Gauss's law and conductors

A conductor is defined by carrying charge that can move through it — in copper roughly 8.5 × 10²⁸ free electrons per cubic metre, about one per atom. Electrostatics means nothing is moving. Put those two together and the macroscopic field inside the conducting material must be exactly zero, because any surviving field would keep pushing that charge and the arrangement would still be changing. The zero is not a convenient approximation; it is the condition that defines the final state. It is reached quickly: the standard estimate for how fast a lump of excess charge inside a metal disperses is ε₀ divided by the conductivity, and for copper at 5.96 × 10⁷ S m⁻¹ that is about 1.5 × 10⁻¹⁹ s. Connect a battery and drive a steady current through the same copper and the field inside it is no longer zero, so every result in this topic is switched off.

This is the one place in the chapter where Gauss's law needs no symmetry at all. Draw a closed surface of any shape you like, provided it lies entirely within the conducting material. E = 0 at every point of it, so ∮ E ⋅ dA = 0, so the enclosed charge is zero. Now move and shrink that surface anywhere in the bulk and the same argument runs again, which means the volume charge density is zero everywhere inside the metal. Excess charge has nowhere left to go but the boundaries. A copper pellet carrying −40 nC keeps those 2.5 × 10¹¹ surplus electrons in a layer a few atoms thick on its outside, not spread through its volume. Notice what has not been proved: Gauss's law fixes the total charge on a surface and never says how that charge is distributed over it. On a sphere symmetry finishes the job; on a lump nothing does.

Two facts about the surface field come free. The tangential component has to vanish, because a tangential field would slide surface charge along the surface forever, so E immediately outside is perpendicular to the conductor. For the magnitude, straddle the surface with a short cylinder — a pillbox — of end area A, one end just inside the metal and one just outside. The inner end contributes nothing, because E = 0 there. The curved side contributes nothing, because E is perpendicular to the surface and therefore parallel to that side. Only the outer end contributes, with flux EA, so Gauss's law reads EA = σA/ε₀ and E = σ/ε₀. With σ = 1.0 × 10⁻⁶ C m⁻² that is 1.13 × 10⁵ V m⁻¹. The relation is local: σ means the density at that patch, and the formula is silent about what value σ takes.

Hollow out a conductor and put nothing in the hole. A Gaussian surface drawn in the metal around the cavity encloses zero net charge, so the cavity wall carries zero net charge. That on its own does not forbid equal and opposite patches on opposite walls, so finish the argument with work. If a field line crossed the cavity it would leave a positive patch and land on a negative one; close the path back through the metal, where E = 0 contributes nothing, and you would have carried a charge around a closed loop with net work done. For an electrostatic field ∮ E ⋅ dl = 0 around every loop, so no such field line exists: the cavity field is zero and the wall charge is zero point by point. Park the hollow conductor in an external field of 1.0 × 10⁵ V m⁻¹ and the cavity still reads zero, because the outer surface charge rearranges to cancel it. That is the Faraday cage.

Now fix a charge qᵢₙ inside the cavity. The Gaussian surface in the metal must still enclose zero, so the wall takes exactly −qᵢₙ, however lopsided the cavity or off-centre the charge. The conductor's own net charge Q cannot change, so the outer surface is left with Q + qᵢₙ. Take Q = −3.0 nC with qᵢₙ = +5.0 nC: the wall holds −5.0 nC and the outer surface +2.0 nC. On a spherical outer boundary that +2.0 nC spreads uniformly, because the wall charge has already absorbed every trace of where qᵢₙ sits and nothing outside can distinguish one direction from another. So at r = 0.20 m from the centre the field is k(2.0 nC)/(0.20 m)² = 4.5 × 10² V m⁻¹ radially outward, and sliding qᵢₙ around the cavity changes nothing out there. Screening runs one way: the cavity is protected from outside, the outside is not protected from the cavity. Grounding the shell breaks that last link by draining the outer charge away.

Gauss's law turns σ into E and E into σ, but on its own it never tells you what σ is on a conductor that is not symmetric. Curvature does, at least qualitatively. Join a 6.0 cm sphere and a 2.0 cm sphere with a long wire and they settle at one potential; each then carries q ∝ R, so σ = q/(4πR²) goes as 1/R and the surface field kq/R² goes as 1/R with it. The small sphere sits at three times the field of the large one, which is why charge leaks from points, why lightning rods are sharpened, and why high-voltage hardware is built from fat rounded shapes. Two further limits are worth naming. E = σ/ε₀ is the field immediately outside the surface, not a metre away, where the whole distribution matters again. And the entire picture is electrostatic: a metal box still screens a 50 Hz field well, but the argument has to be redone, and no conductor screens a field that changes fast enough.