ARCHETYPE: Concentric Shells & Cavity Charges

Lock a point charge inside a metal vault and the vault does bookkeeping on it — instantly, perfectly, and in a way examiners have milked for fifty years. The wall summons exactly $-q$ from the electron sea; the leftover $+q$ is exiled to the outer surface; the field snaps to zero in the metal while surviving in the cavity and reappearing, disguised, outside. Every number in that story is fixed by one move: draw a Gaussian sphere inside the metal and demand zero flux. This is the archetype at the heart of the concentric-shell industry — the most reliable question family built on Gauss's law. CBSE and ISC ask for the induced charges and the field at a marked point (a tidy 3-marker); JEE-Main nests a charged sphere inside a charged shell and asks for the field region by region; JEE-Advanced moves the cavity charge off-centre or earths the outer surface and watches you flinch. Every variant is the same four-line pattern: induced charges by Gauss, outer surface by conservation, field in each gap by counting the charge inside. There is even a founding legend: in 1773, a decade before Coulomb's torsion balance, Henry Cavendish built exactly this setup and used the silence inside as a precision test of the inverse-square law itself. Learn the pattern once and the whole family becomes fill-in-the-blanks.

Key takeaways

The setup: a point charge $q$ in the cavity of a conducting shell (inner radius $a$ , outer $b$ , own charge $Q$ ). Everything follows from $E = 0$ in the metal.
Bookkeeping law one: the cavity wall always takes exactly $-q$ — forced by a Gaussian surface in the metal, where zero field means zero enclosed charge.
Bookkeeping law two: the outer surface carries $Q + q$ — the shell's own charge plus the exile pushed out by the wall's borrowing.
Field by region: $kq/r^2$ in the cavity, $0$ in the metal, $k(Q+q)/r^2$ outside — count every charge closer to the centre than you are.
The amnesia twist: move $q$ off-centre and the wall charge crowds toward it, but the outer surface stays uniform — the exterior never learns where the charge sits.
Key idea: one Gaussian sphere in the metal (zero flux ⇒ wall takes $-q$ ) plus charge conservation (outer takes $Q + q$ ) solves every shell-and-cavity problem ever set.

The core insight

Because the field inside conducting material is zero, any Gaussian surface drawn through the metal encloses zero net charge — pinning $-q$ on the cavity wall, exiling $Q + q$ to the outer surface, and reducing every field question to counting the charge enclosed, region by region.

Where this lives today

The coaxial cable feeding every lab oscilloscope and rooftop antenna is this article rolled into a cylinder: a signal conductor inside a conducting shield whose induced skins keep outside noise out and inside signal in. High-voltage engineering runs the earthed variant at building scale (substation control rooms in grounded enclosures), and chip-fab ESD protocol is cavity bookkeeping with a wrist strap — drain the induced charge continuously, so no vault ever announces itself with a spark. And the founding experiment never stopped running: Cavendish's silent cavity, rebuilt with modern electronics, still delivers the precision test of Coulomb's law — deviations of the exponent from 2 bounded at parts in $10^{16}$ , doubling as the tightest limit on the photon's mass. A JEE examiner asking "find the charge on the inner surface" and a metrologist bounding fundamental physics are reading the same Gaussian surface.

The setup

Geometry first, because in an archetype the geometry is the physics. A conducting shell has inner radius $a$ , outer radius $b$ , and possibly a charge $Q$ of its own; a point charge $q$ sits at the centre of the cavity. Space splits into three regions, with the metal acting as a referee that forces the charge accounts to balance.

Figure

The canonical shell problem — three regions, two induced skins, one referee

Why must the wall respond at all? The charge $q$ throws field lines outward, and field lines cannot enter conducting material — the metal must stay field-free (the doctrine of Electrostatics of Conductors and Shielding). So every line from $q$ must terminate before crossing radius $a$ , and lines terminate only on negative charge: the wall dresses itself with exactly enough, stealing electrons from the metal's sea, while the abandoned positive remainder — barred from the bulk — surfaces at $b$ . The two skins are what keeps the referee's zone silent.

