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Electric Charges and Fields – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 1. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Electric charges exert forces given by Coulomb's law; the electric field and Gauss's law describe these forces around charge distributions.

Charge and force

Charge is quantised (q = ne, e = 1.6 × 10⁻¹⁹ C) and conserved.
Coulomb's law: F = kq₁q₂/r², k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻².
Forces add as vectors (superposition).

Field and flux

Electric field E = F/q₀; for a point charge E = kq/r².
Dipole moment p = q × 2a; torque τ = pE sin θ.
Gauss's law: Φ = q_enclosed/ε₀. Field of infinite line λ/(2πε₀r), infinite sheet σ/(2ε₀).

Key terms

Electric flux: number of field lines through a surface, Φ = E·A
Dipole: two equal and opposite charges separated by a small distance
Coulomb's law: force between two point charges is proportional to the product of charges and inversely proportional to the square of distance

Formulas

F = kq₁q₂/r²
E = kq/r²
τ = pE sin θ
Φ = q/ε₀

Important questions with answers

State Gauss's law in electrostatics.2 marks

The total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed: ∮E·dA = q_enclosed/ε₀.

Derive the electric field due to an infinitely long straight wire with uniform linear charge density λ.3 marks

Take a cylindrical Gaussian surface of radius r and length l coaxial with the wire. By symmetry E is radial and the same on the curved surface; flux through the flat ends is zero. Flux = E × 2πrl = λl/ε₀, so E = λ/(2πε₀r), directed radially outward for positive λ.

Derive the expression for the electric field at a point on the axial line of an electric dipole, and state how it compares with the field on the equatorial line at the same large distance.5 marks

Dipole: charges −q at A and +q at B, separation 2a, dipole moment p = q × 2a from −q to +q. At P on the axis at distance r from the centre (towards +q): E₊ = kq/(r − a)² away from B, E₋ = kq/(r + a)² towards A. Net E = kq[1/(r − a)² − 1/(r + a)²] = kq × 4ar/(r² − a²)² = 2kpr/(r² − a²)². For r ≫ a, E_axial = 2kp/r³ = 2p/(4πε₀r³), along p. On the equatorial line E = kp/r³, opposite to p. So at the same large distance the axial field is twice the equatorial field.

Quick revision

Charge is quantised and conserved.
Field lines start on + and end on −, never cross.
Flux through a closed surface depends only on enclosed charge.

Exam tips

Field inside a conductor is zero in electrostatics; charge resides on the outer surface.
For field lines: they start on +q, end on −q, never cross and are perpendicular to a conductor's surface.

Practice MCQs

SI unit of electric charge is
1. ampere
2. farad
3. volt
4. coulomb
Show answer
Answer: (d) coulomb. Coulomb.
Electric field lines
1. can cross
2. start on negative charges
3. form closed loops always
4. never cross
Show answer
Answer: (d) never cross. Field has one direction at a point.
Two charges repel with force F. If the distance is doubled, the force becomes
1. 2F
2. F/4
3. F/2
4. 4F
Show answer
Answer: (b) F/4. F ∝ 1/r².
Flux through a closed surface enclosing charge q is
1. qε₀
2. zero always
3. q/ε₀
4. q/4πε₀
Show answer
Answer: (c) q/ε₀. Gauss's law.
Two charges repel with force F. If both charges are doubled and distance halved, the force becomes
1. 4F
2. 8F
3. F
4. 16F
Show answer
Answer: (d) 16F. (2 × 2)/(1/2)² = 16.
Field due to an infinite plane sheet of charge density σ is
1. σ/ε₀
2. 2σ/ε₀
3. σ/(2ε₀)
4. σ/(4πε₀)
Show answer
Answer: (c) σ/(2ε₀). Independent of distance.
Torque on a dipole is maximum when the angle with the field is
1. 0°
2. 45°
3. 90°
4. 180°
Show answer
Answer: (c) 90°. sin 90° = 1.
Flux through one face of a cube with charge q at its centre is
1. q/ε₀
2. 6q/ε₀
3. q/4ε₀
4. q/6ε₀
Show answer
Answer: (d) q/6ε₀. Symmetry over six faces.

All chapters of Class 12 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.