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concepts/em/u10/e10-3-capacitance.tex

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+\subsection{Capacitance and Capacitor Geometries}
+
+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.
+
+\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
+\[
+\Delta V=V_{\text{high}}-V_{\text{low}}
+\]
+denote the magnitude of the potential difference between them. The \emph{capacitance} $C$ of the system is defined by
+\[
+C=\frac{Q}{\Delta V}.
+\]
+Capacitance measures how much charge is stored per unit potential difference. Its SI unit is the farad:
+\[
+1\,\mathrm{F}=1\,\mathrm{C/V}.
+\]}
+
+\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
+\[
+C=\frac{Q}{\Delta V}.
+\]
+
+For a vacuum parallel-plate capacitor with plate area $A$ and plate separation $d$, define the surface charge density by
+\[
+\sigma=\frac{Q}{A}.
+\]
+Ignoring edge effects, Gauss's law gives the nearly uniform electric field between the plates:
+\[
+\vec{E}=\frac{\sigma}{\varepsilon_0}\,\hat{n},
+\qquad
+E=\frac{\sigma}{\varepsilon_0}.
+\]
+Using the potential-drop relation with a path across the gap,
+\[
+\Delta V=\left| -\int \vec{E}\cdot d\vec{\ell} \right|=Ed,
+\]
+so
+\[
+\Delta V=\frac{\sigma d}{\varepsilon_0}=\frac{Qd}{\varepsilon_0 A}.
+\]
+Substitute into $C=Q/\Delta V$:
+\[
+C=\frac{Q}{Qd/(\varepsilon_0 A)}=\varepsilon_0\frac{A}{d}.
+\]
+Thus, for an ideal vacuum parallel-plate capacitor,
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+
+Two other useful vacuum results are
+\[
+C_{\text{spherical}}=4\pi\varepsilon_0\frac{ab}{b-a}
+\]
+for concentric spheres of radii $a$ and $b$ with $b>a$, and
+\[
+C_{\text{cylindrical}}=\frac{2\pi\varepsilon_0 L}{\ln(b/a)}
+\]
+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.}
+
+\ex{Illustrative example}{A vacuum parallel-plate capacitor has plate area
+\[
+A=3A_0
+\]
+and separation
+\[
+d=2d_0.
+\]
+If a reference capacitor with area $A_0$ and separation $d_0$ has capacitance $C_0$, then
+\[
+C_0=\varepsilon_0\frac{A_0}{d_0}.
+\]
+For the new capacitor,
+\[
+C=\varepsilon_0\frac{3A_0}{2d_0}=\frac{3}{2}C_0.
+\]
+So increasing plate area increases capacitance, while increasing plate separation decreases it.}
+
+\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.}
+
+\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
+\[
+A=2.0\times 10^{-2}\,\mathrm{m^2}
+\]
+and plate separation
+\[
+d=1.5\times 10^{-3}\,\mathrm{m}.
+\]
+It is connected to a battery that maintains a potential difference
+\[
+\Delta V=12.0\,\mathrm{V}.
+\]
+Take
+\[
+\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
+\]
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the capacitance $C$,
+\item the charge magnitude $Q$ on each plate,
+\item the electric field magnitude $E$ between the plates, and
+\item the surface charge density $\sigma$ on either plate.
+\end{enumerate}}
+
+\sol For part (a), use the parallel-plate formula
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+Substitute the given values:
+\[
+C=(8.85\times 10^{-12})\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}\,\mathrm{F}.
+\]
+First evaluate the geometry factor:
+\[
+\frac{2.0\times 10^{-2}}{1.5\times 10^{-3}}=13.3.
+\]
+So
+\[
+C=(8.85\times 10^{-12})(13.3)\,\mathrm{F}=1.18\times 10^{-10}\,\mathrm{F}.
+\]
+Therefore,
+\[
+C=1.18\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (b), use the definition of capacitance:
+\[
+Q=C\Delta V.
+\]
+Substitute the capacitance and the battery voltage:
+\[
+Q=(1.18\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V}).
+\]
+This gives
+\[
+Q=1.42\times 10^{-9}\,\mathrm{C}.
+\]
+So the plates carry charges $+Q$ and $-Q$ with magnitude
+\[
+Q=1.42\times 10^{-9}\,\mathrm{C}.
+\]
+
+For part (c), the field between ideal parallel plates is approximately uniform, so the potential difference satisfies
+\[
+\Delta V=Ed.
+\]
+Solve for $E$:
+\[
+E=\frac{\Delta V}{d}.
+\]
+Substitute the values:
+\[
+E=\frac{12.0\,\mathrm{V}}{1.5\times 10^{-3}\,\mathrm{m}}=8.0\times 10^3\,\mathrm{V/m}.
+\]
+Since $1\,\mathrm{V/m}=1\,\mathrm{N/C}$,
+\[
+E=8.0\times 10^3\,\mathrm{N/C}.
+\]
+
+For part (d), use
+\[
+\sigma=\frac{Q}{A}.
+\]
+Substitute the charge and area:
+\[
+\sigma=\frac{1.42\times 10^{-9}\,\mathrm{C}}{2.0\times 10^{-2}\,\mathrm{m^2}}.
+\]
+This gives
+\[
+\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
+\]
+
+As a check, the field relation for parallel plates predicts
+\[
+E=\frac{\sigma}{\varepsilon_0}=\frac{7.1\times 10^{-8}}{8.85\times 10^{-12}}\,\mathrm{N/C}
+\approx 8.0\times 10^3\,\mathrm{N/C},
+\]
+which agrees with part (c).
+
+Therefore,
+\[
+C=1.18\times 10^{-10}\,\mathrm{F},
+\qquad
+Q=1.42\times 10^{-9}\,\mathrm{C},
+\]
+\[
+E=8.0\times 10^3\,\mathrm{N/C},
+\qquad
+\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
+\]