Surface Charge and Field at a Conductor

Why are lightning rods sharpened to a point, while the dome of a van de Graaff generator is polished into the smoothest, fattest sphere the budget allows? Both are conductors, both sit at a single potential — yet one is engineered to spray charge into the air and the other to hoard it. The answer lives in the few nanometers where a conductor meets vacuum: the thin sheet of surface charge $\sigma$ , and the electric field it launches. The previous articles established the wholesale facts: $\mathbf{E}=0$ inside a conductor, all charge on the surface, induced films canceling intruding fields. This article zooms in on the retail physics of the surface itself. Three exact statements come out of the pillbox-and-loop machinery of Gauss's law: the field just outside is perpendicular to the surface, its magnitude is $\sigma/\varepsilon_0$ — curiously, twice the field of a free-standing sheet of the same charge — and the surface charge itself feels an outward pull of $\sigma^2/2\varepsilon_0$ per unit area, whatever the sign of $\sigma$ . Every charged conductor is being gently stretched by its own field, like an inflated balloon. The factor of two and the "self-force" question are where students bleed points, and this article dwells on both until they are obvious rather than memorized. The payoff equation, $\sigma = -\varepsilon_0\,\partial V/\partial n$ , is also the bridge to everything that follows: it means that once the E2 chapter teaches you to solve for potentials, the induced charge on every conductor comes free — you will use it to harvest the famous induced-charge maps of the image-charge articles.

Key takeaways

At a conductor's surface in equilibrium the field is purely normal: $E_\parallel = 0$ (from $\nabla\times\mathbf{E}=0$ ), and $E_\perp = \sigma/\varepsilon_0$ just outside (from a Gaussian pillbox).
The conductor value $\sigma/\varepsilon_0$ is twice an isolated sheet's $\sigma/2\varepsilon_0$ : the local patch supplies half, and the rest of the conductor's charge supplies an equal half that adds outside and cancels inside.
A surface element feels only the field of the other charges — the average of the two sides, $\sigma/2\varepsilon_0$ — giving an outward electrostatic pressure $P=\sigma^2/2\varepsilon_0 = \tfrac{1}{2}\varepsilon_0E^2$ , independent of the sign of $\sigma$ .
Because the surface is an equipotential, charge crowds onto regions of small radius: for connected spheres $\sigma \propto 1/R$ — the physics of lightning rods and corona discharge.
$\sigma = -\varepsilon_0\,\partial V/\partial n$ converts any solved potential into a surface-charge map — the harvesting tool of the E2 boundary-value chapter.
KEY IDEA: the conductor surface is where field and charge trade places — $E=\sigma/\varepsilon_0$ head-on, an outward tension $\tfrac{1}{2}\varepsilon_0E^2$ on the skin, and all of it readable from the normal derivative of the potential.

The core insight

Just outside a conductor the field is normal with magnitude $\sigma/\varepsilon_0$ ; the surface charge feels the average field $\sigma/2\varepsilon_0$ , producing an outward pressure $\tfrac{1}{2}\varepsilon_0 E^2$ — and $\sigma=-\varepsilon_0\,\partial V/\partial n$ reads the charge map off any solved potential.

Where this lives today

Equation (3) runs a surprising amount of modern hardware. Electrospray ionization — the technique that won the 2002 chemistry Nobel and feeds most protein mass spectrometers — works by charging a liquid droplet until the outward electrostatic tension $\tfrac{1}{2}\varepsilon_0E^2$ beats surface tension (Rayleigh's 1882 limit) and the droplet bursts into a fine charged mist. Inkjet printers steer droplets by the same competition. At the other extreme of scale, MEMS engineers fight the "pull-in instability": the $\propto E^2$ attraction between a micro-cantilever and its electrode grows faster than the spring restoring force, and past one-third of the gap the electrode snaps shut. The curvature story has its own modern career. Atom-probe tomography deliberately sharpens a specimen to a $\sim 50\,\mathrm{nm}$ tip so that a modest voltage produces fields of $10^{10}\,\mathrm{V/m}$ — enough to field-evaporate individual atoms and reconstruct a material atom by atom. And the boundary condition $\sigma = -\varepsilon_0\,\partial V/\partial n$ is the last line of nearly every electrostatics simulation: finite-element solvers compute $V$ in the vacuum, then differentiate at electrode surfaces to report the charge and force maps that engineers actually asked for.

