Sign In Sign Up
← Back to Physics
AS & A-Level Physics 19 — Capacitors and Time Constants

AS & A-Level Physics 19 — Capacitors and Time Constants

Public

Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 19 of 26: Capacitors and time constants. Concepts, worked applications and misconception checks.

Physics EN A-Level
48 cards
Study this deck on Deckloop

Preview Cards

A sample of cards from this deck.

Example Explainer

A sample of the AI explainer you can generate for cards in this deck.

Defining Capacitance: Charge and Potential Difference

A capacitor is an electrical component designed to store electric charge and, consequently, electrical energy. It typically consists of two conductive plates separated by an insulating material known as a dielectric. When a potential difference is applied across these plates, charge accumulates on them – one plate becomes positively charged, and the other negatively charged, with equal magnitude. Capacitance, symbolised by CC, quantifies a capacitor's ability to store charge. It is formally defined as the ratio of the magnitude of charge, QQ, stored on one plate to the potential difference, VV, across the plates. This relationship is expressed by the formula C=Q/VC = Q/V. The SI unit for capacitance is the Farad (F), where 1 Farad is equivalent to 1 Coulomb per Volt (1 F=1 C V−11 \text{ F} = 1 \text{ C V}^{-1}). A larger capacitance value indicates that more charge can be stored for a given potential difference.

Key points

  • Capacitance (CC) is the ratio of charge (QQ) stored on one plate to the potential difference (VV) across the plates: C=Q/VC = Q/V.
  • The SI unit for capacitance is the Farad (F), where 1 F=1 C V−11 \text{ F} = 1 \text{ C V}^{-1}.
  • Capacitors store electric charge and electrical energy.
  • The insulating material (dielectric) between the plates increases capacitance.

Worked example

Question

A test instrument records these charge–voltage pairs for a component: (0 V, 0 μC), (3.0 V, 12 μC), and (6.0 V, 24 μC). Treat it as an ideal capacitor. Infer its capacitance from the gradient and predict the charge at 9.0 V.

Solution

1. Capacitance is the charge–voltage gradient: C=ΔQ/ΔV=(24−12)/(6−3)=4.0C=\Delta Q/\Delta V=(24-12)/(6-3)=4.0 μC V⁻¹.
2. One μC V⁻¹ is one μF, so C=4.0C=4.0 μF.
3. At 9.0 V, Q=CV=4.0(9.0)=36Q=CV=4.0(9.0)=36 μC. This extrapolation assumes capacitance remains constant.

C = 4.0 μF and Q = 36 μC.

Common pitfalls

  • Confusing the charge stored with the total charge: QQ refers to the magnitude of charge on one plate, not the sum of charges on both plates.
  • Incorrect unit conversions: Always ensure charge is in Coulombs (C) and potential difference in Volts (V) to obtain capacitance in Farads (F).

Prerequisites

  • Understanding electric potential difference is fundamental to defining capacitance.
  • Knowledge of electric current as a flow of charge carriers is necessary to understand charge accumulation.
  • Ability to interpret circuit diagrams containing capacitors is essential for practical application.
  • Basic arithmetic skills are required for calculations involving charge, potential difference, and capacitance.
Further resources