Electrostatics: Charges, Fields & Potential
See how stationary charges create electric fields and potentials, then apply Coulomb's law, Gauss's law and capacitor formulas to JEE Main questions.
The central idea of this chapter
Electrostatics studies charges at rest. Once you know the charge configuration, Coulomb's law and Gauss's law let you find electric field, and potential follows from work done per unit charge.
What should you understand first?
Electrostatics connects charge, field, flux and potential. Most JEE Main problems boil down to choosing between direct Coulomb summation and smart use of Gauss's law or symmetry.
Core concepts
- Properties of charge Quantisation & conservation
- Coulomb's law Point charges
- Electric field & field lines Vector idea
- Electric potential & energy Scalar idea
- Gauss's law & flux Symmetry tool
Useful building blocks
- Point charge F, E, V
- Dipole p = q·2l
- Line / surface charge λ, σ, ρ
- Conductors & shielding E = 0 inside
- Capacitance C = Q/V
Electrostatics formula sheet
Keep track of which quantity is vector (field, force) and which is scalar (charge, potential, energy). Use symmetry to simplify field and flux calculations.
| Topic | Formula / Relation | Meaning or use |
|---|---|---|
| Coulomb's law | F = 1/(4πε₀) · (q₁q₂/r²) | Force between two stationary point charges in vacuum |
| Electric field | E = F/q, for point charge: E = 1/(4πε₀) · q/r² | Force per unit positive test charge |
| Electric potential | V = W/q, for point charge: V = 1/(4πε₀) · q/r | Work done per unit charge in bringing it from infinity |
| Dipole moment | p = q·2l | Magnitude of charge × separation (directed from − to +) |
| Field of dipole (axial) | E = 1/(4πε₀) · 2p/r³ | Along dipole axis, far from dipole |
| Potential of dipole | V = 1/(4πε₀) · (p cosθ / r²) | Potential at point (r,θ) from dipole centre |
| Gauss's law | ∬E·dS = Qenclosed/ε₀ | Net electric flux through closed surface equals enclosed charge divided by ε₀ |
| Infinite plane sheet | E = σ/(2ε₀) | Uniform electric field on each side of sheet |
| Uniformly charged shell | Outside: E = 1/(4πε₀) · Q/r², Inside: E = 0 | Field behaves as if charge at centre; shielding inside |
| Capacitance (parallel plates) | C = ε₀A/d, with dielectric: C = Kε₀A/d | A is plate area, d is separation |
| Energy in capacitor | U = ½ CV² = ½ QV = Q²/(2C) | Electrostatic energy stored in electric field |
| Energy density | u = ½ ε₀E² | Energy per unit volume of electric field |
How to choose the right method
Decide whether direct Coulomb summation or Gauss's law and symmetry will give the field faster, then move from field to potential or energy if required.
Field & potential
- Few point charges? Use Coulomb vector sum
- High symmetry (sphere, sheet, line)? Use Gauss's law
- Potential needed? Use scalar sum V
- Field from potential? E = −dV/dr
- Dipole in field? τ = pE sinθ
Capacitors & energy
- Series combination? 1/C = Σ(1/Cᵢ)
- Parallel combination? C = ΣCᵢ
- Battery connected? V constant
- Isolated system? Q constant
- Energy change asked? Compare ½CV²
How to prepare Electrostatics
Build comfort with vector diagrams for field, then move to potential, energy and capacitor questions.
What to do
- Start with charge properties and Coulomb's law examples
- Practise field diagrams and vector addition for E
- Learn standard results for sheets, lines and shells via Gauss
- Solve potential and energy problems for point charges and dipoles
- Finish with capacitor combinations and dielectric questions
Common mistakes
- Forgetting that inside a conducting shell, electric field is zero
- Mixing up scalar potential with vector field direction
- Applying Gauss's law where symmetry is not sufficient
- Confusing series and parallel rules for capacitors
- Ignoring whether battery remains connected during dielectric insertion
Ready to test Electrostatics?
Revise the electrostatics formula sheet, then solve mixed JEE Main problems on fields, potential and capacitors.
Electrostatics FAQ
Short answers to frequently tested electrostatics ideas.
Coulomb's law gives the electrostatic force between two stationary point charges. It is directly proportional to the product of the charges and inversely proportional to the square of the separation, acting along the line joining them.
Electric field at a point is the force experienced per unit positive test charge placed at that point. For a point charge q, E = 1/(4πε₀) · q/r² radially outward or inward.
Gauss's law states that the total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed by that surface. It is written as ∬E·dS = Qenclosed/ε₀.