The shape of the lesson
By the end of the lesson, learners can
- The radial field: Field lines stream straight out from a positive charge or straight in to a negative one.
- The inverse-square shape: E = Q / (4πε₀r²) shares its form with Coulomb’s law and Newton’s law of gravitation.
- 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
A point charge produces a radial field of strength E = Q / (4πε₀r²), pointing outward for a positive charge and inward for a negative one. Like Coulomb’s law and gravitation, it is an inverse-square field.
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 radiates.. Strip Coulomb’s law of its second charge and you have the field of a point charge itself: radial, and falling away as the inverse square of distance. 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: The radial field | Field lines stream straight out from a positive charge or straight in to a negative one. Project tab 4 of the lab (Field of a point charge) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The inverse-square shape | E = Q / (4πε₀r²) shares its form with Coulomb’s law and Newton’s law of gravitation. | Connect-Extend-Challenge |
| 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, 34 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 The inverse-square shape, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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.
Electric field of a point charge: topic worksheet
3 questions, 34 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.
A point charge Q = +4.0 nC is fixed in a vacuum. This question links Coulomb’s law with the field of a point charge.
(a) A second charge q = +2.0 nC is placed 3.0 cm from Q. Calculate the force on it. [3]
(b) Write the expression for the electric field strength due to a point charge. [1]
(c) Calculate the field strength at 3.0 cm from Q. [2]
(d) Using your answer to (c), show that the force in (a) equals qE. [2]
(e) State and explain how the field strength changes at a distance of 6.0 cm. [2]
(f) State the direction of the field at the point. [1]
A point charge Q = +6.0 nC is in a vacuum. This question links the field of a point charge with electric potential.
(a) State what is meant by the electric potential at a point. [2]
(b) Calculate the potential at 5.0 cm from Q. [2]
(c) Calculate the field strength at 5.0 cm from Q. [2]
(d) State the relation between E and V for a radial field, and verify it at 5.0 cm. [2]
(e) Calculate the work done in moving a +2.0 nC charge from 5.0 cm to 10.0 cm from Q. [3]
Two point charges, A = +5.0 nC and B = −5.0 nC, are fixed 8.0 cm apart in a vacuum. P is the midpoint, as shown in Fig. 10.1. This question uses field strength, Coulomb’s law, the field of a point charge, and potential.
(a) Define electric field strength. [1]
(b) Calculate the force between A and B. [2]
(c) Calculate the field strength at P due to each charge, and the resultant field at P. [4]
(d) Calculate the electric potential at P. [3]
(e) State what your answer to (d) implies about the work done in bringing a charge to P from infinity. [2]
Answer key and marking notes
- (a)F = 8.99 × 10⁹ × (4.0 × 10⁻⁹)(2.0 × 10⁻⁹)/(0.030)² ✓✓ = 8.0 × 10⁻⁵ N ✓
- (b)E = Q/(4πε₀r²) ✓
- (c)E = 8.99 × 10⁹ × 4.0 × 10⁻⁹/(0.030)² ✓ = 4.0 × 10⁴ V m⁻¹ ✓
- (d)qE = 2.0 × 10⁻⁹ × 4.0 × 10⁴ = 8.0 × 10⁻⁵ N ✓, which matches the force in (a) ✓
- (e)It falls to one quarter, 1.0 × 10⁴ V m⁻¹ ✓, because E ∝ 1/r² ✓
- (f)Radially outward from Q (a positive charge) ✓
- (a)The work done per unit positive charge in bringing it from infinity to that point ✓✓
- (b)V = Q/(4πε₀r) = 8.99 × 10⁹ × 6.0 × 10⁻⁹/0.050 ✓ = 1.1 × 10³ V ✓
- (c)E = Q/(4πε₀r²) = 8.99 × 10⁹ × 6.0 × 10⁻⁹/(0.050)² ✓ = 2.2 × 10⁴ V m⁻¹ ✓
- (d)E = −dV/dr; in magnitude E = V/r for a point charge: 2.16 × 10⁴ × 0.050 = 1.1 × 10³ V = V ✓✓
- (e)V at 10 cm = 539 V; ΔV = 539 − 1079 = −540 V ✓; W = qΔV = 2.0 × 10⁻⁹ × (−540) ✓ = −1.1 × 10⁻⁶ J (the field does the work as the like charge moves away) ✓
- (a)The force per unit positive charge at a point ✓
- (b)F = 8.99 × 10⁹ × (5.0 × 10⁻⁹)²/(0.080)² ✓ = 3.5 × 10⁻⁵ N (attractive) ✓
- (c)Each at r = 0.040 m: E = 8.99 × 10⁹ × 5.0 × 10⁻⁹/(0.040)² = 2.8 × 10⁴ V m⁻¹ ✓✓; at P both fields point from A toward B, so they add: resultant = 5.6 × 10⁴ V m⁻¹ ✓✓
- (d)V = ΣQ/(4πε₀r) = 8.99 × 10⁹ × (5.0 × 10⁻⁹/0.040 − 5.0 × 10⁻⁹/0.040) ✓✓ = 0 V ✓
- (e)No net work is needed, since the potential at P is zero ✓✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
The electric field strength a distance r from a point charge Q is:
- AQ / (4πε₀r²)
- BQ / (4πε₀r)
- CQr / (4πε₀)
- DQ² / (4πε₀r²)
The field of a point charge varies with distance as:
- A1 / r²
- B1 / r
- Cr
- Dr²
Halving the distance from a point charge makes the field:
- Afour times as strong
- Btwice as strong
- Chalf as strong
- Dunchanged
The field of a positive point charge points:
- Aradially outward
- Bradially inward
- Cin circles around it
- Dalong one fixed direction
Quiz key
E = Q / (4πε₀r²), the field per unit charge from Coulomb’s law.
It is an inverse-square field: E ∝ 1/r².
E ∝ 1/r², so halving r multiplies the field by four.
Field lines of a positive charge point radially outward, in the direction of the force on a positive test charge.
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.