The master results, before the honest work:

quantity	value	forced by
charge on cavity wall (at $a$ )	$-q$	Gauss's law in the metal
charge on outer surface (at $b$ )	$Q + q$	conservation of the shell's charge
field in the cavity ( $r < a$ )	$kq/r^2$	only $q$ is enclosed
field in the metal ( $a < r < b$ )	$0$	equilibrium doctrine
field outside ( $r > b$ )	$k(Q+q)/r^2$	enclosed total $= q - q + Q + q$

The master solution pattern

Every shell problem starts the same way. Draw a concentric Gaussian sphere inside the metal — any radius between $a$ and $b$ . On it, $E = 0$ everywhere, so the flux through it is zero, and Gauss's law from Gauss's Law does the rest:

q_{\text{enc}} = q + Q_{\text{wall}} = 0 \quad \Rightarrow \quad Q_{\text{wall}} = -q \qquad (1)

where: $Q_{\text{wall}}$ is the total induced charge on the cavity wall. Read the force of this: not "about $-q$ ", not "depends on the shell" — exactly $-q$ , whatever the shell's own charge, whatever the shapes, wherever the charge sits. The shell's own charge $Q$ is conserved, now split between two surfaces:

Q_{\text{wall}} + Q_{\text{outer}} = Q \quad \Rightarrow \quad Q_{\text{outer}} = Q + q \qquad (2)

where: $Q_{\text{outer}}$ sits on the surface at $b$ : the wall borrowed $-q$ , so the ledger dumps the balancing $+q$ outside, on top of whatever $Q$ the shell already owned. A neutral shell shows $+q$ to the world — the vault announces its contents. With the skins pinned, every field question is the counting ritual from Gauss's Law at Work: on a concentric sphere of radius $r$ , symmetry gives $E \cdot 4\pi r^2 = q_{\text{enc}}/\varepsilon_0$ , so

E(r) = \frac{k \, q_{\text{enc}}(r)}{r^2}, \qquad q_{\text{enc}} = \text{sum of ALL charge at radii} < r \qquad (3)

where: $k = 1/4\pi\varepsilon_0 = 9.0 \times 10^9\ \mathrm{N\,m^2/C^2}$ . In the cavity the count is $q$ ; in the metal it is $q - q = 0$ (Gauss agreeing with the doctrine, as it must); outside it is $Q + q$ . One counting rule, three regions, done. And never re-derive the shell's field from scratch: the single-shell result — a uniform spherical skin acts as a point charge at the centre from outside and contributes nothing inside — was established in Gauss's Law at Work; here you spend it, repeatedly and by name.

Derivation AThe two bookkeeping laws from one Gaussian sphere

We derive the master results (1) and (2) — wall charge $-q$ , outer charge $Q + q$ — the way the marking scheme wants: one Gaussian surface, one conservation statement, reasoning spelled out. The strategy is to let the metal's silence do all the work: any surface drawn through field-free material has zero flux, and zero flux means zero enclosed charge. Watch how little the geometry matters.

Step 1. State the doctrine that powers everything: in electrostatic equilibrium the field inside conducting material is zero (Electrostatics of Conductors and Shielding). A statement about the metal, not the cavity.

Step 2. Draw a closed Gaussian surface entirely within the metal, wrapping the cavity. $E = 0$ on all of it, so $\Phi = 0$ , and Gauss's law gives $q_{\text{enc}} = 0$ .

Step 3. Count the enclosure: the cavity charge $q$ plus the induced wall charge $Q_{\text{wall}}$ (the bulk holds none). So $q + Q_{\text{wall}} = 0$ : $Q_{\text{wall}} = -q$ . Why exactly, not approximately: the flux is zero identically — no leftover to argue about. And nothing here used the cavity's shape, the shell's shape, or the charge's position: off-centre charge, dented cavity, cubical shell — the wall total is still exactly minus q.

Step 4. Conserve the shell's own charge, split over its two surfaces: $Q_{\text{wall}} + Q_{\text{outer}} = Q$ , so $Q_{\text{outer}} = Q + q$ .

