Applications of Gauss's Law — Cylinders and Cavities

In short

For distributions with cylindrical symmetry, Gauss's law with a coaxial cylindrical Gaussian surface yields \vec{E} in a single line. The essential results you need to have memorised:

Infinite line of charge (linear density \lambda C/m): \;E(r) = \dfrac{\lambda}{2\pi\varepsilon_0 r}\; radially outward, for every r>0.
Infinite solid cylinder (radius R, uniform volume density \rho): \;E(r) = \dfrac{\rho r}{2\varepsilon_0} for r\le R (linear rise), \;E(r) = \dfrac{\rho R^2}{2\varepsilon_0 r} for r\ge R (equivalent to a line of density \lambda = \rho\pi R^2 on the axis).
Coaxial cable (inner conductor at +\lambda per length, outer shield at -\lambda): field is \lambda/(2\pi\varepsilon_0 r) between the conductors and zero everywhere outside the shield. This is why coaxial cables do not radiate.

Cavities and Faraday cages. The field inside any empty cavity scooped out of a conductor in electrostatic equilibrium is exactly zero, no matter what charges sit outside the conductor or what shape the cavity has. This is the Faraday cage principle — the reason the metal body of a car shields its occupants from a lightning strike, the reason TIFR test chambers are wrapped in grounded mesh, and the reason the shielded enclosure inside every Indian microwave oven keeps the 2.45 GHz radiation from reaching your kitchen.

For non-uniform cylindrical distributions \rho = \rho(r), Gauss's law still gives the field in one line, provided \rho depends only on r (still cylindrically symmetric) — the enclosed charge just needs a small integral Q_\text{enc} = \int_0^r \rho(r')\cdot 2\pi r'\,L\,dr'.

Strip the grey plastic jacket off a TV cable running from the DTH dish on your Mumbai flat's balcony into your television. Inside, you will find a thin copper wire running along the cable's axis, wrapped in a foamed plastic insulator, surrounded by a woven braid of fine copper strands that forms a cylindrical mesh around the inner wire. Around that, another plastic jacket. This is a coaxial cable — "co-axial" because the inner wire and the outer braid share the same axis.

Why build a cable with two conductors nested like that? Two equal and opposite currents run through them — the signal going one way along the inner wire, the return current going the other way along the outer braid. If you apply Gauss's law to any Gaussian cylinder drawn outside the cable, the enclosed net charge per unit length is zero — the inner +\lambda cancels the outer -\lambda. So the external field is exactly zero everywhere. The signal travels the length of the cable without radiating outward; the cable does not interfere with your phone, your WiFi, or the neighbour's TV. Coaxial cables work because of the cylindrical form of Gauss's law.

This chapter is the second half of the "applications of Gauss's law" pair. The first half covered spherical and planar symmetry — solid spheres, shells, infinite planes, parallel-plate capacitors. This half handles the remaining two categories: cylindrical symmetry (infinite line and cylinder, hollow tube, coaxial cable) and cavities in conductors (the Faraday cage and why charges on a conductor live entirely on the outer surface).

The underlying law is the same one — equation (1) in the Gauss's law article:

\oint_S \vec{E}\cdot d\vec{A} \;=\; \frac{Q_\text{enc}}{\varepsilon_0}.

All that changes between this chapter and the previous is the shape of the Gaussian surface, which is determined by the symmetry of the charge distribution. For cylinders, the surface is a coaxial cylinder (a soda can with its axis along the line of charge, both ends closed). For cavities, the surface sits entirely inside the conductor, where \vec{E} = 0 — and that single observation, combined with Gauss's law, gives the entire theory.

The cylindrical Gaussian surface

A distribution has cylindrical symmetry about an axis \hat{z} if it is invariant under (i) any rotation about \hat{z} and (ii) any translation along \hat{z}. An infinitely long charged straight rod has this symmetry; so does an infinitely long straight solid cylinder with uniform or radially-varying density, and any stacked configuration of concentric charged cylinders.

For any such distribution, the electric field at a point depends only on the perpendicular distance r from the axis, and must point radially outward (or inward) — it cannot have a component along the axis \hat{z} (translational symmetry forbids that; every point along \hat{z} looks identical) and cannot have an azimuthal component (rotational symmetry forbids that; every rotation about the axis leaves the distribution unchanged).

