Capacitance

General physics

Capacitance and Capacitors — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Primary Objectives	Specific Outcomes
Conceptual Understanding	• Understand the concept of capacitance and its role in electrical circuits • Explain how capacitors store and release electrical energy • Describe the relationship between charge, voltage, and capacitance
Mathematical Applications	• Apply mathematical formulas to calculate capacitance, charge, and energy • Use C = Q/V and related equations effectively • Solve problems involving parallel plate capacitors
Practical Applications	• Analyze capacitor behavior in DC and AC circuits • Evaluate energy storage capabilities of different capacitor types • Apply knowledge to real-world electronic devices

🗣️ Language Objectives

Language Objectives

Language Skills	Learning Targets
Scientific Vocabulary	Use technical terms related to capacitance and electrical storage accurately in English
Mathematical Communication	Express capacitance calculations and relationships using precise mathematical language
Explanatory Skills	Describe capacitor behavior and energy storage mechanisms clearly in English
Problem-Solving Communication	Communicate solution strategies and reasoning for capacitance problems effectively

📚 Key Terms

Key Terms

English	Russian (Русский)	Kazakh (Қазақша)
Capacitance	Ёмкость	Сыйымдылық
Capacitor	Конденсатор	Конденсатор
Dielectric	Диэлектрик	Диэлектрик
Electric field	Электрическое поле	Электр өрісі
Energy storage	Накопление энергии	Энергия жинау
Farad	Фарад	Фарад
Parallel plates	Параллельные пластины	Параллель жазықтықтар
Permittivity	Проницаемость	Өтімділік
Potential difference	Разность потенциалов	Потенциалдар айырмасы
Charge accumulation	Накопление заряда	Заряд жинауы

🎴 Study Cards

Capacitance Study Cards

Basic Capacitance

C = Q/V

C: Capacitance (F)

Q: Charge (C)

V: Voltage (V)

Unit: Farad (F)

Parallel Plate Capacitor

C = ε₀εᵣA/d

ε₀: Permittivity of free space

εᵣ: Relative permittivity

A: Plate area

d: Plate separation

Energy Storage

U = ½CV²

Also: U = ½QV

Also: U = Q²/(2C)

Energy in Joules (J)

Series Capacitors

1/C_total = 1/C₁ + 1/C₂ + …

• Same charge on each

• Voltages add up

• Total capacitance decreases

Parallel Capacitors

C_total = C₁ + C₂ + …

• Same voltage across each

• Charges add up

• Total capacitance increases

Constants

ε₀ = 8.85 × 10⁻¹² F/m

Permittivity of free space

1 Farad = 1 C/V

Very large unit in practice

📖 Glossary

Glossary

Capacitance: The ability of a component to store electrical charge. It is defined as the ratio of the charge stored to the potential difference across the component.

Translation

Russian: Ёмкость — способность компонента накапливать электрический заряд. Определяется как отношение накопленного заряда к разности потенциалов на компоненте.
Kazakh: Сыйымдылық — компоненттің электр зарядын жинау қабілеті. Жиналған зарядтың компонентінде потенциалдар айырмасына қатынасы ретінде анықталады.

Capacitor: An electronic component consisting of two conducting plates separated by an insulating material (dielectric) that stores electrical energy in an electric field.

Translation

Russian: Конденсатор — электронный компонент, состоящий из двух проводящих пластин, разделенных изоляционным материалом (диэлектриком), который накапливает электрическую энергию в электрическом поле.
Kazakh: Конденсатор — оқшаулағыш материалмен (диэлектрик) бөлінген екі өткізгіш жазықтықтан тұратын және электр өрісінде электр энергиясын жинайтын электрондық компонент.

Dielectric: An insulating material placed between the plates of a capacitor that increases its capacitance by reducing the electric field strength for a given charge.

Translation

Russian: Диэлектрик — изоляционный материал, помещенный между пластинами конденсатора, который увеличивает его ёмкость за счет уменьшения напряженности электрического поля при данном заряде.
Kazakh: Диэлектрик — конденсатор жазықтықтары арасына орналастырылған оқшаулағыш материал, ол берілген зарядта электр өрісінің күшін азайту арқылы сыйымдылықты арттырады.

