Fields at Conductor Surfaces

Derivation

Fields at Conductor Surfaces

D-046 Home PU-102 Threads fields · matter Depends on Gauss's Law (Integral Form) from Coulomb, Irrotationality of the Electrostatic Field

Statement

In electrostatic equilibrium the electric field inside a homogeneous conductor vanishes identically, all net charge resides on its surface as a density σ, the field just outside the surface is purely normal with magnitude E = σ/ε₀, and the conductor (bulk and surface together) is a single equipotential volume.

Why it matters

These four facts are the complete electrostatic boundary condition for any conductor, and they are what make the whole subject of image charges, capacitance, and shielding tractable: once you know a surface is an equipotential and the field meets it at right angles, you can replace the unknown charge distribution by a boundary condition on the potential and solve Laplace's equation instead.

They also underwrite everyday engineering: Faraday cages, coaxial cable, the grounding of instrument chassis, and the field-emission limit of high-voltage electrodes all follow directly from E = 0 inside and E = σ/ε₀ outside.

Assumptions

Static equilibrium (no steady current, ∂/∂t = 0).If charges are still in motion the interior field need not vanish; a current-carrying wire has a nonzero longitudinal E = J/σ_c inside it, and the "field is normal" conclusion fails.

An ideal conductor with freely mobile charge and effectively unlimited supply.Drop perfect mobility and the interior field is only screened to a small residual value over a Debye/Thomas–Fermi length rather than being exactly zero; the surface layer acquires finite thickness.

The surface is smooth on the scale of interest and the charge layer is idealised as a geometric surface density.At an atomically sharp edge or point the "pillbox" construction has no well-defined single normal and the local field diverges; σ is then not slowly varying and the simple σ/ε₀ relation holds only as a coarse-grained average.

Linear, field-independent response (no dielectric breakdown, no field emission).Above ~3 MV/m in air or the material's field-emission threshold the surrounding medium ionises or electrons tunnel out, sourcing new charge that the equilibrium argument does not account for.

Derivation

F = qE ⇒ equilibrium requires E_in = 0

Free carriers feel a force qE; if any macroscopic field remained inside, they would accelerate and rearrange. A static ("no further motion") state is possible only when the net field on every interior charge is zero, so E_in = 0 throughout the bulk. A

ρ_in = ε₀ ∇·E_in = 0

The differential form of the assumed prior result (Gauss's law) reads ∇·E = ρ/ε₀. With E_in = 0 everywhere inside, its divergence is zero, so the interior volume charge density vanishes and any net charge can live only on the boundary as a surface density σ. B

∮E·dl = 0 (prior: curl-free field)

Take a thin rectangular loop straddling the surface, two long sides of length ℓ tangent to it (one just inside, one just outside) and two short ends of vanishing length. The line integral of the electrostatic field around any closed loop is zero. B

(E_out,∥ − E_in,∥)ℓ = 0 ⇒ E_out,∥ = E_in,∥

As the short ends shrink, only the two tangential sides contribute; the loop integral gives the difference of tangential components times ℓ. Since E_in = 0, the tangential field just outside also vanishes: E_∥ = 0. The external field can therefore have only a normal component. B

∮E·dA = Q_enc/ε₀

Now apply the integral Gauss law (prior result) to a short cylindrical "pillbox" of face area A centred on the surface, one face inside the metal and one just outside. B

E_⊥A + 0 + 0 = σA/ε₀

The inner face contributes nothing (E_in = 0); the curved side contributes nothing as its height → 0 and because the outside field is purely normal (Step 4); only the outer face carries flux E_⊥A. The enclosed charge is σA. B

E_⊥ = σ/ε₀, E = (σ/ε₀) n̂

Cancel the common area A (symbols rearranged before any number is inserted) and reinstate the direction: the field points along the outward normal n̂ where σ > 0. A

V(b) − V(a) = −∫_a^b E·dl = 0

For any two points in the bulk the integrand is zero (E_in = 0); for any path along the surface the integrand is zero because E_∥ = 0. Hence V is the same at every interior and surface point: the conductor is one equipotential, and the surface field, being −∇V, must be perpendicular to that equipotential. B

Result

E_in = 0, E_out = (σ/ε₀) n̂, V = const on and in the conductor

Reading. A conductor in equilibrium expels the field from its interior and pushes every excess charge onto its surface. Outside, the field emerges at exactly a right angle with strength set purely by the local surface charge density, and the entire conductor sits at one voltage. The single factor σ/ε₀ — not σ/2ε₀ — already contains the field the sheet makes on itself plus the field of all other charges, which conspire to cancel inside and add outside.

Units check. [σ/ε₀] = (C·m⁻²) / (C²·N⁻¹·m⁻²) = N·C⁻¹ = V·m⁻¹, the correct dimensions of electric field.

