Electrostatics of Conductors & Shielding

In 1836, Michael Faraday walked into a twelve-foot wire-mesh cube in the Royal Institution's lecture theatre, had an assistant charge it with an electrostatic generator until sparks leapt from its skin — and calmly sat inside with his most sensitive instruments. He detected nothing. Not a flicker. The most electrified room in London was, on the inside, the calmest place in the city. That stunt is one corollary of a single sentence you are now fully equipped to unpack: in electrostatic equilibrium, the electric field inside a conductor is zero. Feed that one condition into Gauss's law and an entire doctrine falls out: all excess charge rushes to the outer surface, the field just outside grips the metal at exactly ninety degrees with strength $\sigma/\varepsilon_0$ , charge crowds onto sharp points (the whole design brief of the lightning rod), and any hollow cavity inside the metal is perfectly shielded, whatever storm rages outside. Boards love this topic because it is pure reasoning — "state and explain" marks with no calculation to hide behind; JEE and NEET love it because each property is a one-line MCQ eliminator. This article gives you the property list, the two arguments that generate it, the numbers examiners expect, and the traps (a famous factor of 2 is waiting). By the end, a car in a lightning storm, a phone dying in a lift, and Faraday grinning inside his cube will be the same three lines of physics.

Key takeaways

In electrostatic equilibrium the electric field everywhere inside the material of a conductor is exactly zero — otherwise free electrons would still be moving, and it wouldn't be equilibrium.
Gauss's law then forces every scrap of excess charge to the outer surface: a Gaussian surface drawn anywhere inside the metal encloses zero net charge.
Just outside the surface the field is perpendicular to the metal with magnitude $E = \sigma/\varepsilon_0$ — twice the isolated sheet's value, and examiners test exactly that factor.
Surface charge crowds where the surface curves sharply, so the field is strongest at points and edges — the lightning-rod effect.
A cavity inside a conductor is shielded from all outside static fields (the Faraday cage), but the protection is one-way: a charge inside still makes itself known outside unless the conductor is earthed.
Key idea: set $E = 0$ inside the metal, and every other property — surface-only charge, perpendicular exterior field, point-crowding, shielding — follows from Gauss's law and energy conservation.

The core insight

A conductor in electrostatic equilibrium has zero electric field throughout its material; combining that single fact with Gauss's law generates the entire exam doctrine — charge on the outer surface, a perpendicular surface field of magnitude $\sigma/\varepsilon_0$ , charge crowding at sharp points, and field-free cavities.

Where this lives today

Count the Faraday cages within ten metres of you: the braided shield on every USB and antenna cable, the metal can around your laptop's Wi-Fi module, the perforated screen in the microwave-oven door (holes far smaller than the 12 cm microwaves, so the cage looks solid to them while staying transparent to light), the aircraft fuselage that takes a lightning strike about once a year per plane and delivers its passengers unbothered. MRI scan rooms are welded copper boxes — room-sized cages protecting microvolt spine signals from city radio chatter. Industry runs the other properties too: chip-fab workers wear earthed wrist straps (the grounding cure, draining induced charge before a spark can kill a transistor), and field-emission microscopes sharpen tungsten tips to nanometre radius so modest voltages pull electrons straight out of the metal — the lightning rod's crowding rule, repurposed to image atoms.

The physical picture

A metal is a lattice of fixed positive ions bathed in a sea of free electrons — roughly $10^{29}$ per cubic metre in copper. Switch on an external field and the sea flows: electrons pile up on the face where field lines arrive, leaving exposed positive lattice where they leave. But the pile-up creates its own field, pointing against the applied one inside the metal — the flow weakens the very field that drives it. So the flow is self-terminating: as long as any net field survives inside, electrons keep drifting and the opposing field keeps growing; the drifting stops at exactly one configuration — perfect cancellation everywhere inside the material. Nobody arranges this. It is the only state with nothing left pushing, the way water in a jug stops sloshing only when its surface is level. And it is absurdly fast — about $10^{-19}$ seconds in copper. For anything a board exam calls "electrostatics," equilibrium is instantaneous.