Step 5. Sanity-check the exterior: a Gaussian sphere outside everything encloses $q - q + (Q + q) = Q + q$ . The wall's $-q$ and the cavity's $q$ cancel in every exterior count; only the ledger's bottom line escapes. Verified: limits — $q = 0$ gives an undisturbed shell ( $0$ on the wall, $Q$ outside); $Q = 0$ , $q > 0$ gives the neutral-vault case ( $-q$ wall, $+q$ outside); $Q = -q$ makes the exterior field vanish entirely (the earthed configuration, meeting us in the variants). Matches NCERT's shell discussion and the UG treatment in Induced Charges and Shielding.

Go deeper · Derivation B · 1 minThe field ladder, region by region

Derivation BThe field ladder, region by region

We assemble $E(r)$ for the full canonical setup — charge $q$ at the centre, shell of net charge $Q$ between radii $a$ and $b$ — and check it against every limit we own. The strategy is pure counting: spherical symmetry collapses Gauss's law to $E = k\,q_{\text{enc}}/r^2$ in each region (the ritual from Gauss's Law at Work), so the only work is tallying $q_{\text{enc}}$ three times. The thing to watch is which charges have radius smaller than yours.

Step 1. Cavity, $r < a$ . Enclosed: only $q$ , so $E = kq/r^2$ , outward for positive $q$ . The shell contributes nothing here — its skins are uniform spherical layers at larger radii, and a uniform spherical skin produces zero field inside itself (the shell result, spent by name). Why the occupant is not shielded from itself: shielding protects the cavity from OUTSIDE sources; the occupant is not outside. Its own field fills the cavity at full strength.

Step 2. Metal, $a < r < b$ . Enclosed: $q + (-q) = 0$ , so $E = 0$ — Gauss's count agreeing with the doctrine. Either statement earns the mark; quoting both shows command.

Step 3. Outside, $r > b$ . Enclosed: $q - q + (Q + q) = Q + q$ , so $E = k(Q+q)/r^2$ — a single point charge $Q + q$ at the centre. Positive total points outward, negative inward: state the direction, it is usually a reserved mark.

Step 4. Check the seams: approaching $a$ from inside, $E \to kq/a^2$ (finite), then drops to zero in the metal — the jump is exactly what the wall's negative skin is for. Leaving at $b$ , the field restarts at $k(Q+q)/b^2$ . Verified: with $q = 2.0\ \mu\mathrm{C}$ , $a = 0.10\ \mathrm{m}$ : cavity edge field $kq/a^2 = 1.8 \times 10^6\ \mathrm{N/C}$ ; limits — $Q = 0$ reproduces the naked charge's field outside ( $kq/r^2$ ), $q = 0$ reproduces the plain charged shell, and $Q = -q$ kills the exterior. All three match the single-shell results of Gauss's Law at Work.

The variants examiners actually pose

Variant 1 — the neutral vault (CBSE/NEET favourite). $Q = 0$ : wall $-q$ , outer $+q$ , exterior $kq/r^2$ . The pre-emptable twist: $\sigma_a = -q/4\pi a^2$ and $\sigma_b = +q/4\pi b^2$ differ in magnitude — same coulombs, different areas.

Variant 2 — charged shell plus cavity charge (the 3-marker). The full ladder above. The examiner's favourite catch: $Q_{\text{outer}} = Q$ (forgetting the exile) or $= q$ (forgetting the shell's own charge). Equation (2) carries both terms.

Variant 3 — a sphere inside a shell (JEE-Main staple). Replace the point charge with a charged conducting sphere. Nothing changes: the inner sphere's charge counts as "cavity charge", the shell's inner surface takes its negative, and the field between them is set by the inner sphere alone. Worked in full below.

Variant 4 — the off-centre charge (JEE-Advanced's amnesia test). Slide $q$ toward one side. The wall's $-q$ redistributes — denser on the near side — precisely so the metal stays silent. But the outer surface never hears: the cavity system (charge plus wall) shows zero field to the metal and beyond, so the outer $Q + q$ arranges as if alone — uniformly. Exterior field: still centred, still $k(Q+q)/r^2$ . (The unsymmetric cavity field itself is not asked at HS; the statements are what MCQs test.)

