Electric Charges and Fields Class 12 Notes | CBSE Physics Chapter 1

Electric Charges and Fields is Chapter 1 of CBSE Class 12 Physics. This chapter introduces electrostatics — the study of forces, fields, and potentials arising from static charges. You will learn about Coulomb’s law, electric field, electric field lines, electric flux, and Gauss’s law.

This is a foundational chapter carrying 8–10 marks in board exams. Coulomb’s law numericals, electric field due to various charge distributions, and Gauss’s law applications are heavily tested.

---

Key Concepts

1. Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force in an electric field. There are two types:

• Positive charge: carried by protons
• Negative charge: carried by electrons

SI unit: Coulomb (C). Charge of one electron = 1.6 × 10⁻¹⁹ C

Properties of Electric Charge

• Quantisation: Charge exists only in integral multiples of e (1.6 × 10⁻¹⁹ C). q = ne, where n is an integer.
• Conservation: Total charge in an isolated system is always conserved. Charge can neither be created nor destroyed — only transferred.
• Additivity: Total charge is the algebraic sum of all individual charges.

Methods of Charging

Method	Description
Friction	Rubbing two objects transfers electrons from one to another
Conduction	Touching a charged body to an uncharged conductor transfers charge
Induction	Bringing a charged body near (not touching) an uncharged conductor redistributes charges

---

2. Coulomb’s Law

The force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

• k = 1/(4πε₀) = 9 × 10⁹ N·m²/C²
• ε₀ = 8.854 × 10⁻¹² C²/(N·m²) — permittivity of free space
• Like charges repel; unlike charges attract

In a medium: F = kq₁q₂/(κr²), where κ is the dielectric constant of the medium. Force decreases in a medium.

Vector form: F₁₂ = kq₁q₂r̂₁₂/r² (force on charge 1 due to charge 2, along the line joining them)

Superposition Principle

The net force on a charge due to multiple other charges is the vector sum of the individual forces due to each charge, calculated independently.

---

3. Electric Field

The electric field at a point is the force experienced by a unit positive test charge placed at that point.

E = F/q₀ = kQ/r² (due to a point charge Q)

Unit: N/C or V/m

Electric field is a vector quantity — it points away from positive charges and towards negative charges.

Electric Field Due to Common Configurations

Configuration	Electric Field
Point charge Q	E = kQ/r²
Electric dipole (axial point, far away)	E = 2kp/r³
Electric dipole (equatorial point, far away)	E = kp/r³
Infinite line charge (linear charge density λ)	E = λ/(2πε₀r)
Infinite plane sheet (surface charge density σ)	E = σ/(2ε₀)
Two parallel infinite sheets (+σ and −σ)	E = σ/ε₀ (between), 0 (outside)

---

4. Electric Field Lines

Electric field lines are imaginary curves that represent the direction and strength of the electric field.

Properties

• Start from positive charges, end at negative charges
• Never cross each other
• Closer lines = stronger field
• Perpendicular to the surface of a conductor
• Do not form closed loops (unlike magnetic field lines)
• Continuous curves (no breaks)

---

5. Electric Dipole

An electric dipole consists of two equal and opposite charges (+q and −q) separated by a small distance 2a.

Dipole moment: p = q × 2a (direction: from −q to +q)

Unit: C·m

Dipole in a Uniform Electric Field

• Torque: τ = p × E = pE sin θ (where θ is the angle between p and E)
• Torque is maximum when θ = 90° and zero when θ = 0° or 180°
• Net force = 0 (in a uniform field)
• The dipole tends to align along the field direction

---

6. Electric Flux and Gauss’s Law

Electric flux (Φ) through a surface is the number of electric field lines passing through that surface.

Φ = E · A = EA cos θ

Unit: N·m²/C or V·m

Gauss’s Law

The total electric flux through any closed surface is equal to 1/ε₀ times the total charge enclosed within that surface.

∮ E · dA = q_enclosed / ε₀

Gauss’s law is especially useful for calculating electric fields when there is symmetry (spherical, cylindrical, or planar).

