite signs [1].

(b) At the midpoint (r = 0.10 m from each charge): V₁ = 8.99 × 10⁹ × 3.0 × 10⁻⁶ ÷ 0.10 = 2.70 × 10⁵ V; V₂ = 8.99 × 10⁹ × (−5.0 × 10⁻⁶) ÷ 0.10 = −4.50 × 10⁵ V [2]. V(total) = 2.70 × 10⁵ + (−4.50 × 10⁵) = −1.8 × 10⁵ V [1].

8. (a) Taking logs of Q = Q₀e^(−t/RC) gives ln Q = ln Q₀ − t/RC [1], a straight line with gradient = −1/RC [1], so RC can be found from the reciprocal of the magnitude of the gradient [1].

(b) One time constant is the time for the charge to fall to 1/e, approximately 37%, of its initial value [2].

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Where marks are usually lost

• Applying the inverse square law to a uniform field.
• Combining capacitors like resistors.
• Saying a magnetic field gives a parabolic path.
• Forgetting the ½ in capacitor energy, or being unable to explain it.
• Stating Lenz’s law without linking it to conservation of energy.
• Forgetting the sign of a charge when summing potentials at a point due to multiple charges.
• Confusing the gradient of a ln(Q)-against-t graph with RC itself, rather than −1/RC.