Variant 5 — earth the outer surface. The stranded outer charge drains: $Q_{\text{outer}} \to 0$ , the exterior field dies, and the wall's $-q$ stays exactly where it was — pinned by the cavity charge, not the ground wire. Verdicts only for now; the quantitative "how much flows" machinery is potential, arriving in Sphere Potentials and Earthing.

Variant 6 — storm outside, cavity inside. Apply an external field to the shell: the cavity stays field-free (the cage theorem from Electrostatics of Conductors and Shielding) and the occupant feels nothing. Two protections at once, one in each direction of the wall.

Figure

Field strength versus radius — the three-region ladder

Limiting cases

case	result	meaning
$q = 0$	wall $0$ , outer $Q$ ; plain charged shell	the archetype contains Gauss's Law at Work as a special case
$Q = 0$	wall $-q$ , outer $+q$ ; exterior $kq/r^2$	neutral vault announces its contents
$Q = -q$	outer surface $0$ ; exterior field zero	the "accidentally earthed" configuration
outer earthed	$Q_{\text{outer}} \to 0$ , exterior dies, wall $-q$ survives	grounding edits only the announcing surface
$q$ off-centre	wall redistributes; outer stays uniform	amnesia: outside learns the total, never the position
$a \to b$ (thin shell)	same totals on coincident surfaces	thickness never mattered — only topology
external field applied	cavity and occupant untouched	the cage from Electrostatics of Conductors, running in parallel

Worked examples

Board level — the classic vault with a charged shell (CBSE/ISC pattern, 3 marks)

A conducting shell of inner radius $a = 0.10\ \mathrm{m}$ , outer radius $b = 0.12\ \mathrm{m}$ , carries net charge $+6.0\ \mu\mathrm{C}$ ; a point charge $-2.0\ \mu\mathrm{C}$ sits at its centre. Find (a) the inner-surface charge, (b) the outer-surface charge, (c) the field at $r = 0.20\ \mathrm{m}$ . ( $k = 9.0 \times 10^9$ SI.)

Step 1. Inner surface, by the Gaussian-in-the-metal argument (state it — it carries the method mark): zero flux forces $q + Q_{\text{wall}} = 0$ , so $Q_{\text{wall}} = -(-2.0\ \mu\mathrm{C}) = \boxed{+2.0\ \mu\mathrm{C}}$ . Sign discipline: the wall takes the negative of the cavity charge, which is here itself negative.

Step 2. Outer surface, by conservation: $Q_{\text{outer}} = Q - Q_{\text{wall}} = +6.0 - (+2.0) = \boxed{+4.0\ \mu\mathrm{C}}$ .

Step 3. Field at $r = 0.20\ \mathrm{m}$ (outside everything). Enclosed total: $-2.0 + 2.0 + 4.0 = +4.0\ \mu\mathrm{C}$ — equivalently cavity plus shell, $-2 + 6 = +4\ \mu\mathrm{C}$ .

Step 4. Counting rule: $E = \dfrac{k\,q_{\text{enc}}}{r^2} = \dfrac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{(0.20)^2} = \boxed{9.0 \times 10^5\ \mathrm{N/C}}$ , directed radially outward (positive total). Direction stated = mark secured.

Board level — surface densities for the neutral vault (ISC/NEET pattern)

A point charge $q = +4.0\ \mu\mathrm{C}$ sits at the centre of a neutral conducting shell, $a = 0.05\ \mathrm{m}$ , $b = 0.10\ \mathrm{m}$ . Find both surface charge densities, and the field at $r = 0.02$ and $0.07\ \mathrm{m}$ .

Step 1. Totals first: wall $-4.0\ \mu\mathrm{C}$ , outer $0 + 4.0 = +4.0\ \mu\mathrm{C}$ .

Step 2. Spread each total over its own sphere: $\sigma_a = \dfrac{-4.0 \times 10^{-6}}{4\pi (0.05)^2} = \boxed{-1.3 \times 10^{-4}\ \mathrm{C/m^2}}$ ; $\sigma_b = \dfrac{+4.0 \times 10^{-6}}{4\pi (0.10)^2} = \boxed{+3.2 \times 10^{-5}\ \mathrm{C/m^2}}$ . Same coulombs, four times the area, quarter the density — examiners love asking why $|\sigma_a| \ne |\sigma_b|$ .

