Electrostatics: Coulomb, Gauss and Capacitance
Coulomb's law, electric field of point, line and sheet charges, Gauss's law, electric potential, energy, dielectrics and boundary conditions, and capacitance of common geometries.
Electrostatics deals with charges at rest. The field E and the potential V describe how they push on other charges.
Coulomb's law and electric field
- F = q₁q₂/(4πε r²), with ε = ε₀ε_r and ε₀ = 8.854 × 10⁻¹² F/m (1/(4πε₀) ≈ 9 × 10⁹).
- E = F/q (V/m). Fields of common charge distributions:
| Source | Field |
|---|---|
| Point charge Q | Q/(4πε r²), radial |
| Infinite line charge ρ_L | ρ_L/(2πε r), radial from the line |
| Infinite sheet ρ_S | ρ_S/(2ε), normal to the sheet (independent of distance) |
Gauss's law
∮ D·dS = Q_enclosed, with D = εE. In point form: ∇·D = ρ_v.
It gives the fields above quickly by choosing a surface where D is constant. Inside a conductor in equilibrium, E = 0 and charge sits on the surface.
Potential
- V = −∫E·dl; E = −∇V. For a point charge: V = Q/(4πε r).
- The electrostatic field is conservative: ∮E·dl = 0 (∇ × E = 0).
- Laplace's equation ∇²V = 0 in charge-free regions; Poisson's ∇²V = −ρ_v/ε.
- Energy stored: W = ½ ∫ ε E² dv; energy density ½εE².
Dielectrics and boundaries
- Polarisation increases D: D = ε₀E + P.
- At a boundary between two dielectrics: tangential E is continuous; normal D is continuous (no free surface charge).
- At a conductor surface: E is normal, E_n = ρ_S/ε.
Capacitance
| Geometry | Capacitance |
|---|---|
| Parallel plates | C = εA/d |
| Coaxial cable (per metre) | C = 2πε/ln(b/a) |
| Isolated sphere | C = 4πεa |
| Two-wire line (per metre, d ≫ a) | C = πε/ln(d/a) |
Layered dielectrics in a parallel-plate capacitor act like capacitors in series (layers parallel to plates) or parallel (side by side).
Worked example
Two charges of 1 µC each are 0.3 m apart in air.
- F = 9 × 10⁹ × 10⁻¹²/0.09 = 0.1 N (repulsive).
Practice
A parallel-plate capacitor has A = 0.01 m², d = 1 mm and ε_r = 5. What is its capacitance in pF? (Give 1 decimal place.)
Show answer and explanation
Answer: 442.7C = ε₀ε_rA/d = 8.854 × 10⁻¹² × 5 × 0.01/0.001 = 442.7 pF.
What is the field (V/m) 2 m from an infinite line charge of 4 nC/m in air? Use 1/(4πε₀) = 9 × 10⁹.
Show answer and explanation
Answer: 36E = ρ_L/(2πε₀r) = 2 × 9 × 10⁹ × 4 × 10⁻⁹/2 = 36 V/m.
The field of an infinite sheet of charge at a distance d from it:
Show answer and explanation
Answer: AE = ρ_S/(2ε) for an infinite sheet, the same at every distance.
At the boundary between two perfect dielectrics with no free charge, which components are continuous? (Select all that apply.)
Show answer and explanation
Answer: A, BE_t1 = E_t2 and D_n1 = D_n2. Normal E and tangential D jump by the ratio of permittivities.