Capacitors

A capacitor is a component that stores energy in an electric field between two conducting surfaces. Capacitors are everywhere — from the flash unit in a camera to the smoothing circuits in power supplies and the timing circuits in electronics. Understanding how they store charge and energy is a core part of the Edexcel A-Level specification.

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Capacitance

Capacitance is defined as the charge stored per unit potential difference across the capacitor:

C = Q / V

where:

• C is the capacitance in farads (F)
• Q is the charge stored (C) — this refers to the charge on one plate (the other plate carries −Q)
• V is the potential difference across the capacitor (V)

One farad is an enormous capacitance. In practice, capacitors are measured in smaller units:

Unit	Symbol	Value in Farads
Millifarad	mF	10⁻³ F
Microfarad	μF	10⁻⁶ F
Nanofarad	nF	10⁻⁹ F
Picofarad	pF	10⁻¹² F

A 1 F capacitor is the size of a drinks can. Supercapacitors used in electric vehicles can reach thousands of farads but are very expensive.

Rearranging gives:

• Q = CV (charge stored is proportional to voltage)
• V = Q / C

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Parallel Plate Capacitor

The capacitance of a parallel plate capacitor with a vacuum (or air) between the plates is:

C = ε₀A / d

where:

• ε₀ = 8.85 × 10⁻¹² F m⁻¹ (permittivity of free space)
• A is the area of overlap of the plates (m²)
• d is the separation of the plates (m)

This tells you three ways to increase the capacitance:

1. Increase the plate area A
2. Decrease the plate separation d
3. Insert a dielectric material (which increases ε)

Worked Example 1

A parallel plate capacitor has square plates of side 0.10 m separated by 0.50 mm. Calculate its capacitance.

A = 0.10 × 0.10 = 0.010 m²

C = ε₀A / d = (8.85 × 10⁻¹² × 0.010) / (0.50 × 10⁻³) = 8.85 × 10⁻¹⁴ / 5.0 × 10⁻⁴ = 1.77 × 10⁻¹⁰ F ≈ 180 pF

This small capacitance from relatively large plates shows why practical capacitors use very thin separations, large surface areas (rolled foils), and dielectric materials.

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Energy Stored in a Capacitor

When a capacitor is charged, work is done against the electric field as charge is transferred from one plate to the other. The energy stored is:

E = ½QV = ½CV² = ½Q²/C

These three forms are all equivalent (substitute Q = CV to convert between them). Use whichever is most convenient for the given information.

Choosing the Right Energy Formula

Known quantities	Best formula
Q and V	E = ½QV
C and V	E = ½CV²
Q and C	E = ½Q²/C
C constant, V changes	E = ½CV² (for comparing energies)
Q constant, C changes	E = ½Q²/C (e.g., isolated capacitor with changing d)

Derivation

The energy stored equals the area under the Q–V graph. Since Q = CV (a straight line through the origin), the area is a triangle:

E = ½ × base × height = ½ × V × Q = ½QV

Worked Example 2

A 470 μF capacitor is charged to 12 V. How much energy does it store?

E = ½CV² = ½ × 470 × 10⁻⁶ × 12² = ½ × 470 × 10⁻⁶ × 144 = 0.034 J = 34 mJ

Worked Example 3: Energy in a Camera Flash

A camera flash capacitor (1000 μF, charged to 300 V) discharges through the flash tube in 2.0 ms. Calculate the energy stored and the average power delivered during the flash.

Energy: E = ½CV² = ½ × 1000 × 10⁻⁶ × 300² = 45 J

Average power: P = E / t = 45 / 0.002 = 22 500 W = 22.5 kW

This is why camera flashes are so bright — 45 J is modest, but delivering it in 2 ms gives an average power of 22.5 kW. The capacitor is able to release its stored energy far more quickly than a battery could.

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Capacitors in Parallel

When capacitors are connected in parallel:

• They all have the same voltage across them
• The total charge is the sum of the individual charges
• The total capacitance is:

C_total = C₁ + C₂ + C₃ + ...

Capacitors in parallel add directly — this is the opposite of resistors.

Why?

Each capacitor stores charge Q = CV independently. The total charge Q_total = C₁V + C₂V + C₃V = (C₁ + C₂ + C₃)V, so the effective capacitance is C₁ + C₂ + C₃.

