Physics • Year 12 • Module 6: Electromagnetism • Lesson 2

Trajectories in Electric Fields

Build HSC Band 5–6 extended-response technique: multi-step calculations, experimental design, and evaluation of claims about charged particle motion.

Master · Extended Response

1. Multi-step analysis — mass spectrometer deflection chamber (Band 5–6)

8 marks   Band 5–6

Scenario. In a simplified mass spectrometer, ions are first accelerated from rest through a potential difference Vacc = 2 000 V. They then enter a deflection chamber consisting of two horizontal parallel plates, 8.0 cm long and 1.6 cm apart, with a potential difference of 200 V across them (top plate positive). A detector screen is placed 12.0 cm beyond the exit of the plates. Two ion species are being analysed: singly charged carbon ions (C+, m = 2.00×10−26 kg) and singly charged oxygen ions (O+, m = 2.66×10−26 kg). Both carry q = +1.60×10−19 C.

Q1. Analyse the motion of both ion species and determine which ion strikes the detector screen at a higher point. In your response you must:

  • Calculate the entry speed of each ion into the deflection chamber using the work-energy theorem.
  • Calculate the electric field between the plates and each ion’s electric acceleration.
  • Determine the vertical deflection of each ion at the exit of the plates.
  • Calculate the additional vertical displacement gained by each ion between the plates and the detector screen (treat ions as leaving the plates with horizontal velocity vx and vertical velocity vy, then travelling 12.0 cm in free motion with no field).
  • State the total vertical displacement of each ion at the detector screen and identify which ion hits higher and why.
Plan: (1) v = √(2qVacc/m) for each ion. (2) E = Vplates/d; a = qE/m (different for each ion). (3) tplates = L/vx; y1 = ½atplates². (4) After exit: vy,exit = atplates; t2 = 0.12/vx; y2 = vy,exit×t2. (5) Total y = y1 + y2.

2. Experimental design — measuring charge-to-mass ratio using plate deflection (Band 5–6)

7 marks   Band 5–6

Research question. A student wants to measure the charge-to-mass ratio (q/m) of a charged particle using the deflection it experiences between two parallel plates. The particle source produces particles with unknown speed; the student has access to a vacuum tube apparatus, two parallel plates (adjustable voltage 0–500 V, separable to 1.0–5.0 cm), a velocity selector (crossed E and B fields), a fluorescent screen, a ruler, and a digital voltmeter.

Constraint: The particle’s charge-to-mass ratio should be determined from measured deflection without prior knowledge of q and m separately.

Q2. Design the investigation. Your response must include:

  • A hypothesis stating the expected relationship between deflection and the quantity being varied.
  • Identification of the independent, dependent, and at least two controlled variables.
  • A step-by-step procedure (minimum 4 steps) explaining how to use the deflection data to determine q/m, including the key equation you will use and how you will measure the deflection.
  • An explanation of what the gradient of a relevant linear graph would represent.
  • One limitation of this experimental method and a suggested improvement.
Key equation: y = (qE/m)(L²/2vx²) = (q/m)(VL²/4dVacc) if entry speed comes from acceleration through Vacc. Plotting y vs Vplates gives gradient = (q/m)(L²/4dVacc); rearrange for q/m.
Answers — Do not peek before attempting

Q1 — Sample Band 6 response (8 marks), annotated

Entry speeds. v = √(2qVacc/m). For C+: vC = √(2×1.60×10−19×2000 / 2.00×10−26) = √(3.2×107) = 1.789×104 m/s. For O+: vO = √(3.2×107/2.66×10−26×2.00×10−26×... ) — recalculate: vO = √(2×1.60×10−19×2000 / 2.66×10−26) = √(2.406×107) = 1.551×104 m/s. [1 mark]

Electric field and accelerations. E = Vplates/d = 200/0.016 = 1.25×104 V/m [1]. aC = qE/mC = (1.60×10−19×1.25×104) / 2.00×10−26 = 1.00×1011 m/s². aO = (1.60×10−19×1.25×104) / 2.66×10−26 = 7.52×1010 m/s². [1]

