\subsection{Electric Flux}

This subsection introduces electric flux as a signed measure of how much electric field passes through an oriented surface.

\dfn{Area vector and electric flux}{Let $S$ be a surface broken into small area elements of scalar area $dA$. Let $\hat{n}$ denote a chosen unit normal to a surface element. The corresponding \emph{area vector element} is
\[
d\vec{A}=\hat{n}\,dA.
\]
For an open surface, either choice of normal may be used, but the choice must be kept consistent across the surface. For a closed surface, the standard choice is the outward normal.

Let $\vec{E}$ denote the electric field at each point of the surface. The \emph{electric flux} through the oriented surface is the scalar
\[
\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
\]
Its SI units are $\mathrm{N\cdot m^2/C}$.}

\thm{Surface-integral form and uniform-field special case}{Let $S$ be an oriented surface with area vector element $d\vec{A}=\hat{n}\,dA$, and let $\vec{E}$ be the electric field on that surface. Then the electric flux through $S$ is
\[
\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
\]
This integral adds the component of $\vec{E}$ perpendicular to the surface over the entire surface.

If the surface is flat with area $A$, the field is uniform over it, and $\vec{A}=A\hat{n}$ denotes the surface's area vector, then
\[
\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
\]
where $E=|\vec{E}|$, $\theta$ is the angle between $\vec{E}$ and $\vec{A}$, and $A=|\vec{A}|$. For a closed surface, the same formula is applied piece by piece using outward area vectors on all patches of the surface.}

\ex{Illustrative example}{Let a uniform electric field be
\[
\vec{E}=(300\,\mathrm{N/C})\hat{\imath}.
\]
Let a flat surface have area
\[
A=0.20\,\mathrm{m^2},
\]
and let its area vector make an angle $\theta=60^\circ$ with $\vec{E}$.

Then the flux is
\[
\Phi_E=EA\cos\theta=(300)(0.20)\cos 60^\circ\,\mathrm{N\cdot m^2/C}.
\]
So
\[
\Phi_E=30\,\mathrm{N\cdot m^2/C}.
\]
Because the angle is acute, the flux is positive.}

\nt{Electric flux is not the same thing as electric field magnitude. Flux depends on both the field and the oriented surface. Reversing the chosen normal reverses the sign of $\Phi_E$. A positive flux means the field points generally in the same direction as the chosen area vector, while a negative flux means it points generally opposite that direction. If the field is parallel to the surface, then it is perpendicular to $d\vec{A}$ and the flux is zero even if $|\vec{E}|$ is large.}

\qs{Worked AP-style problem}{A cube of side length $L=0.20\,\mathrm{m}$ is placed in a uniform electric field
\[
\vec{E}=(500\,\mathrm{N/C})\hat{\imath}.
\]
Let the cube's faces be aligned with the coordinate axes. Let the outward area vector of the right face be in the $+\hat{\imath}$ direction, and let the outward area vector of the left face be in the $-\hat{\imath}$ direction.

Find:
\begin{enumerate}[label=(\alph*)]
\item the electric flux through the right face,
\item the electric flux through the left face,
\item the electric flux through any one of the four remaining faces, and
\item the net electric flux through the entire closed cube.
\end{enumerate}}

\sol Let the area of one face be $A$. Since each face is a square of side length $L$,
\[
A=L^2=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}.
\]

For each face of the cube, use
\[
\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
\]
where $\theta$ is the angle between the electric field and that face's outward area vector.

For part (a), the right face has outward area vector in the $+\hat{\imath}$ direction, the same direction as $\vec{E}$. Thus
\[
\theta=0^\circ.
\]
So
\[
\Phi_{E,\mathrm{right}}=EA\cos 0^\circ=(500)(0.040)(1)\,\mathrm{N\cdot m^2/C}.
\]
Therefore,
\[
\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C}.
\]

For part (b), the left face has outward area vector in the $-\hat{\imath}$ direction, opposite the field. Thus
\[
\theta=180^\circ.
\]
So
\[
\Phi_{E,\mathrm{left}}=EA\cos 180^\circ=(500)(0.040)(-1)\,\mathrm{N\cdot m^2/C}.
\]
Therefore,
\[
\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C}.
\]

For part (c), on any of the other four faces, the outward area vector is perpendicular to $\vec{E}$. Thus
\[
\theta=90^\circ,
\]
so
\[
\Phi_E=EA\cos 90^\circ=0.
\]
Therefore, the flux through each of those four faces is
\[
0\,\mathrm{N\cdot m^2/C}.
\]

For part (d), add the fluxes from all six faces:
\[
\Phi_{E,\mathrm{net}}=\Phi_{E,\mathrm{right}}+\Phi_{E,\mathrm{left}}+4(0).
\]
So
\[
\Phi_{E,\mathrm{net}}=20+(-20)=0\,\mathrm{N\cdot m^2/C}.
\]

Therefore,
\[
\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C},
\qquad
\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C},
\]
\[
\Phi_E=0\,\mathrm{N\cdot m^2/C}\text{ for each of the other four faces},
\]
and the net flux through the closed cube is
\[
\Phi_{E,\mathrm{net}}=0.
\]
