18.5 Capacitors and Dielectrics

Capacitors store electric charge and energy by holding opposite charges on two conductors separated by an insulating material (a dielectric). They show up everywhere in circuits, from camera flashes to power supplies, and understanding how they work ties together ideas about electric fields, potential difference, and energy storage.

Capacitor Fundamentals

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Capacitance and energy calculations

Capacitance ($C$) measures how much charge a capacitor can store per volt of potential difference across its plates. The unit is the farad (F), and the defining relationship is:

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

where $Q$ is the magnitude of charge on either plate and $V$ is the voltage between them. The two plates always carry equal and opposite charges (+$Q$ and -$Q$), so you can rearrange to find the charge stored:

$Q = CV$

Energy stored in a capacitor depends on both capacitance and voltage. Two equivalent forms are useful depending on what you're given:

$U = \frac{1}{2}CV^2 \qquad \text{or} \qquad U = \frac{1}{2}\frac{Q^2}{C}$

Energy is measured in joules (J).

• A 10 µF capacitor charged to 5 V stores $U = \frac{1}{2}(10 \times 10^{-6})(5^2) = 125 \text{ µJ}$
• A 100 pF capacitor holding 2 nC of charge has a voltage of $V = \frac{Q}{C} = \frac{2 \times 10^{-9}}{100 \times 10^{-12}} = 20 \text{ V}$

Practice rearranging these equations freely. Exam problems often give you two of the three quantities ($C$, $Q$, $V$) and ask for the third, then follow up with an energy calculation.

Energy density refers to the energy stored per unit volume of the electric field between the plates. This concept becomes more important in later units, but for now just know it exists and depends on the field strength.

Effects of dielectrics on capacitors

A dielectric is an insulating material placed between the plates of a capacitor. Common examples include air, paper, plastic, glass, and ceramic. Dielectrics do three important things:

1. They increase capacitance. Each dielectric material has a dielectric constant ($\kappa$, the Greek letter kappa), which tells you by what factor the capacitance increases compared to a vacuum (or air, which is very close to vacuum):

$C = \kappa C_0$

where $C_0$ is the capacitance without the dielectric. So a capacitor with $\kappa = 2.5$ stores 2.5 times as much charge at the same voltage.

2. They reduce the electric field between the plates. The dielectric partially cancels the applied field through polarization of its molecules. The reduced field is:

$E = \frac{E_0}{\kappa}$

Why does this happen? When the external field is applied, the molecules in the dielectric shift slightly so that their positive ends face the negative plate and vice versa. This creates a small internal field opposing the applied field, weakening the net field inside.

3. They increase the maximum voltage the capacitor can handle. Every dielectric has a dielectric strength, which is the maximum electric field it can withstand before it breaks down and starts conducting. Mica, for example, has a dielectric strength of 100–200 MV/m, which is why it's used in high-voltage capacitors.

Capacitor Geometry and Capacitance

Capacitor geometry vs capacitance

The shape and size of a capacitor determine its capacitance. The general rule across all geometries: more plate area means more capacitance, and less separation between plates means more capacitance. Think of it this way: a larger area gives more room to spread charge, and a smaller gap makes it easier for the plates to "feel" each other's influence.

Parallel plate capacitors are the most common and the easiest to analyze:

$C = \frac{\kappa \varepsilon_0 A}{d}$

where $A$ is the plate area, $d$ is the separation, and $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \text{ F/m}$). Notice that $C$ is directly proportional to $A$ and inversely proportional to $d$. Doubling the area doubles the capacitance; halving the separation also doubles it.

• A parallel plate capacitor with $A = 100 \text{ cm}^2 = 0.01 \text{ m}^2$ and $d = 1 \text{ mm}$ gives $C = \frac{(8.85 \times 10^{-12})(0.01)}{0.001} = 8.85 \text{ pF}$
• Reducing the separation to 0.5 mm doubles the capacitance to 17.7 pF

Cylindrical capacitors (like coaxial cables) depend on the length $l$ and the ratio of outer radius $r_2$ to inner radius $r_1$:

$C = \frac{2\pi \kappa \varepsilon_0 l}{\ln\left(\frac{r_2}{r_1}\right)}$

Spherical capacitors depend on the inner and outer radii:

$C = 4\pi \kappa \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1}$

For an honors course, you should be comfortable with the parallel plate formula and understand the cylindrical and spherical formulas conceptually. The key pattern is the same in all three: capacitance grows with surface area and shrinks with greater separation between conductors.

Electrostatic Induction and Displacement Current

Electrostatic induction occurs when a charged object causes charge to redistribute in a nearby conductor without direct contact. This is the same mechanism behind how dielectrics polarize: the external field pushes charges around, creating regions of induced positive and negative charge.

Displacement current is a concept introduced by Maxwell to explain how current can appear to flow "through" a capacitor in an AC circuit, even though no actual charge crosses the dielectric gap. It's defined as the rate of change of the electric field (or equivalently, the changing electric flux) between the plates. This ensures that the current entering one plate equals the effective current leaving the other, maintaining continuity in the circuit. You'll revisit this idea in more depth when you study electromagnetic waves.