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 macroscopic electric field inside a conductor vanishes, \(\vec{E}_{\text{in}}=\vec{0}\); the bulk and surface together form a single equipotential volume, \(V=\text{const}\); and just outside the surface the field is purely normal with magnitude fixed by the local surface charge density, \(\vec{E}_{\text{out}}=\dfrac{\sigma}{\varepsilon_0}\,\hat{n}\), where \(\hat{n}\) is the outward unit normal. As a corollary the volume charge density inside vanishes and all net charge lives on the surface.

Why it matters

These results are the complete electrostatic boundary condition for any conductor, and they are what make image charges, capacitance, and shielding tractable. Once a surface is known to be an equipotential met by the field at right angles, the unknown charge distribution can be replaced by a Dirichlet condition on the potential, and one solves Laplace's equation \(\nabla^2 V=0\) instead of tracking individual charges.

They also underwrite everyday engineering: Faraday cages, coaxial cable, grounded instrument chassis, and the field-emission limit of high-voltage electrodes all follow directly from \(\vec{E}=\vec{0}\) inside and \(E=\sigma/\varepsilon_0\) outside.

Assumptions

Static equilibrium (no steady current, \(\partial/\partial t=0\)).If charges are still moving, the interior field need not vanish; a current-carrying wire has a nonzero longitudinal \(\vec{E}=\vec{J}/\sigma_c\) inside it and the "field is normal" conclusion fails.

An ideal conductor with freely mobile charge in effectively unlimited supply.Drop perfect mobility and the interior field is only screened to a small residual 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 \(\sigma\).At an atomically sharp edge the pillbox has no single well-defined normal and the local field diverges; \(E=\sigma/\varepsilon_0\) then holds only as a coarse-grained average.

Purely electrostatic conditions, \(\nabla\times\vec{E}=\vec{0}\), with linear field-independent response.A time-varying \(\vec{B}\) makes \(\nabla\times\vec{E}\neq\vec{0}\), so tangential continuity changes and fields penetrate a skin depth; above breakdown the medium ionises and sources new charge the equilibrium argument ignores.

Derivation

\[ \vec{F}=q\vec{E}\ \Longrightarrow\ \vec{E}_{\text{in}}=\vec{0} \]

Free carriers feel a force \(q\vec{E}\); if any macroscopic field remained inside, they would accelerate and rearrange. A static state is possible only when the net field on every interior charge is zero, so \(\vec{E}_{\text{in}}=\vec{0}\) throughout the bulk. A

\[ \rho_{\text{in}}=\varepsilon_0\,\nabla\cdot\vec{E}_{\text{in}}=0 \]

The differential form of the assumed prior result (Gauss's law) is \(\nabla\cdot\vec{E}=\rho/\varepsilon_0\). With \(\vec{E}_{\text{in}}=\vec{0}\) everywhere inside, its divergence is zero, so the interior volume charge density vanishes and any net charge lives only on the boundary as a surface density \(\sigma\). B

\[ \oint \vec{E}\cdot d\vec{\ell}=0 \]

Because \(\nabla\times\vec{E}=\vec{0}\) (prior result), the line integral of the electrostatic field around any closed loop vanishes. Take a thin rectangular loop straddling the surface: long sides of length \(\ell\) tangent to it, one just inside and one just outside, with short ends of vanishing height. B

\[ \left(E_{\parallel}^{\text{out}}-E_{\parallel}^{\text{in}}\right)\ell=0\ \Longrightarrow\ E_{\parallel}^{\text{out}}=0 \]

As the short ends shrink, only the two tangential sides contribute, giving the difference of tangential components times \(\ell\). Since \(E_{\parallel}^{\text{in}}=0\) from Step 1, the tangential field just outside also vanishes; the external field can have only a normal component. B

\[ \oint_S \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0} \]

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, with negligible height. B

\[ E_{\perp}^{\text{out}}A+\underbrace{E_{\perp}^{\text{in}}A}_{=0}+\underbrace{\Phi_{\text{side}}}_{\to 0}=\frac{\sigma A}{\varepsilon_0} \]

The inner face contributes nothing since \(\vec{E}_{\text{in}}=\vec{0}\); the curved side contributes nothing as its height \(\to 0\) and because the outside field is purely normal (Step 4). Only the outer face carries flux, and the enclosed charge on the straddled patch is \(\sigma A\). B

\[ E_{\perp}^{\text{out}}=\frac{\sigma}{\varepsilon_0},\qquad \vec{E}_{\text{out}}=\frac{\sigma}{\varepsilon_0}\,\hat{n} \]