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Capacitors in Series

When capacitors are connected in series:

• They all store the same charge Q
• The total voltage is the sum of the individual voltages
• The total capacitance is:

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + ...

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

Why?

Each capacitor has V = Q/C. The total voltage V_total = Q/C₁ + Q/C₂ + Q/C₃ = Q(1/C₁ + 1/C₂ + 1/C₃). Since V_total = Q/C_total, we get 1/C_total = 1/C₁ + 1/C₂ + 1/C₃.

Worked Example 4

A 100 μF capacitor and a 300 μF capacitor are connected in series. What is the total capacitance?

1/C_total = 1/100 + 1/300 = 3/300 + 1/300 = 4/300

C_total = 300/4 = 75 μF

Summary: Series vs Parallel

Property	Series	Parallel
Same quantity	Charge Q	Voltage V
Splits	Voltage	Charge
Capacitance formula	1/C_T = 1/C₁ + 1/C₂	C_T = C₁ + C₂
Total C compared to individuals	Smaller than smallest	Larger than largest
Analogy	Like resistors in parallel	Like resistors in series

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Dielectrics

A dielectric is an insulating material placed between the plates of a capacitor. It increases the capacitance by a factor called the relative permittivity (or dielectric constant) εᵣ:

C = ε₀εᵣA / d

The dielectric works by becoming polarised in the electric field — positive and negative charges within the dielectric molecules shift slightly, creating an internal field that partially opposes the external field. This effectively reduces the electric field strength for a given charge, allowing more charge to be stored for the same voltage.

Common dielectrics include ceramic, paper, plastic film, and mica. They also prevent the plates from touching, allowing very thin separations.

What Happens When You Insert a Dielectric?

Two scenarios to distinguish:

Capacitor connected to a battery (V constant):

• C increases by factor εᵣ
• Q = CV increases by factor εᵣ
• E = ½CV² increases by factor εᵣ

Capacitor isolated (Q constant):

• C increases by factor εᵣ
• V = Q/C decreases by factor εᵣ
• E = ½Q²/C decreases by factor εᵣ
• The capacitor actually pulls the dielectric in (the energy decreases, so work is done by the system)

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Worked Example 5: Sharing Charge Between Capacitors

A 2200 μF capacitor is charged to 15 V and then disconnected from the supply. It is then connected across an uncharged 1000 μF capacitor. Find the final voltage and the energy lost.

Initial charge: Q = CV = 2200 × 10⁻⁶ × 15 = 0.033 C

After connection, the capacitors are in parallel: C_total = 2200 + 1000 = 3200 μF

Charge is conserved: Q_total = 0.033 C

Final voltage: V = Q / C_total = 0.033 / 3200 × 10⁻⁶ = 10.3 V

Energy before: E₁ = ½ × 2200 × 10⁻⁶ × 15² = 0.2475 J

Energy after: E₂ = ½ × 3200 × 10⁻⁶ × 10.3² = 0.170 J

Energy lost: 0.2475 − 0.170 = 0.078 J

This energy is dissipated as heat in the connecting wires. Whenever charge is shared between capacitors, energy is always lost (unless the capacitors happen to be at the same voltage already).

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Common Exam Mistakes

• Unit errors: Remember that 1 μF = 10⁻⁶ F, not 10⁻³ F. Converting incorrectly gives answers out by factors of 1000.
• Confusing series and parallel rules: Capacitors in series use the reciprocal formula (the opposite of resistors in series).
• Assuming Q is conserved when a battery is connected: Q is only conserved when the capacitor is isolated. If a battery is connected, it supplies whatever charge is needed to maintain V.
• Forgetting the ½ in energy formulas: The energy stored is ½CV², not CV². The ½ comes from the triangular area under the Q-V graph.