Step 3. Field at $r = 0.02\ \mathrm{m}$ (cavity): only $q$ enclosed: $E = \dfrac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{(0.02)^2} = \boxed{9.0 \times 10^7\ \mathrm{N/C}}$ , outward. The vault does not protect the cavity from its own occupant.

Step 4. Field at $r = 0.07\ \mathrm{m}$ (metal): $\boxed{0}$ — by doctrine, or the count $+4 - 4 = 0$ . Give the reason, collect the mark.

Competitive level — a sphere inside a shell, region by region (JEE-Main pattern)

A solid conducting sphere of radius $R_1 = 0.05\ \mathrm{m}$ carrying $+2.0\ \mu\mathrm{C}$ sits at the centre of a concentric conducting shell of inner radius $0.10\ \mathrm{m}$ , outer radius $0.12\ \mathrm{m}$ , carrying net $-5.0\ \mu\mathrm{C}$ . Find (a) the field at $r = 0.08\ \mathrm{m}$ , (b) the charges on the shell's two surfaces, (c) the field at $r = 0.20\ \mathrm{m}$ . ( $k = 9.0 \times 10^9$ SI.)

Step 1. Field in the gap, $r = 0.08\ \mathrm{m}$ : enclosed is the inner sphere's $+2.0\ \mu\mathrm{C}$ only. $E = \dfrac{(9.0 \times 10^9)(2.0 \times 10^{-6})}{(0.08)^2} = \boxed{2.8 \times 10^6\ \mathrm{N/C}}$ , outward. The shell is invisible from the gap.

Step 2. Shell's inner surface: Gaussian sphere in the shell's metal ⇒ encloses zero ⇒ inner surface $= -(+2.0) = \boxed{-2.0\ \mu\mathrm{C}}$ .

Step 3. Shell's outer surface: conservation, $-5.0 = -2.0 + Q_{\text{outer}} \Rightarrow Q_{\text{outer}} = \boxed{-3.0\ \mu\mathrm{C}}$ .

Step 4. Field at $r = 0.20\ \mathrm{m}$ : enclosed total $+2.0 - 5.0 = -3.0\ \mu\mathrm{C}$ . $E = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{(0.20)^2} = \boxed{6.8 \times 10^5\ \mathrm{N/C}}$ , directed inward (negative total). Writing "outward" here is the single most common lost mark in this variant. Timing note: 3 minutes flat. Parts (b) and (c) are ledger lines — if you're summing fields from three surfaces separately, stop and count enclosed charge instead. The one place to slow down: the direction in (c).

Competitive level — off-centre charge and the earthing twist (JEE-Advanced flavour)

A point charge $q = +3.0\ \mu\mathrm{C}$ is fixed off-centre in the cavity of a neutral conducting shell with spherical outer surface of radius $0.15\ \mathrm{m}$ . (a) Judge each statement: (i) the inner-wall charge is $-3.0\ \mu\mathrm{C}$ , distributed non-uniformly; (ii) the outer-surface charge is uniform; (iii) the exterior field points along lines through the charge's actual position. (b) Find the exterior field magnitude at $r = 0.30\ \mathrm{m}$ from the centre. (c) The outer surface is now earthed — what changes, inside and out?

Step 1. (i) TRUE. The total is exactly $-3.0\ \mu\mathrm{C}$ (Gaussian in metal — position never entered the argument), crowded toward the nearby charge so the metal stays field-free: non-uniform.

Step 2. (ii) TRUE — amnesia. The cavity charge and its wall screen show zero net field to the metal and beyond, so the outer $+3.0\ \mu\mathrm{C}$ arranges as if alone on a sphere: uniformly.

Step 3. (iii) FALSE. Uniform outer skin ⇒ exterior field of a point charge at the centre of the shell, not at the charge's hiding spot. The world learns the total, never the address.