Cancel the common area \(A\) (symbols rearranged before any number is inserted) and reinstate direction with the outward normal \(\hat{n}\); for \(\sigma>0\) the field points outward. A

\[ V(\vec{b})-V(\vec{a})=-\int_{\vec{a}}^{\vec{b}}\vec{E}\cdot d\vec{\ell}=0 \]

For any two points in the bulk the integrand is zero (\(\vec{E}_{\text{in}}=\vec{0}\)); for any path along the surface it is zero because \(E_{\parallel}=0\). Hence \(V\) is identical at every interior and surface point: the conductor is one equipotential, and the surface field, being \(-\nabla V\), is perpendicular to it. B

Result

\[ \vec{E}_{\text{in}}=\vec{0},\qquad \vec{E}_{\text{out}}=\frac{\sigma}{\varepsilon_0}\,\hat{n},\qquad V=\text{const on and in the conductor} \]

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 whole conductor sits at one voltage. The single factor \(\sigma/\varepsilon_0\) — not \(\sigma/2\varepsilon_0\) — already contains the field the local sheet makes plus the field of all other charges, which cancel inside and add outside.

Units check. \([\sigma]=\mathrm{C\,m^{-2}}\) and \([\varepsilon_0]=\mathrm{C^2\,N^{-1}\,m^{-2}}\), so \(\left[\dfrac{\sigma}{\varepsilon_0}\right]=\dfrac{\mathrm{C\,m^{-2}}}{\mathrm{C^2\,N^{-1}\,m^{-2}}}=\mathrm{N\,C^{-1}}=\mathrm{V\,m^{-1}}\), the correct dimensions of electric field.

Limiting cases

\(\sigma\to 0\) on a patch: the field there vanishes and neighbouring field lines fan around it — the statement is genuinely local.
Large flat plate, near field: the surface looks like an infinite sheet plus its induced counterpart, recovering the uniform \(\sigma/\varepsilon_0\) of a parallel-plate gap.
Isolated sphere, radius \(R\), charge \(Q\): \(\sigma=Q/4\pi R^2\) gives \(E=Q/4\pi\varepsilon_0 R^2\), matching the point-charge field just outside the surface.
Sharp region (small radius of curvature): charge crowds where curvature is high, so \(\sigma\) and hence \(E\) are largest at points and edges — the lightning-rod limit.
Negative patch: \(\sigma<0\) simply flips the sign, so \(\vec{E}_{\text{out}}\) points inward along \(-\hat{n}\); the derivation is unchanged.

Breaks when

Time dependence / currents. With a driving EMF the interior field is nonzero (\(\vec{E}=\vec{J}/\sigma_c\)) and tangential at a wire's surface; the equilibrium chain collapses. AC fields also penetrate a skin depth \(\delta=\sqrt{2/\mu_0\sigma_c\omega}\), 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 (\(\sim0.05\ \mathrm{nm}\)); in an electrolyte or plasma the Debye length can reach micrometres, so a thin nanostructure never fully screens and \(\vec{E}_{\text{in}}\neq\vec{0}\).
Dielectric breakdown / field emission. Once \(\sigma/\varepsilon_0\) exceeds the breakdown field of the surrounding medium (\(\sim3\times10^{6}\ \mathrm{V\,m^{-1}}\) in air), discharge or electron emission bleeds charge away and no static \(\sigma\) is sustainable.
Superconductors and quantum-scale samples. A superconductor additionally expels magnetic flux (Meissner effect) over the London depth, so it is a perfect diamagnet, not merely a perfect conductor; and a sample a few atoms across has no continuous \(\sigma\) or sharp normal.

Failure modes

Writing \(E=\sigma/2\varepsilon_0\). That is the field of an isolated charge sheet in vacuum. At a conductor the rest of the charge adds a second \(\sigma/2\varepsilon_0\), cancelling inside and doubling outside to the full \(\sigma/\varepsilon_0\).
Claiming charge spreads uniformly. Only on a sphere, by symmetry. On any other shape \(\sigma\) is fixed by the equipotential condition and concentrates at high curvature.
Carrying \(\sigma\) onto an empty cavity wall. A conductor with a charge-free cavity has zero field in the cavity too; students often wrongly place an outer-surface \(\sigma\) on the inner wall.
Confusing \(V=\text{const}\) 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 a tangential surface field.
Asserting the outside field is \(Q/4\pi\varepsilon_0 r^2\) at a non-spherical surface. That Coulomb form holds only for a sphere or far away; locally the correct statement is \(\sigma/\varepsilon_0\).

Discussion

The deep content is that a conductor converts a hard problem — find the charge distribution — into an easy one — solve \(\nabla^2 V=0\) with \(V=\text{const}\) on each conductor. The charges arrange themselves precisely so as to make this boundary condition true, and we never track them individually. The surface density is read off afterward from \(\sigma=\varepsilon_0 E_{\perp}=-\varepsilon_0\,\partial V/\partial n\) at the surface.

