content(warnings): Add W17-W28, X8-X12, N7 — E&M misconceptions, cross-refs, notes

concepts/em/u8/e8-6-gauss-law.tex

@@ -10,8 +10,14 @@ This subsection states Gauss's law and shows how symmetry can reduce a difficult
 \]
 This law is always true. It becomes a practical method for solving for the electric field when the charge distribution has enough symmetry that one can choose a Gaussian surface for which the magnitude $E=|\vec{E}|$ is constant on each flux-contributing part of the surface and the angle between $\vec{E}$ and $d\vec{A}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the flux integral reduces to algebraic terms such as $EA$, $-EA$, or $0$. Common useful cases are spherical, cylindrical, and planar symmetry.}
 
+Gauss's law is one of Maxwell's four equations (in electrostatic form). Its analogue for magnetism is Gauss's law for magnetic fields, $\oint\vec{B}\cdot d\vec{A}=0$ (Section 12.5), which reflects the absence of magnetic monopoles.
+
 \nt{Gauss's law is always true, but it is not always useful for finding $\vec{E}$. In a general asymmetric charge distribution, knowing only the total flux through a closed surface does not tell you the field at each point on that surface. Also, zero net enclosed charge implies zero \emph{net flux}, not necessarily zero field everywhere. The main strategy is therefore: first identify strong symmetry, then choose a Gaussian surface matched to that symmetry.}
 
+\wc{Gauss's law is always true, not just for symmetric cases}{Gauss's law $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$ holds for \emph{any} charge distribution and \emph{any} closed surface. Symmetry only makes it \emph{useful} for calculating $\vec{E}$. Without symmetry, you know the total flux but cannot solve for the field at each point on the surface.}
+
+\wc{Zero net enclosed charge does not mean zero field}{A Gaussian surface enclosing zero net charge has zero \emph{net flux}, but the electric field on the surface need not be zero. External charges can produce nonzero field on the surface, with field lines entering on one side and leaving on the other, yielding zero net flux. $\oint\vec{E}\cdot d\vec{A}=0$ does \emph{not} imply $\vec{E}=0$.}
+
 \pf{Why symmetry reduces the integral}{Let a point charge $Q$ be at the center of a spherical Gaussian surface of radius $r$. By spherical symmetry, the electric field is radial and has the same magnitude $E(r)$ at every point on the sphere. The outward area element $d\vec{A}$ is also radial, so
 \[
 \vec{E}\cdot d\vec{A}=E(r)\,dA