Figure

A metal box in an external field — the induced skin charge cancels everything inside

Three consequences are visible right in the picture. First, no field line ever passes through the metal — lines terminate on the induced negative film and restart from the positive film. Second, lines touching the surface dock at right angles; a slanted arrival would drag surface electrons sideways, contradicting equilibrium. Third — the part Faraday staked his comfort on — hollow the block out, and the cavity inherits the calm. One warning before the properties parade: everything here is about equilibrium, charges at rest. A current-carrying wire is deliberately kept out of equilibrium by a battery — it has an interior field on purpose (the Current Electricity chapter's business). Today's rules apply the instant the pushing stops.

One condition, five properties

Everything descends from the equilibrium condition:

E = 0 \quad \text{(everywhere inside the conducting material)} \qquad (1)

where: $E$ is the total field in the metal — applied plus every induced and excess charge, all summed. Equation (1) is not an approximation about good metals; it is what "electrostatic equilibrium" means. Free charges feel $F = qE$ , so any surviving interior field is still rearranging charge. The settling time in copper is about $10^{-19}\ \mathrm{s}$ — the statement costs nothing and buys everything.

Property 1 — no charge in the bulk. Draw a Gaussian surface entirely inside the metal, any shape. $E = 0$ on all of it, so the flux is zero, and Gauss's law says the enclosed charge is zero. Slide that surface anywhere in the material: no region of bulk holds net charge. Give a conductor an excess $Q$ and there is exactly one address left: the outer surface.

Property 2 — one electrical level. With a field-free interior, moving a test charge between any two points of the conductor takes zero work: the whole body sits at one electrical "level." (Next chapter this gets its proper name, potential; for now, hold the picture of a still lake — one surface height, however lumpy the basin.)

Property 3 — the field just outside. The surface itself carries charge, so the field just outside is not zero: it is perpendicular to the surface, with magnitude

E_{\text{just outside}} = \frac{\sigma}{\varepsilon_0} \qquad (2)

where: $\sigma$ is the local surface charge density in $\mathrm{C/m^2}$ (it varies over a lumpy conductor) and $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/N\,m^2}$ . Read it with suspicion: an isolated sheet with the same $\sigma$ gives $\sigma/2\varepsilon_0$ — you derived that in Gauss's Law at Work. The conductor gives exactly twice the sheet; Derivation A shows where the doubling comes from, and the factor is among the most reliably tested facts in this chapter.

Property 4 — charge crowds at sharp points. On a lumpy conductor, $\sigma$ is largest where the curvature is sharpest, so by equation (2) the field is fiercest just outside tips and edges. When that local field reaches about $3 \times 10^6\ \mathrm{N/C}$ , air ionises and charge leaks away as a faint glow — corona discharge. That is the lightning rod's job description: a sharp, earthed spike that bleeds charge into the air and offers any strike a safe path to ground. Run backwards, the same physics is why high-voltage equipment is compulsively rounded.

Property 5 — the shielded cavity. Hollow the conductor out and the cavity is completely protected from external static fields: $E = 0$ throughout the empty space, whatever the storm outside. This is the Faraday cage; Derivation B proves it with nothing but Gauss's law and energy conservation. Mind the one-way valve: the cage hides the world from its contents, but a charge placed inside still announces itself to the outside — unless the conductor is earthed. The full bookkeeping of that announcement ( $-q$ pulled to the cavity wall, $+q$ pushed outside) is the master pattern of the next module, Concentric Shells and Cavity Charges.

you are asked about	reach for	answer
field or charge inside the metal	equation (1) + Gauss	$E = 0$ , no bulk charge
where excess charge sits	Property 1	outer surface only
field at the surface	equation (2)	$\sigma/\varepsilon_0$ , perpendicular
a hollow region, field applied outside	Property 5	zero — shielded
a charge placed inside a cavity	next module	$-q$ on wall, $+q$ outside
where sparks start	Property 4	sharpest point first

Derivation AFrom one condition to the whole doctrine

We start from the single equilibrium fact $E = 0$ inside the metal and extract the exam properties one by one: surface-only charge, the perpendicular exterior field, and the magnitude $\sigma/\varepsilon_0$ with its famous factor of 2. The strategy: spend Gauss's law on the interior, an equilibrium argument on the direction, and superposition — the sheet result from Gauss's Law at Work — on the magnitude. Watch that no step assumes a shape: the chain holds for spheres, boxes and paperclips alike.

