Class 9 Physics Chapter 1 Electric Charges and Fields: 18 Important Questions with Answers

The CBSE Class 9 Physics syllabus (2024–25 rationalized) places Electric Charges and Fields in Unit II: Electricity. The board exam follows a predictable question distribution: 1-mark MCQs test factual recall (Coulomb's constant k = 9 × 10⁹ N·m²/C², SI unit of charge is coulomb), 2-mark questions demand definition and formula application (e.g., derive electric field due to a point charge), and 5-mark questions require full derivations or multi-step numerical solutions (e.g., calculate the net electric field at a vertex of an equilateral triangle with three point charges). Questions on Gauss's law often appear as "derive" or "prove" problems, testing deeper conceptual understanding. The dipole moment (p = q × d) and electric flux (Φ = E·A·cos θ) are perennially examined. Practising these 18 questions exposes you to all question types, difficulty levels, and likely exam variations. You'll recognize patterns instantly during the exam and avoid time-wasting errors.

**Q1.** The SI unit of electric charge is: (A) Ampere (B) Coulomb (C) Newton (D) Joule **Answer:** (B) Coulomb. Electric charge measures the quantity of electricity. Symbol: C. 1 C = charge transferred when 1 A of current flows for 1 second. **Q2.** According to Coulomb's law, the electrostatic force between two point charges is: (A) Directly proportional to the sum of charges (B) Inversely proportional to the distance (C) Inversely proportional to the square of distance (D) Independent of the medium **Answer:** (C) Inversely proportional to the square of distance. F ∝ 1/r². The force weakens rapidly as charges separate. **Q3.** The electric field strength at a distance r from a point charge q is: (A) E = kq/r (B) E = kq/r² (C) E = kq²/r (D) E = kr/q **Answer:** (B) E = kq/r², where k = 9 × 10⁹ N·m²/C². This is the field intensity formula derived from Coulomb's law divided by test charge. **Q4.** A dipole moment is defined as: (A) p = q × r (B) p = q/d (C) p = q × d (D) p = d/q **Answer:** (C) p = q × d, where q is charge magnitude and d is separation. SI unit: coulomb-metre (C·m). Direction: from negative to positive charge. **Q5.** Electric flux through a closed surface is related to enclosed charge by: (A) Gauss's law (B) Coulomb's law (C) Ohm's law (D) Faraday's law **Answer:** (A) Gauss's law states Φ = Q_enclosed/ε₀. It connects flux through a closed surface to the net charge inside.

**Q1. Define electric field. Write its SI unit.** **Answer:** Electric field is the region around a charged object where another charge experiences an electric force. It is defined as the force per unit test charge: E = F/q. SI unit: N/C (newton per coulomb) or V/m (volt per metre). Direction: along the force on a positive test charge. **Q2. State Coulomb's law in words and write its mathematical form.** **Answer:** Coulomb's law states: The electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, and acts along the line joining them. Mathematical form: F = k(q₁q₂)/r², where k = 9 × 10⁹ N·m²/C², q₁ and q₂ are charges, r is separation. **Q3. What is electric dipole? Give an example.** **Answer:** An electric dipole is a system of two equal and opposite charges separated by a small distance. Example: HCl molecule (H⁺ and Cl⁻ atoms), water molecule (H₂O has a net dipole moment due to asymmetric electron distribution). Dipole moment: p = q × d (directed from −q to +q). **Q4. What is the difference between electric field and electric flux?** **Answer:** Electric field (E) is the force per unit charge at a point; it is a vector quantity with SI unit N/C. Electric flux (Φ) is the total field passing through a surface area; Φ = E·A·cos θ; SI unit: N·m²/C (or V·m). Field is local; flux is a surface integral. **Q5. State Gauss's law.** **Answer:** Gauss's law states: The total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed within the surface. Mathematically: Φ = Q_enclosed/ε₀, where ε₀ = 8.85 × 10⁻¹² C²/(N·m²). It connects field to charge distribution without needing to calculate field at every point.

**Q1. Derive the electric field due to a point charge q at distance r.** **Solution:** Consider a point charge +q. Place a test charge +q₀ at distance r from q. By Coulomb's law, electrostatic force: F = kqq₀/r². Electric field is force per unit test charge: E = F/q₀ = kqq₀/(r²q₀) = kq/r². Direction: radially outward for +q, radially inward for −q. Magnitude: E = kq/r². **Q2. Two point charges q₁ = +2 μC and q₂ = +4 μC are separated by 30 cm. Find the force between them. (k = 9 × 10⁹ N·m²/C²)** **Solution:** Using Coulomb's law: F = kq₁q₂/r² F = (9 × 10⁹ × 2 × 10⁻⁶ × 4 × 10⁻⁶) / (0.30)² F = (9 × 10⁹ × 8 × 10⁻¹²) / 0.09 F = (72 × 10⁻³) / 0.09 = 0.072 / 0.09 = 0.8 N Since both charges are positive, the force is repulsive. **Q3. A uniform electric field of magnitude 10 N/C passes through a surface area of 5 m². The angle between the field and the surface normal is 60°. Calculate the electric flux through the surface.** **Solution:** Electric flux: Φ = E·A·cos θ Φ = 10 × 5 × cos 60° Φ = 50 × 0.5 = 25 N·m²/C The component of field perpendicular to the surface contributes to flux. **Q4. Why is the electric field inside a conductor zero?** **Solution:** In a conductor (e.g., copper), free electrons move until electrostatic equilibrium is reached. At equilibrium, the electric field inside is zero because: (1) any internal field would drive electron motion, contradicting equilibrium; (2) charges redistribute on the surface such that the field inside cancels; (3) by Gauss's law, if E = 0 inside, then no net charge exists in the interior—all charge resides on the surface. This is true for any conductor in electrostatic equilibrium.