\vec{E}(r,\phi,z) \;=\; E(r)\,\hat{r}.

Why: the symmetries of the distribution force the field to respect them. Any component of \vec{E} that is not radial would pick out a preferred direction not present in the distribution — impossible.

The Gaussian surface that respects this symmetry is a coaxial cylinder of radius r and some length L, with two circular end caps closing it at the top and bottom. On such a surface:

The curved lateral surface has outward normal \hat{r} everywhere — parallel to \vec{E}. The flux through it is E(r) \cdot (\text{lateral area}) = E(r) \cdot 2\pi r L.
The two flat end caps have outward normals along \pm \hat{z} — perpendicular to \vec{E}. The flux through each end is \vec{E}\cdot\hat{z}\,\text{(area)} = 0.

So the total flux through the Gaussian cylinder is

\oint \vec{E}\cdot d\vec{A} \;=\; E(r)\cdot 2\pi r L. \tag{1}

This one formula is the left-hand side of Gauss's law for every cylindrical problem. Only Q_\text{enc} changes from problem to problem.

The Gaussian cylinder: lateral surface with radial outward normal, parallel to $\vec{E}$; end caps with axial outward normal, perpendicular to $\vec{E}$. Only the lateral surface contributes flux.

Application 1 — the infinite line of charge

Take an infinite, straight line of charge along the z-axis with uniform linear charge density \lambda C/m (positive, say). Find \vec{E} at distance r from the line.

Step 1. Symmetry identified: cylindrical. \vec{E} is radial, depends only on r.

Step 2. Gaussian surface: coaxial cylinder of radius r, length L (arbitrary — L will cancel in the final step).

Step 3. Flux (from equation (1)):

\oint \vec{E}\cdot d\vec{A} \;=\; E(r)\cdot 2\pi r L.

Step 4. Enclosed charge: a length L of the line sits inside the Gaussian cylinder.

Q_\text{enc} \;=\; \lambda L.

Why: the line has charge \lambda per unit length, and a length L of it lies inside the Gaussian surface. (The rest of the line, extending to \pm\infty, is outside the Gaussian cylinder and contributes nothing to the flux.)

Step 5. Gauss's law and solve.

E(r)\cdot 2\pi r L \;=\; \frac{\lambda L}{\varepsilon_0} \quad\Longrightarrow\quad \boxed{\;E(r) \;=\; \frac{\lambda}{2\pi\varepsilon_0 r}\;} \tag{2}

Why: the L cancels on both sides — as expected, since the problem is translationally invariant along z, the answer cannot depend on our arbitrary choice of L. What remains is the characteristic 1/r falloff of a line source (not 1/r^2 as for a point source, because one dimension of the source is already "infinite").

The field of an infinite line of charge falls off as 1/r, not 1/r^2. This is a general rule: a d-dimensional charge distribution produces a field that falls off as 1/r^{2-d} in the region far from the sources (for d=0, a point: 1/r^2; for d=1, a line: 1/r; for d=2, a plane: constant, i.e., 1/r^0). The dimension of the source determines the geometry of the falloff.

Application 2 — the infinite solid cylinder

Now an infinite solid cylinder of radius R with uniform volume charge density \rho C/m³. This is a useful model for the overhead Delhi Metro traction wire (approximately a uniformly charged cylinder at a known voltage above the grounded track), for a rod used as a charge injector in a Van de Graaff generator at IIT Kharagpur, or for the beam inside an electron gun.

Outside the cylinder: r \ge R.

Gaussian surface: coaxial cylinder of radius r, length L. Enclosed charge: the full cross-section of the cylinder, length L:

Q_\text{enc} \;=\; \rho \cdot \pi R^2 \cdot L.

Gauss's law:

E(r)\cdot 2\pi r L \;=\; \frac{\rho\pi R^2 L}{\varepsilon_0} \quad\Longrightarrow\quad E(r) \;=\; \frac{\rho R^2}{2\varepsilon_0 r} \;=\; \frac{\lambda}{2\pi\varepsilon_0 r},

where \lambda = \rho\pi R^2 is the total charge per unit length of the cylinder.

Why: outside the solid cylinder, the field is identical to that of an infinite line of charge of the same linear density \lambda placed on the axis. This is the cylindrical analogue of the shell-theorem result for spheres — the external field only sees the total charge, not its distribution inside. The R^2 in the numerator and the \pi R^2 in \lambda are exactly the cross-sectional area, encoding the "line of equivalent charge" result.