Physically the two boundary results are two faces of the same coin. "Field is normal" is the tangential (curl) condition; \(E=\sigma/\varepsilon_0\) is the normal (divergence) condition. Every electrostatic interface obeys both — for a dielectric they read \(E_{\parallel}\) continuous and \(\Delta D_{\perp}=\sigma_{\text{free}}\). A conductor is the special case where one side has \(\vec{E}=\vec{0}\), which 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 \(P=\dfrac{\sigma^2}{2\varepsilon_0}=\dfrac{1}{2}\varepsilon_0 E^2\), and the factor \(\tfrac12\) here (versus the full \(\varepsilon_0\) 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 (\(\sigma/\varepsilon_0\)), namely \(\sigma/2\varepsilon_0\). 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 distribution minimises the electrostatic energy at fixed total charge, and a nonzero interior field would signal a lower-energy rearrangement. 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 \(\sigma\) 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 \(\pm\sigma\) on opposite faces. And the surface charge is not "stuck to atoms"; it is a mobile equilibrium that re-forms instantly if you deform the conductor or move a nearby charge.

Worked examples

\[ \textbf{Isolated charged sphere: }R=5.0\ \mathrm{cm},\ Q=2.0\ \mathrm{nC}.\ \text{Find }\sigma\text{ and }E. \]

By spherical symmetry the charge is uniform: \(\sigma=Q/4\pi R^2\), then \(E=\sigma/\varepsilon_0\). Symbols first, numbers after. A

\[ \sigma=\frac{2.0\times10^{-9}}{4\pi(0.050)^2}=\frac{2.0\times10^{-9}}{3.14\times10^{-2}}=6.4\times10^{-8}\ \mathrm{C\,m^{-2}} \]

Here \(4\pi R^2=3.14\times10^{-2}\ \mathrm{m^2}\). A

\[ E=\frac{\sigma}{\varepsilon_0}=\frac{6.4\times10^{-8}}{8.85\times10^{-12}}=7.2\times10^{3}\ \mathrm{V\,m^{-1}} \]

Cross-check with the point-charge form: \(\dfrac{Q}{4\pi\varepsilon_0 R^2}=\dfrac{(8.99\times10^{9})(2.0\times10^{-9})}{(0.050)^2}=7.2\times10^{3}\ \mathrm{V\,m^{-1}}\) — agreement, as required for a sphere. B

\[ \sigma\approx 64\ \mathrm{nC\,m^{-2}},\qquad E\approx 7.2\ \mathrm{kV\,m^{-1}}\ \text{(radially outward)} \]

Reading. Well below air's \(\sim3\ \mathrm{MV\,m^{-1}}\) breakdown, so the charge sits stably.

Units check. \(\mathrm{C\,m^{-2}}/(\mathrm{C^2\,N^{-1}\,m^{-2}})=\mathrm{V\,m^{-1}}\). Correct.

\[ \textbf{Charged plate, one face: }\sigma=1.5\times10^{-6}\ \mathrm{C\,m^{-2}}.\ \text{Find }E\text{ and the outward pressure }P. \]

Locally the boundary law gives \(E=\sigma/\varepsilon_0\) (normal); the Maxwell stress is \(P=\sigma^2/2\varepsilon_0\). B

\[ E=\frac{\sigma}{\varepsilon_0}=\frac{1.5\times10^{-6}}{8.85\times10^{-12}}=1.7\times10^{5}\ \mathrm{V\,m^{-1}} \]

Direct substitution into the boxed result. A

\[ P=\frac{\sigma^2}{2\varepsilon_0}=\frac{(1.5\times10^{-6})^2}{2(8.85\times10^{-12})}=0.13\ \mathrm{Pa} \]

Equivalently \(\tfrac12\varepsilon_0 E^2=\tfrac12(8.85\times10^{-12})(1.7\times10^{5})^2=0.13\ \mathrm{Pa}\), confirming the \(\tfrac12\) "own-field" factor. B

\[ E\approx 1.7\times10^{5}\ \mathrm{V\,m^{-1}},\qquad P\approx 0.13\ \mathrm{Pa}\ \text{(outward)} \]

Reading. The surface is pulled outward whatever the sign of \(\sigma\) — the pressure goes as \(\sigma^2\), a genuine electrostatic tension on the metal.

Units check. \(\mathrm{(C\,m^{-2})^2/(C^2\,N^{-1}\,m^{-2})=N\,m^{-2}=Pa}\). Correct.

