Capacitance

Capacitors are components that store charge and energy in electric fields. They are ubiquitous in electronic circuits and are a major topic in AQA specification 3.7.4. This lesson covers the definition of capacitance, energy storage, the parallel plate model, dielectrics, and capacitor combinations.

Spec mapping (AQA 7408 §3.7.4 — Capacitance): This lesson covers the definition C = Q/V with units of farads, the parallel-plate formula C = ε₀ε_r A/d with the role of permittivity and area/separation dependence, the polarisation mechanism by which a dielectric raises capacitance, series and parallel combinations with their reciprocal/direct addition rules, the derivation of energy stored U = ½CV² = ½QV = Q²/(2C) including the area-under-graph interpretation, and practical applications (camera flash, defibrillator, smoothing). Energy stored and applications are taken to fuller depth in Lesson 6 of this course; charge/discharge dynamics are covered in Lessons 5 and 7. Refer to the official AQA 7408 specification document for the exact wording.

Synoptic links — where this resurfaces in 7408:

• §3.7.3 Electric fields (Lessons 2–3 here) — the parallel-plate derivation uses E = V/d and E = σ/ε₀, both from the §3.7.3 toolkit.
• §3.5 Electricity — capacitors interact with resistors in DC circuits (RC time constant — Lesson 5) and feature in §3.5.1.5 EMF/internal-resistance contexts as smoothing components.
• §3.7.5 Magnetic fields / electromagnetic induction — inductors store energy in magnetic fields (U = ½LI²); capacitors store energy in electric fields (U = ½CV²) — a structural parallel that emerges most clearly in LC resonant circuits (post-spec but flagged in "Going Further").
• §3.4.1.7 Energy — the ½CV² formula is dimensionally identical to ½kx² for a spring (capacitance as "compliance"; voltage as "displacement"); the same energy-as-integral derivation applies.
• §3.2 Quantum physics — the polarisation mechanism in dielectrics is, microscopically, a quantum-mechanical effect of bound-electron displacement; classical εᵣ is the macroscopic average of microscopic dipole moments.
• §3.8.1 Nuclear physics — high-energy capacitor banks drive pulsed magnetic fields used in inertial-confinement fusion experiments and Z-pinch devices (extension reading).

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Definition of Capacitance

Key Definition: Capacitance (C) is the charge stored per unit potential difference across a capacitor.

C = Q/V

where C is the capacitance (F, farads), Q is the charge stored (C), and V is the potential difference (V).

1 farad = 1 coulomb per volt (1 F = 1 C V⁻¹). A 1 F capacitor is extremely large; typical capacitors range from picofarads (pF = 10⁻¹² F) to millifarads (mF = 10⁻³ F).

Prefix	Symbol	Value
picofarad	pF	10⁻¹² F
nanofarad	nF	10⁻⁹ F
microfarad	μF	10⁻⁶ F
millifarad	mF	10⁻³ F

A capacitor consists of two conducting plates (or surfaces) separated by an insulator (the dielectric). When connected to a voltage supply, charge flows onto the plates: +Q on one plate and −Q on the other. The net charge on the capacitor is zero, but we refer to Q as "the charge stored" (meaning the charge on each plate).

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The Parallel Plate Capacitor

For a parallel plate capacitor with vacuum (or air) between the plates:

C = ε₀A/d

where ε₀ = 8.85 × 10⁻¹² F m⁻¹ (permittivity of free space), A is the overlapping area of the plates (m²), and d is the separation between the plates (m).

This shows that capacitance:

• Increases with plate area A (more space to store charge)
• Decreases with plate separation d (a smaller gap means a stronger field for the same voltage, allowing more charge to be held)

Derivation of C = ε₀A/d

Starting from the uniform field between the plates: E = V/d, and the field due to a surface charge density σ = Q/A on a plate: E = σ/ε₀ = Q/(ε₀A).