Inside the cylinder: r < R.

Gaussian surface: coaxial cylinder of radius r, length L, with r<R entirely inside the charged volume. Enclosed charge: only the charge in the narrower cylinder of radius r:

Q_\text{enc} \;=\; \rho \cdot \pi r^2 \cdot L.

Gauss's law:

E(r)\cdot 2\pi r L \;=\; \frac{\rho\pi r^2 L}{\varepsilon_0} \quad\Longrightarrow\quad \boxed{\;E(r) \;=\; \frac{\rho r}{2\varepsilon_0}\;} \tag{3}

Why: inside the cylinder, only the enclosed "core" of charge (up to radius r) contributes flux; the surrounding "shell" of charge at radii r' > r contributes zero flux (by the same argument as for a spherical shell — outside charges cancel in pairs through the Gaussian surface). The field grows linearly with r, from zero at the axis to a maximum \rho R/(2\varepsilon_0) at the surface, where it matches continuously with the outside formula.

The field $E(r)$ for a uniformly charged infinite cylinder rises linearly from zero on the axis to a peak at the surface, then decays as $1/r$ outside, the same profile as an infinite line of charge with equivalent $\lambda$.

Application 3 — the coaxial cable

A coaxial cable consists of an inner cylindrical conductor of radius a carrying charge per unit length +\lambda, and an outer cylindrical shield of radius b (with b > a) carrying -\lambda. Between them: a non-conducting insulator (the foamed plastic). This is the geometry of the cable running from your DTH dish, the coax connecting a cathode-ray oscilloscope to a voltage source in a lab, and the shielded signal lines inside every Indian hospital's MRI room.

There are three regions to handle: inside the inner conductor (r < a), between the two conductors (a < r < b), and outside the outer shield (r > b).

Region 1: r < a — inside the inner conductor.

In electrostatic equilibrium, \vec{E} = 0 inside any conductor. (The inner conductor is a hollow or solid cylinder; the charge +\lambda sits on its outer surface at r = a.) So E = 0 for r < a.

Why: inside any conductor in equilibrium, free charges have already rearranged until no internal field is felt. If \vec{E} were non-zero, free electrons would keep moving — not equilibrium.

Region 2: a < r < b — between the conductors.

Draw a Gaussian cylinder of radius r (with a < r < b) and length L. The enclosed charge is only the charge on the inner conductor's surface:

Q_\text{enc} \;=\; +\lambda L.

Gauss's law:

E(r)\cdot 2\pi r L \;=\; \frac{\lambda L}{\varepsilon_0} \quad\Longrightarrow\quad E(r) \;=\; \frac{\lambda}{2\pi\varepsilon_0 r}.

Why: in this region, the Gaussian surface encloses the inner conductor entirely (charge +\lambda L) but not the outer shield. The field is identical to that of an infinite line of charge — the outer shield might as well not exist for a field point in this annular region.

Region 3: r > b — outside the outer shield.

Now the Gaussian cylinder encloses both the inner conductor and the outer shield. The enclosed charge:

Q_\text{enc} \;=\; +\lambda L + (-\lambda L) \;=\; 0.

Gauss's law:

E(r)\cdot 2\pi r L \;=\; 0 \quad\Longrightarrow\quad E(r) \;=\; 0 \quad\text{for } r > b.

Why: the two equal and opposite line charges cancel in the enclosed total. There is no net charge inside the Gaussian surface; by Gauss's law, the net flux is zero, which by cylindrical symmetry means E = 0 at every point outside the cable.

This is the engineering feature of coaxial design: the cable does not radiate. Signals carried inside the cable do not leak out; external fields (e.g., from someone running a microwave oven next door) do not leak in (the outer shield acts as a Faraday cage for the interior). Radio frequencies, cable TV signals, oscilloscope probes — all use this geometry for the same reason.

Cross-section of a coaxial cable. Inside the inner conductor ($r<a$): zero field. Between the conductors ($a<r<b$): radial field $\lambda/(2\pi\varepsilon_0 r)$. Outside the shield ($r>b$): zero field. The cable is its own electromagnetic containment.

Conductors and cavities — the Faraday cage

The second half of this chapter turns from cylindrical symmetry to a more subtle topic: what happens when you scoop a cavity out of a solid conductor.

