Physics • Year 12 • Module 6: Electromagnetism • Lesson 2
Trajectories in Electric Fields
Apply your understanding of parabolic motion, time of flight, deflection and impact velocity to real calculations and scenario-based problems.
1. Interpret experimental data — varying entry speed
A student fires electrons horizontally into a uniform electric field between two parallel plates. The plates are 8.0 cm long, 2.0 cm apart, and the field strength is 4 000 V/m (directed downward). The table records results for three different entry speeds. 8 marks
| Entry speed vx (m/s) | Time of flight t (s) | Deflection y (m) | Hits plate? | Exit speed v (m/s) |
|---|---|---|---|---|
| 1.0 × 106 | ||||
| 2.0 × 106 | ||||
| 4.0 × 106 |
Use: me = 9.11 × 10−31 kg, qe = 1.60 × 10−19 C. Plates are 2.0 cm apart so the particle hits if |y| > 1.0 cm.
1.1 Calculate the electric acceleration of the electron (show working). Then complete all four missing columns of the table above. 6 marks
1.2 Describe the relationship between entry speed and vertical deflection. Explain this relationship using the formula y = ½at² and t = L/vx. 2 marks
2. Interpret a velocity-time graph — vertical component
An electron enters a uniform electric field horizontally. The graph below shows the vertical component of its velocity (vy) versus time while it is between the plates. 8 marks
Figure 2.1 — Vertical velocity component of an electron in a uniform electric field vs. time. Data collected during passage through the field region only.
2.1 From the graph, determine the gradient of the line and state what physical quantity it represents. Include units. 2 marks
2.2 Use the gradient to calculate the electric field strength E between the plates (use me = 9.11 × 10−31 kg, q = 1.60 × 10−19 C). 2 marks
2.3 If the electron entered at vx = 3.0 × 106 m/s and the plates are 18.0 cm long, use the graph to find the exit vertical velocity and hence calculate the magnitude and direction of the exit velocity. 2 marks
2.4 A student states: “The graph shows that the electron is accelerating, so its kinetic energy is increasing.” Evaluate this claim using the work-energy theorem. 2 marks
3. Compare proton and electron trajectories
Complete the table comparing an electron and a proton both fired horizontally at 2.0 × 106 m/s into the same uniform electric field (E = 3 000 V/m, pointing downward) between plates 6.0 cm long and 2.0 cm apart. 10 marks (1 per cell)
| Quantity | Electron | Proton |
|---|---|---|
| Direction of electric force | ||
| Electric acceleration a (m/s²) | ||
| Time of flight t (s) | ||
| Vertical deflection y (m) | ||
| Hits plate before exit? |
4. Predict and justify — cathode ray tube scenario
In an old television cathode ray tube (CRT), electrons are accelerated from rest through a potential difference of 5 000 V and then enter a set of deflection plates 4.0 cm long and 1.2 cm apart, with 120 V across them.
9 marks
4.1 The electron enters the deflection plates horizontally. Calculate its entry speed using the work-energy theorem (W = qV = ½mv²). Use me = 9.11×10−31 kg, q = 1.60×10−19 C. 3 marks
4.2 Calculate the electric field between the deflection plates and hence the vertical deflection of the electron as it exits the plates. 3 marks
4.3 Explain why a higher accelerating voltage (say 10 000 V instead of 5 000 V) would reduce the deflection of the beam on the screen, even though the electron carries more kinetic energy. 3 marks
Q1 — Electric acceleration and data table
Electric acceleration: a = qE/m = (1.60×10−19 × 4000) / (9.11×10−31) = 7.03×1014 m/s² (upward for the electron, since the field is downward and the electron has negative charge).
Row 1 (vx = 1.0×106 m/s): t = 0.080 / (1.0×106) = 8.0×10−8 s; y = ½×7.03×1014×(8.0×10−8)² = 2.25×10−2 m = 2.25 cm > 1.0 cm → Hits plate. (vy = 5.62×107 m/s; v ≈ 5.62×107 m/s.)
Row 2 (vx = 2.0×106 m/s): t = 4.0×10−8 s; y = ½×7.03×1014×(4.0×10−8)² = 5.62×10−3 m = 5.6 mm < 1.0 cm → Does not hit; vy = 2.81×107 m/s; v = √((2.0×106)² + (2.81×107)²) ≈ 2.82×107 m/s.
Row 3 (vx = 4.0×106 m/s): t = 2.0×10−8 s; y = ½×7.03×1014×(2.0×10−8)² = 1.41×10−3 m = 1.4 mm < 1.0 cm → Does not hit; vy = 1.41×107 m/s; v ≈ 1.41×107 m/s.
1.2 As entry speed doubles, deflection decreases by a factor of 4 (y ∝ 1/vx²). From y = ½a(L/vx)² = aL² / (2vx²), deflection is inversely proportional to the square of the entry speed.
Q2 — Velocity-time graph
2.1 Gradient = (1.20×108 − 0) / (6.0×10−8 − 0) = 2.0×1015 m/s². This represents the electric acceleration a of the electron in the field.
2.2 a = qE/m ⇒ E = am/q = (2.0×1015 × 9.11×10−31) / (1.60×10−19) = 1.14×104 V/m (approximately 1.1×104 V/m).
2.3 t = L/vx = 0.180 / (3.0×106) = 6.0×10−8 s. From the graph at t = 6.0×10−8 s: vy = 1.20×108 m/s. Exit speed: v = √((3.0×106)² + (1.20×108)²) ≈ 1.20×108 m/s; θ = tan−1(1.20×108 / 3.0×106) ≈ 88.6° above horizontal.
2.4 The claim is correct. The electric force does positive work on the electron (force is in the direction of motion for the electron moving toward the positive plate). By the work-energy theorem, W = ΔKE, so the kinetic energy increases. The total speed increases even though vx is constant, because vy grows and the total KE = ½m(vx² + vy²) increases.
Q3 — Electron vs. proton comparison
a(electron) = (1.60×10−19×3000) / (9.11×10−31) = 5.27×1014 m/s² upward (force opposes field).
a(proton) = (1.60×10−19×3000) / (1.67×10−27) = 2.87×1011 m/s² downward.
t = 0.060 / (2.0×106) = 3.0×10−8 s (same for both).
y(electron) = ½×5.27×1014×(3.0×10−8)² = 2.37×10−1 m — this exceeds 1.0 cm, so the electron hits the top plate.
y(proton) = ½×2.87×1011×(3.0×10−8)² = 1.29×10−4 m = 0.13 mm < 1.0 cm → Does not hit.
Note: The electron’s much smaller mass gives it a far larger acceleration and deflection despite having the same charge magnitude.
Q4 — CRT scenario
4.1 Work-energy theorem: qV = ½mv² ⇒ v = √(2qV/m) = √(2 × 1.60×10−19 × 5000 / 9.11×10−31) = √(1.754×1015) = 4.19×107 m/s.
4.2 E = Vplates/d = 120 / 0.012 = 10 000 V/m. a = qE/m = (1.60×10−19×10000) / (9.11×10−31) = 1.757×1015 m/s². t = L/vx = 0.040 / (4.19×107) = 9.55×10−10 s. y = ½×1.757×1015×(9.55×10−10)² = 8.0×10−4 m = 0.80 mm.
4.3 Doubling the accelerating voltage to 10 000 V increases vx by √2 (v ∝ √V). Since y = aL² / (2vx²) and vx² doubles, the deflection y is halved. Although the electron has more kinetic energy, it spends less time between the plates (shorter time of flight), so the electric force has less time to deflect it. Higher speed → shorter t → smaller y.