NCERT Solutions: Class 12 Physics — Chapter 1
Electric Charges and Fields | Step-by-step solutions to all 23 NCERT exercises
Electric Charge
Fundamental property of matter. Like charges repel, unlike attract. Quantized: q = ne (e = 1.6 × 10⁻¹⁹ C, n = integer).
Coulomb's Law
F = k|q₁q₂|/r², where k = 9 × 10⁹ N m² C⁻². Force acts along the line joining the two charges.
Electric Field
E = F/q₀. Field lines go from +ve to –ve charges. Field lines never cross; their density shows field strength.
Gauss's Law
Φ = q_enc/ε₀. Net flux through any closed surface equals enclosed charge divided by ε₀. Powerful for symmetric systems.
Q1. What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air? ⌄
Given: q₁ = 2 × 10⁻⁷ C, q₂ = 3 × 10⁻⁷ C, r = 0.3 m
F = kq₁q₂/r² = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)²
F = 6 × 10⁻³ N (Repulsive)
Q2. The electrostatic force on a sphere of charge 0.4 μC due to another of charge −0.8 μC in air is 0.2 N. (a) What is the distance? (b) Force on second sphere due to first? ⌄
(a) r² = k|q₁||q₂|/F = (9×10⁹ × 0.4×10⁻⁶ × 0.8×10⁻⁶)/0.2 = 0.0144
(a) r = 0.12 m = 12 cm
(b) By Newton's third law, force is equal and opposite.
(b) 0.2 N, Attractive (directed towards q₁)
Q3. Check that the ratio ke²/Gm_em_p is dimensionless. Find its value and state what it signifies. ⌄
ke² → N·m² (numerator). Gm_em_p → N·m² (denominator). Ratio is dimensionless. ✓
Value = (9×10⁹ × (1.6×10⁻¹⁹)²) / (6.67×10⁻¹¹ × 9.1×10⁻³¹ × 1.67×10⁻²⁷)
≈ 2.3 × 10³⁹
Electrostatic force between electron and proton is 2.3 × 10³⁹ times stronger than gravitational force.
Q4. (a) Explain 'electric charge of a body is quantised'. (b) Why can we ignore quantisation for macroscopic charges? ⌄
(a) Charge exists only as q = ne, where e = 1.6×10⁻¹⁹ C and n is an integer. Fractional charges do not exist.
(b) Macroscopic charges (~μC) contain ~10¹³ electrons. The step size e is negligibly small, so charge appears continuous.
Q5. When a glass rod is rubbed with silk, charges appear on both. Explain how this is consistent with conservation of charge. ⌄
Electrons transfer from glass to silk. Glass becomes +q, silk becomes −q. Net charge = +q + (−q) = 0.
Charge is conserved — only transferred, never created or destroyed.
Q6. Charges qA = 2 μC, qB = −5 μC, qC = 2 μC, qD = −5 μC at corners of square ABCD (side 10 cm). Force on 1 μC at centre? ⌄
Centre is equidistant from all corners. qA = qC (opposite corners) → forces cancel. qB = qD (opposite corners) → forces cancel.
Net force = 0
Q7. (a) Why can't electric field lines have sudden breaks? (b) Why do two field lines never cross? ⌄
(a) Field lines trace the continuous path of a positive test charge. A moving charge cannot jump — it moves smoothly, so lines have no breaks.
(b) If two lines crossed, there would be two directions of E at that point — impossible, since E has a unique direction at every point.
Q8. qA = 3 μC, qB = −3 μC, 20 cm apart. (a) Electric field at midpoint O? (b) Force on −1.5 × 10⁻⁹ C placed at O? ⌄
r = 0.1 m. E_A = E_B = (9×10⁹ × 3×10⁻⁶)/(0.1)² = 2.7×10⁶ N/C, both directed A→B.
(a) Net E = 5.4 × 10⁶ N/C, from A to B
(b) F = 8.1 × 10⁻³ N, from B to A (opposite to E; charge is negative)
Q9. System: qA = 2.5×10⁻⁷ C at (0,0,−15 cm), qB = −2.5×10⁻⁷ C at (0,0,+15 cm). Total charge and dipole moment? ⌄
Total charge = 0
p = q × 2l = 2.5×10⁻⁷ × 0.30
p = 7.5 × 10⁻⁸ C·m, along −z axis (from −ve to +ve charge)
Q10. Electric dipole moment 4×10⁻⁹ C·m at 30° with uniform field of 5×10⁴ N C⁻¹. Find the torque. ⌄
τ = pE sinθ = 4×10⁻⁹ × 5×10⁴ × sin30° = 4×10⁻⁹ × 5×10⁴ × 0.5
τ = 10⁻⁴ N·m
Q11. Polythene rubbed with wool has charge −3×10⁻⁷ C. (a) Number of electrons transferred? (b) Is there a mass transfer? ⌄
(a) n = q/e = (3×10⁻⁷)/(1.6×10⁻¹⁹)
(a) n ≈ 1.875 × 10¹² electrons (wool → polythene)
(b) Yes. Mass = n × mₑ = 1.875×10¹² × 9.1×10⁻³¹
(b) ≈ 1.706 × 10⁻¹⁸ kg (negligibly small)
Q12. Two copper spheres 50 cm apart, each with charge 6.5×10⁻⁷ C. (a) Repulsion? (b) If charge doubled and distance halved? ⌄
(a) F = kq²/r² = (9×10⁹ × (6.5×10⁻⁷)²)/(0.5)²
(a) F ≈ 1.52 × 10⁻² N
(b) F' = k(2q)²/(r/2)² = 16F
(b) F' ≈ 0.243 N
Q13. Fig 1.30 shows tracks of three charged particles in a uniform electrostatic field. Signs of charges? Which has highest charge-to-mass ratio? ⌄
Particles deflecting toward the positive plate are negatively charged; toward the negative plate — positively charged.
