Gauss's Law at Work: Wire, Sheet & Shell

In 1766, Benjamin Franklin wrote to Joseph Priestley about something odd: cork balls dangled inside a charged metal cup felt no pull at all. Priestley saw what Franklin didn't — this "no force inside" was the signature Newton had proved for gravity inside a hollow shell, so electricity must obey an inverse-square law too. A null result inside a shell revealed the exponent of the force law. That is the calibre of result this module hands you. The previous module forged Gauss's law; this one weaponises it. There are exactly three charge geometries where the law's flux sum collapses into one line of algebra: the infinitely long wire, the infinite flat sheet, and the uniformly charged shell (or sphere). Master the ritual for each — declare the symmetry, choose the matching Gaussian surface, count the enclosed charge — and you own three formulas that appear on CBSE and ISC papers nearly every year, plus the derivation questions that go with them. More than formulas, you get a pattern language: the three results fall off as $1/r^2$ , $1/r$ and constant, and each exponent is a story about how field lines find room to spread. Read them that way once and you will never mis-remember which formula belongs to which geometry. The Shells & Cavities archetype and the Conductors module build directly on what you derive here.

Key takeaways

Gauss's law becomes a calculating machine in exactly three symmetric geometries: infinite wire, infinite sheet, spherical shell/sphere.
Wire: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ — falls as $1/r$ ; Gaussian surface is a coaxial cylinder.
Sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$ — same at every distance; Gaussian surface is a pillbox straddling the sheet.
Shell: outside, $E = \dfrac{kQ}{r^2}$ exactly as if all charge sat at the centre; inside, $E = 0$ everywhere.
Two parallel sheets with $+\sigma$ and $-\sigma$ : field $\sigma/\varepsilon_0$ between, zero outside — the capacitor's field, free of charge.
The ritual is always the same four beats: symmetry → surface → flux collapse → enclosed charge.
Key idea: symmetry fixes the field's direction and evens out its magnitude, so the flux sum collapses to E × area — and each geometry's falloff power counts how much room its lines have to spread.

The core insight

For the wire, sheet and shell, symmetry makes the field uniform over a matching Gaussian surface, collapsing Gauss's law to $E \times (\text{pierced area}) = q_{\text{enc}}/\varepsilon_0$ — giving $\lambda/2\pi\varepsilon_0 r$ , $\sigma/2\varepsilon_0$ , and $kQ/r^2$ -outside-zero-inside in three lines each.

Where this lives today

The three formulas of this module are working engineering documents. Every coaxial cable spec rests on the cylindrical pattern: the field between core and shield follows $\lambda/2\pi\varepsilon_0 r$ (times a dielectric factor), fiercest at the inner conductor — exactly where insulation is rated. The sheet result runs the parallel-plate geometry inside every capacitor and the gate of every transistor. And the shell's silence remains frontier metrology. The Priestley–Cavendish null test, in modern dress — charge a conducting shell, hunt for any interior field — bounds deviations of Coulomb's exponent from 2 at the $10^{-16}$ level, which doubles as the best laboratory limit on the photon's mass. A result first read off dangling cork balls now guards the foundations of electromagnetism.

The physical picture

Why these three geometries and no others? Because Gauss's law hands you only one number per surface — the total flux — and one number can only determine one unknown. You need the field to have a single, evened-out magnitude over your surface before the law can find it. That evenness is what symmetry supplies. Stand beside an infinite wire: every direction along it looks the same, so the field cannot lean left or right — it must point straight out from the axis, with a strength depending only on distance. Hover above an infinite sheet: no sideways direction is special, so the field points straight away from the sheet. Stand outside a uniform shell: no direction from the centre is special, so the field is radial. In each case symmetry has done nine-tenths of the work before any algebra — it fixed the field's shape, leaving only its size to find.

