Electrostatics – JEE Main Physics Formula Sheet & Class 12 Notes | Coulomb's Law, Electric Field, Potential, Gauss's Law, Capacitors & All Electrostatics Formulas

Coulomb's Law, Electric Field, and Superposition Principle

Why Coulomb's Law and Electric Field Superposition Are the Starting Formulas of All Electrostatics JEE Main Questions

Coulomb's Law (from Aakash PDF — Electrostatics):

The electrostatic force between two point charges q₁ and q₂ separated by distance r in vacuum:

F = kq₁q₂/r² = q₁q₂/(4πε₀r²)

k = 9×10⁹ N·m²/C² (Coulomb's constant). ε₀ = 8.85×10⁻¹² C²/N·m² (permittivity of free space).

Vector form: F₁₂ = kq₁q₂/r² × r̂₁₂ (force on q₁ due to q₂, directed from q₂ to q₁).

Like charges: repulsive. Unlike charges: attractive. Central force (acts along the line joining charges).

In a medium of relative permittivity (dielectric constant) ε_r: F = kq₁q₂/(ε_r r²)

ε_r = C_medium/C_vacuum; for water ε_r ≈ 80 (Coulomb force reduces 80× in water).

Superposition Principle (from Aakash PDF — Electrostatics JEE Main):

The total electric force on a charge due to multiple other charges = vector sum of individual forces:

F_total = ΣFᵢ (each pair acts independently, unaffected by others)

Electric Field (from Aakash PDF — Electrostatics JEE Main):

Electric field at a point = force per unit positive test charge:

E = F/q₀ (direction = direction of force on positive charge)

Due to a point charge q at distance r: E = kq/r² (outward for +q, inward for –q)

Vector: E = (kq/r²) r̂ (r̂ = unit vector from q to field point)

Superposition for field: E_net = ΣEᵢ (vector sum of all individual fields)

SI unit of E: N/C = V/m. Dimensional formula: [MLT⁻³A⁻¹].

Electric field lines: start from +charge, end at –charge; never cross; density ∝ field strength; tangent = field direction. Download the Free PDF for Electrostatics for all Coulomb's law examples for JEE Main.

Electric Dipole — Moment, Field on Axis and Equator, Torque, Potential Energy

Why Dipole Field Formulas (2kp/r³ and kp/r³) and Torque τ=pE sinθ Are the Most-Tested Electrostatics JEE Main Results

Electric Dipole (from Aakash PDF — Electrostatics JEE Main):

Two equal and opposite charges +q and –q separated by distance 2l.

Dipole moment: p = q × 2l (vector, from –q to +q). SI unit: C·m.

Electric Field Due to Dipole (from Aakash PDF — Electrostatics JEE Main):

(1) On axial line (along the dipole axis, at distance r from centre, r>>l):

E_axial = 2kp/r³ (direction: along p, i.e., from –q to +q direction)

(2) On equatorial line (perpendicular bisector of dipole, at distance r from centre, r>>l):

E_equatorial = kp/r³ (direction: antiparallel to p)

Note: E_axial = 2 × E_equatorial at the same distance r.

(3) At general point (at distance r, angle θ from dipole axis):

E = (kp/r³)√(3cos²θ+1); angle β with radius vector: tan β = tanθ/2

Torque on Dipole in Uniform Field (from Aakash PDF — Electrostatics JEE Main):

τ = pE sinθ = p×E

where θ = angle between p and E. Torque tends to align p with E.

Stable equilibrium: θ = 0° (p ∥ E; τ=0; U_min=–pE)

Unstable equilibrium: θ = 180° (p antiparallel to E; τ=0; U_max=+pE)

Maximum torque: θ = 90° (τ_max = pE)

Potential Energy of Dipole (from Aakash PDF — Electrostatics JEE Main):

U = –pE cosθ = –p·E

U = –pE (θ=0°, stable); U = 0 (θ=90°); U = +pE (θ=180°, unstable).

Work to rotate dipole from θ₁ to θ₂: W = pE(cosθ₁–cosθ₂).

Force on dipole in non-uniform field: F = p(dE/dx) [dipole aligns with E and moves toward stronger field if p parallel to E].

Download the Free PDF for Electrostatics for all dipole examples for JEE Main.

Gauss's Law — Statement and All Applications

Why Gauss's Law Applications (Sheet, Line Charge, Sphere) Are the Most Calculation-Efficient Electrostatics JEE Main Tools

Gauss's Law (from Aakash PDF — Electrostatics JEE Main):

The total electric flux through any closed surface (Gaussian surface) equals the net charge enclosed divided by ε₀:

∮E·dA = Q_enc/ε₀

Electric flux: φ = ∮E·dA = ∮E cosα dA (α = angle between E and area vector dA).

For uniform E over area A: φ = EA cosα.

Application 1 — Infinite Line Charge (from Aakash PDF — Electrostatics JEE Main):

Linear charge density λ (C/m). Gaussian surface: cylinder of radius r, length l coaxial with wire.

