CBSE Class 12 Physics: Electrostatics — Important Questions with Answers 2026

Short Answer Questions (2-3 Marks)

• Q: State Coulomb's Law. Define one coulomb of charge.
  Ans: Coulomb's Law: The electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. F = kq₁q₂/r², where k = 9 × 10⁹ Nm²/C². One coulomb is the charge that, when placed 1 m from an equal charge in vacuum, experiences a repulsive force of 9 × 10⁹ N.
• Q: Define electric field. Derive the expression for electric field due to a point charge.
  Ans: Electric field at a point is the force experienced by a unit positive test charge placed at that point. E = F/q₀. For a point charge Q: A test charge q₀ at distance r experiences force F = kQq₀/r². E = F/q₀ = kQ/r² = Q/(4πε₀r²). Direction: radially outward for positive Q, inward for negative Q. Unit: N/C or V/m.
• Q: State and prove Gauss's Law. Use it to find the electric field due to an infinite plane sheet of charge.
  Ans: Gauss's Law: The total electric flux through any closed surface equals 1/ε₀ times the total charge enclosed. φ = ∮E⃗·dA⃗ = q_enclosed/ε₀. For infinite plane sheet (surface charge density σ): Choose a cylindrical Gaussian surface with axis perpendicular to sheet. By symmetry, E is perpendicular to sheet. Flux through two ends: 2EA. Charge enclosed: σA. By Gauss's law: 2EA = σA/ε₀. E = σ/(2ε₀). Field is uniform and independent of distance.
• Q: Derive the expression for capacitance of a parallel plate capacitor.
  Ans: Consider two plates of area A separated by distance d. Charge +Q on one plate, -Q on other. Surface charge density σ = Q/A. Electric field between plates: E = σ/ε₀ = Q/(Aε₀). Potential difference: V = Ed = Qd/(Aε₀). Capacitance C = Q/V = ε₀A/d. With dielectric (K): C = Kε₀A/d. Unit: Farad (F).

Long Answer / Application Questions (4-6 Marks)

• Q: What is an electric dipole? Derive the expression for the electric field at an axial point.
  Ans: An electric dipole consists of two equal and opposite charges (+q, -q) separated by a small distance 2a. Dipole moment p⃗ = q × 2a (from -q to +q). At axial point at distance r from centre: E = E₊ - E₋ = kq/(r-a)² - kq/(r+a)². For r >> a: E_axial = 2kp/r³ = p/(2πε₀r³). Direction: along the dipole moment.
• Q: Three capacitors of 2 μF, 3 μF, and 6 μF are connected (i) in series, (ii) in parallel. Find the equivalent capacitance.
  Ans: (i) Series: 1/C = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 1. C = 1 μF. (ii) Parallel: C = 2 + 3 + 6 = 11 μF. In series, equivalent is less than smallest; in parallel, it is the sum.
• Q: Define equipotential surface. Why is no work done in moving a charge along an equipotential surface?
  Ans: An equipotential surface is a surface where the electric potential is the same at every point. Properties: (1) No work is done moving a charge on it because V is constant, so W = q(V₂-V₁) = 0. (2) Electric field is always perpendicular to the equipotential surface. (3) Two equipotential surfaces never intersect. (4) Closer spacing means stronger field.

Exam Tips for This Chapter

• Revise all definitions and laws from Electrostatics — commonly asked as 1-2 mark questions
• Practice diagrams related to Electrostatics — neat labelled diagrams carry 2-3 marks
• For numericals, always show formula → substitution → answer with correct units
• Previous year analysis shows Electrostatics carries significant marks in the board exam