MP Board Class 12 Physics Ch 1 Electric Charges and Fields

📑 Table of Contents

1. ❓ Multiple Choice Questions
2. ✏️ Very Short Answer Questions
3. 📝 Short Answer Questions (2-3 Marks)
4. 📋 Long Answer Questions (4-5 Marks)
5. 🔢 Numerical Problems
6. ⚡ Gauss’s Law Applications
7. 🧲 Electric Dipole Questions
8. 🎯 Exam Tips & Strategy

Q1: State Coulomb’s law in vector form.

Ans: F = (1/4πε₀) (q₁q₂/r²) r̂ where r̂ is the unit vector from q₁ to q₂. The force is repulsive for like charges and attractive for unlike charges.

Q2: Define electric field intensity. Write its SI unit.

Ans: Electric field intensity at a point is the force experienced by a unit positive charge placed at that point. E = F/q₀. SI unit: N/C or V/m.

Q3: What is an electric dipole? Give an example.

Ans: A pair of equal and opposite point charges separated by a small distance is called an electric dipole. Example: HCl molecule has a dipole moment due to charge separation.

Q4: Define electric flux. Write its SI unit.

Ans: Electric flux through a surface is the dot product of electric field and area vector: Φ_E = E·dS. SI unit: Nm²/C.

Q5: Derive an expression for the electric field due to an electric dipole at a point on its axial line.

Ans: For a dipole with charges +q and -q separated by distance 2a, the electric field at distance r from the center on the axial line is: E_axial = (1/4πε₀) × (2p/r³) where p = q × 2a is the dipole moment. The field direction is from -q to +q along the axis.

Q6: Derive the expression for torque experienced by an electric dipole placed in a uniform electric field.

Ans: Torque τ = p × E = pE sinθ where θ is the angle between dipole moment and electric field. The torque tries to align the dipole with the field. Maximum torque when θ = 90° (τ_max = pE). Minimum torque (zero) when θ = 0° or 180°.

Q7: State and explain Gauss’s law. Write its mathematical form.

Ans: Gauss’s law states that the net electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by that surface. Mathematically: ∮E·dS = q_enclosed/ε₀. This law is valid for any closed surface and is a fundamental law of electrostatics.

Q8: Using Gauss’s law, derive the expression for electric field intensity due to an infinitely long straight uniformly charged wire.

Ans: Consider a cylindrical Gaussian surface of radius r and length L coaxial with the wire. By Gauss’s law: ∮E·dS = q/ε₀. E(2πrL) = λL/ε₀ where λ is linear charge density. Therefore, E = λ/(2πε₀r). The field is radial and varies as 1/r.

Q9: Derive the expression for electric field due to a uniformly charged infinite plane sheet using Gauss’s law.

Ans: Consider a cylindrical Gaussian surface with area A cutting through the sheet. By symmetry, E is perpendicular to the sheet. Flux = 2EA (both ends). Charge enclosed = σA. By Gauss’s law: 2EA = σA/ε₀. Therefore, E = σ/(2ε₀). Note: E is independent of distance from the sheet.

Q10: Derive the expression for electric field due to a uniformly charged thin spherical shell at a point (a) outside, (b) on the surface, and (c) inside the shell using Gauss’s law.

Ans: For a spherical shell of radius R with total charge Q: (a) Outside (r > R): E = Q/(4πε₀r²) — behaves like a point charge at center. (b) On surface (r = R): E = Q/(4πε₀R²). (c) Inside (r < R): E = 0 — no charge enclosed by Gaussian surface.

Q11: Two point charges +10 μC and +20 μC are placed 10 cm apart in air. Find the force between them.

Solution: Using Coulomb’s law: F = (1/4πε₀) × q₁q₂/r² = 9×10⁹ × (10×10⁻⁶ × 20×10⁻⁶)/(0.1)² = 9×10⁹ × 200×10⁻¹²/0.01 = 9×10⁹ × 2×10⁻⁸ = 180 N (repulsive).

Q12: An electric dipole of dipole moment 4×10⁻⁹ Cm is placed in a uniform electric field of 5×10⁴ N/C. Find the torque experienced by the dipole when placed at 30° to the field.

Solution: τ = pE sinθ = (4×10⁻⁹)(5×10⁴)(sin 30°) = 2×10⁻⁴ × 0.5 = 1×10⁻⁴ Nm.

Q13: A sphere of radius 0.1 m has a charge of 1 μC uniformly distributed on its surface. Calculate the electric field at a point (a) 0.2 m from the center and (b) 0.05 m from the center.

Solution: (a) Outside (0.2 m > 0.1 m): E = 9×10⁹ × 1×10⁻⁶/(0.2)² = 9×10³/0.04 = 2.25×10⁵ N/C. (b) Inside (0.05 m < 0.1 m): E = 0 (field inside a uniformly charged shell is zero).

💡 MP Board Important: In the 2027 exam, focus on numerical problems from Gauss’s law applications and electric dipole. At least one numerical (3-5 marks) is guaranteed. Practice drawing electric field line patterns for different charge configurations — this is a common 2-mark question.

📌 Related Resources:
• Class 12 Physics Chapter 2: Electrostatic Potential Important Questions
• Class 12 Physics Chapter 3: Current Electricity Important Questions
• Class 12 Physics Previous Year Papers (2018-2027)