Class 9 Physics Chapter 2: Electrostatic Potential and Capacitance Important Questions (2024-25)

The 2024-25 rationalized CBSE Class 9 Physics syllabus has streamlined Chapter 2 to focus on three high-weightage areas: (1) Electric Potential as a scalar quantity and its relationship to electric field (V = W/q), (2) Equipotential surfaces and why they are always perpendicular to field lines, and (3) Capacitors, dielectrics, and the formula C = ε₀εᵣA/d. In the board exam structure, this chapter typically carries 5–7 marks in the theory paper, distributed across all question types. A typical paper includes: one 1-mark MCQ on definitions (e.g., 'What is an equipotential surface?'), one or two 2-mark numerical problems (e.g., calculating potential at a point given charge and distance), one 3-mark conceptual question (e.g., 'Why does a dielectric increase capacitance?'), and one 5-mark long-answer combining derivation and application (e.g., 'Derive C = ε₀εᵣA/d and explain why capacitance increases with plate area'). Examiners reward students who can distinguish between potential and electric field, apply the parallel-plate capacitor formula, and explain how dielectrics reduce the net electric field. These 18 questions directly target the exam's most frequently asked concepts and question patterns, ensuring you practise exactly what will appear on your test.

**Question 1:** Electric potential at a point is defined as: (A) Force per unit charge (B) Work done per unit charge (C) Charge per unit area (D) Resistance per unit length **Answer:** (B) Work done per unit charge. Electric potential V = W/q, where W is work done by an external agent to bring a unit positive charge from infinity to that point. This is a scalar quantity measured in volts (J/C). **Question 2:** An equipotential surface is one where: (A) Electric field has constant magnitude (B) All points have the same electric potential (C) Electric force always acts tangentially (D) Charge density is uniform **Answer:** (B) All points have the same electric potential. By definition, an equipotential surface is a surface on which every point has the same electric potential. No work is done moving a charge along an equipotential surface because W = q(V₁ − V₂) = 0. **Question 3:** The SI unit of capacitance is: (A) Coulomb (C) (B) Farad (F) (C) Volt (V) (D) Joule (J) **Answer:** (B) Farad (F). Capacitance is defined as C = Q/V, where Q is charge and V is potential difference. One farad = one coulomb per volt (F = C/V). In practice, microfarads (µF) and nanofarads (nF) are common. **Question 4:** When a dielectric material is inserted between the plates of a charged capacitor, the capacitance: (A) Decreases (B) Remains unchanged (C) Increases (D) Becomes zero **Answer:** (C) Increases. A dielectric material has polar molecules or becomes polarized when placed in an electric field. This reduces the net electric field inside the dielectric, which lowers the potential difference across the capacitor plates. Since C = Q/V, a lower V (with Q constant) means higher C. The relationship is C = ε₀εᵣA/d, where εᵣ > 1 is the relative permittivity. **Question 5:** Equipotential surfaces due to a uniformly charged infinite plane sheet are: (A) Concentric spheres (B) Planes parallel to the sheet (C) Circular discs (D) Cylinders **Answer:** (B) Planes parallel to the sheet. For a uniformly charged infinite plane, the electric field is perpendicular to the sheet with constant magnitude at equal distances. Therefore, equipotential surfaces (always perpendicular to field lines) are planes parallel to the charged sheet.