Figure

The symmetry trio — each geometry dictates its field pattern and the Gaussian surface that matches it

The three answers fall off with distance in three different ways, and there is a single idea behind the pattern: how much new room do the field lines get as they travel? A point charge's lines spread over ever-growing spheres — area $\propto r^2$ — so the field dies as $1/r^2$ . A wire's lines spread only sideways, over cylinders — area $\propto r$ — so the field dies as $1/r$ . A sheet's lines leave perpendicular and stay parallel forever — no new room at all — so the field never weakens. Each dimension of source removes one power of falloff. That one sentence is worth three memorised formulas.

The ritual and the three results

Every application runs the same four beats. Beat 1 — symmetry: state the field's direction and dependence in one sentence. Beat 2 — surface: choose the Gaussian surface the field pierces squarely (constant magnitude) or skims (zero contribution). Beat 3 — collapse: write flux $= E \times (\text{pierced area})$ . Beat 4 — count: find $q_{\text{enc}}$ and solve

E \times (\text{pierced area}) = \frac{q_{\text{enc}}}{\varepsilon_0} \qquad (1)

where: the "pierced area" is only the part of the Gaussian surface the field actually crosses head-on — the curved wall of the cylinder, the two caps of the pillbox, the whole concentric sphere. Equation (1) is the previous module's solver mode; what is new here is the professional's checklist of when and how it applies. For the charge count you need the density language from Continuous Charge Distributions: $\lambda$ (charge per length, C/m) for the wire and $\sigma$ (charge per area, C/m²) for the sheet and shell. The three trophies, before the honest work:

E_{\text{wire}} = \frac{\lambda}{2\pi\varepsilon_0 r}, \qquad E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \qquad E_{\text{shell,out}} = \frac{kQ}{r^2} \quad (E_{\text{shell,in}} = 0) \qquad (2)

where: $r$ is the perpendicular distance from the wire's axis (first formula) or the distance from the shell's centre (third); $k = 1/4\pi\varepsilon_0 = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ . Read each through the spreading-room lens: $1/r$ for the wire (lines spread over cylinders), no $r$ at all for the sheet (parallel lines never spread), $1/r^2$ outside the shell (from far away it is just a point charge $Q$ ) — and the shell's silent interior, the result that told Priestley the force law's exponent.

Derivation AField of an infinite line of charge

We want the field at perpendicular distance $r$ from an infinitely long, uniformly charged wire with linear density $\lambda$ — the derivation CBSE and ISC ask for more often than any other in this chapter. The strategy is the four-beat ritual with a coaxial cylinder: the curved wall catches all the flux, the flat caps none. Watch for the arbitrary cylinder length $L$ cancelling at the end — if it doesn't, the surface was drawn wrong.

Step 1. Symmetry. The wire is infinite and uniform, so no point along it and no direction around it is special. The field cannot tilt along the wire (which way would it choose?) and cannot swirl around it — it must point straight away from the axis, with a magnitude depending only on the perpendicular distance $r$ .

Step 2. Choose the surface: an imaginary coaxial cylinder of radius $r$ and length $L$ , closed by two flat caps. On the curved wall, the field pierces every patch head-on with the same magnitude $E$ (every point of the wall is the same distance from the wire). On the caps, the field lies in the cap plane — it skims, contributing zero flux.

Step 3. Collapse the flux. Curved wall area is (circumference) × (length) $= 2\pi r L$ :

\Phi = E \times 2\pi r L + 0

Why the caps really give zero: their normal points along the wire, and the field is perpendicular to the wire everywhere — cosine of 90° kills the contribution.

