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Electric Charges and Fields – Class 12 Physics Chapter 1 Notes & NCERT PDF

Class 12PhysicsChapter 1NCERT book: Physics Part I - Class 12

Notes and a simple summary of Chapter 1 of the NCERT Class 12 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 1: Electric Charges and Fields

Introduction

Chapter 1 opens the study of electrostatics — charges at rest, the forces between them and the fields they create. Everyday observations (a comb attracting paper bits, the crackle of synthetic clothes, lightning) come from the transfer of charge. The chapter builds the whole subject on three pillars: the properties of charge, Coulomb's law, and the electric field, ending with Gauss's law, which turns hard field calculations into short symmetry arguments.

Key Concepts

1.2–1.4 Electric charge and its properties

  • There are two kinds of charge, named positive and negative by Benjamin Franklin; like charges repel, unlike charges attract.
  • Charging by rubbing (friction) transfers electrons; charging by induction rearranges charges without contact.
  • Conductors (metals, human body, earth) let charge move freely; insulators (glass, plastic, nylon) do not. Earthing/grounding shares excess charge with the earth.
  • Additivity: charges add algebraically like real numbers (e.g. +1, +2, −3 and +4 units total +4).
  • Conservation: charge can be transferred but never created or destroyed in an isolated system.
  • Quantisation: every charge is q = ne, with n an integer and e = 1.6 × 10⁻¹⁹ C. At the macroscopic level (µC, mC) quantisation can be ignored because e is so tiny.

    • 1.5 Coulomb's law

    The force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them, acting along the line joining them.

    F = (1/4πε₀) · q₁q₂ / r², with 1/4πε₀ = 9 × 10⁹ N m² C⁻² and ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻². In vector form, the force on q₂ due to q₁ is F₂₁ = k q₁q₂ r̂₂₁/r₂₁², and F₁₂ = −F₂₁ (Newton's third law). The electric force is enormously stronger than gravity: between an electron and a proton the ratio F_e/F_G is about 2.4 × 10³⁹.

    1.6 Forces between multiple charges — superposition

    The force on any charge is the vector sum of the forces due to each other charge taken one at a time; each pair interacts as if the others were absent.

    1.7 Electric field

    E = F/q (force per unit positive test charge); SI unit N C⁻¹. For a point charge, E = kq/r² directed radially (away from +q, towards −q). A charge q placed in a field experiences F = qE. The field is not just a mathematical tool — in time-dependent situations it carries energy and momentum and propagates at the speed of light.

    1.8 Electric field lines

    Curves whose tangent gives the field direction. They start on positive and end on negative charges (or infinity), never cross, never form closed loops in electrostatics, and their density shows field strength.

    1.9 Electric flux

    Δφ = E·ΔS = E ΔS cos θ; SI unit N C⁻¹ m². Flux counts "how many field lines" cross a surface.

    1.10–1.11 Electric dipole

    Two equal and opposite charges ±q separated by 2a; dipole moment p = q × 2a, directed from −q to +q.

  • On the axis (r ≫ a): E = 2kp/r³, along p.
  • On the equatorial plane (r ≫ a): E = kp/r³, opposite to p.
  • The field falls as 1/r³ (faster than a single charge's 1/r²).
  • In a uniform field: net force zero, torque τ = p × E (τ = pE sin θ), which aligns the dipole with the field. In a non-uniform field there is also a net force.
  • Polar molecules (e.g. H₂O) have permanent dipole moments.

    • 1.12 Continuous charge distributions

    Linear density λ (C m⁻¹), surface density σ (C m⁻²), volume density ρ (C m⁻³); the field is found by summing (integrating) contributions of small elements.

    1.13–1.14 Gauss's law and its applications

    The total electric flux through any closed surface equals q_enclosed/ε₀.

    Charges outside the surface contribute zero net flux. With a symmetric Gaussian surface:

  • Infinitely long straight wire (λ): E = λ/(2πε₀r), radial.
  • Infinite plane sheet (σ): E = σ/(2ε₀), independent of distance.
  • Thin spherical shell (radius R, charge q): outside E = kq/r² (as if all charge were at the centre); inside E = 0.