**Question 1:** Define electric potential. Write its SI unit. Calculate the potential at a point 0·1 m away from a charge of 2 × 10⁻⁸ C in vacuum. [Use k = 9 × 10⁹ N⋅m²/C²] **Solution:** Electric potential at a point is the work done per unit positive charge by an external agent to bring a charge from infinity to that point without acceleration. SI unit: Volt (V) = J/C. Using V = kQ/r: V = (9 × 10⁹ × 2 × 10⁻⁸) / 0·1 V = (18 × 10) / 0·1 = 1800 V **Question 2:** Explain why equipotential surfaces are always perpendicular to electric field lines. Give one example. **Solution:** Electric field lines represent the direction of force on a positive test charge. Along an equipotential surface, the potential is constant, so the potential difference dV = 0. Since E = −dV/dr (negative gradient of potential), if dV = 0 in a direction, then E must be perpendicular to that direction. Therefore, equipotential surfaces are always perpendicular to field lines. Example: For a point charge, equipotential surfaces are concentric spheres perpendicular to radial field lines. **Question 3:** The potential difference between two points A and B is 50 V. How much work is done in moving a charge of 5 C from A to B? **Solution:** Work done W = q × ΔV = 5 C × 50 V = 250 J. If the potential at A is higher than at B, the external agent does negative work (the field does positive work on a positive charge moving from higher to lower potential). **Question 4:** What is relative permittivity? How does it affect capacitance? **Solution:** Relative permittivity (εᵣ) is the ratio of the permittivity of a material to the permittivity of free space: εᵣ = ε/ε₀. It is dimensionless and always ≥ 1. From C = ε₀εᵣA/d, when a dielectric with εᵣ > 1 is introduced, the capacitance increases by a factor of εᵣ. For example, if εᵣ = 6, the capacitance becomes 6 times larger. **Question 5:** A parallel-plate capacitor has plate area 100 cm² and plate separation 0·1 m. Calculate its capacitance in vacuum. [ε₀ = 8·85 × 10⁻¹² F/m] **Solution:** Using C = ε₀A/d: C = (8·85 × 10⁻¹² × 100 × 10⁻⁴) / 0·1 C = (8·85 × 10⁻¹² × 10⁻²) / 0·1 C = 8·85 × 10⁻¹⁴ / 0·1 = 8·85 × 10⁻¹³ F = 0·885 pF

**Question 1:** Why does a conductor always have an equipotential surface? Explain how charges distribute on a conductor in an electric field. **Answer:** A conductor has free electrons that move freely inside it. In electrostatic equilibrium (no current flow), the electric field inside a conductor is zero. Since E = −dV/dr, if E = 0 everywhere inside, then dV = 0, meaning the potential is constant throughout the conductor. Therefore, the entire conductor (including its surface) is an equipotential. Charges redistribute on the conductor's surface such that the electric field inside remains zero. Excess positive charges move to the surface if the conductor is positively charged, and the field lines always meet the surface perpendicularly (because the surface is an equipotential, and field lines are perpendicular to equipotentials). The charge density is higher at points with smaller radii of curvature (sharper points), which is why lightning rods are pointed. **Question 2:** A parallel-plate capacitor is charged and then disconnected from the battery. A dielectric slab is then inserted between the plates. What happens to (i) charge Q, (ii) potential difference V, and (iii) capacitance C? Explain each. **Answer:** (i) **Charge Q remains constant** because the capacitor is isolated (disconnected). No charge can flow to or from the plates. (ii) **Potential difference V decreases** because the dielectric reduces the net electric field inside by polarizing. The electric field inside becomes E' = E₀/εᵣ, where E₀ is the initial field. Since V = E·d and d is fixed, V decreases by a factor of εᵣ. (iii) **Capacitance C increases** because C = Q/V, and Q is constant while V decreases. Alternatively, C' = ε₀εᵣA/d = εᵣC₀, showing C increases by factor εᵣ. Energy stored also decreases: U = Q²/(2C) = Q²/(2ε₀εᵣA/d), so U' = U₀/εᵣ. **Question 3:** Two identical conducting spheres A and B have charges +5 µC and −3 µC respectively. When brought into contact, they share charge equally. What is the final charge on each? How does the capacitance of each sphere change? **Answer:** **Final charge on each sphere:** Total charge = 5 + (−3) = 2 µC. When two identical conductors touch, charge distributes equally: Final charge on each = 2/2 = **+1 µC**. **Capacitance change:** For an isolated conducting sphere, C = 4πε₀R, which depends only on radius R, not on charge. Therefore, the capacitance of each sphere **remains unchanged** because their radii do not change during charge redistribution. The capacitance is an intrinsic property of the geometry, independent of the stored charge. **Question 4:** Explain why the potential inside a conductor is constant in electrostatic equilibrium. What does this imply about the electric field inside a conductor? **Answer:** In electrostatic equilibrium, free electrons in a conductor move until the net electric field inside becomes zero. Once E_internal = 0, no force acts on electrons, so they stop moving. Since potential difference dV = −∫E⋅dr, and E = 0 inside the conductor, dV = 0. This means potential is constant everywhere inside. Mathematically, V = kQ/R for a uniformly charged conductor, which is independent of position inside. **Implication:** The electric field inside any conductor in electrostatic equilibrium is always **zero**, regardless of the external field or the amount of charge on it. All the charge resides on the surface, and the surface becomes an equipotential. This principle is used in electrostatic shielding, where a conductor or conducting cage protects the interior from external electric fields.

