Also, this AP Physics C lesson on capacitance notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on capacitance and check yourself with the quiz.
Quick answer: Capacitance is the charge stored per unit potential difference, C = Q/ΔV, measured in farads. Moreover, it depends on geometry: ε₀A/d for parallel plates and 2πε₀L/ln(b/a) for coaxial cylinders. A dielectric of constant κ multiplies C by κ, and the stored energy is U = ½CΔV² = ½ε₀E² per unit volume.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 10
Capacitance notes: the key ideas
What is capacitance and how do you calculate it?
Two conductors that carry equal and opposite charge ±Q have a potential difference ΔV between them. For any fixed shape, ΔV is proportional to Q, because doubling the charge doubles the field everywhere. Also, the constant of proportionality depends only on geometry and on the material between the conductors.
Capacitance is the charge stored per unit potential difference: C = Q/ΔV. In addition, its unit is the farad, 1 F = 1 C/V.
Capacitance formulas
C = Q/ΔV C = ε₀A/d (parallel plates) U = ½QΔV = ½CΔV² = Q²/2C u = ½ε₀E²
Q means the charge on one plate (C), ΔV the potential difference (V), A the plate area (m²), d the gap (m), U the stored energy (J), u the energy density (J/m³), and ε₀ = 8.85 × 10⁻¹² F/m.
The method is always the same: put charge ±Q on the conductors, find E with Gauss’s law, integrate to get ΔV = −∫E·dl, then divide. The geometry fixes the answer.
Data: plates 20 cm × 10 cm, separated by 1.0 mm of air, connected to 12 V. Find C and Q.
Method: A = 0.020 m², so C = ε₀A/d = (8.85 × 10⁻¹²)(0.020)/(1.0 × 10⁻³). Then Q = CΔV.
Answer: C = 177 pF and Q = 2.1 nC.
Data: a coaxial cable has inner radius a = 1.0 mm, outer radius b = 4.0 mm and length L = 2.0 m. Find C.
Method: with λ = Q/L, E = λ/(2πε₀r) and ΔV = (λ/2πε₀) ln(b/a), so C = 2πε₀L/ln(b/a).
Answer: C = 80 pF, since 2πε₀L = 1.112 × 10⁻¹⁰ F and ln 4 = 1.386.
Capacitance of common geometries
| Geometry | Capacitance C |
|---|---|
| Parallel plates (area A, gap d) | ε₀A/d |
| Isolated sphere, radius R | 4πε₀R |
| Concentric spheres, radii a < b | 4πε₀ab/(b − a) |
| Coaxial cylinders, length L | 2πε₀L/ln(b/a) |
| Any of these filled with dielectric κ | κ times the empty value |
What does a dielectric do to a capacitor?
A dielectric is an insulator whose molecules polarise in a field. Its surfaces gain bound charge that opposes the field of the plates, so the net field falls to E₀/κ, where κ is the dielectric constant. For the same charge, ΔV drops by κ, so the capacitance rises by the factor κ.
Energy stored in a capacitor
The field stores the energy, not the plates alone. each cubic metre holds u = ½ε₀E² in vacuum, and u = ½κε₀E² in a dielectric. Moreover, integrating u over the volume reproduces U = Q²/2C.
Data: we charge the 177 pF capacitor to 12 V (Q = 2.12 nC), disconnect it, and then fill it with a dielectric of κ = 4. Find the new V and the energy change.
Method: Q stays fixed, so V = Q/(κC₀) and U = Q²/(2κC₀) = U₀/κ, with U₀ = ½C₀V₀² = 12.7 nJ.
Answer: V = 3.0 V and U falls from 12.7 nJ to 3.2 nJ; the slab pulls in and does work as it enters.
Do not use one rule for every case. With the battery connected, V stays fixed and Q changes. Furthermore, with the battery disconnected, Q stays fixed and V changes. Decide which is constant before you apply a dielectric or change d.
Write down the constant quantity first, V or Q, then express every other change through C = κε₀A/d.
- C = Q/ΔV (1 F = 1 C/V); parallel plates C = ε₀A/d.
- Find C by Gauss’s law for E, then ΔV = −∫E·dl, then C = Q/ΔV.
- U = ½QΔV = ½CΔV² = Q²/2C; energy density u = ½ε₀E².
- Dielectric κ: C = κC₀ and field E = E₀/κ at the same free charge.
- Furthermore, battery connected: V fixed. Battery disconnected: Q fixed.
Capacitors, energy storage and dielectrics: frequently asked questions
Does capacitance depend on charge or voltage?
Moreover, no. Q and ΔV change together, but their ratio depends only on geometry and the material between the conductors. Doubling Q doubles ΔV and leaves C unchanged.
Where is the energy of a capacitor stored?
Also, it is stored in the electric field between the conductors, with density u = ½ε₀E² (or ½κε₀E² in a dielectric). Also, integrating u over the gap gives the same result as ½CΔV².
Why does a dielectric increase capacitance?
Polarisation puts bound charge on the surfaces that cancels part of the field. For the same free charge the field and potential difference drop, so Q/ΔV is larger by the factor κ.
What happens when a dielectric is inserted with the battery still attached?
The voltage stays fixed, so charge Q = κC₀V flows onto the plates and the stored energy rises by the factor κ. The battery supplies the extra energy.
How do capacitors combine?
In parallel the capacitances add, C = C₁ + C₂, because the voltage is shared. In series the reciprocals add, 1/C = 1/C₁ + 1/C₂, because the charge is shared.
