⚡ Capacitance

Storing charge and energy in capacitors — series, parallel and parallel plate arrangements

AQA A-Level Physics 7

📦Define capacitance: C = Q/V

🔗Calculate capacitors in series and parallel

⚡Calculate energy stored: W = ½QV = ½CV²

🏗️Use the parallel plate formula: C = ε₀εᵣA/d

🌐Describe uses of capacitors in circuits

🔬Investigate the charge stored on a capacitor

What is Capacitance?

Capacitor: An electrical component that stores charge. It consists of two conducting plates separated by an insulator (dielectric).

Capacitance (C): The charge stored per unit potential difference across the capacitor.

C = Q / V
C = capacitance (farads, F)
Q = charge stored on one plate (coulombs, C)
V = potential difference across the capacitor (volts, V)

1 F = 1 C V⁻¹ (one farad is very large; practical units are μF, nF, pF)

When a capacitor is connected to a voltage V, charge Q = CV is stored. One plate holds +Q and the other −Q (equal and opposite). The capacitance depends on the physical construction, not the voltage applied.

Parallel plate capacitor:

C = ε₀ ε_r A / d
ε₀ = 8.85 × 10⁻¹² F m⁻¹ (permittivity of free space)
ε_r = relative permittivity of dielectric (dimensionless)
A = area of overlap of the plates (m²)
d = separation of plates (m)

Larger plate area → more charge per volt → higher C. Smaller gap → stronger electric field → higher C. Higher ε_r (e.g. ceramic) → higher C.

Capacitors in Circuits

Capacitors in PARALLEL:
C_total = C₁ + C₂ + C₃ + …

Capacitors in SERIES:
1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + …
(For two: C_total = C₁C₂/(C₁+C₂))

Note: the rules are the OPPOSITE of resistors! Capacitors in parallel add directly; in series they combine reciprocally.

	Series	Parallel
C_total vs individual	Less than smallest	Greater than largest
Voltage	Divides: V = V₁ + V₂	Same across all
Charge	Same on each: Q₁ = Q₂	Divides: Q = Q₁ + Q₂

In a series combination, the same charge Q is stored on each capacitor (the inner plates are isolated, so charge can only rearrange). In parallel, each capacitor has the same voltage and stores charge proportional to its capacitance.

Energy Stored in a Capacitor

As a capacitor charges, the voltage increases. The energy stored equals the area under the Q–V graph (a triangle):

W = ½ Q V = ½ C V² = Q² / (2C)

W = energy stored (joules, J)
Q = charge (C)
V = voltage (V)
C = capacitance (F)

The factor of ½ arises because the voltage builds up from 0 to V during charging — not all charge is transferred at full voltage. Use W = ½CV² when V is known; W = Q²/2C when Q is known; W = ½QV when both are known.

Doubling the voltage quadruples the stored energy (W ∝ V²). Doubling the capacitance doubles the energy. This asymmetry is important when comparing designs.

Uses of Capacitors

Smoothing: In power supply circuits, capacitors smooth the ripple from a rectified AC supply by charging when voltage is high and discharging when it drops.
Energy storage: Camera flashes, defibrillators — large capacitors store energy and release it in a very short time (high power).
Timing circuits: The time constant RC (charging/discharging time) is used to control delays in timers and oscillators.
Signal coupling / filtering: Capacitors block DC but pass AC, allowing AC signals to pass between stages of an amplifier without DC offset.

A 470 μF capacitor is charged to 12 V. Calculate the charge stored and the energy stored.

1Q = CV = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C = 5.64 mC

2W = ½CV² = ½ × 470 × 10⁻⁶ × 12² = ½ × 470 × 10⁻⁶ × 144 = 3.38 × 10⁻² J = 33.8 mJ

Q = 5.64 mC; W = 33.8 mJ

Three capacitors of 10 μF, 20 μF and 30 μF are connected in parallel. Find the total capacitance. Then find the total if connected in series.

1Parallel: C = 10 + 20 + 30 = 60 μF

2Series: 1/C = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60

3C = 60/11 = 5.45 μF

Parallel: C = 60 μF; Series: C = 5.45 μF

A parallel plate capacitor has plates of area 0.025 m² separated by 2.0 mm of air (ε_r = 1). Calculate the capacitance.