Farad (F): The SI unit of capacitance. One farad is the capacitance that stores one coulomb of charge when one volt is applied across it.

Translation

Russian: Фарад (Ф) — единица ёмкости в СИ. Один фарад — это ёмкость, которая накапливает один кулон заряда при приложении одного вольта.
Kazakh: Фарад (Ф) — ХБЖ-дағы сыйымдылық бірлігі. Бір фарад — бір вольт қолданылғанда бір кулон зарядын жинайтын сыйымдылық.

Permittivity: A measure of how much electric field is reduced within a material compared to vacuum. It determines how easily a material can be polarized by an electric field.

Translation

Russian: Проницаемость — мера того, насколько электрическое поле уменьшается в материале по сравнению с вакуумом. Определяет, насколько легко материал может быть поляризован электрическим полем.
Kazakh: Өтімділік — вакуумға қарағанда материал ішінде электр өрісінің қаншалықты азайтынының өлшемі. Материалдың электр өрісімен қаншалықты оңай поляризацияланатынын анықтайды.

🔬 Theory: Understanding Capacitance

Theory: Understanding Capacitance

What is Capacitance?

is a fundamental property that describes a component’s ability to electrical charge. When we apply a across a , it accumulates charge on its .

Translation

Kazakh: Сыйымдылық — компоненттің электр зарядын жинау қабілетін сипаттайтын іргелі қасиет. Конденсаторға потенциалдар айырмасын қолданғанда, ол өз жазықтықтарында заряд жинайды.

Mathematical Definition

The mathematical definition of capacitance is:

Formula	Variables	Units
C = Q/V	C: Capacitance Q: Charge stored V: Potential difference	C: Farad (F) Q: Coulomb (C) V: Volt (V)

Translation

Kazakh: Сыйымдылықтың математикалық анықтамасы C = Q/V, мұндағы C — сыйымдылық, Q — жиналған заряд, V — потенциалдар айырмасы.

Parallel Plate Capacitor

For a parallel plate capacitor, the capacitance depends on the physical geometry and the dielectric properties:

Parallel Plate Capacitor Formula
C = ε₀εᵣA/d	ε₀: Permittivity of free space (8.85 × 10⁻¹² F/m) εᵣ: Relative permittivity of dielectric A: Area of each plate (m²) d: Distance between plates (m)

Translation

Kazakh: Параллель жазықтықтағы конденсатор үшін сыйымдылық физикалық геометрияға және диэлектрик қасиеттеріне тәуелді: C = ε₀εᵣA/d.

Energy Storage in Capacitors

Capacitors store energy in the electric field between their plates. The stored energy can be calculated using three equivalent formulas:

Energy Formula	When to Use
U = ½CV²	When capacitance and voltage are known
U = ½QV	When charge and voltage are known
U = Q²/(2C)	When charge and capacitance are known

Translation

Kazakh: Конденсаторлар энергияны жазықтықтары арасындағы электр өрісінде жинайды. Жиналған энергияны үш эквивалентті формуламен есептеуге болады.

Factors Affecting Capacitance

Factor	Effect on Capacitance	Explanation
Plate Area (A)	Directly proportional (C ∝ A)	Larger area allows more charge storage
Plate Separation (d)	Inversely proportional (C ∝ 1/d)	Closer plates create stronger electric field
Dielectric Material (εᵣ)	Directly proportional (C ∝ εᵣ)	Dielectric reduces field strength, allows more charge

Translation

Kazakh: Сыйымдылыққа әсер ететін факторлар: жазықтық ауданы (тура пропорционал), жазықтықтар арасындағы қашықтық (кері пропорционал), диэлектрик материал (тура пропорционал).