Limiting cases

σ → 0 on a patch: the field there vanishes and neighbouring field lines fan around it — the local statement is genuinely local.
Large flat plate, near field: the surface looks like an infinite sheet-plus-image, recovering the uniform σ/ε₀ of a parallel-plate capacitor gap.
Isolated sphere, radius R, charge Q: σ = Q/4πR² gives E = Q/4πε₀R², matching the point-charge field just outside the surface.
Sharp region (small radius of curvature): charge crowds where curvature is high, so σ and hence E are largest at points and edges — the "lightning-rod" limit.

Breaks when

Time dependence / currents. With a driving EMF the field inside is nonzero (E = ρ_resJ) and tangential at the surface of a wire; the whole equilibrium chain collapses. AC fields also penetrate a skin depth δ = (2/μσ_cω)^1/2, so "field only on the surface" becomes a statement about a finite layer.
Finite screening length. In a real metal the interior field is zero only beyond a Thomas–Fermi length (~0.05 nm); in an electrolyte or plasma the Debye length can be micrometres, so a thin nanostructure never fully screens and E_in ≠ 0.
Dielectric breakdown / field emission. Once σ/ε₀ exceeds the breakdown field of the surrounding medium, discharge or electron emission bleeds charge away and no static σ is sustainable — the equilibrium assumption is violated at the point of interest.
Superconductors and quantum-scale samples. A superconductor expels magnetic flux (Meissner) with its own penetration depth; and when the sample is a few atoms across the notion of a continuous σ and a sharp normal loses meaning.

Failure modes

Writing E = σ/2ε₀. That is the field of an isolated charge sheet in vacuum. At a conductor the far side's charge and the geometry add a second σ/2ε₀, cancelling inside and doubling outside to the full σ/ε₀.
Claiming charge spreads uniformly. Only on a sphere (by symmetry). On any other shape σ is set by the equipotential condition and concentrates at high curvature.
Forgetting the inner pillbox face for a cavity. A conductor with an empty cavity has zero field in the cavity too; students often carry an outer-surface σ into the cavity wall incorrectly.
Confusing V = const with V = 0. The conductor is an equipotential at whatever value the boundary problem fixes, not necessarily ground.
Assuming "field is normal" holds for a current-carrying conductor. It is an electrostatic result; a resistor in a circuit has tangential surface field.

Discussion

The deep content of this derivation is that a conductor converts a hard problem — find the charge distribution — into an easy one — solve Laplace's equation ∇²V = 0 with V = const on each conductor. The charges arrange themselves precisely so as to make this boundary condition true; we never need to track them individually. The surface density is then read off after the fact from σ = ε₀E_⊥ = −ε₀ ∂V/∂n at the surface.

Physically, the two boundary results are two faces of the same coin. "Field is normal" is the tangential (curl) condition; "E = σ/ε₀" is the normal (divergence) condition. Every electrostatic interface obeys both — for a dielectric they read E_∥ continuous and ΔD_⊥ = σ_free. A conductor is just the special case where one side has E = 0, which is what promotes "continuous tangential field" to "zero tangential field" and freezes the surface at one potential.

There is also an elegant energy reading. The outward electrostatic stress on the surface is the Maxwell pressure P = σ²/2ε₀ = ½ε₀E², and the factor of ½ here (versus the full ε₀ in the field law) is exactly the "half the field is your own" bookkeeping: a charge element feels the average of the inside field (0) and the outside field (σ/ε₀), i.e. σ/2ε₀. This resolves the apparent tension between the two "halves".

At a rigorous level the vanishing interior field is not an axiom but the ground state of a Thomson-type variational problem: the equilibrium charge distribution minimises the electrostatic energy subject to fixed total charge, and a nonzero interior field would mean a lower-energy rearrangement exists. In quantum reality the "surface" is the tail of the electron density spilling a fraction of an ångström beyond the ionic lattice (the jellium image plane), and the classical σ is the integral of that spill-out. The classical boundary conditions are the coarse-grained, long-wavelength limit of that microscopic screening — exact to astonishing precision because the Thomas–Fermi length in a good metal is subatomic.

Common misconceptions. The interior field is zero because charges have already moved, not because charge is absent — a neutral conductor still screens an external field by inducing ±σ on opposite faces. And the surface charge is not "stuck to atoms"; it is a mobile equilibrium that instantly re-forms if you deform the conductor or move a nearby charge.

Worked examples

Isolated charged sphere. A metal sphere of radius R = 5.0 cm carries Q = 2.0 nC. Find σ and the surface field.

By spherical symmetry the charge is uniform: σ = Q/4πR², then E = σ/ε₀. Symbols first, numbers after. A

σ = (2.0×10⁻⁹) / [4π(0.050)²] = 6.4×10⁻⁸ C·m⁻²

4πR² = 4π(2.5×10⁻³) = 3.14×10⁻² m². A

E = σ/ε₀ = (6.4×10⁻⁸)/(8.85×10⁻¹²) = 7.2×10³ V·m⁻¹

Cross-check: Q/4πε₀R² = (8.99×10⁹)(2.0×10⁻⁹)/(0.050)² = 7.2×10³ V·m⁻¹. A

σ ≈ 64 nC·m⁻², E ≈ 7.2 kV·m⁻¹ (radially outward)

Reading. Well below air's ~3 MV·m⁻¹ breakdown, so the charge sits stably.