**Question 1:** Derive the expression for capacitance of a parallel-plate capacitor (C = ε₀A/d). Explain how capacitance changes when (a) plate area is doubled, (b) plate separation is halved, and (c) a dielectric with relative permittivity εᵣ = 4 is inserted. **Solution:** **Derivation:** Consider two parallel conducting plates of area A separated by distance d, with charges +Q and −Q. The electric field between the plates is uniform: E = σ/ε₀ = Q/(ε₀A) where σ is surface charge density. The potential difference between the plates is: V = E⋅d = (Q/(ε₀A))⋅d = Qd/(ε₀A) Capacitance is defined as: C = Q/V = Q / [Qd/(ε₀A)] = **ε₀A/d** **(a) When plate area is doubled:** C' = ε₀(2A)/d = 2⋅[ε₀A/d] = **2C** (Capacitance doubles) **(b) When plate separation is halved:** C' = ε₀A/(d/2) = 2⋅[ε₀A/d] = **2C** (Capacitance doubles) **(c) When dielectric (εᵣ = 4) is inserted:** C' = ε₀εᵣA/d = 4⋅[ε₀A/d] = **4C** (Capacitance becomes 4 times) **Explanation:** Dielectric molecules polarize in the external field, creating an opposing internal field. The net field becomes E_net = E₀/εᵣ, reducing potential difference and increasing capacitance. **Question 2:** A 10 µF capacitor is charged to a potential difference of 100 V. Calculate (a) the charge stored, (b) energy stored, (c) the new capacitance if a dielectric of εᵣ = 5 is inserted while connected to the same battery, and (d) the new energy stored. **Solution:** **(a) Charge stored:** Q = C⋅V = 10 × 10⁻⁶ × 100 = **10⁻³ C = 1000 µC = 1 mC** **(b) Energy stored:** U = ½CV² = ½ × 10 × 10⁻⁶ × (100)² U = 5 × 10⁻⁶ × 10⁴ = **5 × 10⁻² J = 0·05 J = 50 mJ** Alternatively, U = Q²/(2C) = (10⁻³)²/(2 × 10 × 10⁻⁶) = 10⁻⁶/(2 × 10⁻⁵) = 0·05 J ✓ **(c) New capacitance with dielectric (εᵣ = 5, capacitor connected to battery):** C' = εᵣ⋅C = 5 × 10 × 10⁻⁶ = **50 × 10⁻⁶ F = 50 µF** (Since capacitor remains connected to battery, V stays 100 V, but C increases.) **(d) New charge and energy stored:** Q' = C'⋅V = 50 × 10⁻⁶ × 100 = **5 × 10⁻³ C = 5 mC** U' = ½C'V² = ½ × 50 × 10⁻⁶ × (100)² U' = 25 × 10⁻⁶ × 10⁴ = **0·25 J = 250 mJ** Alternatively, U' = εᵣ⋅U = 5 × 0·05 = 0·25 J ✓ (Energy increases because the battery supplies additional charge and energy to maintain constant V.) **Question 3:** A parallel-plate capacitor with plate area 200 cm², plate separation 0·5 mm is filled with a dielectric of relative permittivity 6. A potential difference of 1000 V is applied. Calculate: (a) capacitance, (b) charge stored, (c) electric field between plates, and (d) potential difference across the dielectric if only half the space is filled with dielectric. [ε₀ = 8·85 × 10⁻¹² F/m] **Solution:** **(a) Capacitance:** C = ε₀εᵣA/d = (8·85 × 10⁻¹² × 6 × 200 × 10⁻⁴) / (0·5 × 10⁻³) C = (8·85 × 6 × 200 × 10⁻¹⁶) / (0·5 × 10⁻³) C = (10620 × 10⁻¹⁶) / (0·5 × 10⁻³) = (1062 × 10⁻¹⁵) / (5 × 10⁻⁵) C = **2·124 × 10⁻⁹ F ≈ 2·1 nF** (or 2124 pF) **(b) Charge stored:** Q = C⋅V = 2·124 × 10⁻⁹ × 1000 = **2·124 × 10⁻⁶ C ≈ 2·1 µC** **(c) Electric field between plates:** E = V/d = 1000 / (0·5 × 10⁻³) = 1000 / 0·0005 = **2 × 10⁶ V/m** Alternatively, E = σ/(ε₀εᵣ), where σ = Q/A = (2·124 × 10⁻⁶)/(200 × 10⁻⁴) ≈ 1·062 × 10⁻⁴ C/m² **(d) Potential difference with half dielectric:** If half the space (0·25 mm) has dielectric (εᵣ = 6) and half (0·25 mm) is vacuum (εᵣ = 1): Treating as two capacitors in series: C₁ = ε₀εᵣA/(d/2) = (8·85 × 10⁻¹² × 6 × 200 × 10⁻⁴) / (0·25 × 10⁻³) = 2⋅C (doubled from part a) C₂ = ε₀A/(d/2) = (8·85 × 10⁻¹² × 200 × 10⁻⁴) / (0·25 × 10⁻³) = (1/6)⋅C₁ Effective: 1/C_eff = 1/C₁ + 1/C₂ ... (This becomes complex) Simpler: Charge remains Q = 2·124 µC. V = V₁ + V₂ = E₁d₁ + E₂d₂ E₁ = σ/(ε₀εᵣ) = (1·062 × 10⁻⁴)/(8·85 × 10⁻¹² × 6) ≈ 2 × 10⁵ V/m E₂ = σ/ε₀ = (1·062 × 10⁻⁴)/(8·85 × 10⁻¹²) ≈ 1·2 × 10⁶ V/m V = (2 × 10⁵ × 0·25 × 10⁻³) + (1·2 × 10⁶ × 0·25 × 10⁻³) = 50 + 300 = **350 V** (Dielectric region has lower field and lower voltage drop.)

**Case Study: Lightning Protection and Electrostatic Shielding** A researcher is designing a protective enclosure for sensitive electronic equipment in a high-voltage research lab. The lab experiences strong electrostatic fields from nearby high-voltage power transmission lines, with potential differences of up to 10⁶ V across 100 m distances. The researcher proposes using a Faraday cage (conducting mesh enclosure) and also wants to store emergency backup power using capacitors with dielectric-filled spaces. **(a) Explain why a Faraday cage protects equipment inside from external electric fields.** Step 1: In a conductor at electrostatic equilibrium, the electric field inside is zero because free electrons redistribute to cancel any applied field. Step 2: When external electric field is applied to the conducting mesh, charges on the mesh surface redistribute instantly to create an internal field that cancels the external field. Step 3: By Gauss's law, ∮E⋅dA = Q_enclosed/ε₀. Since E = 0 inside and no net charge is enclosed, the external field is completely shielded. Step 4: Equipment inside the cage experiences zero electric field regardless of the external field magnitude. **Answer:** A Faraday cage works because the conductor automatically creates surface charges that generate an opposing electric field, canceling the external field inside. This is electrostatic shielding. **(b) The lab stores energy using a 500 µF capacitor connected to a 1000 V supply. A technician inserts a dielectric (εᵣ = 8) between the plates while the capacitor is connected to the battery. Calculate the increase in energy stored.** Step 1: Initial energy: U₀ = ½CV² = ½ × 500 × 10⁻⁶ × (1000)² = 250 J Step 2: With dielectric, C' = εᵣC = 8 × 500 × 10⁻⁶ = 4000 × 10⁻⁶ F (capacitor remains connected, V = constant) Step 3: New energy: U' = ½C'V² = ½ × 4000 × 10⁻⁶ × (1000)² = 2000 J Step 4: Increase in energy = U' − U₀ = 2000 − 250 = **1750 J** Step 5: This extra energy comes from the battery, which supplies additional charge: ΔQ = Q' − Q₀ = (C' − C)V = (8 − 1)C⋅V = 7CV = 3500 × 10⁻³ C **Answer:** Energy increases by 1750 J (an 8-fold increase overall, from 250 J to 2000 J). **(c) If the same capacitor were disconnected before inserting the dielectric, would the energy increase or decrease? Calculate the final energy.