Capacitors
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
A capacitor is two conducting plates that store equal and opposite charge. Capacitance C = Q/ΔV depends only on the geometry: C = κε0A/d. Bigger plates, closer together, with a dielectric between them, store more charge per volt.
1. Parallel plates
2. Connected or disconnected?
Still connected to the battery: ΔV stays fixed. If C changes, Q changes: Q = CΔV.
Disconnected: the charge has nowhere to go, so Q stays fixed. If C changes, ΔV changes: ΔV = Q/C.
A dielectric (insulator) between the plates multiplies C by κ.
C doesn’t depend on Q or ΔV. A capacitor holding no charge still has the same capacitance.
3. Example, then try it
Air-filled plates, each 0.020 m², are 1.0 mm apart. Find C and the charge stored at 12 V.
C = (8.85 × 10⁻¹²)(0.020) ÷ 0.0010 ≈ 1.8 × 10⁻¹⁰ F (180 pF).
Q = CΔV = (1.77 × 10⁻¹⁰)(12) ≈ 2.1 × 10⁻⁹ C.
A 20 µF capacitor is charged by a 9.0 V battery. Find the charge and stored energy. The battery is then disconnected and the plates are pulled to twice their separation. Find the new C, ΔV and energy.
Charged: Q = (20 × 10⁻⁶)(9.0) = 180 µC; U = ½(20 × 10⁻⁶)(9.0)² = 8.1 × 10⁻⁴ J.
Pulled apart: d doubles, so C halves to 10 µF. Disconnected, Q stays 180 µC.
New voltage: ΔV = Q/C = 180 µC ÷ 10 µF = 18 V.
New energy: U = ½QΔV = ½(180 × 10⁻⁶)(18) = 1.6 × 10⁻³ J, doubled. The extra came from the work you did pulling the attracting plates apart.
Answer: 180 µC and 8.1 × 10⁻⁴ J; then 10 µF, 18 V and 1.6 × 10⁻³ J.
4. Common mistakes
A capacitor stays connected to a battery while a dielectric with κ = 2 is slid between its plates. What happens to the charge on the plates?
The dielectric doubles C. Connected to the battery, ΔV stays fixed, so Q = CΔV doubles.
Ready to practice?
4 questions on Capacitors, each with a hint and an explanation for every wrong answer.