Setting these equal: V/d = Q/(ε₀A)

Rearranging: Q/V = ε₀A/d

Since C = Q/V: C = ε₀A/d ✓

Worked Example 1 — Capacitance of a Parallel Plate Capacitor

Question: A parallel plate capacitor has square plates of side length 20 cm, separated by an air gap of 1.5 mm. Calculate its capacitance.

Solution:

A = (0.20)² = 0.040 m²

C = ε₀A/d = (8.85 × 10⁻¹² × 0.040) / (1.5 × 10⁻³)

C = 3.54 × 10⁻¹³ / 1.5 × 10⁻³

C = 2.36 × 10⁻¹⁰ F = 236 pF

This is a very small capacitance — practical capacitors achieve larger values by using very thin dielectrics and large effective areas (e.g., by rolling the plates into a cylinder).

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Permittivity and Dielectrics

When an insulating material (a dielectric) is placed between the plates, the capacitance increases:

C = εᵣε₀A/d

where εᵣ is the relative permittivity (also called the dielectric constant) of the material. εᵣ is dimensionless and always ≥ 1 (for vacuum, εᵣ = 1).

How Dielectrics Increase Capacitance

1. The electric field between the plates polarises the dielectric molecules — their positive and negative charges are displaced slightly in opposite directions.
2. This creates a field inside the dielectric that opposes the applied field, reducing the net electric field between the plates.
3. A reduced field means a reduced voltage V for the same charge Q.
4. Since C = Q/V, a smaller V for the same Q means a larger C.

Dielectric Material	Relative Permittivity (εᵣ)
Vacuum	1.00
Air	1.0006
Paper	3.5
Mica	5–7
Glass	4–10
Water	80
Barium titanate ceramic	1 000–10 000

Worked Example 2 — Effect of a Dielectric

Question: A capacitor has a capacitance of 470 pF with air between the plates. A sheet of mica (εᵣ = 6.0) is inserted between the plates, completely filling the gap. Calculate the new capacitance.

Solution:

C_new = εᵣ × C_air = 6.0 × 470 × 10⁻¹² = 2.82 × 10⁻⁹ F = 2.82 nF

The capacitance has increased by a factor of 6.

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Energy Stored in a Capacitor

A charged capacitor stores energy in the electric field between its plates. The energy can be calculated using three equivalent expressions:

W = ½QV = ½CV² = Q²/(2C)

Derivation of W = ½QV

As a capacitor charges, the potential difference V increases linearly with charge Q (since V = Q/C). The work done in adding a small charge dQ at potential V is dW = V dQ = (Q/C) dQ.

Integrating from 0 to Q: W = ∫₀ᵠ (Q/C) dQ = Q²/(2C)

Since Q = CV: W = (CV)²/(2C) = ½CV²

And also: W = Q × Q/(2C) × (C/Q) = ... → more simply, W = ½QV

Graphically, W is the area under the Q-V graph (a straight line through the origin with gradient C), which is a triangle with area ½ × base × height = ½QV.

Worked Example 3 — Energy Stored in a Capacitor

Question: A 2200 μF capacitor is charged to 6.0 V. Calculate (a) the charge stored and (b) the energy stored.

Solution:

(a) Q = CV = 2200 × 10⁻⁶ × 6.0 = 1.32 × 10⁻² C = 13.2 mC

(b) W = ½CV² = ½ × 2200 × 10⁻⁶ × (6.0)² = ½ × 2200 × 10⁻⁶ × 36 = 3.96 × 10⁻² J = 39.6 mJ

Check: W = ½QV = ½ × 1.32 × 10⁻² × 6.0 = 3.96 × 10⁻² J ✓

Worked Example 4 — Defibrillator Capacitor

Question: A defibrillator stores 400 J of energy in a capacitor charged to 5000 V. Calculate (a) the capacitance required and (b) the charge stored.

Solution:

(a) W = ½CV² → C = 2W/V² = (2 × 400) / (5000)² = 800 / 2.5 × 10⁷ = 3.2 × 10⁻⁵ F = 32 μF

(b) Q = CV = 3.2 × 10⁻⁵ × 5000 = 0.16 C = 160 mC

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Capacitors in Series and Parallel

Parallel Combination

In parallel, all capacitors have the same voltage V across them, but the total charge Q_total is shared:

Q_total = Q₁ + Q₂ + Q₃ = C₁V + C₂V + C₃V = (C₁ + C₂ + C₃)V

C_total = C₁ + C₂ + C₃ + ...