Step 4. (b) $E = \dfrac{k q}{r^2} = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{(0.30)^2} = \boxed{3.0 \times 10^5\ \mathrm{N/C}}$ , radially outward from the centre.

Step 5. (c) Earthing drains the outer surface: the exterior field collapses to $\boxed{0}$ , while inside nothing changes — the wall's $-3.0\ \mu\mathrm{C}$ is pinned by the cavity charge, and the cavity field never depended on the outer skin. Timing note: 3–4 minutes, mostly reading. Every sub-part is a verdict from the two bookkeeping laws plus amnesia — the only number is one $kq/r^2$ . If a statement has the charge's position affecting anything outside, it is false; train that reflex.

Exam traps

Forgetting the shell's own charge in the outer count. $Q_{\text{outer}} = Q + q$ , both terms, always. Writing $+q$ (ignoring the endowment) or $Q$ (ignoring the exile) are the two halves of the same dropped mark.

Sign slips on the wall. The wall takes the negative of the cavity charge — which is positive when the cavity charge is negative, as in the board example. Read twice, sign once.

Measuring the exterior field from the wrong point. After amnesia, $r$ in $k(Q+q)/r^2$ is measured from the shell's centre — never from an off-centre charge's actual position.

"The cavity is shielded, so E inside is zero." The cage protects the cavity from external sources only; the occupant's own field fills it at full strength, $kq/r^2$ right up to the wall. Zero belongs to the metal.

Earthing confusion. Earthing drains the outer charge only; the wall's $-q$ stays (it has a job). And "earthed" is not "neutral" — an earthed shell around $+q$ carries net $-q$ . Quantitative earthing needs potential: parked until Sphere Potentials and Earthing.

Adding surface fields instead of counting. Never sum three surfaces' fields separately; one Gaussian sphere and an enclosed-charge count is the superposition, pre-digested.

Density arithmetic. Each surface uses its own radius in $\sigma = Q_{\text{surface}}/4\pi r^2$ ; mixing $a$ and $b$ in one line is the classic under-pressure slip.

Direction left unstated. Negative enclosed total ⇒ inward field. Boards reserve a mark for it; MCQ setters build wrong options from the unsigned magnitude.

Origins · 1 minHenry Cavendish and how this was discovered

Henry Cavendish

1731–1810 · London

Cavendish was the strangest great scientist of his century — a millionaire so shy he ordered dinner by leaving notes — and one of the most exact experimenters who ever lived. In 1773 he charged a metal sphere suspended inside two closing hemispheres, connected them briefly with a wire, then opened the apparatus and tested the inner sphere: no charge whatsoever. He read the silence as a signature — only an inverse-square law empties a cavity so perfectly — and bounded the exponent to within about one part in fifty, a decade before Coulomb measured the law directly. Then, being Cavendish, he told almost no one; the work sat in his manuscripts until Maxwell edited and published them in 1879. The same silent-cavity logic, upgraded with electronics, still delivers the tightest tests of Coulomb's law — and the shell problem you just mastered is his apparatus, redrawn for an answer sheet.

History

1755

— Benjamin Franklin reports that a cork ball hanging inside a charged silver can feels no pull toward the walls.

1767

— Joseph Priestley reads Franklin's result the right way: a force-free cavity is the fingerprint of an inverse-square law.

1773

— Cavendish's concentric-sphere experiment turns the silence into a measurement: exponent 2 to within about 2 percent — unpublished for a century.

1843

— Faraday's ice-pail experiment shows the induced charge exactly equals the enclosed charge: equation (1) with Victorian hardware.

1879

— Maxwell publishes Cavendish's electrical manuscripts, revealing how far ahead the silent millionaire had been.

1936

— Plimpton and Lawton's rerun of the cavity test confirms the inverse square to one part in a billion.

Going further

Texts. NCERT Physics Class 12, electrostatics chapters (shell problems and shielding) · HC Verma, Concepts of Physics Vol. 2, Gauss's law exercises · Full story with potential theory: Induced Charges and Shielding (UG deep dive).