Step 4. Count and solve. The cylinder swallows a length $L$ of wire: $q_{\text{enc}} = \lambda L$ . Gauss's law:

E \times 2\pi r L = \frac{\lambda L}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{\lambda}{2\pi\varepsilon_0 r}

The bookkeeping length $L$ cancels — as it must, since an infinite wire offers no way to prefer one $L$ over another. The field falls as $1/r$ : slower than a point charge's $1/r^2$ , because a wire's lines spread over cylinders whose area grows only as the first power of $r$ . Verified: for $\lambda = 5.0\ \mu\mathrm{C/m}$ at $r = 0.20\ \mathrm{m}$ : $E = 2k\lambda/r = 2(9.0\times10^9)(5.0\times10^{-6})/0.20 = 4.5\times10^5\ \mathrm{N/C}$ — matching the worked example below; the $2k\lambda/r$ form is algebraically identical to $\lambda/2\pi\varepsilon_0 r$ .

Go deeper · Derivation B · 1 minField of an infinite sheet of charge

Derivation BField of an infinite sheet of charge

We want the field of an infinite flat sheet with uniform surface density $\sigma$ — the derivation with the famous factor of 2. Strategy: symmetry makes the field perpendicular to the sheet on both sides, and a pillbox (a small closed cylinder) straddling the sheet catches flux through both caps. The 2 is manufactured honestly in Step 3 — watch it arrive.

Step 1. Symmetry. Above an infinite sheet, no sideways direction is special, so the field must point straight away from the sheet (for $\sigma > 0$ ), with equal magnitudes at equal distances on the two sides — the sheet's mirror symmetry demands it.

Step 2. Choose the surface: a pillbox of cap area $A$ , half above and half below the sheet, caps parallel to it. The field pierces both caps head-on and outward; it skims the pillbox's side wall.

Step 3. Collapse the flux — both caps count:

\Phi = EA + EA + 0 = 2EA

Why both are positive: on the top cap, field and outward normal both point up; on the bottom cap, both point down. Each cap has field leaving the box. Nothing cancels — this is where the 2 comes from.

Step 4. Count and solve. The pillbox clips a disc of sheet of area $A$ : $q_{\text{enc}} = \sigma A$ .

2EA = \frac{\sigma A}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{\sigma}{2\varepsilon_0}

The area cancels, and — read it twice — so does the distance. The field is the same one millimetre from the sheet and one metre from it. Parallel lines never dilute: an infinite sheet's lines leave perpendicular and have no room to spread into.

Step 5. The classic superposition payoff. Two parallel sheets, $+\sigma$ and $-\sigma$ : between them the two fields point the same way and add to $\sigma/\varepsilon_0$ ; outside they oppose and cancel to zero. That is the parallel-plate capacitor's field — derived before you ever meet a capacitor. Verified: for $\sigma = 17.7\ \mathrm{nC/m^2}$ : $E = (17.7\times10^{-9})/(2 \times 8.85\times10^{-12}) = 1.0\times10^3\ \mathrm{N/C}$ ; between the paired sheets, double it: $2.0\times10^3\ \mathrm{N/C}$ . Distance-independence holds for real plates at distances small compared to the plate size — the capacitor regime.

Go deeper · Derivation C · 2 minField of a uniformly charged spherical shell — outside and inside

Derivation CField of a uniformly charged spherical shell — outside and inside

We want the shell's field everywhere: the "point-charge impostor" outside and the celebrated zero inside — the shell theorem that took Newton a decade for gravity, plus the null result that let Priestley infer the inverse square. Strategy: spherical symmetry certifies a radial field depending only on $r$ ; concentric Gaussian spheres collapse the flux; outside and inside differ only in the enclosed charge. Watch the whole drama live in $q_{\text{enc}}$ .

Step 1. Symmetry. A uniform shell looks identical from every direction, so its field must be radial with magnitude $E(r)$ — any sideways lean would pick a direction the shell does not possess. This holds inside the shell as well as outside.

Step 2. Choose concentric Gaussian spheres of radius $r$ . On each, the field pierces every patch head-on with one magnitude:

\Phi = E \times 4\pi r^2

Step 3. Outside ( $r > R$ ): the Gaussian sphere swallows the whole shell, $q_{\text{enc}} = Q$ :

E \times 4\pi r^2 = \frac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} = \frac{kQ}{r^2}

Why this deserves its nickname: it is exactly the field of a point charge $Q$ at the centre — not approximately, exactly. From outside, the shell's radius is invisible. Any spherically symmetric charge distribution earns the same verdict.

