CBSE Class 12 Physics Chapter 1 — Electric Charges and Fields: NCERT Notes, Coulomb’s Law, Gauss Theorem and MCQs 2027

Last Updated: April 2026

CBSE Class 12 Physics Chapter 1 — Electric Charges and Fields — is the gateway to the entire Electrostatics unit, which carries 16 of 70 marks (~23%) in the CBSE 2027 board exam. The chapter introduces Coulomb’s Law, electric field, electric flux, and Gauss’s Theorem — concepts you will revisit in Chapter 2 (Potential), Chapter 3 (Current), and Chapter 6 (EM Induction). For students mastering cbse class 12 physics electric charges fields 2027, this comprehensive note covers all NCERT formulas, derivations, 7 worked problems, 20 important questions and 10 MCQs aligned with the CBSE 2026-27 syllabus (no deleted topics).

1. Electric Charge — Properties

• Quantization: q = ne, where n is an integer and e = 1.6 × 10⁻¹⁹ C
• Conservation: Total charge of an isolated system remains constant
• Additivity: Total charge = algebraic sum of all charges (scalar)
• Charge invariance: Independent of velocity of the observer
• Two types: positive (proton) and negative (electron)

1.1 Methods of Charging

• Friction (e.g., glass rod + silk → glass becomes positive)
• Conduction (charge transfer on direct contact)
• Induction (charge separation without contact)

2. Coulomb’s Law

The force between two stationary point charges q₁ and q₂ separated by distance r:

F = (1/4πε₀) · (q₁q₂/r²)

where 1/(4πε₀) = 9 × 10⁹ N·m²/C²; ε₀ = 8.854 × 10⁻¹² C²/(N·m²) is the permittivity of free space.

In a medium of relative permittivity ε_r (dielectric constant K): F = (1/4πε₀ε_r) · (q₁q₂/r²)

2.1 Vector Form

F⃗₁₂ = (1/4πε₀) · (q₁q₂/r²) · r̂₁₂ (force on q₁ due to q₂)

2.2 Principle of Superposition

Net force on a charge = vector sum of forces due to all other charges.

3. Electric Field

Electric field E⃗ at a point = force per unit positive test charge:

E⃗ = F⃗/q₀, units: N/C or V/m

3.1 Field due to a Point Charge

E = (1/4πε₀) · (q/r²), directed radially outward (positive q) or inward (negative q).

3.2 Electric Field Lines — Properties

• Continuous curves from + to – charge
• No two field lines cross (E has unique direction at each point)
• Field-line density ∝ field magnitude
• Always perpendicular to surface of conductor
• No closed loops in electrostatic field (unlike magnetic field)

4. Electric Dipole

Two equal and opposite charges (+q, −q) separated by 2a. Dipole moment p⃗ = q · 2a (from −q to +q).

4.1 Field on Axial Line (end-on)

E_axial = (1/4πε₀) · (2pr/(r² − a²)²) ≈ (1/4πε₀) · (2p/r³) for r >> a

4.2 Field on Equatorial Line (broadside-on)

E_eq = (1/4πε₀) · (p/(r² + a²)^(3/2)) ≈ (1/4πε₀) · (p/r³) for r >> a

Note: E_axial = 2 × E_eq (for r >> a)

4.3 Torque on Dipole in Uniform Field

τ⃗ = p⃗ × E⃗, |τ| = pE sinθ. Maximum at θ=90°, zero at θ=0° or 180°.

5. Continuous Charge Distributions

• Linear charge density: λ = q/L (C/m)
• Surface charge density: σ = q/A (C/m²)
• Volume charge density: ρ = q/V (C/m³)

6. Gauss’s Theorem (Cornerstone of the Chapter)

Total electric flux through a closed surface = (1/ε₀) × charge enclosed.

∮ E⃗ · dA⃗ = q_enc / ε₀

6.1 Applications of Gauss’s Law

7. Master Formula Table (Mandatory)

8. Seven Worked Problems

Problem 1. Two charges q₁ = +4 µC and q₂ = +6 µC are placed 30 cm apart. Find the force between them.
F = 9×10⁹ × (4×10⁻⁶)(6×10⁻⁶)/(0.30)² = 9×10⁹ × 24×10⁻¹² / 0.09 = 2.4 N (repulsive).

Problem 2. An electric dipole has charges ±2 µC separated by 1 cm. Find dipole moment.
p = q × 2a = 2×10⁻⁶ × 0.01 = 2 × 10⁻⁸ C·m.

Problem 3. Field at axial point of dipole 1 m away (r >> a), p = 2×10⁻⁸ C·m.
E = 9×10⁹ × 2 × (2×10⁻⁸)/1³ = 360 N/C.