Problems

A neutral isolated conducting sphere of radius \(R\) is placed in a previously uniform field \(E_0=500\ \mathrm{V\,m^{-1}}\). Find the maximum induced surface charge density and where it occurs.
Solution
The exterior solution is \(V=-E_0\left(r-\dfrac{R^3}{r^2}\right)\cos\theta\). The radial field at the surface is \(E_r(R)=3E_0\cos\theta\), so \(\sigma=\varepsilon_0 E_r=3\varepsilon_0 E_0\cos\theta\), maximal at the poles (\(\theta=0,\pi\)) facing along the field. Numerically \(\sigma_{\max}=3(8.85\times10^{-12})(500)=1.3\times10^{-8}\ \mathrm{C\,m^{-2}}=13\ \mathrm{nC\,m^{-2}}\). The pole field is \(3E_0=1500\ \mathrm{V\,m^{-1}}\), three times the applied field.
A hollow conducting shell (inner radius \(a=2\ \mathrm{cm}\), outer radius \(b=3\ \mathrm{cm}\)) carries net charge \(Q=4\ \mathrm{nC}\), with a point charge \(q=1\ \mathrm{nC}\) at its centre. Find the surface charge on the inner and outer walls.
Solution
The field in the metal is zero, so a Gaussian sphere there encloses zero net charge: the inner wall carries \(-q=-1\ \mathrm{nC}\). Charge conservation on the isolated shell puts \(Q+q=5\ \mathrm{nC}\) on the outer wall. Densities: \(\sigma_{\text{in}}=\dfrac{-1\times10^{-9}}{4\pi(0.02)^2}=-2.0\times10^{-7}\ \mathrm{C\,m^{-2}}\); \(\sigma_{\text{out}}=\dfrac{5\times10^{-9}}{4\pi(0.03)^2}=4.4\times10^{-7}\ \mathrm{C\,m^{-2}}\).
Two parallel conducting plates, area \(A=0.10\ \mathrm{m^2}\) each, carry equal and opposite charges \(\pm Q=\pm20\ \mathrm{nC}\). Find the surface density on the inner faces and the field in the gap, neglecting edge effects.
Solution
Essentially all charge migrates to the inner faces, so \(\sigma=Q/A=\dfrac{20\times10^{-9}}{0.10}=2.0\times10^{-7}\ \mathrm{C\,m^{-2}}\). The gap field is \(E=\sigma/\varepsilon_0=\dfrac{2.0\times10^{-7}}{8.85\times10^{-12}}=2.3\times10^{4}\ \mathrm{V\,m^{-1}}\), directed from the \(+\) to the \(-\) plate. Outside the plates the fields cancel to zero.
A long conducting cylinder of radius \(R=1.0\ \mathrm{mm}\) carries linear charge density \(\lambda=30\ \mathrm{nC\,m^{-1}}\). Find \(\sigma\) on its surface and the field just outside.
Solution
By cylindrical symmetry \(\sigma=\dfrac{\lambda}{2\pi R}=\dfrac{30\times10^{-9}}{2\pi(1.0\times10^{-3})}=4.8\times10^{-6}\ \mathrm{C\,m^{-2}}\). Then \(E=\sigma/\varepsilon_0=\dfrac{4.8\times10^{-6}}{8.85\times10^{-12}}=5.4\times10^{5}\ \mathrm{V\,m^{-1}}\). Cross-check with the line-charge field \(E=\dfrac{\lambda}{2\pi\varepsilon_0 R}=\dfrac{30\times10^{-9}}{2\pi(8.85\times10^{-12})(1.0\times10^{-3})}=5.4\times10^{5}\ \mathrm{V\,m^{-1}}\) — identical.
A conducting sphere of radius \(2.0\ \mathrm{cm}\) is charged until the surface field just reaches the dielectric strength of air, \(E_{\max}=3.0\times10^{6}\ \mathrm{V\,m^{-1}}\). Find the maximum charge and the corresponding potential.
Solution
At breakdown \(\sigma_{\max}=\varepsilon_0 E_{\max}=(8.85\times10^{-12})(3.0\times10^{6})=2.66\times10^{-5}\ \mathrm{C\,m^{-2}}\). Then \(Q_{\max}=\sigma_{\max}\,4\pi R^2=(2.66\times10^{-5})(4\pi)(0.020)^2=1.3\times10^{-7}\ \mathrm{C}=0.13\ \mu\mathrm{C}\). The potential is \(V=E_{\max}R=Q_{\max}/4\pi\varepsilon_0 R=(3.0\times10^{6})(0.020)=6.0\times10^{4}\ \mathrm{V}=60\ \mathrm{kV}\). Note the field, not the potential, sets the limit — small radii break down at modest voltages, which is why sharp points spark first.