What metal does with the charge you give it

Put a conductor in any static arrangement of charge and wait. Free electrons move until they stop moving, and they stop only when the force on them is zero, which means the field inside the metal is zero. That equilibrium statement is the axiom of the topic.

1. $\vec{E} = 0$ inside the conducting material. Not small: exactly zero, everywhere in the metal, once equilibrium is reached.
2. No net charge in the volume. Draw any Gaussian surface entirely inside the metal. $E = 0$ on it, so $\Phi = 0$, so $q_{\text{enc}} = 0$. Excess charge has nowhere to be except a surface.
3. The whole conductor is one equipotential. $V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{\ell} = 0$ along any path through the metal, so every point of it, surface included, sits at the same potential.
4. The field just outside is perpendicular to the surface. A tangential component would push surface charge sideways forever, which contradicts equilibrium.

A pillbox Gaussian surface straddling the surface gets flux only through its outer cap, because the inner cap sits in the field-free metal. That gives

$$E_{\text{just outside}} = \frac{\sigma}{\varepsilon_0},$$

which is twice the $\sigma/2\varepsilon_0$ of an isolated sheet. The doubling is not a new law; it is what happens when the field on one side is forced to zero and the flux that would have gone that way is pushed out the other.

Hollow out a conductor and put a point charge $+q$ in the hole. The field in the cavity is not zero. It is the field of $+q$, distorted near the wall by whatever charge the wall pulled up. Nothing about "zero field in a conductor" applies there, because the cavity is not conductor.

What the axiom does tell you is exactly how much charge the wall must carry. Draw a Gaussian surface that lives entirely inside the metal and wraps the cavity. On it $E = 0$, so the flux is zero, so the enclosed charge is zero:

$$q_{\text{enc}} = q + q_{\text{wall}} = 0 \quad\Longrightarrow\quad q_{\text{wall}} = -q.$$

If the conductor started neutral, that $-q$ came from somewhere: an equal $+q$ is left on the outer surface. Charge conservation, not a separate rule.

Two consequences students routinely miss:

1. The outside still sees $q$. A Gaussian sphere drawn around the whole object encloses $q + (-q) + (+q) = +q$, so flux leaves and there is a field outside. Boxing a charge in metal does not hide it.
2. The outer distribution forgets the details. Move $+q$ off center in a spherical cavity and the wall charge redistributes to compensate, but the outer surface of a spherical conductor stays uniform, so the external field is still $kq/r^2$, radial. The metal launders the interior geometry.

Now empty the cavity and switch on an external field. With no charge inside, the same Gaussian argument gives zero net charge on the wall, and a slightly harder argument (a field line in an empty cavity would have to start and end on the wall, so a closed loop through it would have nonzero $\oint \vec{E}\cdot d\vec{\ell}$) gives $\vec{E} = 0$ throughout the empty cavity. That is real shielding.

Put the two cavity results side by side and the asymmetry is the whole point.

1. Empty cavity, charges outside. Field in the cavity: zero. Outside fields cannot get in. This is why sensitive electronics sit in metal cans.
2. Charge in the cavity, isolated conductor. Field outside: not zero. Induced $+q$ on the outer surface broadcasts the enclosed charge to the world.

Grounding is what closes the second door. Connect the outer surface to earth and charge flows until the conductor sits at $V = 0$; for a shell with $+q$ inside, that means the outer surface ends with no net charge and the exterior field goes to zero. The cavity is unchanged: $+q$ is still there, the wall still carries $-q$, the cavity field is still lively. Grounding silenced the outside, not the inside.

Say it as a slogan and then never trust the slogan alone: a closed conductor blocks fields coming in. It only blocks the news going out if you ground it.

Uniform surface density is a privilege of the sphere, and only of a sphere far from everything else. On a conductor with a bulge, a corner, or a point, charge piles up where the curvature is high.

The crude argument that carries the right scaling: model the object as two conducting spheres of radii $R_1$ and $R_2$ joined by a thin wire. Equal potential gives

$$\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \quad\Longrightarrow\quad \frac{Q_1}{Q_2} = \frac{R_1}{R_2},$$

and since $\sigma = Q/(4\pi R^2)$,

$$\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}, \qquad \frac{E_1}{E_2} = \frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}.$$

Small radius of curvature means large $\sigma$ and a large field just outside. That is the whole physics of the lightning rod: the field at the tip reaches the breakdown strength of air (about $3\times10^6$ V/m) while the field over the blunt parts of the same conductor is still far below it, so the discharge starts at the point.

Two habits follow. First, write $\sigma = Q/A$ only when you have argued for the symmetry. Second, if a question asks where a spark begins, where the field is strongest, or where the charge concentrates, look for the sharpest feature in the picture, not the largest area.