Gauss' Law

Gauss' Law

1. Statement — Electric Flux Form

Gauss's law: total electric flux $Φ_E = ∯_{S} E·n dS$ through any closed surface $S$ equals enclosed charge divided by $ε_0$:

where $E·n = E cosθ$ is normal component. Flux counts net field lines crossing $S$ (outward positive). $Q_{enc}$ includes free + bound if using $D$ form ($∯ D·n dS = Q_{free}$).

2. Why Useful

With symmetry (spherical, cylindrical, planar), $E$ magnitude constant on Gaussian surface and parallel to $n$, so flux = $E·Area$ ⇒ $E$ found without integral. Without symmetry, still true but not helpful for calculation.

3. Applications (exam must-do three)

• Point charge / sphere radius $R$: For $r>R$ (outside), $E·4πr²=Q/ε_0 ⇒ E=Q/(4π ε_0 r²)$ (Coulomb); for $r<R$ with uniform volume charge $ρ$, $Q_{enc}=ρ·4πr³/3 ⇒ E= ρ r/(3ε_0)= Q r/(4π ε_0 R³)$ (linear inside).
• Infinite line charge $λ$: Gaussian cylinder radius $r$ length $l$: flux = $E·2π r l = λ l/ε_0 ⇒ E=λ/(2π ε_0 r)$ (as earlier).
• Infinite plane sheet $σ$: pillbox half in/half out: flux = $E·2A = σA/ε_0 ⇒ E=σ/(2ε_0)$; parallel plate capacitor with $±σ$: between plates fields add ⇒ $σ/ε_0$, outside cancel 0.
• Uniformly charged sphere already; spherical shell $R$: outside same as point, inside $Q_{enc}=0 ⇒ E=0$.

4. Differential Form

(Gauss in differential, one of Maxwell's).

5. Worked Example

(a) Solid sphere charge Q=1 nC, R=0.1 m: E at r=0.05 m (inside) = (9e9·Q·r)/R³ =9e9·1e-9·0.05/0.001=450 V/m ✓ vs outside at 0.2 m: 9e9·1e-9/0.04=225 V/m ✓

(b) Line λ=1 µC/m: E at r=1 cm: 1e-6/(2π·8.85e-12·0.01)=1.8e6 V/m ✓

(c) Why E=0 inside conducting shell? Gaussian surface inside metal encloses no charge ⇒ flux zero ⇒ E=0 (electrostatic shielding).

6. Exam Points

• State $∯ E·n dS=Q/ε_0$ — 3-5 marks, almost certain.
• Derive $E$ for point/sphere/line/sheet/plate via Gauss — 7-10 marks, very frequent (the most common electrostatics derivation).
• Differential form $∇·E=ρ/ε_0$ — 3 marks.
• Explain why symmetry needed — 3 marks.