Step 1. Charge is exiled from the bulk. Take any closed Gaussian surface wholly inside the material. $E = 0$ at every point of it, so $\Phi = 0$ , and Gauss's law gives $q_{\text{enc}} = 0$ . Since this works for every interior surface, no volume of metal holds net charge — excess charge can only live on the boundary. Why the argument stops at the skin: the surface itself is not "inside" — you cannot wrap a Gaussian surface around a patch of skin while staying in field-free metal, so the eviction order does not apply there.

Step 2. The exterior field docks perpendicular. If the field just outside had a component along the surface, it would push the freely-sliding surface electrons — charge still rearranging, contradicting equilibrium. So the parallel component is zero.

Step 3. Set up the magnitude with superposition. Zoom in on a small patch carrying density $\sigma$ ; up close it is a flat sheet, producing $\sigma/2\varepsilon_0$ pointing away on both sides (Gauss's Law at Work). Split the total field near the patch into the patch's own sheet field plus the field of everything else (the rest of the surface charge and any external sources).

Step 4. Use the interior as a probe. Just inside, the total must vanish: the patch contributes $\sigma/2\varepsilon_0$ inward there, so everything-else must supply $\sigma/2\varepsilon_0$ outward to cancel it — and the everything-else field doesn't jump as you cross the thin patch; only the patch's own field flips.

Step 5. Read off the outside. Just outside, patch field and everything-else both point outward:

E_{\text{out}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}

The conductor's other charges, whose whole job was to cancel the patch inside, unavoidably double it outside. That is the factor of 2. Verified: cross-checked on the charged sphere — $\sigma = Q/4\pi R^2$ gives $E = \sigma/\varepsilon_0 = Q/4\pi\varepsilon_0 R^2$ , exactly the Gauss's-law surface field $kQ/R^2$ ; both routes give $2.88 \times 10^6\ \mathrm{N/C}$ for $Q = 3.2\ \mu\mathrm{C}$ , $R = 0.10\ \mathrm{m}$ . Matches the NCERT conductor-properties list.

Go deeper · Derivation B · 1 minWhy an empty cavity is perfectly shielded

Derivation BWhy an empty cavity is perfectly shielded

We prove the Faraday-cage theorem: a hollow cavity inside a conductor has $E = 0$ everywhere, however strong or lopsided the external field. The strategy is proof by contradiction in two moves — Gauss's law first, then a round-trip energy argument. The subtlety worth savouring: Gauss's law alone cannot close the case, because zero net wall charge still permits a plus patch here and a minus patch there with field lines crossing between them. Energy conservation forbids that.

Step 1. Draw a Gaussian surface entirely within the metal, snugly wrapping the cavity. $E = 0$ on all of it, so the flux is zero, and the net charge on the cavity wall is zero. Why this is not yet victory: "net zero" allows equal and opposite patches on opposite walls, with a field threading the cavity between them — an arrangement Gauss's law is blind to.

Step 2. Suppose such a field line exists, running across the cavity from a positive wall patch to a negative one. Carry a positive test charge along it: the field pushes it the whole way, doing positive work — it arrives with more kinetic energy than it started with.

Step 3. Walk the charge back along a return path burrowing through the metal. There $E = 0$ : the return trip costs nothing.

Step 4. Total the round trip: positive work out, zero work back — the charge gains energy every lap from a static arrangement that never changes. Loop it forever and you have a perpetual motion machine. But electrostatic forces are conservative — work around any closed path is zero (the same fact that lets Work and Energy define potential energy at all). Contradiction: the supposed field line cannot exist, and the cavity is field-free — every point, always. Verified: the standard cavity theorem (NCERT's "electrostatic shielding"; Griffiths §2.5 in the UG deep dive); Faraday's 1836 cube and the 1936 Plimpton–Lawton experiment confirm it to better than one part in a billion.

Sharp points, lightning rods and the crowding rule

Why does charge crowd where the surface sharpens? The honest short version: charge arranges itself so the whole body sits at one electrical level (Property 2), and for two spheres of radii $R_1 > R_2$ joined by a long wire — one conductor in two lumps — that balancing act forces the smaller sphere to hold its charge more densely:

\frac{\sigma_2}{\sigma_1} = \frac{R_1}{R_2} \qquad (3)

where: $\sigma_1, \sigma_2$ are the surface densities on the big and small spheres. Halve the radius, double the density — and by equation (2), double the surface field. The full proof needs the potential concept (next chapter); what you carry today is the reading: a bumpy conductor is a family of joined spheres, each bump a sphere of its local curvature radius, so the field outside scales like $1/R$ — gentle over flat regions, fierce at tips, first to reach air's breakdown field at the sharpest point.