Flux through curved surface: E × 2πrl = λl/ε₀

E = λ/(2πε₀r) (radially outward from line; perpendicular to wire)

Application 2 — Infinite Plane Sheet (from Aakash PDF — Electrostatics JEE Main):

Surface charge density σ (C/m²). Gaussian surface: cylindrical pillbox straddling the sheet.

Flux through two flat faces: 2EA = σA/ε₀

E = σ/(2ε₀) (on each side, perpendicular to sheet, independent of distance)

Between two infinite parallel sheets (opposite charges +σ and –σ):

E_between = σ/ε₀ (fields add inside); E_outside = 0 (fields cancel).

Application 3 — Spherical Shell (from Aakash PDF — Electrostatics JEE Main):

Uniformly charged shell, charge Q, radius R.

Outside (r > R): E = kQ/r² (behaves like point charge)

Inside (r < R): E = 0 (no enclosed charge; no field inside)

At surface (r = R): E = kQ/R² = σ/ε₀

Application 4 — Uniformly Charged Solid Sphere (from Aakash PDF — Electrostatics JEE Main):

Volume charge density ρ, total charge Q = (4/3)πR³ρ, radius R.

Outside (r > R): E = kQ/r² (same as shell)

Inside (r < R): enclosed charge = Q(r/R)³; Gaussian sphere radius r:

E = kQr/R³ = ρr/(3ε₀) (increases linearly from 0 at centre to kQ/R² at surface)

At surface (r = R): E = kQ/R² (continuous at surface).

Download the Free PDF for Electrostatics for all Gauss's law examples for JEE Main.

Electric Potential, Potential Difference, and E–V Relationship

Why Electric Potential Formulas and E=–dV/dr Are Essential Electrostatics JEE Main Tools

Electric Potential (from Aakash PDF — Electrostatics JEE Main):

Electric potential V at a point = work done per unit positive charge in bringing a test charge from infinity to that point:

V = W_∞→P / q₀ (reference: V = 0 at infinity)

Due to a point charge q: V = kq/r (scalar; can be positive or negative)

Due to multiple charges: V_total = ΣVᵢ = kΣqᵢ/rᵢ (algebraic sum, not vector)

Electric Potential Due to Dipole (from Aakash PDF — Electrostatics JEE Main):

At distance r, angle θ from dipole axis (r >> l):

V = kp cosθ/r²

On axis (θ=0°): V = kp/r². On equator (θ=90°): V = 0 (potential is zero on equatorial line).

Potential Due to Shell and Sphere (from Aakash PDF — Electrostatics JEE Main):

Conducting shell (charge Q, radius R):

Outside (r>R): V = kQ/r; At surface (r=R): V = kQ/R; Inside (rV = kQ/R (constant = surface value)

Solid insulating sphere (uniform volume charge Q, radius R):

Outside: V = kQ/r; At surface: V = kQ/R; Inside: V = kQ(3R²–r²)/(2R³)

At centre: V_centre = 3kQ/(2R) = (3/2)V_surface

Potential Difference and Work Done (from Aakash PDF — Electrostatics JEE Main):

Potential difference between two points A and B: V_A – V_B

Work done by electric field in moving charge q from A to B:

W_AB = q(V_A – V_B) = qΔV (work done by field = charge × drop in potential)

Work done by external agent = –W_AB = q(V_B – V_A).

E–V Relationship (from Aakash PDF — Electrostatics JEE Main):

In 1D: E = –dV/dr (field = negative gradient of potential)

In 3D: E = –(∂V/∂x î + ∂V/∂y ĵ + ∂V/∂z k̂) = –∇V

Equipotential surfaces: surfaces on which V = constant → E ⊥ equipotential surface (E is perpendicular to every equipotential surface). Work done in moving a charge on an equipotential surface = 0. Closer equipotential lines → stronger field.

Download the Free PDF for Electrostatics for all electric potential examples for JEE Main.

Electric Potential Energy and Energy of a System of Charges

Why PE of Two Charges kq₁q₂/r and System Energy Are Key Electrostatics JEE Main Questions

Electric Potential Energy of Two Point Charges (from Aakash PDF — Electrostatics JEE Main):

U = kq₁q₂/r

U > 0 for like charges (repulsive — energy stored against repulsion). U < 0 for unlike charges (attractive — bound system). U = 0 at r = ∞ (reference).

Potential Energy of a System of Charges (from Aakash PDF — Electrostatics JEE Main):

For n charges, the total PE = sum over all pairs:

U_total = Σ_{all pairs} kqᵢqⱼ/rᵢⱼ

For three charges q₁, q₂, q₃ at pairwise distances r₁₂, r₂₃, r₁₃:

U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃)

Energy to Assemble a System (from Aakash PDF — Electrostatics):

Work done to bring charges from infinity to their final positions = total PE of the configuration. If charges are at vertices of a polygon with same pairwise distance r, all charges q: U = k × C(n,2) × q²/r [C(n,2) = n(n–1)/2 pairs].

