Capacitance and Dielectrics – Study Notes

Capacitance and Dielectrics

Capacitors and Capacitance

A capacitor is a device that stores electric potential energy by separating positive and negative charges onto two nearby surfaces. Capacitors are fundamental components in electric circuits, especially for storing and releasing energy quickly.

• Capacitor Structure: Consists of two conductive plates separated by an insulating material (dielectric).

• Energy Storage: Capacitors store potential energy due to the separation of charges.

• Connection to Battery: When connected to a battery, the voltage across the capacitor equals the battery voltage.

• Charge on Capacitor: The charge stored is proportional to the voltage: $Q = CV$

• Capacitance (C): Measures the ability of a capacitor to store charge per unit voltage. Units: Farads (F).

• Effect of Capacitance: Larger capacitance means more charge can be stored for the same voltage.

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

Example: A 3 F capacitor connected to a 9 V battery stores $Q = (3\,\text{F})(9\,\text{V}) = 27\,\text{C}$ of charge.

Parallel Plate Capacitors

Parallel plate capacitors are the simplest and most common type, consisting of two flat plates of area $A$ separated by a distance $d$.

• Electric Field: The electric field between the plates is uniform and directed from the positive to the negative plate.

• Field Outside: The electric field outside the plates is nearly zero (for large plates and small separation).

• Equipotential Surfaces: Surfaces between the plates where the electric potential is constant and parallel to the plates.

Formulas:

• Electric field between plates: $E = \frac{V}{d} = \frac{Q}{\epsilon_0 A}$

• Capacitance: $C = \epsilon_0 \frac{A}{d}$, where $\epsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}$

Example: For plates of area $5\,\text{cm}^2$ ($5 \times 10^{-4}\,\text{m}^2$), separation $10\,\text{mm}$ ($0.01\,\text{m}$), and $V = 100\,\text{V}$:

• $C = (8.85 \times 10^{-12}) \frac{5 \times 10^{-4}}{0.01} = 4.43 \times 10^{-13}\,\text{F}$

• $Q = CV = (4.43 \times 10^{-13})(100) = 4.43 \times 10^{-11}\,\text{C}$

• $E = \frac{V}{d} = \frac{100}{0.01} = 10,000\,\text{V/m}$

Energy Stored in a Capacitor

When a capacitor is charged, it stores energy in the electric field between its plates. The energy can be calculated in several equivalent ways:

• $U = \frac{1}{2}CV^2$

• $U = \frac{1}{2}QV$

• $U = \frac{Q^2}{2C}$

Energy Density: The energy per unit volume in the electric field is $u = \frac{1}{2}\epsilon_0 E^2$.

Example: For a capacitor with $C = 2\,\text{F}$ and $V = 10\,\text{V}$, $U = \frac{1}{2}(2)(10^2) = 100\,\text{J}$.

Capacitors in Series and Parallel

Capacitors can be combined in circuits to achieve desired total capacitance values.

• Series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$

• Parallel: $C_{eq} = C_1 + C_2 + \cdots$

Key Points:

• In series, all capacitors share the same charge but split the voltage.

• In parallel, all capacitors share the same voltage but split the charge.

Example: Two capacitors, $C_1 = 2\,\text{F}$ and $C_2 = 3\,\text{F}$, in series: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $C_{eq} = 1.2\,\text{F}$.

Dielectrics

A dielectric is an insulating material placed between the plates of a capacitor. It increases the capacitance by reducing the effective electric field.

• Dielectric Constant (\( \kappa \)): $C = \kappa C_0$, where $C_0$ is the capacitance without the dielectric.

• Dielectrics always increase capacitance ($\kappa \geq 1$).

• Electric field inside: $E = \frac{E_0}{\kappa}$

Constant Charge vs. Constant Voltage:

• If the capacitor is disconnected from the battery (constant $Q$), inserting a dielectric decreases $V$.

• If the capacitor remains connected (constant $V$), inserting a dielectric increases $Q$.

Example: A 3 F capacitor connected to a 9 V battery has $Q = 27\,\text{C}$. If a dielectric with $\kappa = 2$ is inserted, $C = 6\,\text{F}$ and $Q = 54\,\text{C}$ (if still connected).

Dielectric Breakdown

Dielectric breakdown occurs when the electric field in the dielectric exceeds a critical value, causing it to become conductive.

• Dielectric Strength: Maximum electric field a dielectric can withstand without breakdown (e.g., air: $3 \times 10^6\,\text{V/m}$).

• Breakdown leads to current flow and possible damage to the capacitor.

• Lightning is a natural example of dielectric breakdown in air.

Example: For air with dielectric strength $3 \times 10^6\,\text{V/m}$ and $V = 100\,\text{V}$, the minimum plate separation is $d = \frac{V}{E_{max}} = \frac{100}{3 \times 10^6} = 3.33 \times 10^{-5}\,\text{m}$.