Overview
Electrostatics studies charges at rest — the forces, fields, potentials, and energy they produce, and how capacitors store charge.
1. Coulomb's Law
F = (1/4πε₀) · q₁q₂/r² , 1/4πε₀ ≈ 9 × 10⁹ N·m²/C²
Force is attractive (unlike) or repulsive (like), along the line joining the charges.
2. Electric Field and Potential
Electric field: E = F/q = (1/4πε₀) q/r² (N/C or V/m)
Potential: V = (1/4πε₀) q/r (volt)
Relation: E = −dV/dr
Potential energy of two charges: U = (1/4πε₀) q₁q₂/r
3. Gauss's Law
Φ = ∮ E·dA = q_enclosed / ε₀
Field of an infinite sheet: E = σ/2ε₀
Field near a conductor: E = σ/ε₀
4. Capacitors
Parallel plate: C = ε₀A/d (with dielectric K: C = Kε₀A/d)
Energy stored: U = ½CV² = ½QV = Q²/2C
Series: 1/C = 1/C₁ + 1/C₂ ...
Parallel: C = C₁ + C₂ ...
Worked Examples
Example 1 — Coulomb's law
Force between 2 μC and 3 μC charges 0.1 m apart.
F = 9×10⁹ × (2×10⁻⁶)(3×10⁻⁶)/(0.1)² = 9×10⁹ × 6×10⁻¹²/0.01 = 5.4 N
Example 2 — Electric field
Field at 0.3 m from a 5 μC charge.
E = 9×10⁹ × 5×10⁻⁶/(0.3)² = 9×10⁹ × 5×10⁻⁶/0.09 = 5 × 10⁵ N/C
Example 3 — Capacitance
Parallel-plate capacitor, A = 0.02 m², d = 1 mm, vacuum.
C = ε₀A/d = (8.85×10⁻¹²)(0.02)/(1×10⁻³) = 1.77 × 10⁻¹⁰ F
Example 4 — Energy stored
Energy in a 2 μF capacitor charged to 100 V.
U = ½CV² = ½ × 2×10⁻⁶ × (100)² = ½ × 2×10⁻⁶ × 10⁴ = 0.01 J
Common Mistakes
- Forgetting to square r in Coulomb's law / field.
- Mixing series and parallel capacitor formulas.
- Treating E and V as the same (E is a vector field, V a scalar).
- Not converting μC and mm to SI units.
Exam Tips & Revision Checklist
- Use 1/4πε₀ ≈ 9 × 10⁹ for quick numericals.
- Capacitors: series adds like resistors-in-parallel and vice-versa.
- Practised: Coulomb force, field, capacitance, stored energy.
Frequently Asked Questions
The force between two point charges is F = (1/4πε₀)·(q₁q₂/r²), acting along the line joining them. The constant 1/4πε₀ ≈ 9 × 10⁹ N·m²/C².