This AP Physics 2 lesson on capacitors notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on capacitors and check yourself with the quiz.
Quick answer: A capacitor stores charge on two plates separated by an insulator, with capacitance C = Q/V in farads. for parallel plates C = κε₀A/d. the stored energy is U = ½CV², and a dielectric of constant κ multiplies the capacitance by κ by reducing the field between the plates.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10
Capacitors notes: the key ideas
What is a capacitor?
A capacitor is two conductors with an insulator between them. When you connect it to a battery, electrons flow onto one plate and leave the other, so the plates hold equal and opposite charges +Q and −Q. Moreover, the battery sets the voltage across the plates. In addition, capacitors sit inside phone touchscreens, camera flashes and radio tuners, where they store charge and energy briefly and release it quickly.
Capacitance is the charge stored per volt, C = Q/V, which we measure in farads (1 F = 1 C/V). Practical capacitors are microfarads (10⁻⁶ F), nanofarads (10⁻⁹ F) or picofarads (10⁻¹² F). Q means the size of the charge on one plate.
How does a parallel-plate capacitor depend on its shape?
Large plates hold more charge, so capacitance grows with area A. Plates that are close together make a strong field for the same voltage and so can hold more charge, so capacitance falls as the gap d grows. Also, this leads to a simple geometric rule.
C = κε₀A / d
C is the capacitance (F), κ the dielectric constant (no unit; κ = 1 for vacuum), ε₀ = 8.85 × 10⁻¹² F/m the permittivity of free space, A the area of one plate (m²) and d the plate separation (m).
Wanjiru builds a capacitor from two square plates of side 10 cm, 1.0 mm apart in air, and connects it to a 12 V battery. Then, find C, Q and the stored energy.
A = 0.10 × 0.10 = 1.0 × 10⁻² m². C = (8.85 × 10⁻¹²)(1.0 × 10⁻²) / (1.0 × 10⁻³) = 8.9 × 10⁻¹¹ F (89 pF). Q = CV = 8.85 × 10⁻¹¹ × 12 = 1.1 × 10⁻⁹ C. U = ½CV² = 0.5 × 8.85 × 10⁻¹¹ × 144 = 6.4 × 10⁻⁹ J.
How much energy does a capacitor store?
To charge a capacitor, you do work to push each extra bit of charge against the voltage already on the plates. The voltage rises in proportion to the charge, so the graph of V against Q is a straight line. Also, the energy equals the area under it, which is a triangle.
U = ½QV = ½CV² = Q² / 2C. Moreover, the electric field between the plates stores the energy. Doubling the voltage on a fixed capacitor makes the energy four times larger.
The flash in Aarav’s camera uses a 330 μF capacitor charged to 300 V. Next, find the energy and the average power if it discharges in 2.0 ms.
U = ½CV² = 0.5 × 330 × 10⁻⁶ × 300² = 15 J. P = U/t = 14.85 / 2.0 × 10⁻³ = 7.4 × 10³ W. A small battery cannot give that power directly, but the capacitor can.
What does a dielectric do?
A dielectric is an insulator that sits between the plates. In addition, its molecules turn in the field, and induced charge forms on its surfaces that opposes the field of the plates. The net field is smaller, so for the same charge the voltage drops, and the capacitance rises by the factor κ.
| Material | Dielectric constant κ |
|---|---|
| Vacuum | 1 (exactly) |
| Air | 1.0006 |
| Polythene | 2.3 |
| Paper | 3.7 |
| Glass | 6 |
| Pure water (20 °C) | 80 |
Capacitors with a dielectric: battery connected or not
What happens next depends on whether the battery stays connected. With the battery connected, V stays fixed, so C rises and so do Q and U. If you isolate the capacitor, Q stays fixed, so C rises but V and U fall.
Capacitors: a worked example with a dielectric
The 89 pF capacitor above has a sheet of paper (κ = 3.7) slid between the plates while the 12 V battery stays connected. Then, find the new C, Q and U.
C = 3.7 × 88.5 = 3.3 × 10² pF. Q = 3.27 × 10⁻¹⁰ × 12 = 3.9 × 10⁻⁹ C. U = ½CV² = 0.5 × 3.27 × 10⁻¹⁰ × 144 = 2.4 × 10⁻⁸ J, which is 3.7 times the earlier energy.
Capacitors: a common mistake and a quick tip
Do not say that a capacitor stores a net charge. The plates carry +Q and −Q, so the total is zero. Also check which quantity stays fixed before inserting a dielectric: the battery fixes V, while disconnection fixes Q.
Before changing a capacitor, write down what stays fixed: V if the battery stays connected, Q if it does not. Then use C = Q/V to find the other quantity.
- Capacitance C = Q/V in farads; one plate holds +Q and the other −Q
- Parallel plates: C = κε₀A/d, with ε₀ = 8.85 × 10⁻¹² F/m
- Energy stored: U = ½QV = ½CV² = Q²/2C
- A dielectric multiplies C by κ; battery connected fixes V, disconnected fixes Q
- Too strong a field makes the dielectric break down and spark
Storing charge and energy between two plates: frequently asked questions
What is a farad?
A farad is one coulomb of stored charge per volt across the capacitor. It is a very large unit, so manufacturers give most capacitors in microfarads, nanofarads or picofarads.
Why does a dielectric increase capacitance?
Its molecules polarise and create a field opposite to the field of the plates. The net field drops, so for the same charge the voltage is smaller, which means a larger Q/V.
Where does a capacitor store its energy?
In the electric field in the space between the plates. A stronger field and a larger volume store more energy, and the energy is U = ½CV².
What happens if you pull apart the plates of a charged, isolated capacitor?
The charge stays the same but C falls, so V rises. Energy increases because you do work pulling the opposite charges apart.
Why do capacitors have a maximum voltage rating?
If the field becomes too strong, the dielectric ionises and conducts, causing a spark or permanent damage. Also, the rating keeps the field below the breakdown value.
