The Physics Question Bank

10 GCSE Physics questions about the electric toothbrush, from Foundation to Higher, each with a full worked answer. Inside the handle, a small motor makes the brush head oscillate hundreds of times every second. The handle is completely sealed so it is waterproof – it charges without any metal contacts, because a coil in the charging stand induces a potential difference in a second coil inside the handle.

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Topics: Waves, Electricity, Energy, Magnetism and electromagnetism

Equations you’ll use

T = 1 ÷ f  |  E = P t  |  Q = I t  |  E = Q V  |  efficiency = useful output energy transfer ÷ total input energy transfer  |  V_p ÷ V_s = n_p ÷ n_s (S only · HT)

Quantity	Symbol	Unit
period	T	seconds (s)
frequency	f	hertz (Hz)
energy transferred	E	joules (J)
power	P	watts (W)
time	t	seconds (s)
charge flow	Q	coulombs (C)
current	I	amperes (A)
potential difference	V	volts (V)
number of turns on a coil	n	no unit

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

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Questions

Q1 (F) The brush head oscillates with a frequency of 260 Hz. Calculate how many oscillations it makes during a 2-minute brush.

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Convert first: 2 minutes = 2 × 60 = 120 s
Idea: 260 Hz means 260 oscillations every second
Calculation: 260 × 120
A: 31 200 oscillations

Q2 (F) Calculate the period of the 260 Hz oscillation, in milliseconds, to 2 significant figures.

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F: T = 1 ÷ f
I: T = 1 ÷ 260
F: T is already the subject
A: T = 0.003 85 s = 3.8 ms (2 s.f.)

Q3 (F) The toothbrush motor has a power of 3.0 W. Calculate the energy it transfers during one 2-minute brush.

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Convert first: 2 minutes = 120 s
F: E = P t
I: E = 3.0 × 120
F: E is already the subject
A: 360 J

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Q4 (F) When fully charged, the toothbrush battery can supply a current of 0.80 A for 1.0 hour. Calculate the charge it stores, in coulombs.

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Convert first: 1.0 hour = 3600 s
F: Q = I t
I: Q = 0.80 × 3600
F: Q is already the subject
A: 2880 C

Q5 (F/H) The battery stores 2880 C at a potential difference of 3.6 V. Calculate the energy stored, in kJ, to 3 significant figures.

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F: E = Q V
I: E = 2880 × 3.6
F: E is already the subject
A: 10 368 J = 10.4 kJ (3 s.f.)

Q6 (F/H) Each brush uses 360 J. The full battery stores 10 368 J. A student brushes twice a day. How many days will a full charge last?

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Step 1 – number of brushes: 10 368 ÷ 360 = 28.8, so 28 complete brushes
Step 2 – days: 28 ÷ 2
A: 14 days (2 weeks)
Round down – there isn’t enough energy for a 29th full brush.

Q7 (F/H) The charging stand takes in 1.2 W and needs 16 hours to fully charge the battery, which then stores 10 368 J. Calculate the efficiency of charging, as a percentage.

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Convert first: 16 hours = 16 × 3600 = 57 600 s
Step 1 – total energy in: E = P t = 1.2 × 57 600 = 69 120 J
F: efficiency = useful output energy ÷ total input energy
I: efficiency = 10 368 ÷ 69 120
F: efficiency is already the subject
A: 0.15 = 15%

Q8 (F/H) Describe the energy transfers when the toothbrush is switched on and used. [3 marks]

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• Energy is transferred from the chemical store of the battery electrically to the motor. (1)
• The motor transfers energy to the kinetic store of the moving brush head. (1)
• Some energy is wasted: it is dissipated to the thermal store of the motor and surroundings and carried away by sound. (1)

Q9 (H · S only · HT) The toothbrush handle is sealed and has no metal contacts. Explain how the charging stand charges the battery, and why the stand must use an alternating current. [6 marks]

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Model answer – marking points:
• The stand contains a coil (primary coil) connected to an alternating supply, often wound around an iron core that pokes up into the handle.
• The alternating current produces a magnetic field that is constantly changing in size and direction.
• The handle contains a second coil (secondary coil); the changing magnetic field passes through it.
• This induces a potential difference across the secondary coil (electromagnetic induction), so a current flows in the handle’s circuit and charges the battery.
• A direct current would produce a steady magnetic field – with no change in the field, no p.d. is induced.
• Because no metal contacts are needed, the handle can be completely sealed, so water cannot get in and cause a short circuit or electric shock.

Q10 (H · S only · HT) Treat the stand and handle as a transformer. The coil in the stand has 150 turns with an alternating p.d. of 12 V across it. The coil in the handle has 50 turns. Calculate the p.d. induced across the handle’s coil. (Assume all the magnetic field passes through both coils.)

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F: V_p ÷ V_s = n_p ÷ n_s
I: 12 ÷ V_s = 150 ÷ 50
F: V_s = 12 × 50 ÷ 150
A: 4.0 V – this is a step-down transformer.

⚠️ Where students lose marks

• Confusing frequency and period – a high frequency means a short period. Always check T = 1 ÷ f gives a small number of seconds.
• Rounding up the number of brushes – 28.8 means only 28 complete brushes.
• Saying the stand “sends current” into the handle. No charge crosses the gap: a changing magnetic field induces a p.d. in the handle’s coil.
• Forgetting to convert hours to seconds before using E = P t to get joules.

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