Last Updated: April 2026
JEE Physics Electrostatics is a foundational chapter for Class 12 Physics and one of the highest-weightage topics in JEE Main and JEE Advanced. Expect 2–3 questions every year from Coulomb’s Law, Electric Field, Potential, Gauss’s Law, and Capacitors. This guide provides comprehensive notes, formulae, and problem-solving strategies for JEE 2027.
Electrostatics Weightage — JEE Main
| Topic | Avg. Questions per Year | Marks |
|---|---|---|
| Coulomb’s Law & Electric Field | 1 | 4 |
| Electric Potential & PE | 1 | 4 |
| Gauss’s Law | 0–1 | 0–4 |
| Capacitance & Dielectrics | 1 | 4 |
| Total | 3–4 | 12–16 |
1. Coulomb’s Law
The electrostatic force between two point charges q₁ and q₂ separated by distance r:
F = kq₁q₂/r² where k = 1/(4πε₀) = 9 × 10⁹ Nm²/C²
- ε₀ (permittivity of free space) = 8.85 × 10⁻¹² C²/Nm²
- Force is along the line joining charges
- Like charges repel, unlike charges attract
- In a medium: F = kq₁q₂/(εᵣ r²) where εᵣ = dielectric constant
- Principle of superposition: Net force = vector sum of all individual forces
2. Electric Field
E = F/q₀ = kQ/r² (due to point charge Q at distance r)
- Electric field is a vector quantity
- Direction: away from +Q, towards −Q
- Field inside a conductor in electrostatic equilibrium = Zero
- Field just outside conductor surface = σ/ε₀ (perpendicular to surface)
- Field due to infinite line charge: E = λ/(2πε₀r)
- Field due to infinite plane sheet: E = σ/(2ε₀)
Electric Field Lines — Properties
- Always start from +charge, end at −charge
- Never cross each other
- Perpendicular to equipotential surfaces
- Density of field lines ∝ magnitude of field
- Number of lines from a charge ∝ magnitude of charge
3. Electric Potential
V = W/q₀ = kQ/r (due to point charge Q)
- Electric potential is a scalar quantity
- SI unit: Volt (V) = J/C
- Potential due to system of charges: V = kΣ(qᵢ/rᵢ)
- Potential inside a conductor = constant (= surface potential)
- Relation: E = −dV/dr (E is negative gradient of V)
- Equipotential surface: surface of constant potential; no work done in moving charge along it
Potential Energy
U = kq₁q₂/r (for two point charges)
- Work done = q(V_A − V_B) = −ΔU
- For positive charge moving from high to low potential: work done by field is positive
4. Gauss’s Law
Φ = Q_enc/ε₀ where Φ = ∮E·dA (electric flux through closed surface)
| Configuration | Electric Field |
|---|---|
| Point charge Q | kQ/r² |
| Uniformly charged sphere (outside, r > R) | kQ/r² |
| Uniformly charged sphere (inside, r < R) | kQr/R³ |
| Infinite line charge (charge density λ) | λ/(2πε₀r) |
| Infinite plane sheet (charge density σ) | σ/(2ε₀) |
| Conducting surface | σ/ε₀ |
5. Capacitance and Capacitors
C = Q/V | SI unit: Farad (F) | Dielectric constant of vacuum = 1
Parallel Plate Capacitor
- C = ε₀A/d (without dielectric)
- C = Kε₀A/d (with dielectric of constant K)
- Energy stored: U = ½CV² = Q²/(2C) = QV/2
- Energy density: u = ½ε₀E²
Capacitors in Combination
- Series: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
- Parallel: C_eq = C₁ + C₂ + C₃
Van de Graaff Generator
Works on the principle of electrostatic induction and the fact that charge resides on outer surface of conductor. Used to accelerate charged particles to high energies. The potential of the outer sphere can reach millions of volts.
JEE Main Previous Year Questions — Electrostatics
| Year | Topic | Type |
|---|---|---|
| 2024 | Capacitor with dielectric slab inserted | Numerical |
| 2023 | Gauss’s Law — flux through hemisphere | MCQ |
| 2022 | Electric potential due to ring | MCQ |
| 2021 | Energy stored in capacitor network | Numerical |
| 2020 | Charge distribution on conductors | MCQ |
Electrostatics Preparation Strategy for JEE 2027
- Understand concepts, don’t memorise blindly — Gauss’s Law is powerful when you choose the right Gaussian surface (spherical/cylindrical/planar)
- Vector vs Scalar — Always remember E is a vector (needs direction), V is scalar (just add with signs)
- Capacitor problems — Master series/parallel combinations, charge redistribution, energy methods
- Superposition principle — Essential for multi-charge problems
- Practice numerical value problems — JEE Main has integer-type questions where exact calculation matters
FAQ — JEE Physics Electrostatics
Is Electrostatics important for JEE Main 2027?
Yes, Electrostatics is very important for JEE Main 2027. It contributes 3–4 questions (12–16 marks) and is foundational for the subsequent chapter of Current Electricity. Topics like Coulomb’s Law, Gauss’s Law, capacitance, and electric potential appear almost every year.
What is the value of Coulomb’s constant k?
Coulomb’s constant k = 1/(4πε₀) = 9 × 10⁹ Nm²/C². The permittivity of free space ε₀ = 8.85 × 10⁻¹² C²/Nm². These values are given in the JEE data sheet but you should memorise them for quick problem-solving.
What is the electric field inside a conductor?
The electric field inside a conductor in electrostatic equilibrium is zero. This is because charges redistribute on the surface until the net field inside becomes zero. Any excess charge on a conductor resides only on its outer surface. The field just outside the conductor surface is σ/ε₀ and perpendicular to the surface.
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