recall and use W = 21QV = 21CV2

Capacitors & Capacitance – A‑Level Physics 9702

⚡️ What is a capacitor? Think of it as a tiny battery that can store electric charge, just like a water tank stores water. The amount of charge it can hold for a given voltage is called its capacitance.

Capacitance (C)

Capacitance is defined as the ratio of the charge (Q) on one plate to the potential difference (V) between the plates:

\$C = \frac{Q}{V}\$

Units: farads (F). 1 F = 1 C V⁻¹.

💡 Analogy: If voltage is the water pressure, capacitance is the size of the tank. A larger tank (higher C) can hold more water (charge) for the same pressure (voltage).

Parallel‑Plate Capacitor

  • Two flat plates of area A separated by distance d.

  • Dielectric constant ε_r (relative permittivity).

  • Capacitance formula:

    \$C = \varepsilon0 \varepsilonr \frac{A}{d}\$

    where ε₀ = 8.85 × 10⁻¹² F m⁻¹.

🧮 Quick example: Two plates 0.02 m² apart, 0.5 mm apart, air dielectric (ε_r ≈ 1).

\$C = 8.85\times10^{-12}\,\frac{0.02}{5\times10^{-4}} \approx 3.5\times10^{-10}\,\text{F} = 0.35\,\text{nF}\$

Energy Stored (W)

The energy stored in a capacitor is the work done to charge it:

\$W = \frac{1}{2} QV = \frac{1}{2} CV^2\$

(Always remember the ½ factor!)

⚡️ Why the ½? Because the voltage across the capacitor increases linearly from 0 to V as it charges, so the average voltage during charging is V/2.

Exam Tips Box

✔️ Remember the key formulas:

  • Capacitance: \$C = Q/V\$

  • Parallel‑plate: \$C = \varepsilon0 \varepsilonr A/d\$

  • Energy: \$W = ½ CV^2\$

✔️ Units check: Farads (F), Coulombs (C), Volts (V), Joules (J).

✔️ Step‑by‑step:

  1. Identify known quantities.

  2. Choose the correct formula.

  3. Insert values and calculate.

  4. Check units and order of magnitude.

✔️ Common pitfalls: Forgetting the ½ in the energy formula; mixing up Q and V when solving for C.

Sample Problem

A parallel‑plate capacitor has plates of area 0.01 m², separated by 2 mm, with a dielectric of ε_r = 2.5. It is charged to 12 V. Calculate:

  1. Capacitance.

  2. Stored energy.

Solution:

  1. Capacitance:

    \$C = 8.85\times10^{-12}\times2.5\times\frac{0.01}{2\times10^{-3}} \approx 1.1\times10^{-10}\,\text{F} = 110\,\text{pF}\$

  2. Energy:

    \$W = \frac{1}{2} C V^2 = 0.5 \times 1.1\times10^{-10} \times 12^2 \approx 7.9\times10^{-9}\,\text{J} = 7.9\,\text{nJ}\$

Summary

• Capacitance measures a capacitor’s ability to store charge.

• For a parallel‑plate capacitor, C increases with plate area and dielectric constant, and decreases with plate separation.

• Energy stored is half the product of capacitance and the square of the voltage.

• Always keep track of units and the ½ factor in energy calculations.

🚀 Good luck on the exam – you’ve got this!