Electricity & Magnetism

Capacitance (Higher Level)

5th Year · 6th Year (Leaving Cert)

✓Define capacitance and state its SI unit.
✓Apply the formula C = Q/V to solve problems involving charge, voltage, and capacitance.
✓Derive and apply the formula for energy stored in a capacitor, E = ½CV².
✓Describe the behaviour of a capacitor in a DC circuit during charging and discharging, including the exponential nature of these processes.
✓Explain the concept of the time constant (τ) for an RC circuit and perform calculations involving it.

Key concepts

Capacitance (C)

Capacitance is a measure of a capacitor's ability to store electric charge. A capacitor is an electrical component designed to store electrical energy in an electric field. It typically consists of two conducting plates separated by an insulating material called a dielectric. The capacitance of a capacitor is defined as the ratio of the magnitude of the charge (Q) stored on either plate to the potential difference (V) between the plates.

C = Q/V

Unit of Capacitance

The SI unit of capacitance is the Farad (F). One Farad is defined as one Coulomb per Volt (1 F = 1 C V⁻¹). The Farad is a very large unit, so practical capacitors often have capacitances in the microfarad (µF = 10⁻⁶ F), nanofarad (nF = 10⁻⁹ F), or picofarad (pF = 10⁻¹² F) range.

Energy Stored in a Capacitor

When a capacitor is charged, work is done to move charge from one plate to the other against the electric field. This work is stored as electrical potential energy within the electric field between the capacitor plates. The energy stored depends on the capacitance and the potential difference across the capacitor.

E = ½CV² = ½QV = ½Q²/C

Capacitor in DC Circuits (Charging)

When a capacitor is connected in series with a resistor (R) to a DC voltage source (V_s), it begins to charge. Current flows, depositing charge on the plates. As charge builds up, the potential difference across the capacitor (V_C) increases, opposing the flow of current. The charging process is exponential, meaning the charge and voltage increase rapidly at first and then more slowly as the capacitor approaches full charge. The current decreases exponentially.

Q(t) = Q₀(1 - e⁻ᵗ/ᴿᶜ) V_C(t) = V_s(1 - e⁻ᵗ/ᴿᶜ) I(t) = I₀e⁻ᵗ/ᴿᶜ

Capacitor in DC Circuits (Discharging)

When a charged capacitor is connected across a resistor (R) without a voltage source, it discharges. Current flows out of the capacitor, and the charge on its plates decreases exponentially. The potential difference across the capacitor (V_C) also decreases exponentially. The current decreases exponentially in the opposite direction to charging.

Q(t) = Q₀e⁻ᵗ/ᴿᶜ V_C(t) = V₀e⁻ᵗ/ᴿᶜ I(t) = I₀e⁻ᵗ/ᴿᶜ

Time Constant (τ)

The time constant (τ) of an RC circuit is a characteristic time that determines the rate at which a capacitor charges or discharges. It is the product of the resistance (R) and the capacitance (C) in the circuit. For a charging capacitor, τ is the time taken for the charge or voltage to reach approximately 63.2% (1 - 1/e) of its maximum value. For a discharging capacitor, τ is the time taken for the charge or voltage to fall to approximately 36.8% (1/e) of its initial value.

τ = RC

Key facts to remember

1Capacitance (C) is defined as the ratio of charge (Q) to potential difference (V): C = Q/V.
2The SI unit of capacitance is the Farad (F), where 1 F = 1 C V⁻¹.
3The energy stored in a capacitor is given by E = ½CV² (or ½QV or ½Q²/C).
4A capacitor blocks direct current (DC) once fully charged but allows alternating current (AC) to pass (after the initial charging phase).
5The time constant (τ) of an RC circuit is τ = RC, and it has units of seconds.
6During charging, the charge and voltage across a capacitor increase exponentially towards their maximum values.
7During discharging, the charge and voltage across a capacitor decrease exponentially from their initial values.
8After one time constant (τ), a charging capacitor reaches approximately 63.2% of its maximum charge/voltage, and a discharging capacitor's charge/voltage falls to approximately 36.8% of its initial value.

Worked examples

Example 1

A capacitor stores 350 µC of charge when connected to a 24 V DC power supply. Calculate its capacitance.