**Q1. State and prove Gauss's law. Mention its applications.** **Solution:** *Statement:* The electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed: Φ_E = Q_enclosed/ε₀. *Proof Outline:* Consider a point charge +q at the centre of a spherical Gaussian surface of radius r. Electric field at distance r: E = kq/r² = q/(4πε₀r²). Flux through the entire sphere: Φ = ∮ E·dA = E × 4πr² = [q/(4πε₀r²)] × 4πr² = q/ε₀. For a non-spherical surface or charge distribution, by superposition and vector calculus, the result remains: Φ = Q_enclosed/ε₀. *Applications:* 1. Finding electric field: For symmetric charge distributions (spherical, cylindrical, planar), choose an appropriate Gaussian surface to simplify calculation. 2. Electric field of infinite uniformly charged plane: E = σ/(2ε₀), where σ is surface charge density. 3. Field inside a conductor: E = 0 (all charge on surface). 4. Field due to infinite wire: E = λ/(2πε₀r), where λ is linear charge density. **Q2. Three point charges, each of magnitude +2 μC, are placed at the three vertices of an equilateral triangle of side 20 cm. Find the net electric field at the centre of the triangle.** **Solution:** Distance from centre to any vertex of an equilateral triangle with side a = 20 cm: r = a/√3 = 0.20/√3 ≈ 0.115 m Electric field due to one charge at the centre: E₁ = kq/r² = (9 × 10⁹ × 2 × 10⁻⁶) / (0.115)² E₁ = 18 × 10³ / 0.0132 ≈ 1.36 × 10⁶ N/C (directed away from that vertex) By symmetry, three field vectors are at 120° to each other. The resultant of three equal vectors at 120° apart is zero: E_net = 0 N/C This is because the vector sum: E₁ + E₂ + E₃ = 0 when magnitudes are equal and angles are symmetric. **Q3. An electric dipole with dipole moment p = 2 × 10⁻²⁹ C·m is placed in a uniform electric field E = 3 × 10⁵ N/C. The angle between the dipole moment and the field is 30°. Calculate: (a) Torque on the dipole, (b) Work done to rotate the dipole from 30° to 0° (alignment).** **Solution:** *(a) Torque:* τ = p × E × sin θ τ = 2 × 10⁻²⁹ × 3 × 10⁵ × sin 30° τ = 6 × 10⁻²⁴ × 0.5 = 3 × 10⁻²⁴ N·m (or 3 × 10⁻²⁴ J/radian) *(b) Work done:* Work by external agent to rotate from θ₁ = 30° to θ₂ = 0°: W = −ΔU = −[U(0°) − U(30°)] U(θ) = −pE cos θ U(0°) = −2 × 10⁻²⁹ × 3 × 10⁵ × cos 0° = −6 × 10⁻²⁴ × 1 = −6 × 10⁻²⁴ J U(30°) = −2 × 10⁻²⁹ × 3 × 10⁵ × cos 30° = −6 × 10⁻²⁴ × (√3/2) ≈ −5.20 × 10⁻²⁴ J W = −[−6 × 10⁻²⁴ − (−5.20 × 10⁻²⁴)] = −(−0.80 × 10⁻²⁴) = 0.80 × 10⁻²⁴ J = 8 × 10⁻²⁵ J (Negative work by the field; external agent must do positive work to align the dipole against the torque's resistance.)

**Case Study: Lightning and Atmospheric Electric Field** During a thunderstorm, the electric field between storm clouds and the Earth's surface can reach 3 × 10⁶ N/C before lightning strikes. A cloud carrying a net charge of −40 C is at height 1000 m above the ground. The ground is approximately neutral. Calculate: (a) The electric field at ground level due to this cloud (treating it as a point charge), (b) The electrostatic force on a person standing on the ground (assume body charge distribution can be modelled as 0.5 C in the upper half), (c) Why does lightning occur, and what initiates the breakdown? **Solution:** *(a) Electric field at ground level:* E = kq/r² = (9 × 10⁹ × 40) / (1000)² E = (3.6 × 10¹¹) / (10⁶) = 3.6 × 10⁵ N/C (This is less than the observed 3 × 10⁶ N/C because real thunderstorms involve larger charge distributions and multiple clouds, but the order of magnitude is correct.) *(b) Force on the person:* Assuming a simplified model where the person's upper body (≈ 0.5 C) is at height ≈ 1.5 m: E_local ≈ kq/r'² = (9 × 10⁹ × 40) / (1000 − 1.5)² ≈ (9 × 10⁹ × 40) / (998.5)² ≈ 3.6 × 10⁵ N/C F = q_body × E = 0.5 × 3.6 × 10⁵ = 1.8 × 10⁵ N (approx., highly simplified) In reality, a person has net charge ≈ 0 (neutrality), but induced charge separation and local field enhancement dominate. *(c) Lightning initiation (Breakdown):* When the electric field reaches the breakdown threshold of air (≈ 3 × 10⁶ N/C), the air becomes ionized. Free electrons and ions in the air are accelerated by the field, collide with neutral molecules, creating an ionization cascade (Townsend avalanche). This forms a conductive path (stepped leader) from cloud to ground. Once the path forms, charge flows in a massive discharge (return stroke), releasing energy as light, heat, and sound (thunder). The initiation condition: **E_applied > E_breakdown of air**.

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