Applications of Gauss’s Law

Charge Distribution	Gaussian Surface	Result
Point charge / Uniformly charged sphere	Concentric sphere	E = kQ/r² (outside); E = 0 (inside hollow sphere); E = kQr/R³ (inside solid sphere)
Infinite line charge	Coaxial cylinder	E = λ/(2πε₀r)
Infinite plane sheet	Pill-box (cylinder perpendicular to sheet)	E = σ/(2ε₀)

---

Important Definitions

Term	Definition
Electric charge	Fundamental property of matter causing it to experience electromagnetic force
Coulomb’s law	F = kq₁q₂/r² — force between two point charges
Electric field	Force per unit positive charge at a point: E = F/q₀
Electric dipole	System of two equal and opposite charges separated by a small distance
Dipole moment	p = q × 2a — measures the strength and direction of a dipole
Electric flux	Number of electric field lines passing through a surface: Φ = EA cos θ
Gauss’s law	Total flux through a closed surface = q_enclosed/ε₀
Dielectric constant	Factor by which a medium reduces the electric force compared to vacuum

---

Solved Examples

Example 1

Two charges of +3 μC and −3 μC are placed 20 cm apart. Find the force between them.

Answer: F = kq₁q₂/r² = (9 × 10⁹)(3 × 10⁻⁶)(3 × 10⁻⁶)/(0.2)² = (9 × 10⁹)(9 × 10⁻¹²)/0.04 = 81 × 10⁻³/0.04 = 2.025 N (attractive, since charges are opposite)

Example 2

An electric dipole with dipole moment 4 × 10⁻⁹ C·m is placed in a uniform electric field of 5 × 10⁴ N/C at 30° to the field. Find the torque.

Answer: τ = pE sin θ = (4 × 10⁻⁹)(5 × 10⁴) sin 30° = 20 × 10⁻⁵ × 0.5 = 10⁻⁴ N·m = 0.1 mN·m

Example 3

A sphere of radius 10 cm encloses a charge of 5 μC. Find the electric flux through the sphere.

Answer: By Gauss’s law: Φ = q/ε₀ = (5 × 10⁻⁶)/(8.854 × 10⁻¹²) = 5.65 × 10⁵ N·m²/C. Note: the flux does not depend on the radius — only on the enclosed charge.

Example 4

Find the electric field at a distance of 20 cm from an infinitely long wire with linear charge density 5 × 10⁻⁶ C/m.

Answer: E = λ/(2πε₀r) = (5 × 10⁻⁶)/(2π × 8.854 × 10⁻¹² × 0.2) = (5 × 10⁻⁶)/(11.13 × 10⁻¹²) = 4.49 × 10⁵ N/C

---

Important Questions for Board Exams

1-Mark Questions

1. State Coulomb’s law in electrostatics.
2. What is the SI unit of electric flux?
3. Why do electric field lines never cross each other?
4. What is the electric field inside a charged hollow conductor?

3-Mark Questions

1. Derive the expression for electric field on the axial line of an electric dipole.
2. State Gauss’s law. Use it to find the electric field due to an infinitely long straight charged wire.
3. Define electric dipole moment. Derive the torque on a dipole in a uniform electric field.

5-Mark Questions

1. State Gauss’s law. Apply it to find the electric field due to (a) a uniformly charged infinite plane sheet, (b) a uniformly charged thin spherical shell.
2. Derive Coulomb’s law from Gauss’s law.

---

Quick Revision Points

• Charge is quantised (q = ne), conserved, and additive
• Coulomb’s law: F = kq₁q₂/r²; k = 9 × 10⁹ N·m²/C²
• Superposition: net force = vector sum of individual forces
• Electric field E = kQ/r² (point charge); direction: away from +, towards −
• Dipole moment p = q × 2a (from −q to +q)
• Torque on dipole: τ = pE sin θ; net force in uniform field = 0
• Gauss’s law: Φ = q_enclosed/ε₀; useful with symmetry
• Infinite plane sheet: E = σ/(2ε₀) — independent of distance
• Inside hollow conductor: E = 0 always

---

Next Chapter: Chapter 2 — Electrostatic Potential and Capacitance