Charged plate, one face. A large flat conducting plate holds a surface density σ = 1.5×10⁻⁶ C·m⁻² on the face in question. Find the field just outside that face and the outward electrostatic pressure.

Locally the boundary law gives E = σ/ε₀ (normal); the Maxwell stress is P = σ²/2ε₀. B

E = σ/ε₀ = (1.5×10⁻⁶)/(8.85×10⁻¹²) = 1.7×10⁵ V·m⁻¹

Direct substitution into the boxed result. A

P = σ²/2ε₀ = (1.5×10⁻⁶)²/[2(8.85×10⁻¹²)] = 0.13 N·m⁻²

Equivalently ½ε₀E² = ½(8.85×10⁻¹²)(1.7×10⁵)² = 0.13 Pa, confirming the ½ ("own field") factor. B

E ≈ 1.7×10⁵ V·m⁻¹, P ≈ 0.13 Pa (outward)

Reading. The surface is pulled outward whatever the sign of σ — the pressure goes as σ², a genuine electrostatic tension on the metal.

Problems

A neutral isolated conducting sphere is placed in a previously uniform field E₀ = 500 V·m⁻¹. What is the maximum induced surface charge density, and where does it occur?
Solution
The exterior solution is V = −E₀(r − R³/r²)cosθ. The radial field at the surface is E_r(R) = 3E₀cosθ, so σ = ε₀E_r = 3ε₀E₀cosθ, maximal at the poles (θ = 0, π) facing along the field. σ_max = 3(8.85×10⁻¹²)(500) = 1.3×10⁻⁸ C·m⁻² = 13 nC·m⁻². The field at the pole is 3E₀ = 1500 V·m⁻¹, three times the applied field.
A hollow conducting shell (inner radius a = 2 cm, outer radius b = 3 cm) carries net charge Q = 4 nC, with a point charge q = 1 nC at its centre. Find the surface charge on the inner and outer walls.
Solution
The field inside the metal (between a and b) is zero, so a Gaussian sphere in the metal encloses zero net charge: the inner wall must carry −q = −1 nC. Charge is conserved on the isolated shell, so the outer wall carries Q + q = 4 + 1 = 5 nC. Densities: σ_in = −1×10⁻⁹/[4π(0.02)²] = −2.0×10⁻⁷ C·m⁻²; σ_out = 5×10⁻⁹/[4π(0.03)²] = 4.4×10⁻⁷ C·m⁻².
Two parallel conducting plates, area A = 0.10 m² each, carry equal and opposite charges ±Q = ±20 nC. Find the field in the gap and the surface density on the inner faces, neglecting edge effects.
Solution
In a parallel-plate capacitor essentially all charge migrates to the inner faces, so on the inner face σ = Q/A = 20×10⁻⁹/0.10 = 2.0×10⁻⁷ C·m⁻². The gap field is E = σ/ε₀ = 2.0×10⁻⁷/8.85×10⁻¹² = 2.3×10⁴ V·m⁻¹, directed from the + to the − plate. Outside the plates the fields cancel to zero.
A long conducting cylinder of radius R = 1.0 mm carries linear charge density λ = 30 nC·m⁻¹. Find σ on its surface and the field just outside.
Solution
By cylindrical symmetry σ = λ/2πR = 30×10⁻⁹/[2π(1.0×10⁻³)] = 4.8×10⁻⁶ C·m⁻². Check via the boundary law: E = σ/ε₀ = 4.8×10⁻⁶/8.85×10⁻¹² = 5.4×10⁵ V·m⁻¹. Compare with the line-charge result E = λ/2πε₀R = (30×10⁻⁹)/[2π(8.85×10⁻¹²)(1.0×10⁻³)] = 5.4×10⁵ V·m⁻¹ — identical, as required.
A conducting sphere is charged until the surface field just reaches the dielectric strength of air, E_max = 3.0×10⁶ V·m⁻¹. If the radius is 2.0 cm, what is the maximum charge it can hold, and what potential does that correspond to?
Solution
At breakdown σ_max = ε₀E_max = (8.85×10⁻¹²)(3.0×10⁶) = 2.66×10⁻⁵ C·m⁻². Then Q_max = σ_max·4πR² = (2.66×10⁻⁵)(4π)(0.020)² = 1.3×10⁻⁷ C = 0.13 µC. The potential is V = E_maxR = Q/4πε₀R = (3.0×10⁶)(0.020) = 6.0×10⁴ V = 60 kV. Note the field, not the potential, sets the limit — small radii break down at modest voltages, which is why sharp points spark first.