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Capacitor Basics and Capacitance

A capacitor is a device that stores electrical energy by separating positive and negative charges on two conducting plates. Capacitance is the measure of a capacitor's ability to store charge per unit voltage.

What is a Capacitor?

Definition

A capacitor consists of two conducting plates separated by an insulating material (dielectric) or vacuum/air. When a voltage is applied, equal and opposite charges accumulate on the plates, creating an electric field between them.

Capacitors are fundamental components in electrical circuits, used for energy storage, filtering, timing, and many other applications. The ability of a capacitor to store charge depends on its geometry and the material between the plates.

Capacitor Structure
Basic capacitor structure with two conducting plates separated by a dielectric.

Capacitance Definition

Fundamental Formula

Capacitance is defined as the ratio of charge stored to the voltage applied:

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

Where:

This relationship shows that:

Parallel Plate Capacitor

Parallel Plate Capacitance

For a parallel plate capacitor with area A and separation d:

$$C = \frac{\epsilon_0 A}{d}$$

Where:

This formula shows that capacitance depends on:

Parallel Plate Capacitor
Parallel plate capacitor with area A and separation d.

Properties of Capacitors

Fundamental Properties

Charge and Voltage Relationship

Charge-Voltage Relationship

The charge stored on a capacitor is directly proportional to the voltage:

$$Q = CV$$

This means:

Electric Field Between Plates

Electric Field Formula

For a parallel plate capacitor, the electric field between plates is:

$$E = \frac{V}{d} = \frac{\sigma}{\epsilon_0}$$

Where σ is the surface charge density on the plates.

This shows that:

Worked Examples

Example 1: Calculating Capacitance

Problem: A parallel plate capacitor has plates of area 0.01 m² separated by 1.0 mm. What is its capacitance?

Solution Steps:

  1. Given: A = 0.01 m², d = 1.0 × 10⁻³ m, ε₀ = 8.85 × 10⁻¹² F/m
  2. Formula: C = ε₀A/d
  3. Substitute: C = (8.85 × 10⁻¹²)(0.01)/(1.0 × 10⁻³)
  4. Calculate: C = 8.85 × 10⁻¹¹ F = 88.5 pF

Answer: The capacitance is 88.5 pF.

Example 2: Charge Storage

Problem: A 100 μF capacitor is charged to 12 V. How much charge is stored on each plate?

Solution Steps:

  1. Given: C = 100 × 10⁻⁶ F, V = 12 V
  2. Formula: Q = CV
  3. Substitute: Q = (100 × 10⁻⁶)(12)
  4. Calculate: Q = 1.2 × 10⁻³ C = 1.2 mC

Answer: Each plate stores 1.2 mC of charge (positive on one plate, negative on the other).

Example 3: Electric Field Between Plates

Problem: A capacitor has plates separated by 2.0 mm with a voltage of 50 V across them. What is the electric field between the plates?

Solution Steps:

  1. Given: V = 50 V, d = 2.0 × 10⁻³ m
  2. Formula: E = V/d
  3. Substitute: E = 50/(2.0 × 10⁻³)
  4. Calculate: E = 2.5 × 10⁴ N/C

Answer: The electric field between the plates is 2.5 × 10⁴ N/C, directed from the positive to negative plate.

Common Mistakes to Avoid

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Practice Problems

Practice Problem 1

Problem: A capacitor stores 5.0 μC of charge when connected to a 10 V battery. What is its capacitance?

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Solution:

  1. Given: Q = 5.0 × 10⁻⁶ C, V = 10 V
  2. Formula: C = Q/V
  3. Substitute: C = (5.0 × 10⁻⁶)/10
  4. Calculate: C = 5.0 × 10⁻⁷ F = 0.5 μF

Answer: 0.5 μF

Practice Problem 2

Problem: A parallel plate capacitor has plates of area 0.05 m² separated by 0.5 mm. What is its capacitance?

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Solution:

  1. Given: A = 0.05 m², d = 0.5 × 10⁻³ m, ε₀ = 8.85 × 10⁻¹² F/m
  2. Formula: C = ε₀A/d
  3. Substitute: C = (8.85 × 10⁻¹²)(0.05)/(0.5 × 10⁻³)
  4. Calculate: C = 8.85 × 10⁻¹⁰ F = 885 pF

Answer: 885 pF

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