Capacitance, energy stored and combining capacitors
A capacitor stores charge, and with it energy, on two plates separated by an insulator. Its capacitance says how much charge it holds for each volt across it, and depends on its size, the gap and the material in between.
Part 1 of 3: Learn it
In short
- C = Q ÷ V, measured in farads; 1 F = 1 C V⁻¹.
- Energy stored E = ½QV = ½CV² = ½Q² ÷ C.
- In parallel, capacitances add; in series, their reciprocals add.
Where this is in your specification
Spec points: AQA 7408 3.7.4.1 to 3.7.4.3, OCR A H556 6.1.1 and 6.1.2, Edexcel 9PH0 Topic 7
| Board | Topic: Capacitors |
|---|---|
| AQA 7408 | 3.7.4 |
| Edexcel 9PH0 | Topic 7 |
| OCR H556 | 6.1 |
| Higher C857 76 | Electricity: capacitors |
Capacitance
A is the area of overlap of the plates, d their separation, ε₀ = 8.85 × 10⁻¹² F m⁻¹ the permittivity of free space, and εᵣ the relative permittivity of the dielectric. A dielectric's molecules become polarised in the field, which lets the plates hold more charge for the same p.d.
Energy stored
A graph of p.d. against charge for a capacitor is a straight line through the origin. The energy stored is the area under it, a triangle, giving E = ½QV. Substituting Q = CV gives the other two forms.
Combining capacitors
| Arrangement | Total capacitance | What is shared |
|---|---|---|
| Parallel | C = C₁ + C₂ + ... | same p.d. across each |
| Series | 1/C = 1/C₁ + 1/C₂ + ... | same charge on each |
These are the opposite way round from resistors. Two equal capacitors in series give half the capacitance of one.
What is the unit of capacitance?
Show the answer
The farad, F.
Part 2 of 3: See it worked
Worked examples
Example 1
A 470 μF capacitor is charged to 9.0 V. Find the charge and the energy stored.
- Q = CV = 470 × 10⁻⁶ × 9.0 = 4.23 × 10⁻³ C
- E = ½CV² = 0.5 × 470 × 10⁻⁶ × 81 = 0.0190 J
Answer: 4.2 mC and 19 mJ.
Example 2
Capacitors of 2.0 μF and 3.0 μF are connected first in series, then in parallel. Find the total capacitance each time.
- Series: 1/C = 1/2 + 1/3 = 5/6, so C = 1.2 μF
- Parallel: C = 2.0 + 3.0 = 5.0 μF
Answer: 1.2 μF in series and 5.0 μF in parallel.
Example 3
Two parallel plates of area 0.010 m² are 1.0 mm apart in air (εᵣ = 1). Find the capacitance.
- C = Aε₀εᵣ ÷ d
- = 0.010 × 8.85 × 10⁻¹² × 1 ÷ 1.0 × 10⁻³
Answer: 8.9 × 10⁻¹¹ F (about 89 pF).
Common mistakes
- Adding series capacitances directly, as you would resistors.
- Forgetting to convert μF to F.
- Writing E = QV, which is the energy supplied by the battery, not the energy stored.
- Leaving the plate separation in mm.
How much energy is stored in a 100 μF capacitor at 12 V?
Show the answer
½ × 100 × 10⁻⁶ × 144 = 7.2 × 10⁻³ J.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
What is the unit of capacitance?
The farad, F.
How much energy is stored in a 100 μF capacitor at 12 V?
½ × 100 × 10⁻⁶ × 144 = 7.2 × 10⁻³ J.
What happens to the capacitance if the gap between the plates is halved?
It doubles, because C is inversely proportional to d.
Jobs that use this
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Diese Lernzettel sind auf Englisch, weil sie britischen Prüfungslehrplänen folgen.
Full lessons and marked practice for this course are coming soon to Brainlag Learn. See courses