Particles 1 & 2: Negative charge. Particle 3: Positive charge.
Particle 3 has the highest charge-to-mass ratio (greatest curvature of track).
Q14. Uniform E = 3×10³ î N/C. (a) Flux through 10 cm square with plane parallel to yz? (b) If normal makes 60° with x-axis? ⌄
A = 0.01 m², E = 3000 N/C
(a) θ = 0°: Φ = EA cos0° = 3000 × 0.01 × 1
(a) Φ = 30 N·m²/C
(b) θ = 60°: Φ = 30 × cos60° = 30 × 0.5
(b) Φ = 15 N·m²/C
Q15. Net flux of uniform E = 3×10³ î N/C through a cube of side 20 cm with faces parallel to coordinate planes? ⌄
Flux entering one face = −EA; flux leaving opposite face = +EA. They cancel. Uniform field → no enclosed charge.
Net flux = 0
Q16. Net outward flux through a black box surface = 8×10³ N m²/C. (a) Net charge inside? (b) If flux = 0, does that mean no charges inside? ⌄
(a) q = Φ × ε₀ = 8×10³ × 8.854×10⁻¹²
(a) q ≈ 7.08 × 10⁻⁸ C = 0.07 μC
(b) No — zero net flux means zero net charge, but equal and opposite charges could still be present inside.
Q17. Point charge +10 μC is 5 cm directly above the centre of a square of side 10 cm (Fig 1.31). Flux through the square? ⌄
Treat the square as one face of a cube of side 10 cm with charge at centre. Total flux = q/ε₀ = 10×10⁻⁶/8.854×10⁻¹² ≈ 1.13×10⁶ N·m²/C
By symmetry, flux divides equally among 6 faces.
Φ through square = 1.13×10⁶/6 ≈ 1.88 × 10⁵ N·m²/C
Q18. Point charge 2 μC at centre of cubic Gaussian surface 9.0 cm on edge. Net electric flux through the surface? ⌄
Φ = q/ε₀ = (2×10⁻⁶)/(8.854×10⁻¹²)
Φ ≈ 2.26 × 10⁵ N·m²/C
Note: Flux depends only on enclosed charge, not on the size or shape of the surface.
Q19. Flux of −1.0×10³ N m²/C through a spherical surface of 10 cm radius. (a) Flux if radius doubled? (b) Magnitude and sign of charge? ⌄
(a) Flux remains −1.0×10³ N·m²/C (depends only on enclosed charge)
(b) q = Φ × ε₀ = −1.0×10³ × 8.854×10⁻¹²
(b) q ≈ −8.85 nC (negative)
Q20. Conducting sphere of radius 10 cm. Electric field 20 cm from centre is 1.5×10³ N/C, pointing radially inward. Net charge on sphere? ⌄
r = 0.2 m, E = 1500 N/C (inward → negative charge)
|q| = Er²/k = (1500 × 0.04)/(9×10⁹)
q = −6.67 × 10⁻⁹ C ≈ −6.67 nC
Q21. Uniformly charged conducting sphere of 2.4 m diameter; surface charge density 80 μC/m². (a) Charge on sphere? (b) Total flux leaving surface? ⌄
R = 1.2 m, A = 4π(1.2)² ≈ 18.09 m²
(a) q = σA = 80×10⁻⁶ × 18.09 ≈ 1.45 × 10⁻³ C = 1.45 mC
(b) Φ = q/ε₀ ≈ 1.63 × 10⁸ N·m²/C
Q22. An infinite line charge produces a field of 9×10⁴ N/C at 2 cm distance. Calculate the linear charge density. ⌄
E = λ/(2πε₀r) → λ = E × 2πε₀r = 9×10⁴ × 2π × 8.854×10⁻¹² × 0.02
λ ≈ 10⁻⁷ C/m = 0.1 μC/m
Q23. Two large parallel plates with opposite surface charge densities of magnitude 17.0×10⁻²² C/m². Find E: (a) outer region of plate 1, (b) outer region of plate 2, (c) between plates. ⌄
σ = 17×10⁻²² C/m². Fields from both plates cancel outside, add between.
(a) E = 0 | (b) E = 0
(c) E = σ/ε₀ = (17×10⁻²²)/(8.854×10⁻¹²)
(c) E ≈ 1.92 × 10⁻¹⁰ N/C
Gauss's Law and field line questions are very frequent in CBSE. Always mention units and direction in numerical answers. For Gauss's Law problems, always identify the symmetry first (spherical / cylindrical / planar) before choosing the Gaussian surface.
Is charge quantised? Explain briefly.
Yes. Charge exists only as q = ±ne, where e = 1.6×10⁻¹⁹ C and n is a positive integer. Fractional multiples of e are not observed. Macroscopic charges appear continuous because n is astronomically large.
Why do electric field lines never cross?
At any intersection, the electric field would point in two directions simultaneously — which is physically impossible since E is a unique vector at every point in space.
What is the significance of Gauss's Law?
Gauss's Law (Φ = q_enc/ε₀) relates flux through any closed surface to the net charge enclosed. It is especially powerful for calculating E when the charge distribution has spherical, cylindrical, or planar symmetry.
Are these solutions updated for the latest CBSE syllabus?
Yes — fully aligned with current NCERT and CBSE 2025–26 guidelines. All solutions follow the step-by-step format recommended by CBSE for maximum marks.
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