Figure

Charge crowds at sharp points — the field outside scales like one over the local radius

Limiting cases

case	result	meaning
isolated charged sphere	$\sigma = Q/4\pi R^2$ , $E_{\text{surface}} = kQ/R^2$	the one shape with an even skin
flat face of a large conductor	$E = \sigma/\varepsilon_0$ outside, $0$ inside	twice the sheet — the tested factor
empty cavity, any outside field	$E = 0$ in the cavity	Faraday cage, unconditional
charge $q$ in the cavity	cavity has a field; metal still doesn't	shielding is one-way — next module
conductor earthed	induced outer charge drains	the cage goes silent both ways
sharper and sharper tip	$E_{\text{tip}} \propto 1/R \to$ breakdown	corona; the lightning rod's principle
current flowing in a wire	$E \ne 0$ inside	not equilibrium — rules don't apply

Common trapThe rubber tyres are what save you inside a car during lightning

Folklore, and worth two marks to demolish properly. A lightning stroke has just jumped kilometres of open air — the best insulator in everyday life; ten centimetres of rubber is not even a speed bump to it. What protects you is the steel body shell: a conductor in near-instant equilibrium, whose skin rearranges so the interior stays field-free while the strike's charge flows over the outside and arcs to ground. You are sitting in Faraday's cube with cup holders. The fine print follows too: a convertible or fibreglass body genuinely loses the protection, and you shouldn't touch the frame while current flows. In an exam, "explain why passengers are safe" wants the cage argument — $E = 0$ in a cavity enclosed by a conductor — not a word about tyres.

Worked examples

Board level — reading one charged sphere every possible way (CBSE/ISC pattern, 3 marks)

A spherical copper shell of radius $R = 0.10\ \mathrm{m}$ carries a charge $Q = +3.2\ \mu\mathrm{C}$ . Find (a) the electric field at a point inside the metal, (b) the surface charge density, and (c) the electric field just outside the surface, checking (c) by two independent routes. Take $k = 9.0 \times 10^9\ \mathrm{N\,m^2/C^2}$ , $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/N\,m^2}$ .

Step 1. Inside the metal: $E = 0$ , by the equilibrium condition — write the reason ("any interior field would drive the free electrons, contradicting equilibrium"); that sentence is typically worth a full mark.

Step 2. Symmetry spreads the charge uniformly: $\sigma = \dfrac{Q}{4\pi R^2} = \dfrac{3.2 \times 10^{-6}}{4\pi (0.10)^2} = \boxed{2.55 \times 10^{-5}\ \mathrm{C/m^2}}$ , about $25.5\ \mu\mathrm{C/m^2}$ .

Step 3. Route one — the conductor-surface formula (2): $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{2.546 \times 10^{-5}}{8.85 \times 10^{-12}} = \boxed{2.9 \times 10^6\ \mathrm{N/C}}$ , radially outward, perpendicular to the surface.

Step 4. Route two — Gauss's law: just outside, the shell acts as a point charge at the centre, $E = \dfrac{kQ}{R^2} = \dfrac{(9.0 \times 10^9)(3.2 \times 10^{-6})}{(0.10)^2} = 2.88 \times 10^6\ \mathrm{N/C}$ . The routes agree (the whisker of mismatch is the rounding in $k$ ): equation (2) is Gauss's law, localised to the skin.

Competitive level — the car under a thundercloud (JEE-Main pattern)

A thundercloud sets up a vertical, uniform field $E_0 = 5.0 \times 10^4\ \mathrm{N/C}$ near the ground. A car with a closed steel body sits beneath it; take the roof as a horizontal conducting surface of area $1.5\ \mathrm{m^2}$ facing the cloud. Find (a) the induced surface charge density on the roof, (b) how many electronic charges per square micrometre that represents, (c) the total induced charge on the roof, and (d) the field inside the cabin. ( $\varepsilon_0 = 8.85 \times 10^{-12}$ , $e = 1.6 \times 10^{-19}\ \mathrm{C}$ .)

