Conductors & Boundary Conditions
- Justify the equilibrium properties of conductors: E = 0 inside, charge on the surface
- Derive the surface field E = σ/ε₀ with a Gaussian pillbox and state the boundary conditions
- Analyze cavities and induced charges using Gauss’s law region by region
Equilibrium forces three properties
A conductor holds mobile charge, and electrostatic equilibrium means that charge has stopped moving. That single condition forces everything else. (1) E = 0 inside: any interior field would push free electrons, contradicting equilibrium. (2) All excess charge sits on the surface: with E = 0 on any interior Gaussian surface, ∮→E·d→A = 0, so Q_enc = 0 everywhere inside — Gauss’s law exiles the charge to the skin. (3) The conductor is an equipotential: with E = 0, the integral −∫→E·d→l between any two interior points vanishes, so V is uniform throughout.
Boundary conditions at the surface
Just outside the surface the field obeys two boundary conditions. Tangential: E_∥ = 0 — a component along the surface would drag surface charge sideways, breaking equilibrium, so the field meets the surface at exactly 90°. Normal: E_⊥ = σ/ε₀, where σ is the local surface charge density. The perpendicular rule also follows from the conductor being an equipotential: field lines always cross equipotential surfaces at right angles.
Use a Gaussian pillbox to derive the field just outside a conductor whose local surface charge density is σ, then evaluate for σ = 8.85 × 10⁻⁸ C/m².
- 1.Choose a tiny cylinder (a “pillbox”) of face area A straddling the surface: one flat face buried in the metal, the other just outside, its axis along the surface normal.
- 2.Flux through the inner face: zero, because E = 0 inside a conductor.
- 3.Flux through the curved side wall: zero, because the outside field is perpendicular to the surface — parallel to the pillbox axis — so it skims along the wall.
- 4.Flux through the outer face: E·A, with E uniform over the tiny face. Enclosed charge: the patch of surface charge, Q_enc = σA.
- 5.Gauss’s law: E·A = σA/ε₀, so E = σ/ε₀. Numerically: E = (8.85 × 10⁻⁸)/(8.85 × 10⁻¹²) = 1.0 × 10⁴ N/C.
Do not confuse E = σ/ε₀ (just outside a conductor) with E = σ/2ε₀ (an isolated sheet of charge). The conductor’s interior field is zero, so all the flux from the surface charge exits through one face of the pillbox instead of splitting both ways — hence the factor of two.
Cavities, induced charge, and shielding
Carve a cavity inside a conductor. If the cavity is empty, the field inside it is zero — the conductor shields its interior from external fields (a Faraday cage). Put a point charge +q in the cavity and Gauss’s law dictates the response: a Gaussian surface drawn through the surrounding metal has E = 0 on it, so it must enclose zero net charge — the cavity wall acquires induced charge −q. If the conductor is neutral overall, charge conservation pushes the balancing +q to the outer surface, and the outside world sees the field of +q as if the shield were not there.
A +2.0 nC point charge sits at the center of a neutral spherical conducting shell (inner radius a, outer radius b). Find the induced surface charges and the field at r = 0.30 m > b.
- 1.Gaussian sphere with a < r < b (inside the metal): E = 0 there, so Q_enc = 0, which requires q_inner = −2.0 nC spread on the cavity wall.
- 2.The shell is neutral: q_inner + q_outer = 0, so q_outer = +2.0 nC on the outside surface.
- 3.For r > b, a Gaussian sphere encloses +2.0 − 2.0 + 2.0 = +2.0 nC. Spherical symmetry gives E·(4πr²) = q/ε₀, so E = kq/r².
- 4.Numerically: E = (8.99 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 18.0/0.090 = 2.0 × 10² N/C, radially outward.
In electrostatic equilibrium, why must the field just outside a conductor be perpendicular to its surface?
A point charge +Q is suspended inside the cavity of a neutral conductor. The total charge on the conductor’s *outer* surface is:
Answer the 2 checkpoints as you read.
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