PE of a Charge in External Field:

U = qV (where V = potential of external field at the location of charge q).

Work done by external agent to move q from A to B: W = q(V_B – V_A).

Self Energy of a Sphere:

For a uniformly charged conducting sphere of charge Q, radius R: self energy = kQ²/2R (energy stored in the electric field).

Download the Free PDF for Electrostatics for all potential energy examples for JEE Main.

Conductors — Properties, Capacitance, Parallel Plate, Combinations, Energy

Why Capacitor Energy U=½CV² and Combination Formulas Are the Most Directly Tested Electrostatics JEE Main Results

Properties of Conductors (from Aakash PDF — Electrostatics JEE Main):

(1) E = 0 inside a conductor in electrostatic equilibrium.

(2) V = constant throughout the conductor (same potential everywhere).

(3) All charge resides on the outer surface of the conductor.

(4) At the surface: E = σ/ε₀ (perpendicular to surface; directed outward for +σ).

(5) Electric field lines are perpendicular to conductor surface.

(6) Charge distributes more densely at sharp points → lightning conductors.

Capacitance (from Aakash PDF — Electrostatics JEE Main):

Capacitance C = Q/V. SI unit: Farad (F). 1 μF = 10⁻⁶ F; 1 nF = 10⁻⁹ F; 1 pF = 10⁻¹² F.

Parallel plate capacitor: Two parallel plates, area A, separation d, in vacuum:

C = ε₀A/d (increases with area, decreases with separation)

Spherical capacitor: Inner radius a, outer radius b (b>a):

C = 4πε₀ab/(b–a)

Isolated sphere: let b→∞: C = 4πε₀R (capacity depends only on radius)

Capacitors in Series (from Aakash PDF — Electrostatics JEE Main):

Same charge Q on each; voltages add: V = V₁+V₂+...

1/C_eff = 1/C₁ + 1/C₂ + ... (C_eff < smallest C)

For two equal capacitors C in series: C_eff = C/2. Voltage across each: V₁/V₂ = C₂/C₁.

Capacitors in Parallel (from Aakash PDF — Electrostatics JEE Main):

Same voltage V; charges add: Q = Q₁+Q₂+...

C_eff = C₁ + C₂ + ... (C_eff > largest C)

For two equal capacitors C in parallel: C_eff = 2C. Charge on each: Q₁/Q₂ = C₁/C₂.

Energy Stored in Capacitor (from Aakash PDF — Electrostatics JEE Main):

U = ½CV² = Q²/2C = QV/2

Energy density (energy per unit volume between plates): u = ½ε₀E²

For parallel plate: U = (½ε₀E²) × (Ad) = ½ε₀E²×Volume.

Force between plates: F = Q²/(2ε₀A) = σ²A/(2ε₀) (attractive for opposite plates).

Download the Free PDF for Electrostatics for all capacitor examples for JEE Main.

Dielectrics — Constant κ, Effect on Capacitor, Partial Slab, Energy Changes

Why Dielectric Constant and Capacitor with Dielectric Are Direct Electrostatics JEE Main Questions

Dielectric in a Capacitor (from Aakash PDF — Electrostatics JEE Main):

When a dielectric of constant κ fills the space between the plates:

C = κε₀A/d = κC₀ (capacitance increases by factor κ)

For a dielectric: ε = κε₀ (permittivity); κ = 1 for vacuum/air, κ > 1 for all insulators.

Two Cases When Dielectric Is Inserted (from Aakash PDF — Electrostatics JEE Main):

Case 1 — Capacitor connected to battery (V = constant):

V unchanged → Q increases (Q = CV → Q' = κQ₀); E between plates = V/d unchanged? Wait — E=V/d, V same, d same → E same. Actually charge increases: Q'=κQ₀; E=V/d (unchanged); C=κC₀. Energy: U'=½κC₀V² = κU₀ (increases — battery supplies extra energy).

Case 2 — Isolated capacitor (Q = constant):

Q unchanged → V decreases (V' = Q/C' = Q/(κC₀) = V₀/κ); E' = V'/d = E₀/κ (decreases); C' = κC₀. Energy: U' = Q²/2C' = Q²/(2κC₀) = U₀/κ (DECREASES — energy goes into polarising the dielectric).

Partial Dielectric Slab (from Aakash PDF — Electrostatics JEE Main):

A slab of dielectric constant κ, thickness t, inserted between plates (total separation d):

C = ε₀A/(d–t+t/κ)

Effective reduction in separation: the slab of thickness t behaves like air gap of t/κ.

d_eff = d – t + t/κ = d – t(1–1/κ). For conducting slab (κ→∞): d_eff = d–t.

Capacitors with Slabs as Series/Parallel (from Aakash PDF — Electrostatics JEE Main):

Dielectric slab dividing a capacitor into two sections: treat as two capacitors in SERIES (same charge).

Vertical dielectric filling half area: treat as two capacitors in PARALLEL (same voltage).

Download the Free PDF for Electrostatics for all dielectric examples for JEE Main.