The shape of the lesson
By the end of the lesson, learners can
- Reading energy off a graph: Vary C and V and watch the shaded triangle under the V-Q line.
- Three forms, one energy: Because the voltage rises from zero to V as the charge builds up, the average voltage is ½V and the energy is ½QV, not QV.
- Capacitance: C = Q/V. In parallel C = C₁ + C₂; in series 1/C = 1/C₁ + 1/C₂.
- Energy: W = ½QV = ½CV² = ½Q²/C, the area under the V-Q graph.
- Discharge: x = x₀ e⁻ᵗּᵣᶜ; the time constant τ = RC is the time to fall to 1/e (about 37%).
One picture every idea returns to
The energy stored in a capacitor is the area under the graph of voltage against charge, a triangle, so W = ½QV. Using Q = CV this is also ½CV² or ½Q²/C.
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, The energy in the field.. Charging a capacitor does work against the charge already there, and that work is stored as energy. 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: Reading energy off a graph | Vary C and V and watch the shaded triangle under the V-Q line. Project tab 2 of the lab (Energy stored) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Three forms, one energy | Because the voltage rises from zero to V as the charge builds up, the average voltage is ½V and the energy is ½QV, not QV. | Think-Pair-Share |
| 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 (5 questions, 57 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 Three forms, one energy, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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.
Energy stored in a capacitor: topic worksheet
5 questions, 57 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 2200 µF capacitor is charged to a potential difference of 9.0 V. Fig. 2.1 shows how the voltage across a capacitor varies with the charge stored.
(a) Calculate the charge stored on the capacitor. [2]
(b) Calculate the energy stored, using W = ½CV². [2]
(c) Show that this is equal to ½QV. [2]
(d) The capacitor is instead charged to 18 V. State and explain the new energy stored. [2]
(e) Using Fig. 2.1, explain why the energy is ½QV and not QV. [2]
(f) The capacitor is then fully discharged through a resistor. State what happens to the stored energy. [2]
(g) Write a third expression for the energy in terms of Q and C. [1]
A 4.7 μF capacitor is charged to 9.0 V. Parts (a) and (b) use capacitance; the rest use the energy stored.
(a) Calculate the charge stored. [2]
(b) State what happens to the charge if the capacitor is disconnected from the supply. [1]
(c) Calculate the energy stored. [2]
(d) Calculate the energy stored if the voltage were doubled to 18 V. [2]
(e) State and explain the factor by which the stored energy increases when the voltage is doubled. [3]
Fig. 7.1 shows how the charge on a capacitor varies with the p.d. across it.
(a) Write three equivalent expressions for the energy stored in a capacitor. [2]
(b) Explain, with reference to the Q-V graph, why the energy is ½QV and not QV. [3]
(c) A 100 μF capacitor is charged to 50 V. Calculate the energy stored. [2]
(d) Calculate the charge stored. [2]
(e) The capacitor is discharged through a flash lamp in 2.0 ms. Calculate the average power delivered. [2]
A 10 μF capacitor charged to 12 V is connected across an uncharged 5.0 μF capacitor. This question links capacitance with the stored energy.
(a) Calculate the initial charge on the 10 μF capacitor. [2]
(b) After connection the charge is shared. Calculate the common p.d. across the two capacitors. [3]
(c) Calculate the total energy stored before and after the connection. [3]
(d) Calculate the energy that appears to be lost, and state where it goes. [2]
(e) State why the charge is conserved but the energy is not. [1]
A 470 μF capacitor is charged to 6.0 V and then discharged through a 10 kΩ resistor. This question uses all three lessons of the topic: capacitance, energy, and discharge.
