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concepts/em/u8/e8-5-electric-flux.tex

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+\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
+\[
+\Phi_{E,\mathrm{net}}=0.
+\]