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Key Points

• Capacitance C = Q / V, measured in farads (F)
• Parallel plate capacitor: C = ε₀A / d
• Energy stored: E = ½QV = ½CV² = ½Q²/C
• In parallel: C_total = C₁ + C₂ + ... (capacitances add)
• In series: 1/C_total = 1/C₁ + 1/C₂ + ... (reciprocals add)
• Dielectrics increase capacitance by the factor εᵣ
• When sharing charge between capacitors, charge is conserved but energy is always lost

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A-Level Deep Dive: Capacitors

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields covers capacitance $C = Q/V$C=Q/V; the parallel-plate formula $C = \varepsilon_0 \varepsilon_r A/d$C=ε0​εr​A/d; energy stored in the equivalent forms $E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \tfrac{1}{2}Q^2/C$E=21​QV=21​CV2=21​Q2/C; series and parallel combinations; and the role of dielectrics (refer to the official Pearson Edexcel specification document for exact wording). Capacitance sits inside Topic 7 alongside electric fields, magnetic flux density and electromagnetic induction, and feeds into the time-dependent charge/discharge content. Capacitors are examined predominantly on 9PH0 Paper 2, with practical-skills crossover into Paper 3 through the required time-constant practical. The Edexcel formula booklet does list $C = Q/V$C=Q/V, the energy forms and the parallel-plate result; it does not give the series/parallel combination rules — these must be reasoned from charge conservation (series) and voltage equality (parallel).

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Worked example with full mark scheme

Question (8 marks):

A capacitor of capacitance $C = 220\ \mu\text{F}$C=220 μF is charged through a fixed resistor of resistance $R = 47\ \text{k}\Omega$R=47 kΩ from a $12.0\ \text{V}$12.0 V supply. The capacitor is initially uncharged.

(a) Calculate the time constant of the charging circuit and state what it physically represents. (2)

(b) Calculate the charge stored on the capacitor at $t = 5.0\ \text{s}$t=5.0 s after the circuit is closed. (3)

(c) Calculate the energy stored in the capacitor when it is fully charged, and the total energy delivered by the supply during the charging process. Comment on the difference. (3)

Solution with mark scheme:

(a) Step 1 — apply $\tau = RC$τ=RC.

$\tau = RC = (47 \times 10^3)(220 \times 10^{-6}) = 10.34\ \text{s} \approx 10\ \text{s}$τ=RC=(47×103)(220×10−6)=10.34 s≈10 s

M1 — correct substitution into $\tau = RC$τ=RC with both quantities in SI units. A common slip is to leave $C$C in microfarads, producing an answer in $\text{k}\Omega\cdot\mu\text{F}$kΩ⋅μF that is numerically right but dimensionally meaningless.

B1 — physical interpretation: $\tau$τ is the time taken for the charge on the capacitor to rise to $(1 - 1/e) \approx 63\%$(1−1/e)≈63% of its final value during charging (or equivalently, the time to fall to $1/e \approx 37\%$1/e≈37% of its initial value during discharge). Stating only "the time constant" without a numerical or proportional interpretation forfeits this mark.

(b) Step 1 — write the charging equation.

$Q(t) = Q_0\left(1 - e^{-t/RC}\right) \quad \text{with}\ Q_0 = CV_0$Q(t)=Q0​(1−e−t/RC)with Q0​=CV0​

M1 — correct charging form. Candidates who write the discharge form $Q = Q_0 e^{-t/RC}$Q=Q0​e−t/RC here lose all subsequent marks for this sub-part; the question explicitly states the capacitor starts uncharged.

Step 2 — substitute.

$Q_0 = CV_0 = (220 \times 10^{-6})(12.0) = 2.64 \times 10^{-3}\ \text{C}$Q0​=CV0​=(220×10−6)(12.0)=2.64×10−3 C.

$t/RC = 5.0/10.34 = 0.4836$t/RC=5.0/10.34=0.4836.

$Q(5.0) = 2.64 \times 10^{-3}\left(1 - e^{-0.4836}\right) = 2.64 \times 10^{-3}(1 - 0.6166) = 2.64 \times 10^{-3}(0.3834)$Q(5.0)=2.64×10−3(1−e−0.4836)=2.64×10−3(1−0.6166)=2.64×10−3(0.3834)

M1 — correct evaluation of the exponential. Examiners accept any value within reasonable rounding — typically $e^{-0.48}$e−0.48 to $e^{-0.49}$e−0.49 — provided the working is shown.

$Q(5.0) \approx 1.01 \times 10^{-3}\ \text{C} \approx 1.0\ \text{mC}$Q(5.0)≈1.01×10−3 C≈1.0 mC

A1 — final answer to two significant figures, matching the precision of the input data.

(c) Step 1 — energy stored when fully charged.

