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Capacitor Energy Storage

Capacitors store electrical energy in the form of an electric field between their plates. Understanding how capacitors store and release energy is crucial for many applications in electronics and electrical engineering.

Energy Storage Formula

Capacitor Energy Formula

$$E = \frac{1}{2}CV^2$$

Where:

Alternative Forms of the Energy Equation

Different Ways to Express Capacitor Energy

The energy stored in a capacitor can be expressed in several equivalent forms:

Form 1: In terms of Capacitance and Voltage

$$E = \frac{1}{2}CV^2$$

This is the most common form, useful when you know the capacitance and voltage.

Form 2: In terms of Charge and Voltage

$$E = \frac{1}{2}QV$$

Since Q = CV, this form is useful when you know the charge and voltage.

Form 3: In terms of Charge and Capacitance

$$E = \frac{Q^2}{2C}$$

This form is useful when you know the charge and capacitance but not the voltage.

Form 4: In terms of Electric Field

$$E = \frac{1}{2}\epsilon_0 E^2 \cdot Ad$$

Where ε₀ is the permittivity of free space, E is the electric field strength, A is the plate area, and d is the plate separation.

Converting Between Forms

To convert between these forms, use these relationships:

Derivation of the Energy Formula

Let's derive why the energy stored in a capacitor is given by \(E = \frac{1}{2}CV^2\):

Derivation Steps:

  1. Work done to charge: To charge a capacitor, we must move charge against the electric field
  2. Incremental work: \(dW = V \, dq\) where \(V\) is the voltage and \(dq\) is a small amount of charge
  3. Voltage relationship: \(V = \frac{q}{C}\) (from \(Q = CV\))
  4. Substitute: \(dW = \frac{q}{C} \, dq\)
  5. Integrate: \(W = \int_0^Q \frac{q}{C} \, dq = \frac{1}{C} \int_0^Q q \, dq\)
  6. Result: \(W = \frac{1}{C} \cdot \frac{Q^2}{2} = \frac{Q^2}{2C}\)
  7. Final form: Since \(Q = CV\), we get \(E = \frac{1}{2}CV^2\)

Physical Intuition

Think of charging a capacitor like filling a water tank:

Energy in Series and Parallel Capacitors

Series Capacitors

Total Energy:

$$E_{total} = \frac{1}{2}C_{eq}V^2$$

Where \(C_{eq}\) is the equivalent capacitance of the series combination.

Individual Capacitor Energy:

$$E_i = \frac{1}{2}C_iV_i^2$$

Where \(V_i\) is the voltage across each capacitor.

Parallel Capacitors

Total Energy:

$$E_{total} = \frac{1}{2}C_{eq}V^2$$

Where \(C_{eq}\) is the equivalent capacitance of the parallel combination.

Individual Capacitor Energy:

$$E_i = \frac{1}{2}C_iV^2$$

All capacitors have the same voltage \(V\) in parallel.

Worked Examples

Example 1: Single Capacitor Energy

Problem: A 10 μF capacitor is charged to 12 V. How much energy is stored?

Solution Steps:

  1. Given: C = 10 μF = 10 × 10⁻⁶ F, V = 12 V
  2. Formula: E = ½ × C × V²
  3. Substitute: E = ½ × (10 × 10⁻⁶) × (12)²
  4. Calculate: E = ½ × 10⁻⁵ × 144 = 7.2 × 10⁻⁴ J
  5. Convert: E = 0.72 mJ

Answer: The capacitor stores 0.72 millijoules of energy.

Example 1b: Using Different Energy Forms

Problem: A 10 μF capacitor is charged to 12 V. Calculate the energy using different forms of the equation.

Solution using different forms:

  1. Form 1 (CV²): E = ½ × (10 × 10⁻⁶) × (12)² = 0.72 mJ
  2. Form 2 (QV): Q = CV = 120 μC, E = ½ × (120 × 10⁻⁶) × 12 = 0.72 mJ
  3. Form 3 (Q²/2C): Q = 120 μC, E = (120 × 10⁻⁶)² / (2 × 10 × 10⁻⁶) = 0.72 mJ

Note: All forms give the same result, showing they are equivalent!

Example 2: Series Capacitors Energy

Problem: Two capacitors of 2 μF and 4 μF are connected in series and charged to 24 V. Find the total energy stored.

Solution Steps:

  1. Step 1 - Equivalent capacitance: \(\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\)
  2. Result: C_eq = 1.33 μF
  3. Step 2 - Total energy: E = ½ × C_eq × V²
  4. Substitute: E = ½ × (1.33 × 10⁻⁶) × (24)²
  5. Calculate: E = 0.384 mJ

Answer: The total energy stored is 0.384 millijoules.

Example 3: Parallel Capacitors Energy

Problem: Three capacitors of 1 μF, 2 μF, and 3 μF are connected in parallel and charged to 6 V. Find the total energy stored.

Solution Steps:

  1. Step 1 - Equivalent capacitance: C_eq = 1 + 2 + 3 = 6 μF
  2. Step 2 - Total energy: E = ½ × C_eq × V²
  3. Substitute: E = ½ × (6 × 10⁻⁶) × (6)²
  4. Calculate: E = 0.108 mJ

Answer: The total energy stored is 0.108 millijoules.

Interactive Energy Calculator

Calculate Capacitor Energy

Capacitor Parameters

10.0 μF
12 V

Energy Results

Applications of Capacitor Energy

Energy Storage Systems

Electronic Applications

High-Energy Applications

Energy Density and Efficiency

Energy Density Comparison

Storage Type Energy Density (J/kg) Advantages
Capacitors 10⁴ - 10⁵ Fast charge/discharge, long life
Batteries 10⁵ - 10⁶ High energy density, stable voltage
Supercapacitors 10⁴ - 10⁵ Very fast charge/discharge

Common Mistakes to Avoid

⚠️ Common Errors

Practice Problems

Practice Problem 1

Problem: A 5 μF capacitor stores 0.5 mJ of energy. What is the voltage across it?

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Solution:

  1. Given: C = 5 μF = 5 × 10⁻⁶ F, E = 0.5 mJ = 5 × 10⁻⁴ J
  2. Formula: E = ½CV² → V² = 2E/C
  3. Substitute: V² = 2 × (5 × 10⁻⁴) / (5 × 10⁻⁶) = 200
  4. Calculate: V = √200 = 14.14 V

Answer: 14.14 V

Practice Problem 2

Problem: Two capacitors of 3 μF and 6 μF are connected in parallel and charged to 9 V. What is the total energy stored?

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Solution:

  1. Step 1 - Equivalent capacitance: C_eq = 3 + 6 = 9 μF
  2. Step 2 - Total energy: E = ½ × C_eq × V²
  3. Substitute: E = ½ × (9 × 10⁻⁶) × (9)²
  4. Calculate: E = 0.3645 mJ

Answer: 0.365 mJ

Key Concepts Summary

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