Origins · 1 minCharles-Augustin de Coulomb and how this was discovered

Charles-Augustin de Coulomb

1736–1806 · Paris, Académie des Sciences

Coulomb came to electricity as a military engineer, after nine years fortifying Martinique wrecked his health and taught him more about soil mechanics and friction than most academics learned in a lifetime. Precision measurement was his signature. The torsion balance — a fiber's twist converting attonewton-scale forces into readable angles — gave him the inverse-square law in 1785; less celebrated, but central to this article, are the later memoirs in which he crawled over charged conductors with a "proof plane," a small insulated disk touched to the surface to sample the local charge. He established experimentally that charge lives on the surface, that its density varies with the body's shape, and that it crowds toward edges and points — the facts that $E=\sigma/\varepsilon_0$ and the equipotential constraint would later explain. The result $E=\sigma/\varepsilon_0$ at a conductor still carries his name in the older literature: Coulomb's theorem. Poisson, a generation later, turned Coulomb's charge maps into the first boundary-value calculations — the mathematics this chapter now inherits.

History

1767

— Priestley infers the inverse-square law from Franklin's observation that a charged vessel holds no charge, and no field, inside

1785

— Coulomb's torsion-balance memoirs begin; later installments map surface charge on conductors with the proof plane

1812

— Poisson computes the charge distribution on conductors (spheres, ellipsoids) — the first solved boundary-value problems of electrostatics

1873

— Maxwell's Treatise derives the surface tension

\tfrac{1}{2}\varepsilon_0E^2

as one face of the general field stress

1882

— Rayleigh finds the charge limit at which electrostatic tension bursts a droplet — the working principle of modern electrospray

The physical picture

Stand (in imagination) on the surface of a charged conductor, on a patch of charge small enough to look flat. Your patch, like any sheet of charge, pushes outward on both of its sides with $\sigma/2\varepsilon_0$ . But you are not alone: every other charge on the conductor also contributes a field where you stand — smooth, continuous, unbothered by your little patch. Equilibrium fixes that contribution completely. Just beneath your feet the total field must vanish, so the others must supply exactly $+\sigma/2\varepsilon_0$ outward to cancel your patch's inward pull. Step just above the surface and nothing about the others has changed — their $\sigma/2\varepsilon_0$ now adds to your patch's $\sigma/2\varepsilon_0$ . Zero inside, $\sigma/\varepsilon_0$ outside: the factor of two is the rest of the conductor voting with the patch on one side and against it on the other.

Figure

The factor of two — patch field plus everything else

The same picture answers the self-force puzzle. Your patch cannot pull on itself — its own field points away from it on both sides symmetrically, and Newton's third law forbids a net bootstrap. The only field that moves it is the smooth green one belonging to everyone else, $\sigma/2\varepsilon_0$ . Multiply by the patch's charge per area and the force per area is $\sigma^2/2\varepsilon_0$ — proportional to $\sigma^2$ , hence outward whether the conductor is positive or negative. Every charged conductor is under tension, its own field trying to peel its skin off. Why must the field arrive head-on in the first place? Because free charges can slide along a surface as easily as through the bulk. Any tangential field component would drag them sideways until it didn't — the same self-terminating flow that empties the interior. Equilibrium leaves only the perpendicular component standing, which is another way of saying the surface is an equipotential: field lines cross equipotentials at right angles. The equipotential condition has a geometric sting. A small sphere at potential $V$ needs more charge per area than a big one ( $V = \sigma R/\varepsilon_0$ for a sphere, so $\sigma = \varepsilon_0 V/R$ ). Weld a knob of small radius onto a large conductor and the knob, forced to the common potential, crowds charge onto itself at density $\propto 1/R$ ; its surface field $E = V/R$ soars. Sharpen the knob to a point and the field wins the arms race with the surrounding air's dielectric strength — corona discharge. That is the whole design theory of the lightning rod, and, run in reverse, of the van de Graaff dome.

