Capacitance
Key idea: Use C = Q/V and the V–Q graph area to solve capacitance and energy stored in a capacitor questions (A Level Physics).
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The core idea
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Learning objectives
- Apply capacitance and capacitor-energy relationships.
1. Definitions (Must Know)
A. Capacitor
A capacitor is two conductors separated by an insulator. It stores charge and energy.
B. Capacitance, C
Capacitance, C, is charge stored per unit potential difference:
C = Q/V
Unit: farad (F), where 1 F = 1 C V⁻¹.
Here Q is the magnitude of the charge on either plate. The plates carry equal and opposite charges, + Q and -Q.
2. Key Ideas (What Earns Marks)
- Rearrangements:
- Q = CV, V = Q/C
- Energy stored in a capacitor is the area under the V–Q graph.
- For a capacitor with constant C, V = Q/C is a straight line through the origin, so the area is a triangle:
- U = (1/2)QV = Q²/2C = (1/2)CV²
V–Q graph for a capacitor (constant C)
A straight-line V–Q graph through the origin; the gradient is 1/C and the area under the line represents the energy stored.
Scroll across the graph to read all labels.
View figure data
| Charge, Q (mC) | V = Q/C (example gradient) |
|---|---|
| 0 | 0 |
| 1 | 2 |
| 2 | 4 |
| 3 | 6 |
| 4 | 8 |
| 5 | 10 |
During charging, V rises from 0 to V, so the average potential difference is V/2. That’s why U = (1/2)QV, not QV.
3. Detailed Explanations
A. What capacitance measures
Capacitance tells you how much charge a capacitor can store for a given potential difference:
- larger C means more charge stored for the same V (since Q = CV).
B. Why energy is the area under the V–Q graph
When you add a small charge dQ to a capacitor at potential V, the work done is:
dW = V dQ
If C is constant, V increases linearly with Q from 0 to V. So the V–Q graph is a straight line and the area under it is a triangle:
U = (1/2)QV = (1/2)CV²
4. Common Mistakes
- Forgetting to convert μF and nF to farads.
- Using U = QV instead of U = (1/2)QV for a capacitor (the voltage rises from 0 to V as it charges).
- Mixing Q in coulombs with “number of electrons” (need Q = Ne if converting).
5. Exam Tips
- If C and V are given: use U = (1/2)CV².
- If C and Q are given: use U = Q²/2C.
- If Q and V are given: use U = (1/2)QV.
- Quick unit check: F·V² = (C/V)·V² = C·V = J.
6. Worked Examples
Modelled example 1
Find Q and U from C and V
Problem
Study the worked solution
Find charge
Method
Use Q = CV.Reason
Capacitance is charge stored per unit p.d.Working
Q = (220 × 10⁻⁶)(12) = 2.64 × 10⁻³ CFind energy
Method
Use U = (1/2)CV².Reason
The capacitor p.d. rises from zero during charging.Working
U = (1/2)(220 × 10⁻⁶)(12)² = 1.58 × 10⁻² J
Guided practice 2
Find C and U from Q and V
Problem
Try this before viewing the solution
Hints
Hint 1: match given quantities
View solution step by step
Find capacitance
Method
Divide charge by p.d.Reason
C = Q/V.Working
C = (6.0 × 10⁻⁶)/200 = 3.0 × 10⁻⁸ F = 30 nFFind energy
Method
Use the given charge and p.d.Reason
U = (1/2)QV avoids an unnecessary substitution.Working
U = (1/2)(6.0 × 10⁻⁶)(200) = 6.0 × 10⁻⁴ J
Common misconception 3
Use a straight-line V–Q graph (triangle area)
Learner claim
Try this before viewing the solution
View solution step by step
Find capacitance
Method
Use C = Q/V.Reason
The endpoint gives the constant capacitance.Working
C = (4.0 × 10⁻³)/8.0 = 5.0 × 10⁻⁴ FFind triangular area
Method
Use U = (1/2)QV.Reason
Potential rises linearly from zero as charge accumulates.Working
U = (1/2)(4.0 × 10⁻³)(8.0) = 1.6 × 10⁻² J
Examiner practice 4
Find energy from Q and C
Examination question
Try this before viewing the solution
View solution step by step
Choose the direct form
1 markMethod
Use U = Q²/(2C).Reason
Charge and capacitance are given.Working
U = Q²/2CConvert and substitute
1 markMethod
Use C = 4.0 × 10⁻⁶ F.Reason
Energy formula requires SI units.Working
U = ((3.0 × 10⁻⁵)²)/(2(4.0 × 10⁻⁶))Evaluate
1 markMethod
Obtain 1.13 × 10⁻⁴ J.Reason
Charge is squared.Working
U = 1.13 × 10⁻⁴ J.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark formula, conversion and result.
Challenge 5
Finding voltage from energy and capacitance
Independent transfer
Try this before viewing the solution
Hints
Hint 1: voltage is squared
View solution step by step
Rearrange
Method
Use V = square root of (2U/C).Reason
Voltage appears squared in the energy relation.Working
V = square root of (2U/C)Evaluate
Method
Obtain 30 V.Reason
Use C = 2.0 × 10⁻⁶ F.Working
V = square root of ((2(9.0 × 10⁻⁴))/(2.0 × 10⁻⁶)) = 30 VInterpret the root
Method
State the requested p.d. magnitude as positive.Reason
The energy determines V²; polarity would require separately labelled terminals.Working
|V| = 30 V.
7. Mind Stretchers
Mind stretcher 1: Explain why U = (1/2)QV and not U = QVExtension
Show Answer
As the capacitor charges, the potential difference increases from 0 to V.
So the average potential during charging is V/2, and: U = Q(V/2) = (1/2)QV
Mind stretcher 2: Unit check for 1/2 CV²Extension
Show that 1/2 CV² has units of joules.
Show Answer
C has unit farad (F) and 1 F = 1 C V⁻¹.
So: CV² = (C V⁻¹)V² = C V = J since 1 V = 1 J C⁻¹.
Mind stretcher 3: Optional (Enrichment)Extension
A. Calculus form (same idea)
The “area under the V–Q graph” can also be written as:
U = ∫₀^Q V dQ
If C is constant, V = Q/C and this gives U = Q²/2C, which is the same as (1/2)QV.
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027