    • Worked Examples

  • Example 1.1 (quantisation): 10⁹ electrons leave a body every second. Charge of one electron = 1.6 × 10⁻¹⁹ C, so 1 C needs n = 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons. Time = 6.25 × 10¹⁸ / 10⁹ = 6.25 × 10⁹ s ≈ 198 years — showing how large a coulomb is.
  • Coulomb force: two 1 µC charges 10 cm apart: F = 9 × 10⁹ × (10⁻⁶)² / (0.1)² = 0.9 N (repulsive).
  • Example 1.7 (motion in a field): in E = 2.0 × 10⁴ N C⁻¹, an electron falling 1.5 cm takes t = √(2h m/(eE)) ≈ 2.9 × 10⁻⁹ s, while a proton takes ≈ 1.3 × 10⁻⁷ s — the heavier particle accelerates far less, and gravity is negligible in both cases.
  • Example 1.9 (dipole): ±10 µC, 5.0 mm apart, so p = 5 × 10⁻⁸ C m. At 15 cm on the axis, E ≈ 2kp/r³ ≈ 2.67 × 10⁵ N C⁻¹; at 15 cm on the equatorial line, E ≈ kp/r³ ≈ 1.33 × 10⁵ N C⁻¹ — exactly half.
  • Gauss's law reasoning: flux through a closed surface around a charge q is q/ε₀ regardless of the surface's shape or size; a charge outside gives zero net flux.

    • Formula / Key-Facts Sheet

  • e = 1.6 × 10⁻¹⁹ C; q = ne
  • F = k q₁q₂/r², k = 1/4πε₀ = 9 × 10⁹ N m² C⁻²
  • E = F/q; point charge E = kq/r²
  • φ = E·S = ES cos θ; closed surface φ = q_enc/ε₀
  • p = q(2a); axial E = 2kp/r³; equatorial E = kp/r³
  • τ = pE sin θ (τ = p × E)
  • Wire E = λ/2πε₀r; sheet E = σ/2ε₀; shell: outside kq/r², inside 0

    • Board Exam Focus

  • Numericals: Coulomb force between charges, superposition at a vertex or centroid, field at a point due to several charges, dipole field and torque, flux through a surface.
  • Derivations (3–5 marks): field on the axis or equatorial line of a dipole; torque on a dipole; fields of a wire, sheet and shell using Gauss's law.
  • Answer outline — field due to a sheet: choose a cylindrical Gaussian surface through the sheet → flux only through the two flat faces = 2EA → enclosed charge σA → 2EA = σA/ε₀ → E = σ/2ε₀.
  • Answer outline — why field lines never cross: at the crossing point the field would have two directions, which is impossible.
  • Conceptual: why charge is quantised yet treated as continuous; why the field inside a charged shell is zero.

    • Key Terms

    Electric charge (विद्युत आवेश), Conductor (चालक), Insulator (विद्युतरोधी), Induction (प्रेरण), Quantisation (क्वांटमीकरण), Coulomb's law (कूलॉम का नियम), Permittivity (परावैद्युतांक/विद्युतशीलता), Electric field (विद्युत क्षेत्र), Field lines (क्षेत्र रेखाएँ), Electric flux (विद्युत फ्लक्स), Dipole (द्विध्रुव), Dipole moment (द्विध्रुव आघूर्ण), Torque (बल आघूर्ण), Gauss's law (गाउस का नियम), Surface charge density (पृष्ठ आवेश घनत्व)

    Common Mistakes

  • Forgetting that force and field are vectors — adding magnitudes instead of components.
  • Writing the dipole moment direction from + to − (it is from −q to +q).
  • Using E = σ/ε₀ for an isolated sheet (correct is σ/2ε₀).
  • Thinking the flux through a closed surface depends on charges outside it.
  • Forgetting to convert µC → C and cm → m before substituting.

    • 💡 Key Learning Points

    • ✓Properties of charge: additivity, conservation and quantisation (q = ne)
    • ✓Coulomb's law in scalar and vector form, and the superposition principle
    • ✓Electric field, field lines and electric flux
    • ✓Electric dipole: moment, axial/equatorial fields, torque in a uniform field
    • ✓Gauss's law and its use for wire, plane sheet and spherical shell

    👨‍🏫 Teaching Tips

    • →Start with rubbing a comb and a glass rod to show two kinds of charge and induction
    • →Draw field-line patterns for single charges, dipoles and like charges side by side
    • →Work Example 1.1 to show how huge one coulomb is
    • →Teach superposition with an equilateral triangle problem using components
    • →Derive the dipole field step by step, then use the r ≫ a approximation
    • →Choose Gaussian surfaces by symmetry: cylinder for wire/sheet, sphere for shell
    • →Stress unit conversion (µC, cm) before every numerical

    📋 Assessment Questions

    1. Can students state and apply Coulomb's law with vectors?
    2. Can they derive the axial and equatorial fields of a dipole?
    3. Can they choose a correct Gaussian surface and derive the field of a wire, sheet or shell?
    4. Can they explain why the field inside a charged shell is zero?

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    Keep learning

    All chapters in Physics Part I - Class 12

    1. 1. Electric Charges and Fields
    2. 2. Electrostatic Potential and Capacitance
    3. 3. Current Electricity
    4. 4. Moving Charges and Magnetism
    5. 5. Magnetism and Matter
    6. 6. Electromagnetic Induction
    7. 7. Alternating Current
    8. 8. Electromagnetic Waves