Two foundational facts about conductors in equilibrium

Before the cavity theorem, establish the two facts about conductors that make it work:

Fact 1. \vec{E} = 0 everywhere inside the bulk of a conductor in electrostatic equilibrium.

Why: if \vec{E} were non-zero inside, the free electrons in the conductor (which are not bound to any particular atom in a metal) would experience a force and move. They would continue moving until the field vanished — i.e., until equilibrium was restored. The equilibrium state, by definition, has \vec{E} = 0 in the bulk.

Fact 2. Any net charge on an isolated conductor resides on its outer surface.

Why: apply Gauss's law to any closed surface drawn entirely within the bulk of the conductor. On this surface, \vec{E} = 0 (by Fact 1), so the flux is zero, so the enclosed charge is zero. Shrink the surface to any size; the enclosed charge is still zero. Hence there is no net charge anywhere in the bulk — all charge lives on surfaces.

The cavity theorem

Now scoop a cavity out of a conductor — a hollow region inside the conductor, with no charges placed inside the cavity. Claim: the electric field inside the cavity is exactly zero, regardless of whatever charges sit outside the conductor.

A closed Gaussian surface $S$ drawn entirely inside the conductor (just outside the cavity). On $S$, $\vec{E}=0$. So the flux through $S$ is zero, so the net charge enclosed by $S$ (which is the charge inside the cavity plus any charge on the cavity wall) is zero. With no charges placed inside the cavity, the wall must have zero net charge too — and some careful argument (below) then shows the field inside the cavity is zero everywhere.

Proof.

Step 1. Draw a Gaussian surface S entirely in the bulk of the conductor, enclosing the cavity.

Step 2. By Fact 1, \vec{E}=0 on every point of S, so \oint_S\vec{E}\cdot d\vec{A}=0.

Step 3. By Gauss's law, Q_\text{enc}=0. The enclosed region consists of (i) the cavity interior (with no charges, by assumption) and (ii) the inner surface of the cavity (where induced surface charge may live). So the total induced charge on the inner cavity wall is zero.

Why: the cavity contents are zero (by assumption), so whatever charge S encloses must be on the inner wall. That total is zero.

Step 4. Could the induced charge on the wall be locally non-zero (positive in some places, negative in others) while summing to zero? In principle yes — but that would produce a non-zero field inside the cavity, and by a clever argument (next) that is impossible.

Step 5. Assume for contradiction that there is some field inside the cavity. A field line inside the cavity must start on a positive surface charge on the wall and end on a negative surface charge on the wall (there are no charges inside the cavity to begin or end on). Consider the path along such a field line, extended outside the cavity wall. The line integral \int \vec{E}\cdot d\vec{l} along the field line (inside the cavity) is strictly positive (field is along the direction of integration). To close the path, you go back to the starting charge through the conductor bulk, where \vec{E}=0. The total closed-loop integral is therefore positive.

But the electrostatic field is conservative: \oint \vec{E}\cdot d\vec{l} = 0 for any closed path. Contradiction.

Why: the electrostatic field is the gradient of a potential (\vec{E} = -\nabla V), and line integrals of a gradient around closed paths are zero. If \oint\vec{E}\cdot d\vec{l} were non-zero, the field would not be conservative and a potential V would not exist — violating the fundamental properties of electrostatic fields. So our assumption of a non-zero field inside the cavity must be false.

Step 6. Therefore \vec{E}=0 everywhere inside the cavity. The induced charge on the inner wall is not just zero in total; it is zero everywhere.

Conclusion. \vec{E} = 0 inside any empty cavity in a conductor, regardless of what charges sit outside. The cavity is electrostatically shielded by the surrounding conductor.

Why this matters — the Faraday cage

This theorem — that a conducting shell of any shape shields its interior from external electrostatic fields — is called the Faraday cage principle, after the experiments Michael Faraday did in the 1830s, in which he stood inside a metal box while his assistants applied huge electrostatic voltages to the outside. Faraday felt nothing; the electroscopes inside the box registered nothing.

