191 lines
5.1 KiB
TeX
191 lines
5.1 KiB
TeX
\subsection{Capacitance and Capacitor Geometries}
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This subsection defines capacitance, shows how capacitor geometry controls it, and derives the parallel-plate result from Gauss's law and the field-potential relation.
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\dfn{Capacitor and capacitance}{A \emph{capacitor} is a system of two conductors that can hold equal and opposite charges. Let the conductors carry charges $+Q$ and $-Q$, and let
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\[
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\Delta V=V_{\text{high}}-V_{\text{low}}
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\]
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denote the magnitude of the potential difference between them. The \emph{capacitance} $C$ of the system is defined by
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\[
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C=\frac{Q}{\Delta V}.
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\]
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Capacitance measures how much charge is stored per unit potential difference. Its SI unit is the farad:
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\[
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1\,\mathrm{F}=1\,\mathrm{C/V}.
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\]}
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\thm{Capacitance formula and common geometries}{Let a capacitor carry plate charges $\pm Q$, and let $\Delta V$ be the magnitude of the potential difference between its conductors. Then
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\[
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C=\frac{Q}{\Delta V}.
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\]
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For a vacuum parallel-plate capacitor with plate area $A$ and plate separation $d$, define the surface charge density by
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\[
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\sigma=\frac{Q}{A}.
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\]
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Ignoring edge effects, Gauss's law gives the nearly uniform electric field between the plates:
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\[
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\vec{E}=\frac{\sigma}{\varepsilon_0}\,\hat{n},
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\qquad
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E=\frac{\sigma}{\varepsilon_0}.
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\]
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Using the potential-drop relation with a path across the gap,
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\[
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\Delta V=\left| -\int \vec{E}\cdot d\vec{\ell} \right|=Ed,
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\]
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so
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\[
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\Delta V=\frac{\sigma d}{\varepsilon_0}=\frac{Qd}{\varepsilon_0 A}.
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\]
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Substitute into $C=Q/\Delta V$:
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\[
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C=\frac{Q}{Qd/(\varepsilon_0 A)}=\varepsilon_0\frac{A}{d}.
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\]
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Thus, for an ideal vacuum parallel-plate capacitor,
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\[
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C=\varepsilon_0\frac{A}{d}.
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\]
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Two other useful vacuum results are
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\[
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C_{\text{spherical}}=4\pi\varepsilon_0\frac{ab}{b-a}
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\]
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for concentric spheres of radii $a$ and $b$ with $b>a$, and
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\[
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C_{\text{cylindrical}}=\frac{2\pi\varepsilon_0 L}{\ln(b/a)}
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\]
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for coaxial cylinders of length $L$ and radii $a$ and $b$ with $b>a$. In every case, capacitance is determined by geometry and the material between the conductors.}
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\ex{Illustrative example}{A vacuum parallel-plate capacitor has plate area
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\[
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A=3A_0
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\]
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and separation
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\[
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d=2d_0.
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\]
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If a reference capacitor with area $A_0$ and separation $d_0$ has capacitance $C_0$, then
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\[
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C_0=\varepsilon_0\frac{A_0}{d_0}.
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\]
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For the new capacitor,
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\[
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C=\varepsilon_0\frac{3A_0}{2d_0}=\frac{3}{2}C_0.
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\]
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So increasing plate area increases capacitance, while increasing plate separation decreases it.}
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\nt{For an ideal capacitor, $C$ is a property of the physical setup, not of the momentary charge or battery setting. In a vacuum parallel-plate capacitor, changing $A$ or $d$ changes $C$, and inserting a dielectric would also change $C$. But if the same capacitor is connected to a larger battery, then $\Delta V$ increases and $Q$ increases proportionally, so the ratio $Q/\Delta V$ stays the same.}
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\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
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\[
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A=2.0\times 10^{-2}\,\mathrm{m^2}
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\]
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and plate separation
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\[
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d=1.5\times 10^{-3}\,\mathrm{m}.
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\]
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It is connected to a battery that maintains a potential difference
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\[
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\Delta V=12.0\,\mathrm{V}.
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\]
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Take
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\[
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\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
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\]
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the capacitance $C$,
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\item the charge magnitude $Q$ on each plate,
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\item the electric field magnitude $E$ between the plates, and
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\item the surface charge density $\sigma$ on either plate.
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\end{enumerate}}
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\sol For part (a), use the parallel-plate formula
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\[
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C=\varepsilon_0\frac{A}{d}.
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\]
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Substitute the given values:
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\[
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C=(8.85\times 10^{-12})\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}\,\mathrm{F}.
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\]
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First evaluate the geometry factor:
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\[
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\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}=13.3.
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\]
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So
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\[
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C=(8.85\times 10^{-12})(13.3)\,\mathrm{F}=1.18\times 10^{-10}\,\mathrm{F}.
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\]
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Therefore,
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\[
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C=1.18\times 10^{-10}\,\mathrm{F}.
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\]
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For part (b), use the definition of capacitance:
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\[
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Q=C\Delta V.
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\]
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Substitute the capacitance and the battery voltage:
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\[
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Q=(1.18\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V}).
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\]
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This gives
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\[
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Q=1.42\times 10^{-9}\,\mathrm{C}.
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\]
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So the plates carry charges $+Q$ and $-Q$ with magnitude
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\[
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Q=1.42\times 10^{-9}\,\mathrm{C}.
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\]
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For part (c), the field between ideal parallel plates is approximately uniform, so the potential difference satisfies
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\[
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\Delta V=Ed.
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\]
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Solve for $E$:
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\[
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E=\frac{\Delta V}{d}.
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\]
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Substitute the values:
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\[
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E=\frac{12.0\,\mathrm{V}}{1.5\times 10^{-3}\,\mathrm{m}}=8.0\times 10^3\,\mathrm{V/m}.
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\]
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Since $1\,\mathrm{V/m}=1\,\mathrm{N/C}$,
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\[
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E=8.0\times 10^3\,\mathrm{N/C}.
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\]
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For part (d), use
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\[
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\sigma=\frac{Q}{A}.
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\]
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Substitute the charge and area:
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\[
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\sigma=\frac{1.42\times 10^{-9}\,\mathrm{C}}{2.0\times 10^{-2}\,\mathrm{m^2}}.
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\]
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This gives
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\[
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\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
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\]
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As a check, the field relation for parallel plates predicts
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\[
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E=\frac{\sigma}{\varepsilon_0}=\frac{7.1\times 10^{-8}}{8.85\times 10^{-12}}\,\mathrm{N/C}
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\approx 8.0\times 10^3\,\mathrm{N/C},
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\]
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which agrees with part (c).
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Therefore,
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\[
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C=1.18\times 10^{-10}\,\mathrm{F},
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\qquad
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Q=1.42\times 10^{-9}\,\mathrm{C},
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\]
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\[
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E=8.0\times 10^3\,\mathrm{N/C},
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\qquad
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\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
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\]
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