Electric Charges and Fields

Why This Chapter Matters

Electrostatics is the first chapter of Class 12 Physics and one of the highest-weightage sections — 8-10 marks in boards and similar in JEE. Coulomb's law, electric field, Gauss's law, and potential are all tested.

Core Concepts

1. Electric Charge

Properties: quantised (q = ne, n = integer, e = 1.6×10⁻¹⁹C), conserved (net charge constant), additive.

Two types: positive (+) and negative (-). Like charges repel, unlike attract.

2. Coulomb's Law

F = kq₁q₂/r² = q₁q₂/4πε₀r²

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

Vector form: Force is along line joining charges.

In medium: F = q₁q₂/4πεr² where ε = ε₀εᵣ (εᵣ = relative permittivity/dielectric constant).

F_medium = F_vacuum/εᵣ.

Superposition: Net force = vector sum of individual forces.

3. Electric Field

E = F/q₀ (force per unit positive test charge).

Due to point charge: E = kQ/r² (direction: radially outward for +Q).

Field lines:

4. Electric Flux and Gauss's Law

Electric flux: Φ = E⃗·A⃗ = EA cosθ. Unit: N·m²/C.

Gauss's Law: ∮E⃗·dA⃗ = Q_enclosed/ε₀.

Applications:

Infinite line charge (λ C/m): E = λ/2πε₀r.

Infinite sheet (σ C/m²): E = σ/2ε₀ (uniform, both sides).

Spherical shell (charge Q, radius R):

Outside (r>R): E = kQ/r² (same as point charge).

Inside (r

Solid sphere (uniform charge density):

Outside: E = kQ/r².

Inside: E = kQr/R³ (proportional to r).

5. Electric Potential

V = W/q₀ (work done to bring unit + charge from ∞ to point).

Due to point charge: V = kQ/r (scalar, can be +/-).

Relation: E = -dV/dr, E⃗ = -∇V.

Equipotential surfaces: V = constant. E⊥ equipotential. No work done moving charge on equipotential.

For point charge: equipotentials are spheres.

Potential energy of system:

Two charges: U = kq₁q₂/r.

Three charges: U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃).

Board Examples

Q1: Two charges +4μC and -2μC separated by 0.5m. Find E and V at midpoint.

At midpoint (0.25m from each):

E₁ = k×4×10⁻⁶/(0.25)² = 9×10⁹×4×10⁻⁶/0.0625 = 576000 N/C (toward -2μC)

E₂ = k×2×10⁻⁶/(0.25)² = 288000 N/C (also toward -2μC, away from +4μC, so same direction)

Net E = 576000+288000 = 864000 N/C.

V₁ = k×4×10⁻⁶/0.25 = 144000 V. V₂ = k×(-2×10⁻⁶)/0.25 = -72000 V.

Net V = 144000-72000 = 72000 V = 72 kV.

PYQs (CBSE)

CBSE 2023: Define electric flux. Write Gauss's law. Using Gauss's law, derive E for uniformly charged infinite plane sheet.

Electric flux = total number of field lines passing through a surface = ∮E⃗·dA⃗.

Gauss's law: ∮E⃗·dA⃗ = Q_enc/ε₀.

For sheet: Gaussian surface = cylinder. E passes through both flat faces (area A each).

Φ = 2EA = σA/ε₀ → E = σ/2ε₀.

CBSE 2022: What is the work done in moving a charge of 2C from one point to another on an equipotential surface of 10V?

W = q×ΔV = 2×0 = 0 J (potential difference on same equipotential = 0).

Revision Notes

COULOMB: F=kq₁q₂/r², k=9×10⁹, In medium: divide by εᵣ
ELECTRIC FIELD: E=F/q₀, Point charge E=kQ/r²
SUPERPOSITION applies to both F and E (vector sum)

GAUSS'S LAW: ∮E·dA = Q_enc/ε₀
Line charge: E=λ/2πε₀r | Sheet: E=σ/2ε₀
Outside sphere: E=kQ/r² | Inside hollow sphere: E=0
Inside solid sphere: E=kQr/R³

POTENTIAL: V=kQ/r (scalar sum for multiple charges)
E=-dV/dr | Equipotential: V=const, E⊥surface
Work on equipotential = 0
PE of two charges: U=kq₁q₂/r