Every Faraday cage in daily life uses this theorem:

The metal body of a Maruti Suzuki Alto or a Hyundai i20 acts as a Faraday cage during a thunderstorm: if lightning strikes the car, the enormous charge flows around the car's outer shell and to the ground, leaving the occupants inside untouched. (Keep the windows rolled up; a partial Faraday cage has partial shielding.)
The shielded enclosure of an MRI scanner at AIIMS or Manipal Hospital blocks external radiofrequency noise from corrupting the imaging sequence. The cage has a carefully designed mesh spacing so that radio frequencies (up to the operating MRI frequency, \sim 100 MHz) are strongly attenuated.
The perforated metal grid inside the glass door of every Indian microwave oven is a Faraday cage for 2.45 GHz radiation. The grid holes are much smaller than the wavelength (\lambda = c/f = 12 cm), so microwaves cannot escape, but visible light (\lambda \sim 500 nm) passes through easily.
Cleanrooms at TIFR and VECC Kolkata conducting sensitive electrostatic experiments use grounded metal cages to prevent stray charges (from lab personnel, clothing, or nearby equipment) from reaching the experimental apparatus inside.

The theorem is remarkably strong: it says nothing about how thick the conductor must be, what material it is made of, or what shape it is. Any closed conductor shields its interior. In practice, conductors are not perfect (finite conductivity means time-varying fields can penetrate; a real metal mesh has holes), so the shielding is imperfect for high-frequency fields. But for static fields, the theorem is exact.

Worked examples

Example 1: Field of a charged overhead metro wire

The overhead traction wire of the Delhi Metro carries a total charge of about 4.5 \times 10^{-9} coulombs per metre of length when energised at 25 kV above the grounded rails. Model the wire as an infinitely long straight line charge with \lambda = 4.5 nC/m, and find the electric field 0.50 m below the wire.

The wire is modelled as a line charge; the field at a distance $r$ below it points radially (downward, toward the ground).

Step 1. Apply equation (2) for the infinite line of charge.

E(r) \;=\; \frac{\lambda}{2\pi\varepsilon_0 r}.

Why: the wire is much longer than our distance of observation (kilometres long, vs. 0.5 m away), so the infinite-line approximation is excellent. Cylindrical symmetry applies.

Step 2. Insert numbers. Use \varepsilon_0 = 8.854\times 10^{-12} C²/(N·m²).

E \;=\; \frac{4.5\times 10^{-9}}{2\pi (8.854\times 10^{-12})(0.50)} \;=\; \frac{4.5\times 10^{-9}}{2.78\times 10^{-11}} \;\approx\; 162 \;\text{V/m}.

Why: compute the denominator (2\pi\varepsilon_0 \cdot r = 2\pi \cdot 8.854\times 10^{-12} \cdot 0.50 = 2.78\times 10^{-11}), then divide. The answer is in V/m (equivalently N/C), the SI unit of electric field.

Step 3. Sanity check.

Fair-weather atmospheric field near Earth's surface is about 100 V/m. The field right under a 25 kV traction wire is a few times that — reasonable. If you were 5 m away (standing on the platform), the field would be 10\times smaller, about 16 V/m — safely comparable to fair-weather background.

Why: the 1/r falloff lets you estimate the field at different distances quickly. A factor-of-10 increase in r gives a factor-of-10 decrease in E.

Result: \boxed{E \approx 162\;\text{V/m}}, pointing radially away from the wire (and in particular, downward at the observer's location).

What this shows: The cylindrical line-charge formula is a good model whenever the wire is long compared to the distance at which you are measuring. The key signature is the 1/r falloff — not the 1/r^2 of a point charge. This is why moving twice as far from a power line halves the field, not quarters it.

Example 2: Charged non-conducting cylinder — inside and outside

A long non-conducting glass cylinder of radius R = 2.0 cm carries a uniform volume charge density \rho = +5.0\,\mu\text{C/m}^3. Find the electric field at (a) r = 1.0 cm (inside the cylinder) and (b) r = 5.0 cm (outside).

Two field points: $P_1$ inside the charged cylinder (at $r=R/2$) and $P_2$ outside (at $r=2.5R$).

(a) Inside, r = 1.0 cm. Use equation (3): E = \rho r/(2\varepsilon_0).

E \;=\; \frac{(5.0\times 10^{-6})(0.010)}{2(8.854\times 10^{-12})} \;=\; \frac{5.0\times 10^{-8}}{1.77\times 10^{-11}} \;\approx\; 2.82\times 10^{3} \;\text{V/m}.

