137 lines
4.7 KiB
TeX
137 lines
4.7 KiB
TeX
\subsection{Electric Flux}
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This subsection introduces electric flux as a signed measure of how much electric field passes through an oriented surface.
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\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
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\[
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d\vec{A}=\hat{n}\,dA.
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\]
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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.
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Let $\vec{E}$ denote the electric field at each point of the surface. The \emph{electric flux} through the oriented surface is the scalar
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\[
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\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
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\]
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Its SI units are $\mathrm{N\cdot m^2/C}$.}
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\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
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\[
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\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
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\]
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This integral adds the component of $\vec{E}$ perpendicular to the surface over the entire surface.
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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
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\[
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\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
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\]
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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.}
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\ex{Illustrative example}{Let a uniform electric field be
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\[
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\vec{E}=(300\,\mathrm{N/C})\hat{\imath}.
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\]
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Let a flat surface have area
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\[
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A=0.20\,\mathrm{m^2},
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\]
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and let its area vector make an angle $\theta=60^\circ$ with $\vec{E}$.
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Then the flux is
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\[
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\Phi_E=EA\cos\theta=(300)(0.20)\cos 60^\circ\,\mathrm{N\cdot m^2/C}.
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\]
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So
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\[
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\Phi_E=30\,\mathrm{N\cdot m^2/C}.
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\]
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Because the angle is acute, the flux is positive.}
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\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.}
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\qs{Worked AP-style problem}{A cube of side length $L=0.20\,\mathrm{m}$ is placed in a uniform electric field
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\[
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\vec{E}=(500\,\mathrm{N/C})\hat{\imath}.
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\]
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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.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the electric flux through the right face,
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\item the electric flux through the left face,
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\item the electric flux through any one of the four remaining faces, and
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\item the net electric flux through the entire closed cube.
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\end{enumerate}}
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\sol Let the area of one face be $A$. Since each face is a square of side length $L$,
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\[
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A=L^2=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}.
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\]
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For each face of the cube, use
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\[
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\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
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\]
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where $\theta$ is the angle between the electric field and that face's outward area vector.
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For part (a), the right face has outward area vector in the $+\hat{\imath}$ direction, the same direction as $\vec{E}$. Thus
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\[
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\theta=0^\circ.
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\]
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So
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\[
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\Phi_{E,\mathrm{right}}=EA\cos 0^\circ=(500)(0.040)(1)\,\mathrm{N\cdot m^2/C}.
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\]
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Therefore,
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\[
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\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C}.
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\]
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For part (b), the left face has outward area vector in the $-\hat{\imath}$ direction, opposite the field. Thus
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\[
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\theta=180^\circ.
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\]
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So
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\[
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\Phi_{E,\mathrm{left}}=EA\cos 180^\circ=(500)(0.040)(-1)\,\mathrm{N\cdot m^2/C}.
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\]
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Therefore,
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\[
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\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C}.
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\]
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For part (c), on any of the other four faces, the outward area vector is perpendicular to $\vec{E}$. Thus
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\[
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\theta=90^\circ,
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\]
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so
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\[
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\Phi_E=EA\cos 90^\circ=0.
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\]
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Therefore, the flux through each of those four faces is
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\[
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0\,\mathrm{N\cdot m^2/C}.
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\]
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For part (d), add the fluxes from all six faces:
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\[
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\Phi_{E,\mathrm{net}}=\Phi_{E,\mathrm{right}}+\Phi_{E,\mathrm{left}}+4(0).
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\]
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So
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\[
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\Phi_{E,\mathrm{net}}=20+(-20)=0\,\mathrm{N\cdot m^2/C}.
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\]
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Therefore,
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\[
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\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C},
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\qquad
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\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C},
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\]
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\[
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\Phi_E=0\,\mathrm{N\cdot m^2/C}\text{ for each of the other four faces},
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\]
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and the net flux through the closed cube is
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\[
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\Phi_{E,\mathrm{net}}=0.
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\]
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