Step 4. Inside ( $r < R$ ): the Gaussian sphere encloses nothing — all the charge sits farther out on the shell itself. So $q_{\text{enc}} = 0$ :

E \times 4\pi r^2 = 0 \quad\Rightarrow\quad E = 0 \quad\text{everywhere inside}

Why zero and not merely small: both premises matter. Zero enclosed charge makes the flux zero; spherical symmetry makes the field one constant number over the Gaussian sphere; a constant times a nonzero area equalling zero forces the constant to vanish. Physically, the near side of the shell pulls harder but the far side has more charge in view — and for an exact inverse-square law the two effects cancel perfectly at every interior point.

Step 5. For a solid uniformly charged sphere (a JEE favourite), only Step 4's count changes: a Gaussian sphere at $r < R$ encloses the fraction $(r/R)^3$ of the charge, giving $E = kQr/R^3$ — a linear ramp from zero at the centre to a peak at the surface, handing over to $kQ/r^2$ outside. The branches agree at $r = R$ . Verified: for $Q = 4.0\ \mu\mathrm{C}$ , $R = 10\ \mathrm{cm}$ : outside at $r = 30\ \mathrm{cm}$ , $E = (9.0\times10^9)(4.0\times10^{-6})/0.09 = 4.0\times10^5\ \mathrm{N/C}$ ; inside, $0$ . Solid-sphere check ( $Q = 80\ \mathrm{nC}$ , $R = 10\ \mathrm{cm}$ ): $E = 3.6\times10^4$ at $r = 5\ \mathrm{cm}$ , $7.2\times10^4$ at the surface, $1.8\times10^4$ at $20\ \mathrm{cm}$ — linear then inverse-square, meeting at the peak.

Figure

The shell's field profile — silence inside, point-charge impostor outside; the solid sphere adds a linear ramp

Which formula answers which problem

you face	symmetry	Gaussian surface	result	falloff
long straight wire, cable	cylindrical	coaxial cylinder	$\lambda/2\pi\varepsilon_0 r$	$1/r$
large flat sheet, plate	planar	straddling pillbox	$\sigma/2\varepsilon_0$	constant
shell, hollow sphere, outside	spherical	concentric sphere	$kQ/r^2$	$1/r^2$
shell, inside	spherical	concentric sphere	$0$	—
solid sphere, inside	spherical	concentric sphere	$kQr/R^3$	$\propto r$
two plates $+\sigma$ , $-\sigma$	superpose sheets	—	$\sigma/\varepsilon_0$ between, $0$ outside	capacitor
finite rod, ring, disc, cube	none usable	—	back to Coulomb/superposition	—

Limiting cases

case	result	meaning
very far from any of the three	$E \to kQ_{\text{total}}/r^2$	from far away, everything is a point charge
very close to any charged surface	sheet-like behaviour	up close, every surface looks flat and infinite
shell interior, any point	$E = 0$	the Priestley/Cavendish null signature
solid sphere centre	$E = 0$	equal pulls from all sides
wire at doubled distance	$E$ halves	$1/r$ , not $1/r^2$ — cylinders, not spheres
sheet at doubled distance	$E$ unchanged	parallel lines never spread

Common trapThe formulas are universal — just plug into whichever looks closest

Each formula is welded to its symmetry, and transplanting one is the chapter's costliest error. The wire formula holds only while the wire looks effectively infinite — near the ends, or far from a short rod, the field tilts and the formula lies. The sheet formula likewise: valid close to a large plate, dead wrong at distances comparable to the plate's size (where the plate shrinks toward a point charge and $1/r^2$ takes over). And the shell's interior zero is for a uniform shell's own field — an external charge's field passes into a non-conducting shell untouched (a conducting shell is a different, better story — the next module's). Before every plug-in, ask: does my geometry actually have this symmetry, at this distance?