Deflection inside plates. tC = 0.080/1.789×104 = 4.47×10−6 s; yC,1 = ½×1.00×1011×(4.47×10−6)² = 9.99×10−2 m ≈ 10.0 cm. But plate separation is only 1.6 cm (half-gap = 0.8 cm), so the C+ ion strikes the top plate before exiting. [1] For O+: tO = 0.080/1.551×104 = 5.16×10−6 s; yO,1 = ½×7.52×1010×(5.16×10−6)² = 1.00×10−1 m ≈ 10.0 cm — also strikes top plate. Both ions hit the plate; the lighter C+ hits it sooner (larger acceleration). [1]

Examiner’s note: This scenario is intentionally designed so both ions strike the plates, testing whether students check the hit-plate condition. Award credit for correctly identifying the plate-strike condition and explaining which ion hits first. A fully credited response identifies C+ as hitting first (larger a, same vx roughly similar), with clear mathematical support.

Marking criteria (8 marks): 1 = correct entry speed calculation for both ions with working; 1 = correct E field; 1 = correct accelerations for both ions; 1 = correct time of flight and deflection for at least one ion; 1 = identifies that one or both ions strike the plate (checks |y| vs d/2); 1 = determines which ion hits first or higher; 1 = correct additional deflection beyond plates for ions that exit (if applicable, or correctly states plate strike occurs); 1 = final conclusion with correct physical reasoning (C+ has larger q/m → larger a → deflects more/hits sooner).

Q2 — Sample Band 6 response (7 marks), annotated

Hypothesis: If the plate voltage Vplates is increased while all other variables are held constant, the vertical deflection y of the particle on the screen will increase proportionally, because y ∝ E ∝ Vplates from the formula y = (qEL²)/(2mvx²). [1]

Variables: IV = plate voltage Vplates (varied from 50 to 450 V in steps of 50 V); DV = vertical deflection y of the particle spot on the screen (measured with ruler against the fluorescent screen); Controlled: particle entry speed vx (fixed by velocity selector at one setting), plate separation d (fixed), plate length L (fixed). [1]

Procedure: (1) Use the velocity selector (crossed E and B fields) to select a known particle speed vx = Esel/Bsel; record this value. (2) Set plate separation d and plate length L; record with a ruler. (3) Apply a known voltage Vplates across the plates using the digital voltmeter; observe the deflected spot on the fluorescent screen and measure deflection y from the undeflected position. (4) Repeat for at least 8 values of Vplates from 50 V to 450 V; record y for each. (5) Plot y vs Vplates; determine gradient; use q/m = gradient × (4dVacc)/L² (if entry speed is from Vacc) or q/m = gradient × (2vx² d)/(L²) (if velocity selector is used). [1]

Gradient interpretation: Gradient of y vs Vplates = qL²/(2mvx²d) = (q/m) × L²/(2vx²d). Since L, vx and d are known constants, q/m = gradient × 2vx²d/L². [1]

Limitation: The screen deflection y may be difficult to measure precisely if the particle beam has finite width, introducing uncertainty. Also, stray electric fields near the plate edges can cause fringing effects that distort the uniform-field assumption. [1]

Improvement: Use a camera and image analysis software to measure spot position to 0.1 mm precision. Alternatively, extend the plate length to increase deflection and improve the signal-to-noise ratio for y measurements. [1]

Justification of design: This experiment correctly isolates the proportionality y ∝ Vplates by holding vx constant via the velocity selector, allowing q/m to be determined from the slope without needing to know q and m individually. [1]

Marking criteria (7 marks): 1 = testable hypothesis with correct proportionality stated; 1 = IV, DV, and two controlled variables named; 1 = four or more clear procedural steps including how y is measured; 1 = correct key equation stated and manipulated for q/m; 1 = gradient of linear graph correctly identified with physical meaning; 1 = one valid limitation stated with physical reasoning; 1 = one improvement that would genuinely reduce the identified limitation.

HSCScience • Physics • Year 12 • Module 6 • Lesson 2 • Worksheet 3 (Master)