A charged particle between parallel plates

8 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

Parallel plates are 0.020 m apart. The left plate is at +600 V and the right plate at 0 V. A particle of charge −2.0 × 10^-9 C and mass 3.0 × 10^-6 kg is released from rest halfway between them. Ignore gravity, drag and edge effects. Right is positive. The table specifies the ideal field model.

Potential across the gap
Distance right from left plate / m	Potential / V
0	600
0.01	300
0.02	0

Part (a)

Determine the electric field magnitude from the table. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / V m⁻¹

Working

Part (b)

Calculate the signed horizontal electric force on the particle. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / N

Working

Part (c)

Calculate the horizontal velocity after 0.010 s, before the particle reaches a plate. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻¹

Working

Part (d)

Explain why the particle moves towards the higher potential plate. [2 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 3.0 × 10⁴ V m⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use |ΔV/Δx| = 600/0.020. (1 method mark)
Obtain 3.0 × 10⁴ V m⁻¹. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: This is potential difference, not field strength. Divide by plate separation.

Part (b) marking points

Final answer: -6.0 × 10^-5 N. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Field is positive/right; use signed F = qE. (1 method mark)
Obtain −6.0 × 10^-5 N. (1 accuracy mark)

Use the field from part (a) in F = qE with the negative charge sign.

Common errors: This loses the negative test-charge sign.

Part (c) marking points

Final answer: -0.20 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use a = F/m = −20 m s⁻². (1 method mark)
Use v = at to obtain −0.20 m s⁻¹. (1 accuracy mark)

Use the signed force from part (b) in v = Ft/m from rest. Follow through numerical errors, but the velocity must be leftward for this charge and field.

Common errors: The motion is left; right was defined as positive.

Part (d) marking points

Its negative charge makes its force opposite to the field. (1 independent mark)
Its electric potential energy qV decreases as V increases for negative q. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Common errors: All charges move towards lower potential: that applies to a positive charge released in this model.

Report a problem with this case

Related skills

fields practice

Scope and reference sources

AQA 3.5 and selected 3.7.3/3.7.4; OCR A Modules 4/6 (selected). These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: charged-drop · version 1.1.0 · seed 0. Choose another case