Based on detailed analysis of last 5 years' papers. Perfect for 2026 Boards prep!
| Question Example | Type/Marks | Years Repeated | Notes |
|---|---|---|---|
| Using Gauss's law, derive electric field due to infinitely long straight uniformly charged wire (or infinite plane sheet / spherical shell). | Derivation (3-4 marks) | 2021, 2022, 2023, 2024, 2025 | Repeated 5x; Gaussian surface cylinder/plane/sphere → flux = E×area = Q_enclosed/ε₀ → E = λ/(2πε₀r) or σ/(2ε₀) or Q/(4πε₀r²). |
| Electric dipole moment p is placed in uniform electric field E. Find torque on dipole and potential energy. Also find field on axial & equatorial line. | Short Answer (3 marks) | 2021 Term 2, 2022, 2023, 2024 | Repeated 4x; τ = pE sinθ, U = -p·E = -pE cosθ; E_axial = 2p/(4πε₀r³), E_equatorial = -p/(4πε₀r³). |
| Two point charges +q and -q are placed at distance 2a. Find electric field at point on perpendicular bisector at distance r from centre. | Numerical/Short Answer (2-3 marks) | 2022, 2023, 2024, 2025 | Repeated 4x; Dipole field on equatorial line: E = p/(4πε₀r³) towards -q (p = q×2a). |
| Assertion: Electric field lines never form closed loops. Reason: Electrostatic field is conservative. | Assertion-Reason (1 mark) | 2023, 2024, 2025 | Repeated 3x; Both true, reason explains (work zero in closed path). |
| Calculate electric flux through a cube of side a placed with one face perpendicular to uniform electric field E (or Gaussian surface variants). | Short Answer (2 marks) | 2021 Term 1, 2022, 2023, 2025 | Repeated 4x; Total flux = 0 (no charge inside) or E a² through one face if field along normal. |
| Force between two charges q1 and q2 separated by r is F. If separation becomes r/2 in a medium of dielectric constant K, find new force. | Numerical (2 marks) | 2022, 2024 | Repeated 2x; F' = K (q1 q2)/(4πε₀ (r/2)²) = 4K F (in medium force decreases by K). |
| MCQ: The SI unit of electric flux is: (a) N m² C⁻¹ (b) V m (c) both (d) none | MCQ (1 mark) | 2021 Term 1, 2023, 2024 | Repeated 3x; Answer (c) both (N m²/C = V m). |
| Derive expression for electric field due to dipole at axial point. | Derivation (3 marks) | 2023, 2025 | Repeated 2x; E = (1/(4πε₀)) (2p / r³) along p for r >> a. |
| Case-based: Given charge distribution (line/sheet/sphere), apply Gauss's law to find field/flux at given point. | Case-Based (4 marks) | 2023, 2025 | Repeated 2x; Choose symmetric Gaussian surface → apply Gauss's law. |
| Two charges +q and +4q are separated by distance d. Find point where electric field is zero. | Short Answer (2 marks) | 2021 Term 2, 2024 | Repeated 2x; Outside on line joining, closer to smaller charge: distance from +q = d/3 towards +4q. |
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