Provenance. Standard Gauss's-law shell doctrine, cross-checked against NCERT and the UG source above. A verifier should recompute: board 1 — wall $+2.0$ , outer $+4.0\ \mu\mathrm{C}$ , $E(0.20) = 9.0 \times 10^5$ N/C out; board 2 — $\sigma_a = -1.27 \times 10^{-4}$ , $\sigma_b = +3.18 \times 10^{-5}\ \mathrm{C/m^2}$ , $E(0.02) = 9.0 \times 10^7$ , $E(0.07) = 0$ ; competitive 1 — shell skins $-2.0/-3.0\ \mu\mathrm{C}$ , $E(0.08) = 2.8 \times 10^6$ out, $E(0.20) = 6.8 \times 10^5$ N/C in; competitive 2 — $E(0.30) = 3.0 \times 10^5$ N/C, zero after earthing. Cavendish 1773: exponent 2 within $\sim 1/50$ ; Plimpton–Lawton 1936: $\sim 10^{-9}$ .

Prerequisites. Gauss's Law at Work: Wire, Sheet and Shell · Electrostatics of Conductors and Shielding. Leads to. Sphere Potentials and Earthing (coming in the next chapter) · Induced Charges and Shielding (UG deep dive).

Why is the induced wall charge exactly −q and not a little less?

Because the flux through a Gaussian surface in the metal is identically zero — not small, zero — so the enclosed charge must cancel to the last electron. There is no efficiency factor, no dependence on the conductor's quality or the shell's thickness. If the wall held only $-0.99q$ , one percent of the field lines from $q$ would survive into the metal, push electrons, and the rearrangement would continue until they too were terminated. The exactness is the equilibrium condition wearing Gauss's law as a uniform.

If no field crosses the metal, how does the outside world find out about the cavity charge at all?

Through conservation, not transmission. The wall's $-q$ had to come from somewhere: it was separated out of the neutral metal, leaving $+q$ behind, and that orphan's only address is the outer surface — where it radiates a perfectly ordinary field. No field line from the cavity ever reaches outside; the message is carried by the ledger. That is why earthing censors it so cleanly: give the orphan an escape route and the outside hears nothing, while the cavity's story continues unchanged.

Does the wall charge follow the cavity charge around if I move it?

Continuously and essentially instantly — the metal re-settles in about $10^{-19}$ s. Bring $q$ near one side and the wall's negative charge crowds toward it, always in exactly the arrangement that keeps the metal field-free; the total stays $-q$ throughout. The outer surface never budges: the cavity-plus-wall system shows zero field to the rest of the world, so the outer skin has no news to respond to. Wave the charge around inside like a lantern — the exterior field stands perfectly still.

When the shell is earthed, which charge leaves, and why not the wall's?

The outer surface's charge drains, because it is the only charge with both a reason to leave (mutual repulsion, no duty) and a route (the ground wire touches the outside). The wall's $-q$ has a job it cannot abandon: terminating the cavity charge's field lines so the metal stays silent. Remove any part of it and a field would penetrate the metal, dragging electrons back within $10^{-19}$ s. So earthing performs surgery on exactly one surface. The quantitative version — how much flows when geometries get complicated — is a potential calculation, coming in Sphere Potentials and Earthing.

Is the charge inside the cavity protected from fields outside the shell?

Yes — the Faraday-cage theorem from Electrostatics of Conductors and Shielding runs simultaneously with everything here. An external storm rearranges charge on the outer surface only; the cavity and its occupant feel nothing. The wall is a two-directional referee — it terminates the occupant's lines (inward duty) while the outer skin absorbs the world's lines (outward duty) — and the jobs never interfere, so you can superpose the answers: cavity = occupant's own field; metal = zero; outside = ledger total plus any external field.

What if there are two cavities, each with its own charge?

Each wall does its own bookkeeping: cavity 1's wall takes $-q_1$ , cavity 2's takes $-q_2$ — the Gaussian argument runs separately around each. Conservation piles the combined exile onto the single outer surface: $Q_{\text{outer}} = Q + q_1 + q_2$ . Outside, the world sees the grand total, centred (for a spherical outer surface), with no hint of how the charge is divided or where the cavities sit. More rooms in the vault, same ledger.