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: $$C = \varepsilon_0 \varepsilon_r \frac{A}{d}$$ 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: $C = \varepsilon_0 \varepsilon_r A/d$
• 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!