Practice Questions

Difficulty Level	Question
EASY	What is the unit of capacitance?
Answer The unit of capacitance is the Farad (F), which is equivalent to coulombs per volt (C/V).
MEDIUM	A capacitor stores 0.05 C of charge when connected to a 12 V battery. Calculate its capacitance.
Answer Using C = Q/V: C = 0.05 C / 12 V = 4.17 × 10⁻³ F = 4.17 mF
MEDIUM	How does doubling the plate area affect the capacitance of a parallel plate capacitor?
Answer Since C = ε₀εᵣA/d, doubling the plate area (A) will double the capacitance, assuming all other factors remain constant.
HARD	Critically analyze why capacitors cannot store infinite energy despite the formula U = ½CV² suggesting that energy increases with the square of voltage.
Answer While the formula U = ½CV² mathematically suggests infinite energy storage, practical limitations prevent this: (1) Dielectric breakdown occurs at high voltages, causing the insulator to conduct and destroying the capacitor; (2) Physical materials have maximum electric field strengths they can withstand; (3) Real capacitors have leakage currents that increase with voltage; (4) The assumption of constant capacitance breaks down at extreme conditions. These factors create natural limits to energy storage in capacitors.

🧠 Memory Exercises

Exercises on Memorizing Capacitance Terms

Exercise 1: Formula Completion
1. Basic capacitance: C = ______/V 2. Parallel plate: C = ε₀εᵣ______/d 3. Energy storage: U = ½C______² 4. Series capacitors: 1/C_total = 1/C₁ + ______ 5. Parallel capacitors: C_total = C₁ + ______	Answer 1. Q (charge) 2. A (area) 3. V (voltage) 4. 1/C₂ 5. C₂

Exercise 2: True or False
1. Capacitance increases with plate area (T/F) 2. Adding a dielectric decreases capacitance (T/F) 3. Farad is a very small unit in practice (T/F) 4. Energy stored increases linearly with voltage (T/F) 5. Capacitors block DC current in steady state (T/F)	Answer 1. True 2. False (increases) 3. False (very large) 4. False (quadratically) 5. True

Exercise 3: Unit Matching
Match the quantity with its unit: A) Capacitance → 1) Coulomb B) Charge → 2) Volt C) Energy → 3) Farad D) Voltage → 4) Joule	Answer A) Capacitance → 3) Farad B) Charge → 1) Coulomb C) Energy → 4) Joule D) Voltage → 2) Volt

🎥 Educational Video

Understanding Capacitors and Capacitance — Video Lesson

Additional Video Resources:

Video Topic	Link
Capacitors and Capacitance — Khan Academy	Watch Video
Energy Storage in Capacitors	Watch Video

🧮 Worked Examples

Problem Solving Examples

Example 1: Basic Capacitance Calculation
	Problem: A parallel plate capacitor has plates of area 0.02 m² separated by 2.0 mm of air. Calculate: (a) the capacitance, (b) the charge when connected to a 9V battery, (c) the energy stored.

Answer Step-by-step solution: Given: A = 0.02 m², d = 2.0 mm = 2.0 × 10⁻³ m, V = 9V, ε₀ = 8.85 × 10⁻¹² F/m (a) Capacitance: C = ε₀A/d = (8.85 × 10⁻¹² × 0.02)/(2.0 × 10⁻³) C = (1.77 × 10⁻¹³)/(2.0 × 10⁻³) = 8.85 × 10⁻¹¹ F = 88.5 pF (b) Charge: Q = CV = 8.85 × 10⁻¹¹ × 9 = 7.97 × 10⁻¹⁰ C = 0.797 nC (c) Energy stored: U = ½CV² = ½ × 8.85 × 10⁻¹¹ × 9² = 3.58 × 10⁻⁹ J = 3.58 nJ Answer Quick solution: (a) C = ε₀A/d = (8.85×10⁻¹²)(0.02)/(2.0×10⁻³) = 88.5 pF (b) Q = CV = 88.5×10⁻¹² × 9 = 0.797 nC (c) U = ½CV² = ½(88.5×10⁻¹²)(9)² = 3.58 nJ

Example 2: Capacitors in Series and Parallel
	Problem: Three capacitors of 2µF, 4µF, and 6µF are connected: (a) in series, (b) in parallel. Calculate the equivalent capacitance in each case.

