Conductors and electric potential

A metal carries roughly 10²⁸ mobile electrons per cubic metre. Put it in a field and they drift, pile up on the surfaces, and build a field of their own that opposes the applied one; the drift stops only when the total field inside the material is zero. That is what electrostatic equilibrium means, and it is fast: the ohmic estimate of the settling time is τ = ε₀ρ, which for copper with ρ = 1.7 × 10⁻⁸ Ω m gives 1.5 × 10⁻¹⁹ s. That figure sits well below the ~10⁻¹⁴ s between electron collisions, where the ohmic model stops being trustworthy, but either estimate finishes long before any measurement starts. Two consequences follow from Gauss's law. Any closed surface drawn entirely within the metal encloses zero flux, so it encloses zero net charge: excess charge can only sit on surfaces. And the assumption is doing real work — a copper wire carrying a current density of 1.0 × 10⁶ A m⁻² has E = ρJ = 0.017 V m⁻¹ inside it, small but not zero.

Potential difference is a line integral of the field, V(b) − V(a) = −∫E⋅dl. Choose any path from a to b that stays in the metal and the integrand is zero everywhere along it, so V(b) = V(a). This is a statement about the volume, not just the skin: the centre of a solid conductor, its outer surface, and the wall of any cavity in it all share one value of V. Nothing about that value is special: it is set by the body's charge and by everything else in the room. Outside, the consequence is geometric. The field just beyond the surface must be perpendicular to it, because a tangential component would push the surface charge sideways and equilibrium would not have been reached; equivalently, moving along the surface would change V and the surface would not be an equipotential. So field lines meet a conductor at 90° and the equipotential surfaces of the surrounding space wrap it, with the conductor supplying one contour for free.

Take a short cylinder straddling the surface, one flat face of area A just outside, the other just inside the metal. The inner face contributes nothing because E = 0 there, the side wall contributes nothing because E is perpendicular to the surface, and Gauss's law leaves EA = σA/ε₀, so E = σ/ε₀ just outside. This is not the σ/2ε₀ of an isolated charged sheet: the local patch supplies σ/2ε₀, and all the other charge supplies a second σ/2ε₀ that adds outside and cancels inside. Written generally, the electrostatic boundary conditions say that across any charged surface the tangential field is continuous, the normal field jumps by σ/ε₀, and V is continuous. On a conductor they read E∥ = 0 and E⊥ = σ/ε₀. For the 15 cm sphere with 20 nC, σ = Q/(4πR²) = 2.0 × 10⁻⁸/0.283 = 7.1 × 10⁻⁸ C m⁻², and σ/ε₀ = 8.0 × 10³ V m⁻¹, matching kQ/R² exactly.

Hollow the conductor out, leaving an empty cavity. Its entire wall sits at one potential V₀, and inside the empty region V obeys Laplace's equation, which admits no maximum or minimum in the interior of a region — so V is pinned to V₀ everywhere in the cavity and E = 0 there. The field-line version is quicker: a line would have to run from one part of the wall to another, and closing the loop through the metal would give a non-zero ∮E⋅dl. Two results drop out. The cavity wall carries no charge, and no arrangement of external charge, however large, produces a field in the hole. That is the Faraday cage, and it works one way only. Place q inside the cavity and a Gaussian surface in the metal, where E = 0, forces −q onto the cavity wall; conservation puts Qₙₑₜ + q on the outer surface, spread according to the outer shape alone. Sliding q around inside the cavity changes nothing measurable outside.

A ground connection is a conducting path to a reservoir large enough that its potential is unaffected by what flows, and that potential is defined as zero. Grounding therefore does not 'drain the charge' as a rule; it lets charge move until V = 0, and how much moves depends on the geometry. Ground the isolated 15 cm sphere sitting at 1.2 × 10³ V and all 20 nC leaves, giving Q = 0. Now take a neutral shell with q = +4.0 nC inside its cavity: the wall holds −4.0 nC, the outer surface +4.0 nC, and the shell is at some positive potential. Ground the outer surface and that +4.0 nC flows to earth. The shell and everything outside it are now at V = 0 with no external field at all — but the cavity is untouched, the wall still holds −4.0 nC, and the field inside the hole is exactly what it was. Grounding a shield kills its field outside; it cannot shield the inside from a charge already in there.

Join two distant spheres, radii R₁ and R₂, with a thin wire and they become one conductor at one potential: kQ₁/R₁ = kQ₂/R₂, so charge shares out as Q ∝ R, surface density as σ ∝ 1/R, and surface field as E = V/R ∝ 1/R. With R₁ = 6.0 cm, R₂ = 2.0 cm and 8.0 nC in total, Q₁ = 6.0 nC and Q₂ = 2.0 nC, both spheres sit at 8.99 × 10² V, and the surface fields are 1.5 × 10⁴ V m⁻¹ and 4.5 × 10⁴ V m⁻¹. The small sphere holds a quarter of the charge and three times the field. Two spheres on a wire are a caricature of one curved surface, but the trend survives: on a body at a single potential, σ and E peak where the radius of curvature is smallest. Dry air breaks down near 3 × 10⁶ V m⁻¹, and V_break = E_break R for an isolated sphere, so a 1.0 mm ball electrode ionises the air around it at only 3.0 × 10³ V while a 15 cm sphere holds 4.5 × 10⁵ V. Hence smooth, fat high-voltage terminals — and hence a lightning rod is a spike.