Worked examples

Board level — deriving to order, then the NCERT wire

(a) Using Gauss's law, derive the expression for the electric field due to an infinitely long straight uniformly charged wire. (b) An infinite line charge produces a field of $9\times10^4\ \mathrm{N/C}$ at a distance of $2\ \mathrm{cm}$ . Calculate the linear charge density. (CBSE pattern — the derivation is the marks; the numerical is NCERT's own.)

Step 1. For part (a), reproduce the four beats of the first derivation above — symmetry sentence, coaxial cylinder, caps-skim-wall-pierces, $q_{\text{enc}} = \lambda L$ — arriving at $E = \lambda/2\pi\varepsilon_0 r$ . In board marking, the symmetry sentence and the diagram of the cylinder carry dedicated marks; the algebra alone does not fetch full credit.

Step 2. Part (b): solve the formula for $\lambda$ :

\lambda = 2\pi\varepsilon_0\, r E = \frac{rE}{2k}

Step 3. Substitute in SI ( $r = 0.02\ \mathrm{m}$ ):

\lambda = \frac{(0.02)(9\times10^4)}{2(9.0\times10^9)} = \boxed{1.0\times10^{-7}\ \mathrm{C/m} = 0.1\ \mu\mathrm{C/m}}

Step 4. Sense check: modest field, centimetre distance — a tenth of a microcoulomb per metre is a plausible laboratory-scale density. (If you get $10^7$ , the cm→m conversion was dropped.)

Competitive level — inside and outside a charged sphere (JEE-Main pattern)

A solid non-conducting sphere of radius $R = 10\ \mathrm{cm}$ carries charge $Q = 80\ \mathrm{nC}$ distributed uniformly through its volume. Find the electric field at (a) $r = 5\ \mathrm{cm}$ , (b) the surface, (c) $r = 20\ \mathrm{cm}$ . (d) At what interior radius does the field equal its value at $20\ \mathrm{cm}$ ?

Step 1. Precompute the reusable constant: $kQ = (9.0\times10^9)(80\times10^{-9}) = 720\ \mathrm{N\,m^2/C}$ .

Step 2. (b) Surface first — it anchors everything: $E_R = kQ/R^2 = 720/(0.10)^2 = \boxed{7.2\times10^4\ \mathrm{N/C}}$ . This is the profile's peak.

Step 3. (a) Inside, the linear ramp: $E = kQr/R^3 = E_R \times (r/R) = 7.2\times10^4 \times 0.5 = \boxed{3.6\times10^4\ \mathrm{N/C}}$ . Half the radius, half the surface field — the ramp makes interior values pure ratio work.

Step 4. (c) Outside, the impostor: $E = kQ/r^2 = 720/(0.20)^2 = \boxed{1.8\times10^4\ \mathrm{N/C}}$ — one quarter of the surface value at double the distance, as $1/r^2$ demands.

Step 5. (d) Set the ramp equal to $1.8\times10^4$ : $E_R(r/R) = E_R/4 \Rightarrow r = R/4 = \boxed{2.5\ \mathrm{cm}}$ . Every field value below the peak is reached twice — once on the ramp, once on the tail; JEE loves this double-crossing. Timing note: 3 minutes for all four parts. The efficiency move is Step 1's $kQ$ plus ratio reasoning — after the surface value, parts (a), (c), (d) are one-line proportions, no fresh substitutions.

Exam traps

The factor of 2 family. Isolated sheet: $\sigma/2\varepsilon_0$ . Between oppositely charged parallel plates: $\sigma/\varepsilon_0$ . Just outside a charged conductor: $\sigma/\varepsilon_0$ (the next module derives why). Three different questions, three different answers — misfiling one costs the whole numerical. Anchor: one lonely sheet splits its flux both ways.