Figure

Curvature and charge — one potential, very different fields

The boundary conditions at the skin

Everything quantitative descends from two small geometric constructions at the surface — a loop and a pillbox — plus the interior facts from Conductors in Electrostatic Equilibrium. The results, stated first:

E_\parallel = 0, \qquad \mathbf{E}_{\text{out}} = \frac{\sigma}{\varepsilon_0}\,\hat{\mathbf{n}} \qquad (1)

where: $\hat{\mathbf{n}}$ — the outward unit normal; $\sigma$ — the local surface charge density, which generally varies from point to point. Equation (1) is local: it ties the field at each point of the surface to the charge density at that same point, no matter how lumpy the far side of the conductor is. It is the reason a measured field map just outside any electrode is also a charge map. Since $\mathbf{E} = -\nabla V$ and only the normal component survives, the charge map can be read directly from the potential:

\sigma = -\varepsilon_0\,\frac{\partial V}{\partial n} \qquad (2)

where: $\partial V/\partial n = \nabla V\cdot\hat{\mathbf{n}}$ — the outward normal derivative evaluated at the surface. This innocent-looking line is the workhorse of the entire E2 chapter: solve a boundary-value problem for $V$ by any method — images, series, relaxation — then differentiate once, and the induced charge on every conductor appears. When the image-charge articles produce their celebrated $\sigma$ maps, equation (2) is doing the harvesting. Finally the mechanical consequence — the force per unit area on the surface:

\mathbf{f} = \frac{\sigma^2}{2\varepsilon_0}\,\hat{\mathbf{n}}, \qquad P = \frac{\sigma^2}{2\varepsilon_0} = \tfrac{1}{2}\varepsilon_0 E^2 \qquad (3)

where: $E$ — the field magnitude just outside. The last form is worth staring at: $\tfrac{1}{2}\varepsilon_0E^2$ is exactly the field's energy density just outside the surface. Pressure equals energy density — a coincidence here, a theorem (the Maxwell stress tensor) in the conservation-laws chapter. Numerically it is feeble: even at air's breakdown field the tension is $\sim 40\,\mathrm{Pa}$ , four ten-thousandths of an atmosphere.

Derivation AThe two boundary conditions from a loop and a pillbox

We derive both halves of equation (1) with the standard surgical tools: an Amperian-style rectangle for the tangential component and a Gaussian pillbox for the normal one. The strategy is to straddle the surface with each construction, shrink the straddling dimension to zero, and let the interior facts ( $\mathbf{E}=0$ in the metal) kill half the terms. Watch where the surface charge enters: it feeds the pillbox but not the loop — which is precisely why the two components behave so differently.

Step 1. Tangential: lay a thin rectangular loop of length $\ell$ across the surface, long sides parallel to it — one just outside, one just inside the metal. The electrostatic field is curl-free, so $\oint\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}=0$ around the loop.

Step 2. Shrink the short sides to zero; they contribute nothing. The inside long side lies in field-free metal and contributes nothing either. What remains is $E_\parallel^{\text{out}}\,\ell = 0$ , so

E_\parallel^{\text{out}} = 0.

Why no $\sigma$ here: surface charge produces no circulation; a loop integral never sees a charge sheet. Tangential continuity is a curl fact, not a Gauss fact — with zero inside, continuity means zero outside.

Step 3. Normal: stand a squat pillbox of cap area $A$ across the surface, one cap in the vacuum, one in the metal. Gauss's law:

\oint\mathbf{E}\cdot\mathrm{d}\mathbf{a} = E_\perp^{\text{out}}A - E_\perp^{\text{in}}A + (\text{sides}) = \frac{\sigma A}{\varepsilon_0}.

Step 4. Shrink the height: the side contribution vanishes. The inner cap sits in the metal, where $E_\perp^{\text{in}}=0$ . Only the outer cap survives:

E_\perp^{\text{out}} = \frac{\sigma}{\varepsilon_0}.

Why the whole jump lands outside: across any charged sheet the normal field jumps by $\sigma/\varepsilon_0$ — that much is universal. A conductor is the special case where one side of the jump is clamped to zero, so the exterior inherits the entire discontinuity. Verified: Griffiths Eqs. 2.49 and 2.23; the general two-sided jump conditions (for non-conductor interfaces) are the business of Boundary Conditions on E and V, and their potential-language form is developed for the E2 chapter in Boundary Conditions for the Potential.