1C = ε₀ ε_r A / d = (8.85 × 10⁻¹²) × 1 × 0.025 / (2.0 × 10⁻³)

2C = 8.85 × 10⁻¹² × 0.025 / 0.002 = 2.21 × 10⁻¹³ / 0.002 = 1.11 × 10⁻¹⁰ F = 111 pF

C = 111 pF

A 100 μF capacitor and a 200 μF capacitor are connected in series to a 9 V supply. Find the total capacitance, total charge, and voltage across each capacitor.

11/C = 1/100 + 1/200 = 2/200 + 1/200 = 3/200 → C = 200/3 = 66.7 μF

2Q = CV = 66.7 × 10⁻⁶ × 9 = 6.0 × 10⁻⁴ C = 600 μC (same on each capacitor in series)

3V₁₀₀ = Q/C₁ = 600/100 = 6.0 V; V₂₀₀ = Q/C₂ = 600/200 = 3.0 V

4Check: 6.0 + 3.0 = 9.0 V ✓

C = 66.7 μF; Q = 600 μC; V across 100 μF = 6.0 V; V across 200 μF = 3.0 V

1. A 50 μF capacitor stores 2.0 × 10⁻³ C of charge. What is the potential difference across it?

V = Q/C = 2.0×10⁻³ / 50×10⁻⁶ = 2.0×10⁻³ / 5.0×10⁻⁵ = 40 V.

2. How does the energy stored in a capacitor change when the voltage across it is doubled (capacitance unchanged)?

W = ½CV². If V doubles: W ∝ V² → W × 4. Energy quadruples.

3. Two capacitors of 4 μF and 12 μF are connected in parallel. What is the combined capacitance?

Parallel: C = C₁ + C₂ = 4 + 12 = 16 μF. (Unlike resistors in parallel, capacitors in parallel simply add.)

4. A capacitor has plates of area A separated by distance d. If d is halved and A is doubled, how does the capacitance change?

C = ε₀A/d. If A → 2A and d → d/2: C_new = ε₀(2A)/(d/2) = 4ε₀A/d = 4C. Capacitance quadruples.

5. A 220 μF capacitor is charged to 15 V. Calculate the energy stored in mJ.

1. A 10 μF capacitor charged to 200 V is connected in parallel with an uncharged 40 μF capacitor. Find the final voltage across both capacitors and the energy lost in the process. Explain where the energy goes.

2. A defibrillator stores 360 J of energy in a capacitor charged to 5000 V. Calculate: (a) the capacitance needed; (b) the charge stored; (c) the average power delivered if the pulse lasts 4.0 ms.

3. Three capacitors are connected: C₁ = 6 μF and C₂ = 3 μF in series, and this combination is connected in parallel with C₃ = 4 μF. The whole arrangement is connected to 12 V. Find the total capacitance, total charge from the supply, and the voltage across C₁.

Required Practical Investigating Charge Stored on a Capacitor

Objective: Verify C = Q/V by measuring the charge stored on capacitors of different values at different voltages.

Equipment

Selection of capacitors (e.g. 100 μF, 220 μF, 470 μF)
Variable DC supply (0–10 V)
Coulombmeter (or high-sensitivity galvanometer to integrate current)
Voltmeter
Switch and connecting leads

Method

Connect the capacitor to the supply via a switch. Connect the voltmeter in parallel.
Charge the capacitor to a set voltage (e.g. 2 V). Disconnect from supply. Connect the coulombmeter to measure Q.
Repeat for different voltages (2, 4, 6, 8, 10 V). Plot Q vs V — expect a straight line through the origin with gradient = C.
Repeat for different capacitors to verify the relationship.

Analysis

V / V	Q / μC (expected for 100 μF)
2	200
4	400
6	600
8	800

Gradient of Q vs V = C. Energy stored = area under Q–V graph = ½QV.

Safety

Do not exceed the rated working voltage of the capacitor (printed on the component). Electrolytic capacitors must be connected with the correct polarity — reversed connection can cause failure or explosion.