Enclosed charge, and when it hands you E

For any closed surface,

$$\oint \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}.$$

Two things about this equation are read wrong more often than they are read right.

1. $q_{\text{enc}}$ is the net charge inside, signs included. A $+q$ and a $-q$ both inside give zero, and a charge outside contributes nothing to $q_{\text{enc}}$.
2. The $\vec{E}$ under the integral is the TOTAL field, produced by every charge in the universe, outside charges included. An outside charge sends field lines in one side of the surface and out the other, so its NET flux is zero. Its field on the surface is not zero anywhere.

So "only enclosed charge matters" is a statement about the flux total, never about the field at a point. Put a point charge $q$ inside a sphere and a second charge $Q$ just outside it: the net flux is $q/\varepsilon_0$, unchanged by $Q$, while the field at every point on that sphere has a piece from $Q$ in it. Dropping $Q$ from a field calculation because Gauss's law does not count it is the first error of the topic.

The surface itself is imagined. Nothing physical happens when you draw it, move it, or resize it. It is bookkeeping around a region, and the field at any point is whatever the charges make it, whether or not a surface passes through that point.

To go from flux to field, you need the integral to collapse to $E \times (\text{area})$. That happens only when, on the part of the surface that carries flux, $E$ has the same magnitude everywhere and is parallel to the normal, and the rest of the surface carries no flux at all. Three geometries deliver it:

1. Spherical symmetry: a concentric Gaussian sphere. $E(4\pi r^2) = q_{\text{enc}}/\varepsilon_0$.
2. Cylindrical symmetry (an infinite line or cylinder): a coaxial Gaussian cylinder of length $L$. The end caps carry nothing because $\vec{E}$ lies in their planes; $E(2\pi r L) = q_{\text{enc}}/\varepsilon_0$.
3. Planar symmetry (an infinite sheet): a pillbox through the sheet. The curved wall carries nothing; the two caps carry $E A$ each.

Where symmetry is missing, the law still holds and still gives the flux, but it stops there. Around a point charge draw a cube instead of a sphere: the flux is still $q/\varepsilon_0$, and by symmetry each face carries $q/6\varepsilon_0$. But on a face the field varies in magnitude and crosses at every angle, so writing $E \times L^2 = q/6\varepsilon_0$ and solving for $E$ produces a number that is the field at no point on the face. For the field at a point, use the surface $E$ is constant on, which is the sphere, or fall back on Coulomb's law directly.

Same for a dipole inside a Gaussian sphere: $q_{\text{enc}} = 0$, so the net flux is zero, and that is all. The field on that sphere is nonzero, and it varies, so no value of $E$ comes out.

Both cases use a concentric Gaussian sphere of radius $r$; what changes is $q_{\text{enc}}$.

Thin uniform shell of radius $R$ and charge $Q$. For $r < R$ the sphere encloses nothing, so $E = 0$ everywhere inside. For $r > R$ it encloses all of $Q$: $E = kQ/r^2$, exactly as if $Q$ were a point at the center. The field jumps from $0$ to $kQ/R^2$ at the shell.

Uniform solid insulating sphere of radius $R$ and charge $Q$. Outside, the same $kQ/r^2$. Inside, a sphere of radius $r$ encloses the fraction of the volume it holds, $(r/R)^3$ of $Q$:

$$E(4\pi r^2) = \frac{Q\,(r/R)^3}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{kQ\,r}{R^3}, \qquad r \le R.$$

So the field is zero at the center, grows linearly to $kQ/R^2$ at the surface, and falls as $1/r^2$ beyond. Its maximum is at the surface, not the center. Two errors are common here: carrying the shell's $E = 0$ into the solid sphere, and using the total $Q$ inside the solid sphere instead of the enclosed fraction, which makes the field blow up toward the center instead of vanishing there.

The test is always the same question: what does my Gaussian sphere actually enclose?

Infinite line, charge per length $\lambda$. The coaxial cylinder of radius $r$ and length $L$ encloses $\lambda L$:

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

The $L$ cancels, as it must: the field cannot depend on how long a piece of imagined cylinder you drew. Doubling $r$ halves the field. This is $1/r$, not $1/r^2$.

Infinite sheet, charge per area $\sigma$. A pillbox of cap area $A$ straddling the sheet encloses $\sigma A$ and has flux out of BOTH caps:

$$2EA = \frac{\sigma A}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{\sigma}{2\varepsilon_0}.$$

No $r$ appears. The field is the same at any distance from an infinite sheet, because the geometry has no length scale for the field to fade over. Students bolt a $1/r^2$ onto this result on the grounds that all fields should weaken with distance; the sheet is the case where the ideal geometry says otherwise.

The factor-of-two partner: just outside the surface of a conductor with local surface density $\sigma$, the field is $\sigma/\varepsilon_0$. Same pillbox, but now the cap inside the conductor sits in zero field, so all the flux leaves through one cap: $EA = \sigma A/\varepsilon_0$. A free-standing sheet radiates both ways and gets $\sigma/2\varepsilon_0$; a conductor's surface radiates one way and gets $\sigma/\varepsilon_0$. Mixing them up is exactly a factor of two.