Go deeper · Derivation B · 1 minThe force on the skin — why the average field, and why always outward

Derivation BThe force on the skin — why the average field, and why always outward

We want the force per unit area on the conductor's surface charge. The trap is to write $f = \sigma E = \sigma^2/\varepsilon_0$ using the full exterior field — overcounting by exactly a factor of two, because that field includes the patch's own contribution, and no object accelerates itself with its own field. The honest route decomposes the field at the patch into "self" and "other," discards the self-force, and finds that the acting field is the average of the fields on the two sides of the sheet — a rule worth keeping, since it holds for any charged sheet, conductor or not.

Step 1. Zoom in until the patch is a locally flat sheet of density $\sigma$ . Its own field is $\pm\sigma/2\varepsilon_0\,\hat{\mathbf{n}}$ — outward on both faces, antisymmetric through the sheet. Acting on itself, it integrates to zero net force.

Step 2. Write the total field on the two sides as self plus other, with $\mathbf{E}_{\text{other}}$ smooth across the infinitesimal patch:

\mathbf{E}_{\text{out}} = \mathbf{E}_{\text{other}} + \frac{\sigma}{2\varepsilon_0}\hat{\mathbf{n}}, \qquad \mathbf{E}_{\text{in}} = \mathbf{E}_{\text{other}} - \frac{\sigma}{2\varepsilon_0}\hat{\mathbf{n}}.

Adding and halving isolates the actor:

\mathbf{E}_{\text{other}} = \tfrac{1}{2}\left(\mathbf{E}_{\text{out}} + \mathbf{E}_{\text{in}}\right).

Why the average rule is general: nothing in this step used the conductor — the force on any surface charge layer is its density times the mean of the fields on its two sides.

Step 3. Specialize to the conductor: $\mathbf{E}_{\text{in}}=0$ and $\mathbf{E}_{\text{out}}=(\sigma/\varepsilon_0)\hat{\mathbf{n}}$ , so $\mathbf{E}_{\text{other}} = (\sigma/2\varepsilon_0)\hat{\mathbf{n}}$ , and

\mathbf{f} = \sigma\,\mathbf{E}_{\text{other}} = \frac{\sigma^2}{2\varepsilon_0}\,\hat{\mathbf{n}} = \tfrac{1}{2}\varepsilon_0 E^2\,\hat{\mathbf{n}}.

Step 4. Read the sign: $\sigma^2$ cannot be negative, so the force is along $+\hat{\mathbf{n}}$ — outward — for positive and negative conductors. The field does not push the surface around; it tries to pull the surface apart. A conductor's skin is under electrostatic tension equal to the energy density camped just outside it. Verified: Griffiths §2.5.3 (Eq. 2.51); equals the Maxwell stress tensor's normal-normal component $T_{nn}=\tfrac{1}{2}\varepsilon_0E^2$ , Jackson §1.11 — the stress-tensor derivation arrives in the conservation-laws chapter.

Limiting cases

Case	Result	Meaning
Isolated free sheet of charge	$E = \sigma/2\varepsilon_0$ both sides	No conductor backstage; the field is half the conductor value
Conductor surface	$E = \sigma/\varepsilon_0$ outside, $0$ inside	The rest of the conductor's charge doubles outside, cancels inside
Flat regions of a lumpy conductor	smaller $\sigma$ , smaller $E$	Equipotential constraint starves flat regions of charge
Radius of curvature $R\to 0$ (points, edges)	$\sigma \sim \varepsilon_0 V/R \to$ large	Field diverges at sharp features — corona, breakdown, lightning rods
Force on the skin	$P = \tfrac{1}{2}\varepsilon_0E^2$ , outward for either sign	Tension, not pressure; equals the exterior field energy density
At air breakdown $E = 3\times10^6\,\mathrm{V/m}$	$P \approx 40\,\mathrm{Pa}$	Electrostatic stresses are mechanically tiny at everyday scales

Worked examples

1 · How much charge can a 5-cm sphere hold? (confidence builder)

A conducting sphere of radius $R = 5.0\,\mathrm{cm}$ in dry air carries $Q = 1.0\,\mu\mathrm{C}$ . Find $\sigma$ , the surface field, and the electrostatic tension on the skin — and decide whether the sphere can keep its charge.

