Last Updated: April 2026
Electrostatics is one of the highest-weight chapters in JEE Main Physics, contributing 3–4 questions per paper (12–16 marks). In JEE Main 2025, 4 out of 4 electrostatics questions were formula-application type. Mastering Coulombs law, electric field, potential, capacitors, and Gauss law is non-negotiable for a 150+ score in Physics.
JEE Main Physics Chapter-wise Weightage 2025
| Chapter | Avg Questions | Marks | Priority |
|---|---|---|---|
| Electrostatics and Capacitors | 3–4 | 12–16 | HIGHEST |
| Current Electricity | 3–4 | 12–16 | HIGHEST |
| Mechanics (all topics) | 8–10 | 32–40 | HIGHEST |
| Optics | 3–4 | 12–16 | HIGH |
| Modern Physics | 2–3 | 8–12 | HIGH |
| Thermodynamics | 2–3 | 8–12 | MEDIUM |
| Magnetism | 2–3 | 8–12 | MEDIUM |
| Waves | 2 | 8 | MEDIUM |
Electrostatics — Core Concepts and Formulas
| Concept | Formula | Key Points |
|---|---|---|
| Coulombs Law | F = kq1q2/r² (k = 9×10⁹ Nm²/C²) | Force along line joining charges |
| Electric Field | E = kq/r² ; F = qE | Superposition applies |
| Electric Potential | V = kq/r ; W = q(V1-V2) | Scalar quantity |
| Potential Energy | U = kq1q2/r | System property |
| Electric Flux | phi = E.A.cosθ | Scalar product |
| Gauss Law | phi = Q_enc/ε₀ | For symmetric charge distributions |
| Capacitance (parallel plate) | C = ε₀A/d | C increases with dielectric |
| Energy stored in capacitor | U = (1/2)CV² = Q²/2C | Energy in electric field |
Gauss Law Applications — JEE Favourites
- Infinite line charge: E = λ/(2πε₀r) — cylindrical Gaussian surface
- Infinite plane sheet: E = σ/(2ε₀) — pillbox Gaussian surface
- Solid sphere (inside): E = Qr/(4πε₀R³) — proportional to r
- Solid sphere (outside): E = kQ/r² — same as point charge
- Hollow sphere (inside): E = 0 — no charge enclosed
Capacitor Combinations — Quick Reference
| Configuration | Equivalent Capacitance | Voltage | Charge |
|---|---|---|---|
| Series | 1/Ceq = 1/C1 + 1/C2 | Divides (V = V1 + V2) | Same on each |
| Parallel | Ceq = C1 + C2 | Same on each | Divides (Q = Q1 + Q2) |
JEE Main Electrostatics — 5 Practice Problems
- Two charges +3μC and -3μC are placed 10 cm apart. Find the electric field at the midpoint.
Solution: Each charge contributes E = kq/r² = 9×10⁹ x 3×10⁻⁶ / (0.05)² = 1.08×10⁷ N/C. Both fields point in same direction at midpoint. Total E = 2.16×10⁷ N/C - A parallel plate capacitor has plates of area 100 cm² separated by 2mm. Find capacitance.
C = ε₀A/d = 8.85×10⁻¹² x 100×10⁻⁴ / 2×10⁻³ = 4.425×10⁻¹² F = 4.4 pF - Two capacitors 4μF and 6μF are connected in series across 100V. Find charge on each.
Ceq = (4×6)/(4+6) = 2.4μF. Q = CeqV = 2.4×10⁻⁶ x 100 = 240μC. Same charge on each in series.
Frequently Asked Questions — JEE Electrostatics
How many questions from Electrostatics come in JEE Main?
JEE Main typically has 3–4 questions from Electrostatics (Class 12 Chapter 1 and 2 combined). With 4 marks each, this chapter alone can contribute 12–16 marks — roughly 8-10% of total Physics marks. It is consistently the highest-scoring chapter relative to preparation time invested.
Is Gauss Law important for JEE Main?
Yes — Gauss Law is tested directly in JEE Main almost every year. You must know its applications for 4 standard configurations: infinite line charge, infinite plane sheet, solid sphere (inside and outside), and hollow sphere. These are application questions, not derivation-based.
What is the difference between electric potential and electric potential energy?
Electric Potential (V) is a property of a point in space due to a charge — it equals the work done per unit charge to bring a test charge from infinity to that point. Electric Potential Energy (U) is the energy of a system of charges — U = qV for a charge q at potential V. Potential is a scalar field; potential energy is the energy stored in the configuration.
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