Step 1. Field lines from the cloud land perpendicular on the roof and must terminate there (none may enter the metal). Just outside a conductor $E = \sigma/\varepsilon_0$ , and that total field is the given $E_0$ , so the induced density has magnitude $\sigma = \varepsilon_0 E_0 = (8.85 \times 10^{-12})(5.0 \times 10^4) = \boxed{4.4 \times 10^{-7}\ \mathrm{C/m^2}}$ — about $0.44\ \mu\mathrm{C/m^2}$ .

Step 2. In electrons per area: $\dfrac{\sigma}{e} = \dfrac{4.43 \times 10^{-7}}{1.6 \times 10^{-19}} = 2.8 \times 10^{12}\ \mathrm{per\ m^2} = \boxed{2.8\ \text{electrons per}\ \mu\mathrm{m^2}}$ — a pre-lightning field cancelled by three electrons per square micron. Conductors shield with absurdly little effort.

Step 3. Total induced charge on the roof: $q = \sigma A = (4.43 \times 10^{-7})(1.5) = \boxed{6.6 \times 10^{-7}\ \mathrm{C}} \approx 0.66\ \mu\mathrm{C}$ .

Step 4. Inside the cabin: $\boxed{E = 0}$ — a cavity wrapped in conductor, Derivation B. No calculation, full marks for the reason. Timing note: 2–2.5 minutes. Parts (a)–(c) are one formula each; the only thinking is recognising that $\sigma = \varepsilon_0 E_0$ (no factor 2 — the conductor relation, not the sheet relation). Part (d) should take five seconds; if you started computing, you missed the doctrine.

Exam traps

The factor of 2, both ways. Just outside a conductor $E = \sigma/\varepsilon_0$ ; an isolated sheet gives $\sigma/2\varepsilon_0$ . Swapping them is the classic error here. Anchor: the conductor's other charges cancel the sheet's field inside, so they double it outside.

" $E = 0$ inside" claimed in the wrong place. The theorem covers the metal and empty cavities. A cavity holding a charge has a real field in it (the surrounding metal is still field-free), and a current-carrying wire is not in equilibrium at all. Quote the condition with its jurisdiction.

Shielding direction reversed. The cage always protects inside from outside; it hides its contents only when earthed. MCQs with a charge inside an unearthed shell are testing exactly this — the exterior field is not zero.

Perpendicularity stated without reason. "Field lines meet the conductor normally" earns the mark only with the because: a parallel component would slide the free surface charges.

Where the charge sits. Outer surface, always, whatever the shape — never "through the volume". A solid sphere and a hollow shell of equal radius and charge have identical outside worlds.

Sign and unit slips. $\sigma$ in $\mathrm{C/m^2}$ ; field points away from positive skin, toward negative. Use the constants your paper states ( $k = 9.0 \times 10^9$ , $\varepsilon_0 = 8.85 \times 10^{-12}$ are the board defaults) and never mix precisions mid-solution.

Lightning-rod mechanism. Not "the rod attracts the lightning away". Marks want: sharp tip → high local field (crowding rule) → corona leakage, plus an earthed path if the strike lands.

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · Royal Institution, London

Faraday came to physics by way of a bookbindery — an apprentice who read the encyclopedias he was stitching, talked his way into Humphry Davy's lab as a bottle-washer, and became the greatest experimentalist of his century without ever learning calculus. The cage experiment of 1836 was pure Faraday: rather than argue theory, he built a twelve-foot wire-mesh cube, charged it until sparks flew from its skin, and moved in. "I went into the cube and lived in it," he wrote, "and using lighted candles, electrometers, and all other tests of electrical states, I could not find the least influence upon them." His 1843 ice-pail experiment then quantified induction itself with a pewter pail and a gold-leaf electroscope. Maxwell later said his own equations were largely Faraday's pictures translated into mathematics.

History

1729

— Stephen Gray discovers conduction: electrical "virtue" flows along some materials and not others, splitting the world into conductors and insulators.

1755

— Benjamin Franklin notices a cork ball hanging inside a charged silver can feels no force — the first recorded hint of cavity shielding.

1767

— Joseph Priestley, prompted by Franklin's report, infers that the electric force must follow an inverse-square law — only that law gives a force-free cavity.

1836

— Faraday charges a twelve-foot cube and sits inside it with electroscopes: the cage works, dramatically and in person.

1936

— Plimpton and Lawton re-run the cavity test with electronics, confirming the physics behind shielding to one part in a billion.

