\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,
\[
C=1.18\times 10^{-10}\,\mathrm{F},
\qquad
Q=1.42\times 10^{-9}\,\mathrm{C},
\]
\[
E=8.0\times 10^3\,\mathrm{N/C},
\qquad
\sigma=7.1\times 10^{-8}\,\mathrm{C/m^2}.
\]