Lesson Overview

Unit 6 explains electric charges, electric forces, electric fields, electric potential, electric flux, Gauss’s law, and capacitors. These form the foundation of all electrical and electronic systems.

1. Core Concepts (Short Notes)

6.1 Coulomb’s Law

• Force between two point charges:F = (1 / 4πϵ) × (q₁q₂ / r²)

• Directly proportional to product of charges.

• Inversely proportional to square of distance.

• Acts along the line joining charges.

6.2 Electric Field (E)

• Region where charge experiences a force.

• E = F / q

• Due to point charge: E = (1 / 4πϵ) × (Q / r²).

• Direction: away from +Q, toward –Q.

6.3 Electric Potential (V)

• Work done per unit charge to bring a charge from infinity.

• V = (1 / 4πϵ) × (Q / r).

• Potential difference (V = W/q).

• Unit: volt (V).

6.4 Electric Potential Energy (U)

• U = qV.

• For two charges: U = (1 / 4πϵ) × (q₁q₂ / r).

6.5 Electric Flux (Φₑ) & Gauss’s Law

• Electric flux: Φ = EA cosθ.

• Gauss’s Law:Φ = Q / ϵ.

• Used to calculate E around symmetric charge distributions.

6.6 Capacitors

• Capacitance (C): ability to store charge.

• C = Q / V.

• Parallel plate capacitor:C = ϵA / d.

• Energy stored:U = ½CV² = ½QV = ½Q²/C.

2. Detailed Notes for Each Section

6.1 Coulomb’s Law

Expression:

F = (1 / 4πϵ₀) × (q₁q₂ / r²).

Permittivity (ϵ)

• ϵ₀: permittivity of free space.

• ϵ = ϵ₀ϵᵣ (ϵᵣ = relative permittivity).

Vector Form:

6.2 Electric Field

Definition

E = F/q.

Due to a point charge:

E = (1 / 4πϵ) × (Q / r²).

Electric Field Lines

Properties:

• Start on positive, end on negative.

• Never intersect.

• Density shows strength.

Common Field Patterns:

• Isolated charge.

• Dipole.

• Parallel plates.

6.3 Electric Potential (V)

Definition

V = W/q.

Expression for a point charge:

V = (1 / 4πϵ) × (Q / r).

Potential Difference & Work Done

W = Vq.

Equipotential Surfaces

• Perpendicular to electric field lines.

• No work done moving along an equipotential.

6.4 Electric Potential Energy

Formula:

U = (1 / 4πϵ) × (q₁q₂ / r).

Potential Gradient:

E = – dV/dr.

6.5 Electric Flux & Gauss’s Law

Flux:

Φ = EA cosθ.

Gauss’s Law:

Φ = Q / ϵ.

Applications:

• Spherical shell:

  • Inside: E = 0.

  • Outside: E = (1 / 4πϵ) × (Q / r²).

• Infinite sheet:E = σ / (2ϵ).

• Infinite line of charge:E = λ / (2πϵr).

6.6 Capacitors

Definition:

C = Q/V.

Parallel Plate Capacitor:

C = ϵA/d.

Effect of Dielectric:

C increases → C = kϵ₀A/d.

Combination of Capacitors:

• Parallel: Cₑ = C₁ + C₂ + ...

• Series: 1/Cₑ = 1/C₁ + 1/C₂ + ...

Energy Stored:

U = ½CV².U = ½Q²/C.U = ½QV.

3. Formula Summary (Unit 6)

• F = (1 / 4πϵ) × (q₁q₂ / r²)

• E = F/q

• E = (1 / 4πϵ) × (Q / r²)

• V = (1 / 4πϵ) × (Q / r)

• U = qV

• Φ = EA cosθ

• Q = ϵΦ

• C = Q/V

• C = ϵA/d

• U = ½CV²

4. Common Mistakes to Avoid

• Forgetting permittivity (ϵ) in equations.

• Using radius (r) incorrectly in Gauss’s law problems.

• Mixing series and parallel formulas.

• Confusing potential (V) with potential energy (U).

• Forgetting that electric field lines never cross.

5. Exam Tips

• Identify symmetry before applying Gauss’s Law.

• Show field-line diagrams clearly.

• Always convert cm → m in capacitor questions.

• Mention assumptions (e.g., uniform field between plates).

• Use potential gradient for tough E-field problems.

6. Quick Revision Table