= Electric flux
{title2=$\Phi_E=\int_S E\cdot n\,dS$}
{wiki}
<Electric flux> measures how much <electric field> crosses a chosen <oriented surface>, with a sign distinguishing the two crossing directions. A strong <electric field> running along a surface contributes nothing to its <electric flux>: only the <normal component> counts. This lets <Gauss's law> turn a potentially complicated <electric field> into one number recording the net enclosed <electric charge>.
For a surface $S$ with chosen <unit normal> $\mathbf n$, the <electric flux> is the <surface integral>
$$
\Phi_E(S)=\int_S\mathbf E\cdot\mathbf n\,dA=\int_S\mathbf E\cdot d\mathbf S,\qquad d\mathbf S=\mathbf n\,dA.
$$
Here $dA$ is positive <surface area>, while $d\mathbf S$ is the <vector area element>. The <dot product> makes <electric flux> a signed <scalar>, rather than a <vector> like the <electric field>. Its <SI> unit is $\mathrm{N\,m^2/C}$: <electric field> measured using the <newton (unit)> per <coulomb>, multiplied by <surface area> in square metres. Reversing the <orientation of a surface> replaces $\mathbf n$ by $-\mathbf n$ and reverses the <electric flux>. Calling it <electric flux> does not imply that <electric charge> is moving through the surface. https://www.feynmanlectures.caltech.edu/II_03.html#Ch3-S2[Feynman lectures II.3, §3–2: flux of a vector field]{ref}.
For an open patch in a <plane>, with <surface area> $A$ and a constant <electric field>, the <unit normal> is constant and the <surface integral> simplifies to
$$
\Phi_E=\mathbf E\cdot(A\mathbf n)=EA\cos\theta,\qquad E=|\mathbf E|.
$$
The <angle> $\theta$ is between the <electric field> and the chosen <unit normal>, not between the <electric field> and the <plane>. The <cosine> describes the signed projected <surface area> seen by the <electric field>. Thus <electric flux> is positive for $0\leq\theta<90^\circ$, zero at $90^\circ$, and negative for $90^\circ<\theta\leq180^\circ$. https://openstax.org/books/university-physics-volume-2/pages/6-1-electric-flux[Ling, Sanny and Moebs, University Physics II, §6.1: electric flux]{ref}.
For example, take $E=100\,\mathrm{N/C}$ and $A=0.02\,\mathrm{m^2}$, so $EA=2\,\mathrm{N\,m^2/C}$. Three choices of <angle> give
$$
\begin{aligned}
\Phi_E(0^\circ)&=2\cos0^\circ=2\,\mathrm{N\,m^2/C},\\
\Phi_E(60^\circ)&=2\cos60^\circ=1\,\mathrm{N\,m^2/C},\\
\Phi_E(120^\circ)&=2\cos120^\circ=-1\,\mathrm{N\,m^2/C}.
\end{aligned}
$$
Keeping the $60^\circ$ patch fixed and reversing its <unit normal> gives $120^\circ$ and changes $+1$ to $-1\,\mathrm{N\,m^2/C}$. The <electric field> and physical patch have not changed. If the <electric field> is instead tangent to the patch, its <dot product> with the <unit normal> vanishes: <electric flux> is zero even though $E=100\,\mathrm{N/C}$.
\Image[/open-surface-electric-flux.png]
{title=Three open patches in a <plane> in a constant <electric field>: the <angle> to the chosen <unit normal> gives <electric fluxes> of 2, 1 and minus 1 newton square metres per <coulomb>}
For a <closed surface> enclosing a volume $V$, choose the outward <unit normal>. <Gauss's law> gives
$$
\Phi_E(\partial V)=\oint_{\partial V}\mathbf E\cdot\mathbf n\,dA=\frac{Q_{\mathrm{enclosed}}}{\varepsilon_0}.
$$
Here $\varepsilon_0$ is the <vacuum permittivity>, and $Q_{\mathrm{enclosed}}$ is the signed total <electric charge> inside. The <electric field> in the <surface integral> includes all sources, including external <electric charges>. Assume no ideal <point charge> lies on the integration surface. Zero enclosed <electric charge> implies zero total <electric flux>; it does \b[not] imply zero <electric field> everywhere on the <closed surface> or inside it. Positive and negative contributions to the <surface integral> can cancel. https://openstax.org/books/university-physics-volume-2/pages/6-2-explaining-gausss-law[Ling, Sanny and Moebs, University Physics II, §6.2: outward flux and enclosed charge]{ref}.
To see the cancellation with sources inside, place <point charges> $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\,\mathrm m$, and both lie inside the <sphere> of radius $R=1\,\mathrm m$ centered at the origin. This separated pair has the <electric dipole> geometry. Its enclosed <electric charge> is $q-q=0$, so its total outward <electric flux> is zero. At the point $P=(0,0,R)$ on the <sphere>, <Coulomb's law> and the <superposition principle> give the individual <electric fields>
$$
\mathbf E_+(P)=\frac{kq(a,0,R)}{(R^2+a^2)^{3/2}},\qquad
\mathbf E_-(P)=\frac{kq(a,0,-R)}{(R^2+a^2)^{3/2}},\qquad k=\frac{1}{4\pi\varepsilon_0}.
$$
The $z$ components cancel and the positive $x$ components add:
$$
\mathbf E(P)=\frac{2kqa}{(R^2+a^2)^{3/2}}\widehat{\mathbf x}\approx0.527244\frac{kq}{R^2}\widehat{\mathbf x}\ne\mathbf0.
$$
For the stipulated approximation $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$ and $q=1\,\mathrm{nC}$, this is about $4.740\,\mathrm{N/C}$ along positive $x$. At $P$, the outward <unit normal> is $\widehat{\mathbf z}$, so this nonzero <electric field> is tangent to the <sphere>. At other points of the <sphere>, its <normal component> can be positive or negative. The source-field formula applies away from the two <point charges>, where the <electric field> has zero <divergence>; it is singular at the sources. https://www.feynmanlectures.caltech.edu/II_04.html#Ch4-S2[Feynman lectures II.4, §§4–1 and 4–2: electrostatic sources and superposition]{ref}.
\Image[/zero-net-flux-nonzero-field.png]
{title=Cross-section of the three-dimensional <electric field> of two opposite <point charges> inside a unit-radius <sphere>: zero total <electric flux> coexists with a nonzero tangential <electric field> at the top}
The diagram shows <streamlines> of the three-dimensional <electric field> in the section $y=0$. The circle is a cross-section of the <sphere>, not the integration surface: <Gauss's law> uses the entire three-dimensional <surface integral>. A two-dimensional <line integral> of outward field around that circle would be a different quantity, with units $\mathrm{N\,m/C}$ rather than the <electric flux> units $\mathrm{N\,m^2/C}$.
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