Wrong falloff transplant. Writing $kλ/r^2$ for a wire or expecting the sheet's field to weaken with distance. Recite the spreading-room story: point $1/r^2$ , wire $1/r$ , sheet constant.

Applying the shell's interior zero to a solid sphere. Inside a shell: zero. Inside a solid sphere: $kQr/R^3$ , growing linearly. The word "hollow"/"solid" in the question is load-bearing — underline it.

Forgetting the caps or the second cap. In derivations, CBSE explicitly awards the statement that cap flux vanishes (wire) or that both caps contribute (sheet). Write those sentences.

Densities mixed up. $\lambda$ is C/m, $\sigma$ is C/m². Convert the given quantity to the density the formula wants before substituting — a wire problem stated via total charge and length needs $\lambda = Q/L$ first.

Constants discipline. $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ and $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/N\,m^2}$ are the accepted board values; JEE supplies what it wants used. The handy hybrid for wires: $E = 2k\lambda/r$ .

Origins · 1 minJoseph Priestley and how this was discovered

Joseph Priestley

1733–1804 · Leeds and Birmingham

Priestley — dissenting minister, discoverer of oxygen, Benjamin Franklin's scientific correspondent — pulled the inverse-square law of electricity out of a null result. In 1766, at Franklin's suggestion, he repeated the charged-cup experiment: cork balls suspended inside a charged metal vessel felt no force whatsoever. Priestley recognised the fingerprint. Newton had proved a hollow shell of matter exerts no gravity inside it — a cancellation that works only for an exact inverse square. If charge in a cup exerts no force inside, then electricity, "like gravity, obeys the law of the inverse square of the distances." He published the inference in 1767, eighteen years before Coulomb's torsion balance measured the law directly — making the shell's silent interior, the very result you derived on this page, the first evidence for the equation that opens your syllabus.

History

1687

— Newton's shell theorem: no gravity inside a uniform shell — the template for every null test to come

1766

— Franklin's charged-cup observation; Priestley infers the inverse-square law from the force-free interior (published 1767)

1773

— Cavendish's concentric-sphere experiment sharpens the null test of the exponent to 2 ± 0.02, unpublished until 1879

1813

— Gauss proves the flux–source theorem that turns all such null results into three-line derivations

1936

— Plimpton and Lawton's shell experiment pins the exponent to within one part in a billion

Going further

Texts. NCERT Physics Part 1 (Class XII), Ch. 1 §1.15 — all three derivations in board-expected form · H.C. Verma, Concepts of Physics Part 2, Ch. 30 · Halliday, Resnick & Walker Ch. 23 · Gauss's Law Applications and Symmetry (UG deep dive) for the full symmetry arguments and the coaxial-cable design worked end to end.

Provenance. All results standard (NCERT §1.15; Griffiths Examples 2.1, 2.3, 2.5 downshifted). A verifier should recompute: $\lambda = rE/2k = (0.02)(9\times10^4)/(1.8\times10^{10}) = 1.0\times10^{-7}\ \mathrm{C/m}$ ; $E = 2k\lambda/r = 4.5\times10^5\ \mathrm{N/C}$ for $5\ \mu\mathrm{C/m}$ at $20\ \mathrm{cm}$ ; $\sigma/2\varepsilon_0 = 1.0\times10^3\ \mathrm{N/C}$ for $\sigma = 17.7\ \mathrm{nC/m^2}$ ; shell $E = 4.0\times10^5\ \mathrm{N/C}$ for $4\ \mu\mathrm{C}$ at $30\ \mathrm{cm}$ ; solid-sphere trio $3.6/7.2/1.8\times10^4\ \mathrm{N/C}$ and the matching radius $R/4$ .

Prerequisites. Gauss's Law · Continuous Charge Distributions. Leads to. Concentric Shells & Cavity Charges · Symmetry Kills Options · Gauss's Law Applications and Symmetry (UG deep dive).

