The shape of the lesson
By the end of the lesson, learners can
- A uniform field, and a parabola: Set the plate voltage and separation to control E = V/d, then watch a positive charge curve toward the negative plate.
- Why the path is a parabola: The field, and so the force qE, is the same everywhere between the plates.
- Uniform field: E = V / d; the force on a charge is F = qE, and a charge crossing the field follows a parabola.
- Point charges: Coulomb's law F = Q₁Q₂ / (4πε₀r²) and field E = Q / (4πε₀r²) are both inverse-square.
- Potential: V = Q / (4πε₀r) falls off as 1/r; the work to move charge q is W = qΔV, and E = −dV/dr.
One picture every idea returns to
Between two parallel plates a distance d apart with potential difference V, the field is uniform with strength E = V / d. A charge fired across it has constant velocity along the plates and constant acceleration across them, so it follows a parabola.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A field that is the same everywhere.. Parallel plates make a field that is constant in size and direction. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: A uniform field, and a parabola | Set the plate voltage and separation to control E = V/d, then watch a positive charge curve toward the negative plate. Project tab 2 of the lab (Uniform fields) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Why the path is a parabola | The field, and so the force qE, is the same everywhere between the plates. | Corners |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (3 questions, 35 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why the path is a parabola, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Uniform electric fields: topic worksheet
3 questions, 35 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
Two horizontal parallel plates are 2.0 cm apart with a potential difference of 600 V between them. An electron enters midway between the plates, moving parallel to them at 6.0 × 10⁷ m s⁻¹. The plates are 6.0 cm long. Fig. 1.1 shows the arrangement.
(a) Calculate the electric field strength between the plates. [2]
(b) Calculate the magnitude of the electric force on the electron. [2]
(c) State the direction of this force relative to the field, and explain why. [1]
(d) Explain why the path of the electron between the plates is parabolic. [2]
(e) Calculate the time the electron spends between the plates. [2]
(f) Calculate the vertical deflection of the electron as it leaves the plates. [3]
(g) State the effect on the deflection of doubling the plate voltage. [1]
Two parallel plates 4.0 cm apart have a potential difference of 2000 V across them.
(a) Calculate the electric field strength between the plates. [2]
(b) State the direction and nature of the field between the plates. [2]
(c) Calculate the force on an electron between the plates. [2]
(d) Calculate the acceleration of the electron. [2]
(e) Calculate the work done on the electron as it moves from one plate to the other. [2]
(f) State what happens to this energy. [1]
An electron enters the uniform field between two horizontal plates moving horizontally at 3.0 × 10⁷ m s⁻¹, as shown in Fig. 6.1. The field region is 5.0 cm long and the field strength is 2.0 × 10⁴ V m⁻¹.
(a) State the shape of the path in the field and explain why. [2]
(b) Calculate the vertical force and the vertical acceleration of the electron. [3]
(c) Calculate the time the electron spends in the field. [2]
(d) Calculate the vertical deflection as it leaves the field. [3]
(e) State the direction of the deflection. [1]
Answer key and marking notes
- (a)E = V / d = 600 / 0.020 ✓ = 3.0 × 10⁴ V m⁻¹ ✓
- (b)F = qE = 1.60 × 10⁻¹⁹ × 3.0 × 10⁴ ✓ = 4.8 × 10⁻¹⁵ N ✓
- (c)Opposite to the field (toward the positive plate), because the electron carries a negative charge ✓
- (d)The horizontal velocity is constant while the uniform field gives a constant vertical acceleration ✓; constant velocity in one direction and constant acceleration at right angles to it produce a parabola, just like a projectile ✓
- (e)t = L / v = 0.060 / (6.0 × 10⁷) ✓ = 1.0 × 10⁻⁹ s ✓
- (f)a = F / m = 4.8 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 5.3 × 10¹⁵ m s⁻² ✓; y = ½at² = 0.5 × 5.27 × 10¹⁵ × (1.0 × 10⁻⁹)² ✓ = 2.6 × 10⁻³ m ✓
- (g)The deflection doubles, since y ∝ a ∝ E ∝ V ✓
- (a)E = V/d = 2000 / 0.040 ✓ = 5.0 × 10⁴ V m⁻¹ ✓
- (b)Uniform ✓, directed from the positive plate to the negative plate ✓
- (c)F = qE = 1.60 × 10⁻¹⁹ × 5.0 × 10⁴ ✓ = 8.0 × 10⁻¹⁵ N ✓
- (d)a = F/m = 8.0 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ✓ = 8.8 × 10¹⁵ m s⁻² ✓
- (e)W = qV = 1.60 × 10⁻¹⁹ × 2000 ✓ = 3.2 × 10⁻¹⁶ J ✓
- (f)It becomes kinetic energy of the electron ✓
- (a)Parabolic ✓: the horizontal velocity is constant while a constant vertical force gives constant vertical acceleration (like a projectile) ✓
- (b)F = qE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N ✓; a = F/m = 3.2 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ✓ = 3.5 × 10¹⁵ m s⁻² ✓
- (c)t = L/v = 0.050 / 3.0 × 10⁷ ✓ = 1.7 × 10⁻⁹ s ✓
- (d)y = ½at² = ½ × 3.51 × 10¹⁵ × (1.67 × 10⁻⁹)² ✓✓ = 4.9 × 10⁻³ m (about 4.9 mm) ✓
- (e)Toward the positive plate ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
Between parallel plates a distance d apart with a potential difference V, the field strength is:
- AV / d
- BVd
- Cd / V
- DV / d²
The electric field between two parallel plates is:
- Auniform: the same strength and direction everywhere
- Bstrongest near the plates
- Cradial
- Dzero in the middle
A charged particle fired parallel to the plates follows:
- Aa parabolic path
- Ba circular path
- Ca straight line at constant speed
- Da spiral
Keeping the voltage fixed and doubling the plate separation makes the field:
- Ahalf as strong
- Btwice as strong
- Cfour times as strong
- Dunchanged
Quiz key
For a uniform field, E = V/d, in volts per metre.
Parallel plates produce a uniform field, shown by evenly spaced parallel field lines.
Constant velocity along the plates plus constant acceleration across them gives a parabola.
E = V/d, so for fixed V the field is inversely proportional to d.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.