Going further

Texts. NCERT Physics Class 12, electrostatics chapters — the conductor-properties list and shielding discussion are the board-exam source · HC Verma, Concepts of Physics Vol. 2, electric field and Gauss's law chapters · Full UG treatment: Electrostatics of Conductors (UG deep dive) and Induced Charges and Shielding (UG deep dive).

Provenance. Standard conductor doctrine, cross-checked against NCERT's property list and the UG sources above. A verifier should recompute: sphere $R = 0.10$ m, $Q = 3.2\ \mu\mathrm{C}$ : $\sigma = 2.55 \times 10^{-5}\ \mathrm{C/m^2}$ , $E = \sigma/\varepsilon_0 = 2.88 \times 10^6\ \mathrm{N/C} = kQ/R^2$ ✓; car roof at $E_0 = 5.0 \times 10^4$ N/C: $\sigma = \varepsilon_0 E_0 = 4.43 \times 10^{-7}\ \mathrm{C/m^2}$ , $2.8\ e/\mu\mathrm{m^2}$ , $q = \sigma A = 0.66\ \mu\mathrm{C}$ on $1.5\ \mathrm{m^2}$ ; copper relaxation $1.5 \times 10^{-19}$ s; air breakdown $3 \times 10^6$ N/C.

Prerequisites. Gauss's Law · Charging Bodies: Conductors, Insulators and Induction. Leads to. Concentric Shells and Cavity Charges (the archetype gym built on this doctrine) · Capacitance and the Parallel-Plate Capacitor (coming in the next chapter).

If the field inside a conductor is zero, how does a wire ever carry a current?

Because a current-carrying wire is not in electrostatic equilibrium — that is the entire answer. Equilibrium means the free charges have finished rearranging; $E = 0$ is the signature of that finished state. A battery's job is to prevent the finish: it keeps pumping charge so a small field survives inside the wire and keeps driving electrons along it. Disconnect the battery and the wire settles in about $10^{-19}$ seconds — and this article takes over again. The current-carrying picture is built properly in the Current Electricity chapter.

Does a Faraday cage block magnetic fields and gravity too?

No, and the reasons are instructive. Electric shielding works by response: the metal has mobile charges of both signs that rearrange until they cancel the intruder. Gravity has only one sign of "charge" — mass — and nothing mobile to rearrange, so nothing shields it. Static magnetic fields pass through copper almost untouched (no magnetic charges to rearrange either); you redirect those with high-permeability alloys instead. Rapidly changing fields are different — the cage fights them with induced currents, which is why the same mesh that ignores a magnet blocks a phone signal.

Why exactly perpendicular? Could the field not be almost perpendicular?

Exactly, in equilibrium. Split the surface field into perpendicular and parallel parts. Surface electrons cannot leave the metal (the perpendicular part just presses them against the boundary), but they slide along it freely — so even a tiny parallel component starts a surface current, and the arrangement isn't final. Charges shuffle until the parallel part is dead everywhere; what survives is purely perpendicular. Exam answers need that mechanism, not just the fact.

How thick is the "surface" where the charge sits?

An atom or two. The classical idealisation — charge on a mathematical surface of zero thickness — corresponds to a layer about $10^{-10}$ m deep where the electron sea adjusts its density. For anything you will calculate or measure with school apparatus, zero thickness is exact. It's a reminder of what "conductor" means as a model: infinitely responsive charge, instantly settled, infinitely thin skin — and real metals honour it because $10^{29}$ electrons per cubic metre is, in practice, infinite responsiveness.

How do the electrons know exactly where to go to cancel the field?

They don't — they just fall downhill until there is no downhill left. Wherever cancellation is incomplete, a residual field survives and keeps pushing charge; the flow stops only when the residual is zero everywhere. So the final arrangement is automatically the perfect one, the way water finds a flat surface without solving any equations — and that is also why the result never depends on the conductor's shape: whatever the geometry, the stopping condition is the same, $E = 0$ inside.

My phone loses signal in a lift — is that this article?

Mostly yes. A lift car is a rough Faraday cage: a metal box whose induced charges terminate incoming fields. The wrinkle is that phone signals are oscillating waves, not static fields, so the full story involves how the cage responds at that frequency and how big its openings are compared to the wavelength — a lift with glass panels leaks more than a sealed one. The static theorem here is the foundation; the frequency fine print arrives with electromagnetic waves at the end of this track.