IIdentify the given values: Charge, Q = 350 µC = 350 × 10⁻⁶ C; Voltage, V = 24 V.

IIRecall the formula for capacitance: C = Q/V.

IIISubstitute the values into the formula and calculate: C = (350 × 10⁻⁶ C) / (24 V).

IVC = 0.000014583 F.

Answer

The capacitance is 1.46 × 10⁻⁵ F or 14.6 µF (to 3 significant figures).

Always convert microcoulombs (µC) to coulombs (C) before calculation.

Example 2

A 22 µF capacitor is charged to a potential difference of 100 V. Calculate the energy stored in the capacitor.

IIdentify the given values: Capacitance, C = 22 µF = 22 × 10⁻⁶ F; Voltage, V = 100 V.

IIRecall the formula for energy stored in a capacitor: E = ½CV².

IIISubstitute the values into the formula and calculate: E = ½ × (22 × 10⁻⁶ F) × (100 V)².

IVE = ½ × (22 × 10⁻⁶) × 10000.

VE = 0.11 J.

Answer

The energy stored in the capacitor is 0.11 J.

Ensure units are in Farads and Volts for the energy calculation to yield Joules.

Example 3

A 100 µF capacitor is connected in series with a 1.5 kΩ resistor and a 12 V DC power supply.

I(a) Calculate the time constant of the circuit.

IIIdentify given values: C = 100 µF = 100 × 10⁻⁶ F; R = 1.5 kΩ = 1.5 × 10³ Ω.

IIIRecall the formula for time constant: τ = RC.

IVSubstitute values: τ = (1.5 × 10³ Ω) × (100 × 10⁻⁶ F) = 0.15 s.

V(b) Calculate the voltage across the capacitor after one time constant during charging.

VIAt t = τ, the voltage across the capacitor is V_C(τ) = V_s(1 - e⁻¹).

VIIV_C(τ) = 12 V × (1 - e⁻¹) = 12 V × (1 - 0.36788) = 12 V × 0.63212 = 7.585 V.

VIII(c) Calculate the charge on the capacitor after 0.2 seconds during charging.

9First, calculate the maximum charge Q₀ = CV_s = (100 × 10⁻⁶ F) × (12 V) = 1.2 × 10⁻³ C.

10Use the charging formula: Q(t) = Q₀(1 - e⁻ᵗ/ᴿᶜ).

11Q(0.2 s) = (1.2 × 10⁻³ C) × (1 - e⁻⁰·²/⁰·¹⁵).

12Q(0.2 s) = (1.2 × 10⁻³ C) × (1 - e⁻¹·³³³).

13Q(0.2 s) = (1.2 × 10⁻³ C) × (1 - 0.2636) = (1.2 × 10⁻³ C) × 0.7364 = 8.8368 × 10⁻⁴ C.

Answer

(a) The time constant is 0.15 s. (b) The voltage across the capacitor after one time constant is 7.59 V (to 3 s.f.). (c) The charge on the capacitor after 0.2 seconds is 8.84 × 10⁻⁴ C (to 3 s.f.).

Remember to use the correct exponential formula for charging (1 - e⁻ᵗ/ᴿᶜ) and discharging (e⁻ᵗ/ᴿᶜ).

Common mistakes

✗Failing to convert units (e.g., µF to F, kΩ to Ω, µC to C) before performing calculations.
✗Confusing the different forms of the energy stored formula (e.g., using E = ½QV when C and V are given, or vice versa).
✗Incorrectly applying the exponential charging and discharging formulas (e.g., using Q₀e⁻ᵗ/ᴿᶜ for charging instead of Q₀(1 - e⁻ᵗ/ᴿᶜ)).
✗Misinterpreting the meaning or significance of the time constant (τ).
✗Assuming that capacitors charge or discharge linearly with time, rather than exponentially.

Exam tips

★Always write down the formula you intend to use before substituting values; this can earn marks even if there's a calculation error.
★Pay meticulous attention to units. Convert all values to SI units (Farads, Ohms, Volts, Coulombs, Seconds) at the beginning of a problem.
★For problems involving exponential functions, ensure your calculator is set up correctly for natural logarithms (e^x).
★Practice the derivation of the energy stored formula (E = ½CV²), as it is a common exam question.
★Clearly label all variables and show all steps in your worked solutions to maximise potential marks.

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