** Step 1: With capacitor disconnected, charge Q = CV₀ = 500 × 10⁻⁶ × 1000 = 0·5 C remains constant. Step 2: After inserting dielectric: C' = 8C = 4000 × 10⁻⁶ F Step 3: New voltage: V' = Q/C' = 0·5 / (4000 × 10⁻⁶) = 125 V Step 4: Initial energy: U₀ = ½CV² = ½ × 500 × 10⁻⁶ × (1000)² = 250 J Step 5: Final energy: U' = Q²/(2C') = (0·5)² / (2 × 4000 × 10⁻⁶) = 0·25 / (8 × 10⁻³) = 31·25 J Step 6: Energy **decreases** by 250 − 31·25 = **218·75 J** **Answer:** Energy decreases to 31·25 J (or by 218·75 J). The dielectric absorbs energy as it polarizes; the electric field does work on the dielectric molecules, reducing stored energy when the capacitor is isolated. **(d) Write a brief recommendation for the lab technician on which scenario (dielectric inserted with battery connected or disconnected) is better for energy storage, with reasoning.** **Recommendation:** **Insert the dielectric while the capacitor is connected to the battery.** Reason: When connected, the battery supplies additional charge and energy as the dielectric increases capacitance. Energy stored increases from 250 J to 2000 J (8-fold). When disconnected, inserting the dielectric actually reduces stored energy to 31·25 J because the potential difference drops dramatically (from 1000 V to 125 V). For a backup power supply, the connected scenario stores significantly more energy and is the practical choice. The extra energy cost (1750 J from the battery) is offset by the much larger storage capacity.

CBSETUTOR.ai's AI tutor is specifically designed to master Class 9 NCERT Physics through daily adaptive practice on exactly these question types and difficulty bands. Here's how the platform works: **Adaptive Question Generation:** The AI generates unlimited variations of 1-mark MCQs, 2-mark calculations, and 3-mark conceptual questions on Electrostatic Potential and Capacitance—ensuring you never see the exact same problem twice but always within the 2024-25 CBSE syllabus. **Intelligent Error Diagnosis:** When you attempt a question, the AI detects whether you've made a conceptual error (e.g., confusing potential with field) or a calculation mistake (e.g., unit conversion). It immediately offers a mini-lesson and similar practice problems to reinforce. **Spaced Repetition Scheduling:** The platform tracks which topics you struggle with and resurfaces them at scientifically-proven intervals (2 days, 1 week, 2 weeks) to move knowledge from short-term to long-term memory. For instance, if you get equipotential-surface questions wrong, the AI schedules 3–4 variations over the next 10 days. **Exam-Pattern Practice Drills:** Twice weekly, the AI generates full-chapter mock tests that mirror the exact 2026-27 CBSE board distribution: 20% MCQs (1 mark each), 30% short-answer (2 marks), 30% medium-answer (3 marks), 20% long-answer (5 marks). After each drill, you receive a detailed performance report breaking down marks by topic (Potential vs. Capacitors vs. Dielectrics). **Real-Time Feedback on Board-Specific Writing:** For 5-mark long-answer questions, the AI evaluates not just correctness but also structure: Did you define the concept first? Did you derive the formula? Did you apply it to a real example? This mirrors how CBSE examiners score. **Parent & Student Dashboards:** Parents can see weekly progress: % of correct answers, topics mastered, topics needing review, and estimated board exam score based on current performance. Start a 3-day free trial at cbsetutor.ai to unlock unlimited Class 9 Physics drills, personalized feedback, and real-time doubt resolution via AI—no ads, no distractions, just exam-focused learning.