📘 Chapter 29: Capacitance of a Parallel Plate Capacitor with Dielectric Medium (Class XII)

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🔷 1. Introduction

A parallel plate capacitor becomes more efficient when a dielectric medium is inserted between its plates. This modification significantly increases its capacitance and energy storage capability.

This chapter explains how dielectric materials influence capacitance and electric fields.

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🔷 2. Basic Parallel Plate Capacitor

Without dielectric, capacitance is:

C₀ = ε₀ A / d

• A = area of plates
• d = separation
• ε₀ = permittivity of free space

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🔷 3. With Dielectric Medium

When a dielectric is inserted, capacitance becomes:

C = K ε₀ A / d

Where:

• K = dielectric constant

Capacitance increases K times

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📦 4. Important Results (Must Remember)

• C₀ = ε₀A/d (without dielectric)
• C = Kε₀A/d (with dielectric)
• K > 1 for all dielectrics
• Capacitance increases
• Electric field decreases

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🔷 5. Effect on Electric Field

Electric field without dielectric:

E₀ = σ / ε₀

With dielectric:

E = E₀ / K

Thus, dielectric reduces electric field inside capacitor.

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🔷 6. Effect on Potential Difference

• Potential difference decreases
• Charge remains constant (if battery disconnected)

Thus:

V decreases → C increases

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🔷 7. Polarisation Mechanism

When dielectric is placed:

• Dipoles are induced
• Bound charges appear on surfaces
• Internal field opposes external field

This is called polarisation

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🔷 8. Energy Stored with Dielectric

Energy stored:

U = ½ C V²

• Energy increases when capacitor is connected to battery
• Energy decreases if isolated

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🔷 9. Different Cases

🔹 Case 1: Battery Connected

• Voltage constant
• Charge increases
• Energy increases

🔹 Case 2: Battery Disconnected

• Charge constant
• Voltage decreases
• Energy decreases

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🔷 10. Physical Interpretation

Dielectric reduces the effective electric field inside capacitor, allowing more charge to be stored for the same voltage. This enhances capacitance.

Dielectrics improve energy storage capability

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🧠 11. Solved Numerical Problems

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🔹 Q1

Find capacitance with dielectric K = 5, A = 1 m², d = 0.01 m.

Solution:

C = Kε₀A/d

= 5 × 8.85×10⁻¹² / 0.01

C = 4.425 × 10⁻⁹ F

Answer: 4.425 × 10⁻⁹ Farad

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🔹 Q2

What happens to electric field when dielectric is inserted?

Answer:

It decreases

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🔹 Q3

What is role of dielectric constant?

Answer:

It measures increase in capacitance.

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🔹 Q4

Which case increases energy?

Answer:

Battery connected case.

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🔹 Q5

Why capacitance increases?

Answer:

Due to reduction in electric field.

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🔷 12. Advanced Conceptual Insight

Dielectric materials play a vital role in modern capacitors, energy storage systems, and electronic devices. Their ability to enhance capacitance makes them essential in circuit design.

Advanced materials like ceramics and polymers are widely used as dielectrics.

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🔷 13. Applications

• Capacitors in electronics
• Energy storage devices
• Insulation systems
• Communication circuits

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🔷 14. Summary

Insertion of dielectric in a capacitor increases its capacitance by reducing the internal electric field. This enhances its ability to store charge and energy efficiently.

✨ End of Chapter 29: Dielectric Capacitors ✨