Step 1. By symmetry the charge spreads uniformly:

\sigma = \frac{Q}{4\pi R^2} = \frac{1.0\times10^{-6}}{4\pi(0.050)^2}\,\mathrm{C/m^2} = 3.2\times10^{-5}\,\mathrm{C/m^2}.

Step 2. The surface field follows from the boundary condition:

E = \frac{\sigma}{\varepsilon_0} = \frac{3.18\times10^{-5}}{8.854\times10^{-12}}\,\mathrm{V/m} = 3.6\times10^{6}\,\mathrm{V/m}.

Step 3. That exceeds the dielectric strength of dry air, $E_{\text{bd}} \approx 3\times10^{6}\,\mathrm{V/m}$ — the surrounding air ionizes and the sphere cannot hold this charge. The maximum is

Q_{\max} = 4\pi\varepsilon_0 R^2 E_{\text{bd}} = 4\pi(8.854\times10^{-12})(0.050)^2(3\times10^{6})\,\mathrm{C} \approx 0.83\,\mu\mathrm{C}.

Step 4. The tension on the skin at breakdown:

P = \tfrac{1}{2}\varepsilon_0 E_{\text{bd}}^2 = \tfrac{1}{2}(8.854\times10^{-12})(3\times10^{6})^2\,\mathrm{Pa} \approx 40\,\mathrm{Pa}.

A palm-sized sphere in air tops out near $0.8\,\mu\mathrm{C}$ , its skin tugged outward by a mere 40 pascals. Microcoulombs are the natural currency of electrostatics — this is why.

2 · The dome and the needle (exam level)

A van de Graaff dome of radius $R_1 = 15\,\mathrm{cm}$ is charged toward its limit in air. A stray needle point of tip radius $R_2 = 0.50\,\mathrm{mm}$ sticks out of a fitting on the dome, electrically connected to it. Find the dome's maximum potential if it were smooth, then the field at the needle tip at that potential, and the potential at which the needle starts corona.

Step 1. For an isolated sphere, $V = ER$ at the surface. A smooth dome could reach

V_{\max} = E_{\text{bd}}R_1 = (3\times10^{6})(0.15)\,\mathrm{V} = 4.5\times10^{5}\,\mathrm{V} = 450\,\mathrm{kV}.

Step 2. Model the tip as a small sphere at the common potential $V$ . Its surface field is $E_2 = V/R_2$ . At $V = 450\,\mathrm{kV}$ :

E_2 = \frac{4.5\times10^{5}}{5.0\times10^{-4}}\,\mathrm{V/m} = 9.0\times10^{8}\,\mathrm{V/m}

— three hundred times the breakdown field. The needle would be in violent corona long before the dome approached its own limit.

Step 3. Corona at the tip begins when $E_2 = E_{\text{bd}}$ :

V_{\text{corona}} = E_{\text{bd}}R_2 = (3\times10^{6})(5.0\times10^{-4})\,\mathrm{V} = 1.5\,\mathrm{kV}.

Step 4. Compare the two surface charge densities at that moment: $\sigma \propto E$ , so $\sigma_2/\sigma_1 = E_2/E_1 = R_1/R_2 = 300$ . One stray needle caps a 450 kV machine at 1.5 kV — a factor-of-300 sabotage by curvature. Every seam, bolt, and burr on high-voltage hardware is polished for exactly this reason — and a lightning rod is this "sabotage" deployed on purpose.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.5.1–2.5.3 (boundary conditions and force on a conductor) · Jackson, Classical Electrodynamics, 3rd ed., §1.5, §2.3 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 3 · Zangwill, Modern Electrodynamics, Ch. 5.

Provenance. Equations are standard textbook results: $E=\sigma/\varepsilon_0$ (Griffiths Eq. 2.49, Jackson §1.5), $P=\sigma^2/2\varepsilon_0=\tfrac12\varepsilon_0E^2$ (Griffiths Eq. 2.51). A verifier should reconfirm the pillbox/superposition derivations and recompute the numbers: ratio $R_1/R_2=5$ in the spheres example; pressure $\tfrac12(8.854\times10^{-12})(3\times10^6)^2=39.8\,\mathrm{Pa}$ at air breakdown; image-charge density $\sigma=-qd/[2\pi(r^2+d^2)^{3/2}]$ (Griffiths Eq. 3.10) integrating to $-q$ . Constant $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ and air dielectric strength $\approx3\times10^6\,\mathrm{V/m}$ are standard values.