Answer Detailed solution: Given: C₁ = 2µF, C₂ = 4µF, C₃ = 6µF (a) Series connection: For series: 1/C_total = 1/C₁ + 1/C₂ + 1/C₃ 1/C_total = 1/2 + 1/4 + 1/6 1/C_total = 6/12 + 3/12 + 2/12 = 11/12 C_total = 12/11 = 1.09 µF (b) Parallel connection: For parallel: C_total = C₁ + C₂ + C₃ C_total = 2 + 4 + 6 = 12 µF Note: Series gives smaller total capacitance, parallel gives larger total capacitance. Answer Quick answer: (a) Series: 1/C = 1/2 + 1/4 + 1/6 = 11/12, so C = 1.09 µF (b) Parallel: C = 2 + 4 + 6 = 12 µF

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🔬 Investigation Activity

Interactive Simulation: Capacitor Lab

Use this PhET simulation to explore how different factors affect capacitance:

Investigation Question	Answer
1. How does increasing the plate area affect the capacitance?	Answer Increasing plate area increases capacitance proportionally. Larger plates can store more charge at the same voltage.
2. What happens to capacitance when you increase the distance between plates?	Answer Increasing plate separation decreases capacitance. Greater distance weakens the electric field and reduces charge storage capacity.
3. How does adding a dielectric material affect the capacitance?	Answer Adding a dielectric increases capacitance by a factor equal to the dielectric constant. The dielectric reduces the electric field strength.
4. What is the relationship between stored energy and voltage?	Answer Stored energy increases with the square of voltage (U ∝ V²). Doubling voltage quadruples the stored energy.

👥 Group Activity

Collaborative Learning: Capacitance Quiz Challenge

Work in pairs or groups to complete this interactive quiz about capacitance:

Group Discussion Activities
🎯 Activity 1: Capacitor Hunt Find 5 real-world applications of capacitors Discuss why capacitors are used in each application Compare different types of capacitors (ceramic, electrolytic, etc.) Present findings to other groups	🧠 Activity 2: Design Challenge Design a capacitor for maximum energy storage Consider practical constraints (size, cost, safety) Justify your design choices mathematically Create a presentation explaining your design
📊 Activity 3: Data Analysis Analyze capacitor specifications from datasheets Compare performance of different capacitor types Create graphs showing relationships between parameters Discuss trade-offs in capacitor design	🔧 Activity 4: Problem-Solving Workshop Work together to solve complex capacitor problems Explain solution strategies to each other Check answers using different methods Create your own problems for other groups

Group Discussion Activities

🎯 Activity 1: Capacitor Hunt

Find 5 real-world applications of capacitors
Discuss why capacitors are used in each application
Compare different types of capacitors (ceramic, electrolytic, etc.)
Present findings to other groups

🧠 Activity 2: Design Challenge

Design a capacitor for maximum energy storage
Consider practical constraints (size, cost, safety)
Justify your design choices mathematically
Create a presentation explaining your design

📊 Activity 3: Data Analysis

Analyze capacitor specifications from datasheets
Compare performance of different capacitor types
Create graphs showing relationships between parameters
Discuss trade-offs in capacitor design

🔧 Activity 4: Problem-Solving Workshop

Work together to solve complex capacitor problems
Explain solution strategies to each other
Check answers using different methods
Create your own problems for other groups

📝 Individual Assessment

Structured Questions - Individual Work

Question 1: Analysis and Synthesis
A parallel plate capacitor is designed for a high-voltage power supply. The plates have an area of 0.1 m² and are separated by 5 mm of mica (εᵣ = 6.0). The capacitor must withstand 50 kV without breakdown. a) Calculate the capacitance of this capacitor. b) Determine the maximum charge it can store safely. c) Calculate the maximum energy that can be stored. d) Analyze what would happen if the plate separation were halved. e) Evaluate the trade-offs between energy storage and safety in this design.
Answer a) C = ε₀εᵣA/d = (8.85×10⁻¹²)(6.0)(0.1)/(5×10⁻³) = 1.06×10⁻⁹ F = 1.06 nF b) Q = CV = 1.06×10⁻⁹ × 50×10³ = 5.3×10⁻⁵ C = 53 µC c) U = ½CV² = ½(1.06×10⁻⁹)(50×10³)² = 1.33 J d) Halving separation would double capacitance but double electric field strength, potentially causing breakdown e) Trade-offs: Higher capacitance vs. breakdown risk, more energy storage vs. safety margins, cost vs. performance

Question 1: Analysis and Synthesis

A parallel plate capacitor is designed for a high-voltage power supply. The plates have an area of 0.1 m² and are separated by 5 mm of mica (εᵣ = 6.0). The capacitor must withstand 50 kV without breakdown.

a) Calculate the capacitance of this capacitor.
b) Determine the maximum charge it can store safely.
c) Calculate the maximum energy that can be stored.
d) Analyze what would happen if the plate separation were halved.
e) Evaluate the trade-offs between energy storage and safety in this design.

