Paper 2 June 2025 Foundation tier, Question 10 11 marks
Figure 19 shows two metal plates.
The plates are charged by a very high voltage (potential difference).
Draw, on Figure 19, the shape and direction of the electric field between the two plates.
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Figure 20 shows a device that removes small particles of soot from sooty air.
There is a large potential difference between the metal tube and the metal rod in the centre of the tube.
This large potential difference causes the soot particles to gain a negative charge as they pass between the rod and the tube.
The soot particles have
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Explain why, in Figure 20, the charged soot particles move towards the metal tube.
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When the soot particles reach the metal tube, the charge on them changes.
State how the charge on them changes.
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The metal tube does not become charged when the soot particles reach it.
Explain why the metal tube does not become charged.
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There is a current of 2.5 mA in a circuit.
Calculate the amount of charge that moves past a point in the circuit in 80 seconds.
State the unit of charge.
If you’re taking GCSE (9–1) Combined Science or GCSE (9–1) Physics, you will need these equations:
HT = higher tier
| distance travelled = average speed × time | |
| acceleration = change in velocity ÷ time taken | a v t u |
| force = mass × acceleration | F = m × a |
| weight = mass × gravitational field strength | W = m × g |
| HT momentum = mass × velocity | p = m × v |
| change in gravitational potential energy = mass × gravitational field strength × change in vertical height | ∆GPE = m × g × ∆h |
| kinetic energy = 1/2 × mass × (speed)2 | KE 21 m v2 |
| efficiency = useful energy transferred by the devicetotal energy supplied to the device | |
| wave speed = frequency × wavelength | v = f × λ |
| wave speed = distance ÷ time | v |
| work done = force × distance moved in the direction of the force | E = F × d |
| power = work done ÷ time taken | P |
| energy transferred = charge moved × potential difference | E = Q × V |
| charge = current × time | Q = I × t |
| potential difference = current × resistance | V = I × R |
| power = energy transferred ÷ time taken | P |
| electrical power = current × potential difference | P = I × V |
| electrical power = (current)2 × resistance | P = I2 × R |
| density = mass ÷ volume |
force exerted on a spring = spring constant × extension F = k × x
| (final velocity)2 – (initial velocity)2 = 2 × acceleration × distance | v2 − u2 = 2 × a × x |
| HT force = change in momentum ÷ time | F mv t mu |
| energy transferred = current × potential difference × time | E = I × V × t |
| HT force on a conductor at right angles to a magnetic fieldcarrying a current = magnetic flux density × current × length | F = B × I × l |
| For transformers with 100% efficiency, potential difference across primary coil × current in primary coil = potential difference across secondary coil × current in secondary coil | VP × IP = VS × IS |
| change in thermal energy = mass × specific heat capacity × change in temperature | ∆Q = m × c × ∆θ |
| thermal energy for a change of state = mass × specific latent heat | Q = m × L |
| energy transferred in stretching = 0.5 × spring constant × (extension)2 | E 21 k x2 |
If you’re taking GCSE (9–1) Physics, you also need these extra equations:
| moment of a force = force × distance normal to the direction of the force | |
| pressure = force normal to surface ÷ area of surface | P |
| HT potential difference across secondary coilpotential difference across primary coil | VVpS |
| to calculate pressure or volume for gases of fixed mass at constant temperature | P1 × V1 = P2 × V2 |
| HT pressure due to a column of liquid liquid × gravitational field strength= height of column × density of | P = h × ρ × g |
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