Why: inside the cylinder, the field grows linearly with r — doubling r doubles E. At the axis (r=0), the field is zero by symmetry; at the surface (r=R), it reaches its maximum.

(b) Outside, r = 5.0 cm. Use the external formula: E = \rho R^2/(2\varepsilon_0 r).

E \;=\; \frac{(5.0\times 10^{-6})(0.020)^2}{2(8.854\times 10^{-12})(0.050)} \;=\; \frac{(5.0\times 10^{-6})(4\times 10^{-4})}{8.85\times 10^{-13}}

E \;=\; \frac{2.0\times 10^{-9}}{8.85\times 10^{-13}} \;\approx\; 2.26\times 10^{3} \;\text{V/m}.

Why: outside the cylinder, the field falls off as 1/r. Equivalently, treat the whole cylinder as a line of charge with \lambda = \rho\pi R^2 = (5\times 10^{-6})(π)(0.02)^2 = 6.28\times 10^{-9} C/m, and use E = \lambda/(2\pi\varepsilon_0 r). Same answer — and the equivalence is a useful cross-check.

Quick sanity check at the surface. At r = R = 2 cm, the inside formula gives E = \rho R/(2\varepsilon_0) = (5\times 10^{-6})(0.02)/(1.77\times 10^{-11}) = 5.65 \times 10^3 V/m. The outside formula gives E = \rho R^2/(2\varepsilon_0 R) = \rho R/(2\varepsilon_0) = 5.65\times 10^3 V/m. The two match exactly at the boundary, as they should.

Why: E is a continuous function across the boundary of a volume charge distribution (only surface charges can produce a discontinuity; a bulk density makes a smooth profile).

Result: E(1\text{ cm}) \approx 2.8 \times 10^3 V/m. E(5\text{ cm}) \approx 2.3 \times 10^3 V/m.

What this shows: Inside the cylinder, the field scales linearly with position; outside, it decays as 1/r. The peak occurs at the surface. The same two-region behaviour (linear inside, 1/r outside) that governs any uniformly charged cylindrical distribution.

Example 3: A cavity inside a charged conductor

A solid conducting sphere of radius R carries a total charge Q. Inside it, an off-centre spherical cavity of radius a has been hollowed out. No charges are placed inside the cavity. Find (a) the field inside the cavity, (b) the field in the conducting bulk, and (c) the field outside the sphere at a distance r > R.

A cavity of any shape inside a charged conductor has no field within it. All the charge lives on the conductor's outer surface. The external field depends only on the total charge, not on the cavity's size or position.

(a) Inside the cavity. By the cavity theorem, \vec{E} = 0 everywhere inside the cavity, regardless of the cavity's shape, position, or what charges sit outside the conductor.

Why: this is the Faraday cage theorem, proved in the section above. The surrounding conductor shields the cavity from any external field and, since no charges are inside the cavity, there are no interior sources to produce a field either. Both contributions are zero, so the field is zero.

(b) Inside the bulk of the conductor. \vec{E} = 0, by Fact 1 above (any conductor in electrostatic equilibrium has zero bulk field).

Why: free electrons rearrange until the bulk field is zero. This is the definition of equilibrium.

(c) Outside the sphere, r > R. By Gauss's law with a Gaussian sphere of radius r concentric with the conductor, Q_\text{enc} = Q (all the charge lives on the outer surface, inside the Gaussian sphere). By spherical symmetry outside the conductor — even with an off-centre cavity, as long as the cavity is completely inside the outer sphere — the exterior field is E = Q/(4\pi\varepsilon_0 r^2), identical to that of a point charge Q at the centre of the outer sphere.

Why: the charge on the outer surface of the conductor arranges itself uniformly in the simplest case (when there is no external field imposed on the conductor); when there is an external field or an internal cavity, the surface distribution can be non-uniform, but for a cavity entirely interior to the sphere, the outside multipole moments remain those of a uniformly charged sphere — the cavity can shift the distribution near itself, but cannot produce multipoles visible from outside. For a non-spherical conductor this would need more care; for our spherical conductor, the external field is that of a centred point charge Q.

Result: \vec{E} = 0 inside the cavity, \vec{E} = 0 in the conducting bulk, and |\vec{E}| = Q/(4\pi\varepsilon_0 r^2) outside.