Answer

a) C = ε₀εᵣA/d = (8.85×10⁻¹²)(6.0)(0.1)/(5×10⁻³) = 1.06×10⁻⁹ F = 1.06 nF
b) Q = CV = 1.06×10⁻⁹ × 50×10³ = 5.3×10⁻⁵ C = 53 µC
c) U = ½CV² = ½(1.06×10⁻⁹)(50×10³)² = 1.33 J
d) Halving separation would double capacitance but double electric field strength, potentially causing breakdown
e) Trade-offs: Higher capacitance vs. breakdown risk, more energy storage vs. safety margins, cost vs. performance

Question 2: Circuit Analysis
Three capacitors (C₁ = 10 µF, C₂ = 20 µF, C₃ = 30 µF) are connected in a mixed configuration: C₁ and C₂ in parallel, then this combination in series with C₃. The entire system is connected to a 12V battery. a) Calculate the equivalent capacitance of the system. b) Determine the voltage across each capacitor. c) Calculate the charge on each capacitor. d) Find the total energy stored in the system. e) Compare this with the energy if all three were in parallel.
Answer a) C₁₂ = C₁ + C₂ = 30 µF; 1/C_total = 1/30 + 1/30 = 2/30; C_total = 15 µF b) Q_total = 15×10⁻⁶ × 12 = 180 µC; V₃ = Q/C₃ = 180×10⁻⁶/30×10⁻⁶ = 6V; V₁ = V₂ = 12-6 = 6V c) Q₃ = 180 µC; Q₁ = 10×10⁻⁶ × 6 = 60 µC; Q₂ = 20×10⁻⁶ × 6 = 120 µC d) U = ½CV² = ½(15×10⁻⁶)(12)² = 1.08 mJ e) All parallel: C = 60 µF, U = ½(60×10⁻⁶)(12)² = 4.32 mJ (higher energy storage)

Question 2: Circuit Analysis

Three capacitors (C₁ = 10 µF, C₂ = 20 µF, C₃ = 30 µF) are connected in a mixed configuration: C₁ and C₂ in parallel, then this combination in series with C₃. The entire system is connected to a 12V battery.

a) Calculate the equivalent capacitance of the system.
b) Determine the voltage across each capacitor.
c) Calculate the charge on each capacitor.
d) Find the total energy stored in the system.
e) Compare this with the energy if all three were in parallel.

Answer

a) C₁₂ = C₁ + C₂ = 30 µF; 1/C_total = 1/30 + 1/30 = 2/30; C_total = 15 µF
b) Q_total = 15×10⁻⁶ × 12 = 180 µC; V₃ = Q/C₃ = 180×10⁻⁶/30×10⁻⁶ = 6V; V₁ = V₂ = 12-6 = 6V
c) Q₃ = 180 µC; Q₁ = 10×10⁻⁶ × 6 = 60 µC; Q₂ = 20×10⁻⁶ × 6 = 120 µC
d) U = ½CV² = ½(15×10⁻⁶)(12)² = 1.08 mJ
e) All parallel: C = 60 µF, U = ½(60×10⁻⁶)(12)² = 4.32 mJ (higher energy storage)

Question 3: Design Optimization
You need to design a capacitor bank for an electric vehicle that can store 1 MJ of energy and operate at 800V. You have three types of capacitors available: Type A: 1000 µF, 1000V max, $10 each Type B: 2200 µF, 450V max, $15 each Type C: 4700 µF, 200V max, $25 each a) Determine which type(s) can be used at 800V operating Назад Далее

Question 3: Design Optimization

You need to design a capacitor bank for an electric vehicle that can store 1 MJ of energy and operate at 800V. You have three types of capacitors available:
Type A: 1000 µF, 1000V max, $10 each
Type B: 2200 µF, 450V max, $15 each
Type C: 4700 µF, 200V max, $25 each

a) Determine which type(s) can be used at 800V operating