Why exactly three geometries? Why can't a clever surface handle, say, a charged cube?

Because the law provides one equation, and one equation finds one unknown. For a cube, no closed surface exists on which the cube's field has a single constant magnitude — near faces, edges and corners the field differs in strength and direction, so the flux sum contains an unknown pattern, not an unknown number. No cleverness in surface-drawing fixes that; the symmetry belongs to the source, not to your surface. Sphere-like, wire-like, sheet-like — that is the complete menu, and recognising "not on the menu" instantly is itself an exam skill.

How can the sheet's field really not weaken with distance? That seems unphysical.

For a truly infinite sheet, walk away and the view is unchanged — nothing exists to measure your distance against, so the field has nothing to depend on. The lines leave perpendicular and stay parallel; parallel lines never spread; no spreading, no weakening. The idealisation earns its keep as a close-up approximation: near a real plate, the edges are effectively at infinity and $\sigma/2\varepsilon_0$ is excellent — precisely the regime inside a capacitor. Far away, the same plate looks like a point charge and $1/r^2$ resumes. Near view: sheet formula; far view: point-charge formula.

Where does the factor of 2 in the sheet formula come from, and why does it vanish between capacitor plates?

It comes from the pillbox catching flux through both caps: a lonely sheet pushes field out of both its faces, so its budget $\sigma A/\varepsilon_0$ is split two ways — $\sigma/2\varepsilon_0$ per side. Between two plates carrying $+\sigma$ and $-\sigma$ , each plate still contributes its $\sigma/2\varepsilon_0$ , but in the gap the two contributions point the same way (away from +, toward −) and add: $\sigma/\varepsilon_0$ . Outside the pair they oppose and cancel. No formula was repealed — superposition just did its job. Tallying "how many half-fields point which way" resolves every plate configuration an exam can throw.

Is the field inside a shell zero even if I sit very close to the inner wall?

Yes — spectacularly, yes. A millimetre from the wall, the nearby patch of charge pulls hard; but the entire far side, though weaker per unit charge, has vastly more charge in view, and for an exact inverse square the two cancel perfectly at every interior point, wall-hugging included. This is not approximate — it is so sensitive to the exponent being exactly 2 that experiments use this very configuration to test Coulomb's law. No interior field has ever been found.

Does the shell's E = 0 inside still hold if there are charges outside the shell?

Careful — this is where the fine print lives. The shell's own field is zero inside, always. But Gauss's law never cancels the fields of outside charges (only their net flux), so an external charge's field passes straight through a charged non-conducting shell and is alive inside it. The total interior field is then the outsider's field plus the shell's zero. A conducting shell is different: its mobile charges rearrange to cancel external fields inside — genuine shielding, with a different mechanism. That story, and why it makes your car a lightning shelter, is the next module's.

For the wire, why must the Gaussian cylinder be coaxial? Could I use an off-centre cylinder or a sphere?

The law would still be true on those surfaces — flux would still equal $q_{\text{enc}}/\varepsilon_0$ — but useless, because the field would no longer have one magnitude over the surface. On an off-centre cylinder, patches near the wire see a stronger field than distant patches; on a sphere, the field isn't even perpendicular to the surface. Only the coaxial cylinder matches the source's symmetry: every point of its curved wall is equivalent. The Gaussian surface never creates the collapse — it inherits it from the symmetry.

The solid sphere's field grows with r inside. Shouldn't the field always decrease as I move away from charge?

Moving outward inside the sphere, you are not moving away from the charge — more of it accumulates beneath you. Enclosed charge grows as $r^3$ , the Gaussian sphere's area only as $r^2$ ; the ratio grows as $r$ . At the centre, equal pulls cancel: zero. At the surface, everything is finally beneath you, and the ordinary $1/r^2$ decline begins. The surface peak is the crossover between "gaining charge" and "losing proximity."