What this shows: The cavity theorem gives you free reason to place sensitive equipment inside conducting enclosures. Any shape of cavity, any external field — the inside is field-free. This is the physics behind every shielded instrument at TIFR, every MRI room at AIIMS, and every Maruti's passenger compartment during a lightning strike.

Common confusions

"A long line charge's field falls off as 1/r^2." No — for an infinite line, it falls off as 1/r. The 1/r^2 applies to point charges. A finite-length line charge has a more complicated field that interpolates between the two: 1/r^2 at distances much larger than the line's length (where the line looks like a point), and 1/r at distances much smaller than the line's length (where the line looks infinite).
"Gauss's law fails for a finite cylinder." The law holds for any closed surface; what fails is the computational shortcut. A finite cylinder lacks translational symmetry along its axis, so |\vec{E}| is not constant on a coaxial Gaussian cylinder, and you cannot pull it out of the integral. The full calculation requires direct Coulomb integration.
"Inside a hollow conductor, the field is zero because the charges on the outer surface cancel at each interior point." The cancellation is the mechanism, but the cleaner proof is the Gauss's law argument above: any interior Gaussian surface has zero enclosed charge, zero flux, and (with the symmetry of a neutral interior) zero field. The conceptual point is that the field is zero everywhere inside, not by accident, but by a theorem.
"If I place a charge inside the cavity, then what?" Then the cavity is no longer "empty," and the cavity theorem does not directly apply. The charge inside the cavity induces an equal and opposite charge on the inner wall of the cavity (by Gauss's law applied to a Gaussian surface in the bulk), and an additional charge on the outer surface of the conductor to keep the total of the conductor plus induced charges constant. The field inside the cavity is now non-zero (it is the sum of the field from the interior charge and the field from the induced wall charge — together, they give a specific field pattern determined by the cavity's geometry).
"Gauss's law proves a non-conducting hollow cylinder also shields its interior." No! The cavity theorem requires the cage to be a conductor, because Fact 1 (\vec{E}=0 in the bulk) uses the fact that free charges rearrange to kill the bulk field. A non-conducting hollow cylinder does not have this property; it transmits external fields into its interior. Only metals (and, approximately, ionic solutions) can form Faraday cages.
"A metal car shields its occupants from a lightning strike because the car has insulating tires." Partly true but not the main reason. The tires provide no pathway to ground, so once the car is struck, the charge distributes around the outer shell and, being a conductor, the shell keeps its interior field-free — this is the Faraday cage. The charge eventually drains off when the tires finally break down or when the car touches a grounded object (a puddle, a wet road). The insulating tires delay the draining but the Faraday cage is what protects the occupants while the charge is on the car.

If you came here to handle cylindrical distributions, coaxial cables, and Faraday cages, you have what you need. What follows is for readers interested in non-uniform cylindrical distributions, the shielding of time-varying fields, and a surprising consequence for biology.

Non-uniform cylindrical distributions

What if the volume density depends on r? As long as \rho = \rho(r) only (still cylindrically symmetric — no \phi or z dependence), Gauss's law still delivers the field in one line. The trick is just a small integral to find the enclosed charge.

For a Gaussian cylinder of radius r and length L inside the charge distribution, the enclosed charge is

Q_\text{enc} \;=\; \int_0^r \rho(r')\cdot 2\pi r'\,L\,dr',

where 2\pi r'\,L\,dr' is the volume of a thin cylindrical shell of radius r' and thickness dr'.

Gauss's law:

E(r) \;=\; \frac{Q_\text{enc}/L}{2\pi\varepsilon_0 r} \;=\; \frac{1}{\varepsilon_0 r}\int_0^r \rho(r')\,r'\,dr'.

Example: a cylindrical beam of charge with \rho(r) = \rho_0 e^{-r/a} (a Gaussian-like radial profile, roughly the charge density inside the space-charge region of an electron gun). The enclosed charge at radius r:

Q_\text{enc}/L \;=\; 2\pi \int_0^r \rho_0 e^{-r'/a}\,r'\,dr' \;=\; 2\pi\rho_0 a^2\left[1 - \left(1 + \frac{r}{a}\right)e^{-r/a}\right],

using integration by parts. The field is then (Q_\text{enc}/L)/(2\pi\varepsilon_0 r) — in one line.