(a) Calculate the initial charge stored. [2]
(b) Calculate the initial energy stored. [2]
(c) Calculate the time constant of the discharge. [2]
(d) Calculate the p.d. across the capacitor after 3.0 s. [3]
(e) Calculate the energy still stored at that time. [3]
Answer key and marking notes
- (a)Q = CV = 2200 × 10⁻⁶ × 9.0 ✓ = 2.0 × 10⁻² C ✓
- (b)W = ½CV² = 0.5 × 2200 × 10⁻⁶ × 9.0² ✓ = 8.9 × 10⁻² J ✓
- (c)½QV = 0.5 × 1.98 × 10⁻² × 9.0 ✓ = 8.9 × 10⁻² J, the same value ✓
- (d)Four times as large, about 0.36 J ✓; W ∝ V², so doubling V multiplies the energy by four ✓
- (e)The charge builds from 0 to Q while the voltage rises from 0 to V ✓; the energy is the area under the line, a triangle of area ½QV, since the average voltage is ½V ✓
- (f)It is dissipated in the resistor, transferred to thermal energy (heat) ✓✓
- (g)W = ½Q²/C ✓
- (a)Q = CV = 4.7 × 10⁻⁶ × 9.0 ✓ = 4.2 × 10⁻⁵ C ✓
- (b)The charge remains stored on the capacitor ✓
- (c)W = ½CV² = ½ × 4.7 × 10⁻⁶ × 9.0² ✓ = 1.9 × 10⁻⁴ J ✓
- (d)W = ½ × 4.7 × 10⁻⁶ × 18² ✓ = 7.6 × 10⁻⁴ J ✓
- (e)It increases by a factor of 4 ✓✓, because the energy is proportional to V² ✓
- (a)W = ½QV = ½CV² = ½Q²/C ✓✓
- (b)As charge is added the p.d. rises, so the average p.d. during charging is ½V ✓; the energy is the area under the Q-V line, a triangle of area ½QV ✓✓
- (c)W = ½CV² = ½ × 100 × 10⁻⁶ × 50² ✓ = 0.13 J ✓
- (d)Q = CV = 100 × 10⁻⁶ × 50 ✓ = 5.0 × 10⁻³ C ✓
- (e)P = W/t = 0.125 / 2.0 × 10⁻³ ✓ = 63 W ✓
- (a)Q = CV = 10 × 10⁻⁶ × 12 ✓ = 1.2 × 10⁻⁴ C ✓
- (b)Total capacitance = 15 μF; V = Q/C = 1.2 × 10⁻⁴ / 15 × 10⁻⁶ ✓✓ = 8.0 V ✓
- (c)Before: W = ½CV² = ½ × 10 × 10⁻⁶ × 12² = 7.2 × 10⁻⁴ J ✓; after: ½ × 15 × 10⁻⁶ × 8.0² = 4.8 × 10⁻⁴ J ✓✓
- (d)Energy lost = 7.2 − 4.8 = 2.4 × 10⁻⁴ J ✓; dissipated as heat (and a little radiation) in the connecting wires ✓
- (e)Charge cannot be created or destroyed (conservation of charge), but energy is dissipated as the charge redistributes ✓
- (a)Q = CV = 470 × 10⁻⁶ × 6.0 ✓ = 2.8 × 10⁻³ C ✓
- (b)W = ½CV² = ½ × 470 × 10⁻⁶ × 6.0² ✓ = 8.5 × 10⁻³ J ✓
- (c)τ = RC = 10 × 10³ × 470 × 10⁻⁶ ✓ = 4.7 s ✓
- (d)V = 6.0 e^(−3.0/4.7) = 6.0 × e^(−0.638) ✓✓ = 3.2 V ✓
- (e)W = ½CV² = ½ × 470 × 10⁻⁶ × 3.17² ✓✓ = 2.4 × 10⁻³ J ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
The energy stored in a capacitor equals the area under a graph of:
- Avoltage against charge
- Bcharge against time
- Ccurrent against voltage
- Dpower against time
The energy stored in a capacitor can be written as:
- A½QV = ½CV² = ½Q²/C
- BQV
- CCV²
- DQ²/C
Doubling the voltage on a capacitor changes the energy stored by a factor of:
- A4
- B2
- C½
- D8
The factor of ½ in W = ½QV arises because:
- Athe voltage rises from zero to V as the charge builds up
- Bthe charge is halved
- Conly half the plates are used
- Dthe capacitor is half full
Quiz key
The area under the V-Q line is the work done charging the capacitor, the stored energy.
All three forms are equal, obtained by substituting Q = CV into ½QV.
W ∝ V², so doubling V multiplies the energy by four.
The energy is the area of a triangle: the average voltage during charging is ½V.
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.