📘 Chapter 7: Electric Field (Class XII)

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🔷 1. Introduction

The concept of electric field is central to electrostatics. Instead of thinking that charges directly exert force at a distance, we introduce the idea that a charge creates a field in space, and another charge placed in this region experiences force.

This field-based description is more powerful and forms the foundation of classical and modern electromagnetism.

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🔷 2. Definition of Electric Field

Electric field at a point is defined as the force experienced per unit positive test charge placed at that point.

E⃗ = F⃗ / q₀

Where:

• E⃗ = electric field
• F⃗ = force on test charge
• q₀ = small positive test charge

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🔷 3. Electric Field Due to Point Charge

For a point charge q:

E = (1 / 4πε₀) (q / r²)

Direction:

• Outward for positive charge
• Inward for negative charge

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🔷 4. Nature of Electric Field

• Vector quantity
• Exists even in vacuum
• Independent of test charge magnitude
• Depends only on source charges

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📦 5. Important Results (Must Remember)

• Definition: E⃗ = F⃗ / q₀
• Point charge field: E = kq / r²
• Superposition applies: E⃗net = Σ E⃗i
• Field independent of test charge
• Unit: N/C or V/m

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🔷 6. Electric Field Lines

Electric field lines are imaginary lines used to represent the electric field visually.

• Start from positive charge and end at negative charge
• Never intersect each other
• Denser lines → stronger field
• Tangent gives direction of field

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🔷 7. Physical Interpretation

Electric field represents a force field in space. It allows us to describe interaction without direct contact between charges.

Thus, force becomes:

F⃗ = qE⃗

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🔷 8. Electric Field for Continuous Distribution

For distributed charges:

E⃗ = ∫ dE⃗ = ∫ (1 / 4πε₀) (dq / r²) r̂

This requires integration and symmetry considerations.

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🔷 9. Superposition Principle in Field

If multiple charges are present:

E⃗net = E⃗₁ + E⃗₂ + E⃗₃ + ...

Each field is calculated independently and added vectorially.

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🧠 10. Solved Numerical Problems

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🔹 Q1

Find electric field at 0.2 m from a charge of 5 μC.

Solution:

E = kq / r²

E = 9×10⁹ × 5×10⁻⁶ / (0.2)²

E = 45×10³ / 0.04

E = 1.125 × 10⁶ N/C

Answer: 1.125 × 10⁶ N/C

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🔹 Q2

Why is electric field independent of test charge?

Answer:

Because E is defined as F/q₀, and force is directly proportional to test charge. Hence q₀ cancels out.

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🔹 Q3

What is electric field at midpoint of two equal positive charges?

Solution:

Fields due to both charges are equal and opposite.

Net field = 0

Answer: Zero

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🔹 Q4

If distance from charge is doubled, how does field change?

Solution:

E ∝ 1/r² → becomes 1/4

Answer: Field becomes one-fourth

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🔷 11. Advanced Conceptual Insight

Electric field is not just a mathematical tool but a physical entity that stores energy and momentum. In modern physics, it is part of the electromagnetic field described by Maxwell’s equations.

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🔷 12. Applications

• Capacitor design
• Particle acceleration
• Electrostatic shielding
• Medical imaging devices

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🔷 13. Summary

Electric field is a fundamental concept that replaces action-at-a-distance with a field-based interaction model. It simplifies analysis of complex charge systems and forms the backbone of electromagnetism.

✨ End of Chapter 7: Electric Field ✨