$E_\text{cap} = \tfrac{1}{2}CV_0^2 = \tfrac{1}{2}(220 \times 10^{-6})(12.0)^2 = \tfrac{1}{2}(220 \times 10^{-6})(144) = 1.58 \times 10^{-2}\ \text{J} \approx 16\ \text{mJ}$Ecap​=21​CV02​=21​(220×10−6)(12.0)2=21​(220×10−6)(144)=1.58×10−2 J≈16 mJ

M1 — correct use of $E = \tfrac{1}{2}CV^2$E=21​CV2.

Step 2 — total energy delivered by the supply.

The supply delivers charge $Q_0$Q0​ at constant emf $V_0$V0​ throughout, so

$E_\text{supply} = Q_0 V_0 = (2.64 \times 10^{-3})(12.0) = 3.17 \times 10^{-2}\ \text{J} \approx 32\ \text{mJ}$Esupply​=Q0​V0​=(2.64×10−3)(12.0)=3.17×10−2 J≈32 mJ

A1 — correct, with the recognition that the supply does work $QV$QV rather than $\tfrac{1}{2}QV$21​QV.

Step 3 — comment.

Exactly half of the energy delivered by the supply is dissipated as heat in the resistor during charging; only half ends up stored on the capacitor. This 50:50 split is independent of $R$R and $C$C — it follows from integrating $P = I^2 R$P=I2R along the charging curve.

B1 — explicit comment that 50% of the supply energy is dissipated, with a brief justification.

Total: 8 marks (M4 A2 B2).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A $1000\ \mu\text{F}$1000 μF capacitor is charged to $25\ \text{V}$25 V and then disconnected from the supply. It is then connected in parallel with an uncharged $500\ \mu\text{F}$500 μF capacitor.

(a) Calculate the common potential difference across the two capacitors after they are connected. (3)

(b) Show that energy is lost in the process and calculate the magnitude of this loss. State where this energy goes. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — initial charge: $Q_0 = C_1 V_0 = (1000 \times 10^{-6})(25) = 2.5 \times 10^{-2}\ \text{C}$Q0​=C1​V0​=(1000×10−6)(25)=2.5×10−2 C.
• M1 (AO2.1) — apply conservation of charge to the isolated combined system: total charge after $=$= total charge before $=$= $2.5 \times 10^{-2}\ \text{C}$2.5×10−2 C, distributed across $C_\text{total} = 1000 + 500 = 1500\ \mu\text{F}$Ctotal​=1000+500=1500 μF.
• A1 (AO1.1b) — final voltage $V_\text{f} = Q_0/C_\text{total} = 2.5 \times 10^{-2}/1.5 \times 10^{-3} = 16.7\ \text{V}$Vf​=Q0​/Ctotal​=2.5×10−2/1.5×10−3=16.7 V.

(b)

• M1 (AO2.1) — initial energy stored: $E_i = \tfrac{1}{2}C_1 V_0^2 = 0.5 \times 1000 \times 10^{-6} \times 25^2 = 0.3125\ \text{J} \approx 0.31\ \text{J}$Ei​=21​C1​V02​=0.5×1000×10−6×252=0.3125 J≈0.31 J.
• M1 (AO1.1b) — final energy stored: $E_f = \tfrac{1}{2}C_\text{total} V_\text{f}^2 = 0.5 \times 1500 \times 10^{-6} \times 16.7^2 \approx 0.21\ \text{J}$Ef​=21​Ctotal​Vf2​=0.5×1500×10−6×16.72≈0.21 J.
• A1 (AO3.2a) — energy lost $\Delta E = 0.31 - 0.21 = 0.10\ \text{J}$ΔE=0.31−0.21=0.10 J, dissipated as heat in the connecting wires (and as a small amount of electromagnetic radiation), arising because charge redistribution is necessarily irreversible when the two capacitors start at different voltages.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Paper 2 routinely expects candidates to combine charge conservation with energy accounting in one multi-step question — the AO3 mark is the explicit recognition that energy redistribution between capacitors is always lossy.

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Synoptic links

Connects to:

1. Topic 7 — Electric fields: $C = \varepsilon_0 \varepsilon_r A/d$C=ε0​εr​A/d is derived from the uniform-field result $E = V/d$E=V/d combined with the surface-charge relationship $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$E=σ/ε0​=Q/(ε0​A). The capacitor is the parallel-plate field problem dressed in a circuit element.