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JEE · NEET Physics

Class 12 · Chapter 1

Electric Charges & Fields

Overview, notes, short notes, formula sheet, daily practice problems, previous year questions, and videos for this chapter — all in one place.

Electric Charges & Fields Formula Sheet

8 formulas across 3 topics in Electric Charges & Fields.

1 min read

Updated 2026-07-05 · v1.0.0

Coulomb's Law & Electric Field

Coulomb's law is the electrostatic twin of gravitation — inverse square, but with two signs and vastly stronger. Superpose forces or fields as vectors, always.

F = (1/4πε₀) q₁q₂/r² = k q₁q₂/r²

Coulomb's law

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
kCoulomb constant 1/4πε₀k = 9 × 10⁹ N·m²/C²N·m²/C²[ML³T⁻⁴A⁻²]
ε₀permittivity of free space8.85 × 10⁻¹² C²/(N·m²)C²/(N·m²)[M⁻¹L⁻³T⁴A²]
q₁, q₂the two point chargesC[AT]
rseparation between chargesm[L]

Valid when

  • Point charges (or spherical charges, centre-to-centre)
  • In a medium of dielectric constant K, divide by K

Common mistakes

  • Coulomb force obeys superposition — the force between two charges is unaffected by a third; add the separate forces as vectors

E = F/q₀ = (1/4πε₀) q/r²

Electric field of a point charge

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
Eelectric field strengthN/C (= V/m)[MLT⁻³A⁻¹]
q₀small positive test chargeC[AT]
qsource chargeC[AT]

Valid when

  • Field points AWAY from positive charge, TOWARD negative

Common mistakes

  • E is defined in the LIMIT of a vanishing test charge, so the test charge does not disturb the source

Worth remembering

  • Field lines start on positive, end on negative charge; they never cross, and their density shows field strength
  • Electrostatic force is ~10³⁹ times stronger than gravity between two protons

Electric Dipole

A dipole is two equal and opposite charges a short distance apart. Its field falls as 1/r³ — faster than a point charge — and it feels torque in a uniform field.

p = q(2a), directed from −q to +q

Electric dipole moment

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
pdipole momentC·m[ATL]
2aseparation between the chargesm[L]

Valid when

  • Convention: points from negative to positive charge

E(axial) = (1/4πε₀) 2p/r³; E(equatorial) = (1/4πε₀) p/r³

Dipole field — axial and equatorial

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Superposing the fields of +q and −q for r ≫ a

Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
rdistance from the dipole centrem[L]
pdipole momentC·m[ATL]

Valid when

  • Short dipole approximation r ≫ a
  • Axial field is TWICE the equatorial field at the same distance, and points along p; equatorial points opposite to p

Common mistakes

  • Dipole field falls as 1/r³, faster than a point charge's 1/r² — the two charges nearly cancel at large distance

τ = pE sinθ; U = −pE cosθ

Torque and energy of a dipole in a uniform field

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
θangle between p and Erad (or °)[M⁰L⁰T⁰]
Eexternal uniform fieldN/C[MLT⁻³A⁻¹]

Valid when

  • Uniform field → net force zero, only torque
  • Stable equilibrium at θ = 0 (p along E), unstable at θ = 180°

Common mistakes

  • In a NON-uniform field a dipole also feels a net FORCE, not just torque

Worth remembering

  • Work to rotate a dipole from θ₁ to θ₂: W = pE(cosθ₁ − cosθ₂); flipping from aligned to anti-aligned costs 2pE

Gauss's Law & Field of Symmetric Bodies

Gauss's law turns hard integrals into one-line answers whenever symmetry lets you choose the right closed surface.

Φ = ∮ E·dA = q(enc)/ε₀

Gauss's law

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
Φelectric flux through the closed surfaceN·m²/C (= V·m)[ML³T⁻³A⁻¹]
q(enc)net charge enclosedC[AT]

Valid when

  • Flux depends ONLY on enclosed charge — external charges contribute zero NET flux
  • The field E in the integral is due to ALL charges, inside and out

Common mistakes

  • A charge at the centre of a cube gives flux q/ε₀ total, so q/6ε₀ through each face — but only by symmetry; a charge at a corner needs q/8 enclosed

Outside: E = kq/r²; On surface: kq/R²; Inside conductor: 0

Field of a charged sphere

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Gauss's law with a concentric spherical surface

Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
Rradius of the spherem[L]
rdistance from centrem[L]

Valid when

  • Outside: behaves as if all charge were at the centre
  • Solid NON-conductor (uniform charge): inside E = kqr/R³ (grows linearly from centre)

Common mistakes

  • Conductor vs uniformly-charged insulator differ INSIDE: conductor has E = 0, insulator has E ∝ r

Line: E = λ/(2πε₀r); Sheet: E = σ/(2ε₀); Conductor surface: σ/ε₀

Field of a line charge and a sheet

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Variables used in this formula, with units and dimensions
SymbolMeaningUnitDimension
λlinear charge densityC/m[ATL⁻¹]
σsurface charge densityC/m²[ATL⁻²]

Valid when

  • Infinite line: E ∝ 1/r; infinite sheet: E is UNIFORM (independent of distance)
  • Field just outside a charged conductor is σ/ε₀, twice that of an isolated sheet

Common mistakes

  • Between two oppositely charged parallel sheets E = σ/ε₀; outside them the field is zero

Worth remembering

  • Flux through an open surface uses the same E·dA, but only a CLOSED surface gives q(enc)/ε₀

NEET asks flux-through-a-face and sphere-field-region questions; JEE Advanced hides Gauss inside cavity and superposition problems (charge in an off-centre cavity).

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