Capacitors in parallel add directly — the total capacitance is greater than any individual capacitance.

Series Combination

In series, all capacitors store the same charge Q (charge displaced from one plate moves onto the next), but the voltages add:

V_total = V₁ + V₂ + V₃ = Q/C₁ + Q/C₂ + Q/C₃

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + ...

The total capacitance in series is always less than the smallest individual capacitance.

Exam Tip: The formulae for capacitors in series and parallel are the reverse of those for resistors. Capacitors in parallel add directly (like resistors in series). Capacitors in series combine reciprocally (like resistors in parallel). A common trick to remember: the combination that makes more "space" for charge (parallel) gives a larger total capacitance.

Worked Example 5 — Capacitor Combinations

Question: Three capacitors of 10 μF, 22 μF, and 47 μF are connected in (a) parallel and (b) series. Calculate the total capacitance in each case.

Solution:

(a) Parallel: C_total = 10 + 22 + 47 = 79 μF

(b) Series: 1/C_total = 1/10 + 1/22 + 1/47 = 0.100 + 0.04545 + 0.02128 = 0.1667 C_total = 1/0.1667 = 6.0 μF

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Practical Applications of Capacitors

• Camera flash: A capacitor is charged slowly from a battery (over a second or two), then discharged rapidly through the flash tube (over milliseconds), delivering a brief, intense burst of energy at peak powers far above what the battery alone could supply.
• Defibrillators: A capacitor is charged to a high voltage and discharged through the patient's chest to restart the heart. The capacitor delivers a controlled pulse of energy (typically 200–400 J in modern biphasic defibrillators).
• Smoothing in power supplies: Capacitors smooth the pulsating output of a rectifier, reducing ripple in a DC supply. During peaks, the capacitor charges; during troughs, it discharges, maintaining a more constant output voltage. The ripple amplitude is set by the load current, the smoothing capacitance, and the mains frequency.
• Timing circuits: The predictable charge/discharge behaviour of RC circuits is used in timing applications such as windscreen wiper delays, oscillators, and digital signal debouncing networks.
• Touchscreens (capacitive sensing): Modern smartphone touchscreens detect the change in capacitance when a fingertip approaches the screen surface — the human body, being mostly water, acts as a finite-impedance ground that perturbs the local capacitance map.

Worked Example 6 — Capacitors in a Mixed Network

Question: Three capacitors are connected as follows: C₁ = 4.0 μF and C₂ = 6.0 μF are in series, and this series combination is in parallel with C₃ = 8.0 μF. The combination is connected across a 12 V supply. Calculate (a) the equivalent capacitance of the network; (b) the total charge stored; (c) the voltage across C₁ alone; (d) the total energy stored.

Solution:

(a) Series combination of C₁ and C₂: 1/C_series = 1/4.0 + 1/6.0 = 0.250 + 0.1667 = 0.4167 C_series = 1/0.4167 = 2.40 μF

Parallel with C₃: C_total = C_series + C₃ = 2.40 + 8.0 = 10.4 μF

(b) Q_total = C_total × V = 10.4 × 10⁻⁶ × 12 = 1.25 × 10⁻⁴ C ≈ 125 μC

(c) The series branch (C₁ and C₂) sees the full 12 V across the two together. Its charge is: Q_series_branch = C_series × V = 2.40 × 10⁻⁶ × 12 = 2.88 × 10⁻⁵ C = 28.8 μC

This same charge sits on each of C₁ and C₂ (because they are in series). So: V across C₁ = Q / C₁ = 28.8 × 10⁻⁶ / 4.0 × 10⁻⁶ = 7.2 V

(Check: V across C₂ = 28.8 / 6.0 = 4.8 V; V_C₁ + V_C₂ = 12.0 V ✓.)