For an exact analytical answer, you need an integrable \rho(r). For more complex profiles, the integral is numerical. But the structure is always the same: Gauss's law reduces the field computation to a one-dimensional radial integral.

Shielding of time-varying fields (AC shielding)

The Faraday cage theorem as stated applies strictly to electrostatic fields. For time-varying fields — a radio wave, a pulsed laser, the electromagnetic field of a mobile phone — the situation is different.

A time-varying external field induces currents in the shield (not just surface charges). The currents themselves produce a magnetic field that, combined with the electric field, must satisfy Maxwell's equations. For a good conductor (like copper or aluminium), the induced currents cancel the external field at the shield's surface, and the field decays exponentially as it penetrates — this is the skin effect. The penetration depth (the skin depth \delta) scales as \delta \propto 1/\sqrt{\omega\sigma}, where \omega is the angular frequency and \sigma the conductivity. Higher frequency means shallower penetration; at microwave frequencies, the skin depth in copper is less than a micrometre.

This is why the perforated grid in a microwave-oven door works: the holes are much smaller than the 12-cm wavelength of the 2.45 GHz radiation. The field inside the oven induces currents that flow around the holes on the outer surface of the mesh, and those currents radiate nothing significant through holes this small. The oven's interior looks like an electrically closed cavity to the microwaves, even though it looks open to visible light.

Why cell membranes are not Faraday cages (and why that matters)

The lipid bilayer of a biological cell is about 5 nm thick and is an insulator, not a conductor. So the Faraday cage theorem does not apply — the cell's interior is not shielded from external electric fields.

This is why electric fields can influence biological cells. Pulsed high-field electroporation — a standard technique at NCBS Bangalore and IISER Pune for introducing DNA into cells — uses fields of \sim 10 kV/m for microseconds to induce transient pores in the membrane. The ATM machine in your pocket also uses electric fields to move ions across channels in cell membranes (that is how nerves fire). If cells were Faraday cages, none of this would be possible.

The Faraday cage principle is specifically about conductors. Any material with mobile free charges (metals, ionic solutions, semiconductors) can form a cage. Materials without free charges (insulators, vacuum, the lipid bilayer) cannot.

The non-conducting hollow cylinder — a non-trivial problem

What happens inside a non-conducting hollow cylinder with surface charge density \sigma? Gauss's law, applied to an interior Gaussian cylinder, shows that the enclosed charge is zero (no charge inside the cylinder's walls), so the net flux is zero. But this alone does not prove the field is zero — it only gives you a single scalar equation.

To conclude that \vec{E}=0 inside, you need an additional argument: cylindrical symmetry implies \vec{E} is radial and depends only on r. For an interior Gaussian cylinder at any radius r<R, the flux is E(r)\cdot 2\pi r L = 0, which forces E(r)=0 for all r<R. So yes, a uniformly charged infinite non-conducting hollow cylinder also has zero field inside — the cylindrical symmetry is doing all the work, not the conductor property.

This is a surprise: for this specific geometry (hollow infinite cylinder with uniform surface charge), even non-conductors shield their interior. But this is an artefact of the cylindrical symmetry and the infinite length. A non-conducting spherical shell with non-uniform charge does not shield its interior, because the charge can be arranged asymmetrically and the symmetry argument fails.

Historical note: Faraday's own experiment

In 1836, Michael Faraday built a room-sized metal cage, stood inside with sensitive electroscopes, and had his assistants apply enormous voltages (from a large electrostatic generator) to the outside. He reported that the electroscopes inside registered nothing — no matter what was done outside. This was the definitive experimental test of the cavity theorem.

The modern version runs daily in TIFR's shielded labs, in the CERN accelerator control rooms, and in the MRI bay of any hospital. The result is always the same: \vec{E} = 0 inside, no matter what storms rage outside.

Where this leads next

Gauss's Law — the parent law, derived from Coulomb's law, that this chapter applies to cylinders and cavities.
Applications of Gauss's Law — Spheres and Planes — the companion chapter covering spherical and planar symmetry, the other two symmetric cases.
Electric Flux — the surface integral \oint \vec{E}\cdot d\vec{A} that Gauss's law is built on.
Electric Potential and Potential Difference — the scalar field whose gradient gives \vec{E}, the cleaner energy approach that complements the field approach.
Coulomb's Law — the underlying force law from which Gauss's law follows; the fallback when symmetry is not available.