Builds on: Conductors in Electrostatic Equilibrium · Induced Charge and Electrostatic Shielding · Gauss's Law in Integral Form · The Curl of the Electrostatic Field · The Electrostatic Potential

Leads to: Boundary Conditions for the Potential · Method of Images · Image Charge for a Grounded Plane

How can the boundary condition be local when the charges elsewhere obviously contribute to the field?

They contribute, but their contribution is smooth at your patch — no jump. The discontinuity in the normal field across the surface is produced entirely by the local sheet, and on a conductor the interior side of the jump is clamped at zero. So the exterior value $\sigma/\varepsilon_0$ is pinned to the local density even though half of it is "supplied" by distant charges. The distant charges are not optional extras; the equilibrium arrangement forces them to supply exactly the right smooth field to zero the interior everywhere.

If the patch's own field is σ/2ε₀ on each side, why doesn't it contribute to the force on itself?

Its field pushes the top of the patch outward and the bottom of the patch... also outward — but the sources feeling those pushes are the same charges, and by Newton's third law the internal pushes and pulls of a rigid patch on itself cancel identically. A system cannot accelerate itself with its own internal forces. What survives is the external field, which is smooth across the patch and equals the average of the total field's two side-values — a rule that works for any charged sheet anywhere.

Why does charge concentrate at sharp points if the whole conductor is at one potential?

Precisely because it is at one potential. A region of small curvature radius $R$ resembles a small sphere, and a small sphere needs a large $\sigma$ ( $= \varepsilon_0 V/R$ ) to hold itself at potential $V$ . The two-connected-spheres model makes it quantitative: joined by a wire, spheres satisfy $\sigma_1 R_1 \approx \sigma_2 R_2$ , so density and field scale like $1/R$ . It's not that charge "likes" points — the equipotential constraint leaves it no choice.

Is the 40 Pa electrostatic tension ever mechanically important?

At everyday scales almost never — 40 Pa won't dent a soda can. It matters when the competing forces are also small or the fields are huge: surface tension of microdroplets (electrospray bursts them), the spring constants of MEMS cantilevers (pull-in collapse), liquid-metal ion sources, and field-evaporation tips where $E \sim 10^{10}\,\mathrm{V/m}$ makes $P \sim \tfrac{1}{2}\varepsilon_0E^2 \sim 4\times10^{8}\,\mathrm{Pa}$ — enough to rip individual atoms off a tungsten needle.

Does E = σ/ε₀ hold on a charged insulator too?

No. The two ingredients that produced it — interior field clamped to zero and charge free to form an equipotential — are conductor privileges. On an insulator the charge is stuck where it was put, the interior field need not vanish, and the field just outside is whatever superposition delivers; the only universal statement is the jump condition, that the normal component changes by $\sigma/\varepsilon_0$ across the sheet while the tangential component is continuous. Those general two-sided conditions are treated in Boundary Conditions on E and V.

The pressure equals the energy density just outside. Coincidence?

No — it is the simplest appearance of a deep identity. The electromagnetic field carries momentum and exerts stresses described by the Maxwell stress tensor; along the field direction the stress is a tension of magnitude $\tfrac{1}{2}\varepsilon_0E^2$ , numerically the energy density. At a conductor the field is purely normal, so the surface feels exactly that tension. The full stress-tensor machinery, including the sideways pressure $-\tfrac{1}{2}\varepsilon_0E^2$ across field lines, arrives in the conservation-laws chapter.

Since σ = −ε₀ ∂V/∂n needs the solved potential, isn't it useless until I can solve boundary-value problems?

That's exactly the right reading of where you are in the chapter — and why the E2 articles come next. For high-symmetry cases you already can (sphere, plane, cylinder via Gauss). For everything else, equation (2) is a promissory note: the moment Laplace's equation, uniqueness, and images give you $V$ , one derivative pays out the full induced-charge map. The image-charge articles cash the note within this chapter — their $\sigma(\theta)$ and $\sigma(s)$ maps are equation (2) applied to a cleverly guessed potential.