Problem 4. Charge of 5 µC is placed at the centre of a cube of side 10 cm. Find flux through one face.
Total flux = q/ε₀ = 5×10⁻⁶/(8.854×10⁻¹²) = 5.65×10⁵ N·m²/C. Per face = 9.42×10⁴ N·m²/C.

Problem 5. Field due to infinite sheet σ = 1.77×10⁻¹¹ C/m².
E = σ/(2ε₀) = 1.77×10⁻¹¹/(2×8.854×10⁻¹²) = 1 N/C.

Problem 6. Torque on a dipole p = 4×10⁻⁹ C·m at 30° to a field E = 5×10⁴ N/C.
τ = pE sinθ = 4×10⁻⁹ × 5×10⁴ × 0.5 = 10⁻⁴ N·m.

Problem 7. Field at 5 cm outside a hollow sphere of radius 3 cm carrying 6 µC.
r = 5+3 = 8 cm = 0.08 m. E = 9×10⁹ × 6×10⁻⁶/(0.08)² = 8.44×10⁶ N/C.

9. 20 Important Questions with Concise Answers

1. Define electric flux. Φ = E⃗·A⃗ = ∫E·dA — count of field lines through a surface (scalar; SI unit: N·m²/C).
2. State Coulomb’s Law. F = (1/4πε₀)(q₁q₂/r²); along the line joining charges.
3. State Gauss’s Theorem. ∮ E·dA = q_enc/ε₀.
4. Why are field lines perpendicular to a conductor’s surface? Tangential component would set up surface currents, contradicting electrostatic equilibrium.
5. Why is electric field zero inside a hollow conductor? By Gauss’s law — no enclosed charge in interior cavity.
6. Differentiate axial and equatorial dipole field. E_axial = 2p/(4πε₀r³); E_eq = p/(4πε₀r³); axial is twice equatorial and parallel to p; equatorial is anti-parallel.
7. Define dielectric constant K. K = ε/ε₀ = ratio of force in vacuum to force in medium for same charges.
8. Why can two field lines never intersect? Otherwise E would have two directions at the intersection — unphysical.
9. Define linear charge density. λ = dq/dl (C/m).
10. What is the SI unit of electric dipole moment? Coulomb·metre (C·m).
11. Torque on a dipole in uniform field. τ⃗ = p⃗ × E⃗.
12. Field due to infinite line charge. E = λ/(2πε₀r).
13. Why is electric field independent of size of Gaussian surface? Field depends only on enclosed charge, not on surface chosen.
14. Difference between charging by induction and conduction. Induction = no contact, charge separation; conduction = direct contact, charge transfer.
15. What is meant by quantization of charge? q = ne (n integer); e = 1.6×10⁻¹⁹ C.
16. State principle of superposition. Net force = vector sum of pairwise forces.
17. Why a charge inside a Gaussian surface contributes flux but outside doesn’t? External charge produces equal incoming and outgoing flux that cancel.
18. Equipotential surface. Surface where V is constant; E ⊥ to it.
19. Field at centre of a uniformly charged ring. Zero (by symmetry).
20. Field of two parallel sheets +σ and −σ. Inside: σ/ε₀; Outside: 0.

10. Internal Resources

Practice with our Chapter 2 — Electrostatic Potential and Capacitance. Browse Class 12 Physics courses, the CBSE 2027 hub and our free resources library.

11. FAQ

Q1. How important is Chapter 1 for CBSE Class 12 Physics 2027?

Electrostatics (Chapters 1+2) carry 16 of 70 marks (~23%) in the CBSE 2027 paper, with at least one 5-mark and one 3-mark question typical.

Q2. Are derivations from Gauss’s Theorem in the syllabus?

Yes — derivations for infinite line charge, infinite sheet, and uniformly charged spherical shell are in the CBSE 2026-27 syllabus.

Q3. What is the difference between electric field and electric force?

Force F is on a charge; field E exists due to other charges, defined as F/q. Field is independent of test charge magnitude.

Q4. How do you tackle a question on field due to a continuous charge distribution?

Identify symmetry → choose appropriate Gaussian surface → apply ∮E·dA = q_enc/ε₀ → solve for E.

Q5. Is the principle of superposition important for boards?

Yes — every multi-charge problem uses it. NCERT example 1.6 directly tests it.

Quiz — Electric Charges and Fields — 10 MCQs

Quiz data error: Syntax error

Conclusion & CTA

Master Chapter 1 with NCERT examples + 7 worked problems above + 10 PYQ MCQs. Then advance to Chapter 2 (Potential & Capacitance). Need video lectures, daily PYQ practice, and answer-checking AI? Join Ready For Boards Class 12 Physics 2027 program.