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concepts/em/u10/e10-4-capacitor-energy.tex

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+\subsection{Energy Stored in Capacitors and Fields}
+
+This subsection develops the standard formulas for capacitor energy and connects them to the electric-field energy density in vacuum.
+
+\dfn{Capacitor energy and the field-energy viewpoint}{Let a capacitor have capacitance $C$, let the magnitude of the charge on either conductor be $Q$, and let
+\[
+\Delta V=V_{\text{high}}-V_{\text{low}}
+\]
+denote the magnitude of the potential difference between the conductors. The \emph{energy stored in the capacitor}, denoted $U_C$, is the electric potential energy gained while the capacitor is charged from $0$ to $Q$.
+
+In the field viewpoint, let $u_E$ denote the \emph{electric-field energy density}, measured in joules per cubic meter. For a vacuum region with electric-field magnitude $E$, the stored energy can be regarded as distributed through the field-filled volume.}
+
+\thm{Equivalent capacitor-energy formulas and vacuum field-energy density}{Let a capacitor have capacitance $C$, plate charges $\pm Q$, and potential-difference magnitude $\Delta V$. Then the stored energy can be written in any of the equivalent forms
+\[
+U_C=\frac12 Q\Delta V=\frac12 C(\Delta V)^2=\frac{Q^2}{2C}.
+\]
+For a vacuum region in which the electric-field magnitude is $E$, the electric-field energy density is
+\[
+u_E=\frac12 \varepsilon_0 E^2.
+\]
+For an ideal vacuum parallel-plate capacitor of plate area $A$ and separation $d$, the field is approximately uniform, so
+\[
+U_C=u_E(Ad).
+\]}
+
+\nt{The energy grows quadratically with charge or voltage because charging is cumulative. For a capacitor with fixed $C$, the potential difference is not constant while it charges: it rises in proportion to the accumulated charge, since $\Delta V=Q/C$. That means later bits of charge are harder to add than earlier ones. The average potential difference during charging is therefore half the final value, which is why $U_C=\tfrac12 Q\Delta V$ and why doubling $Q$ or $\Delta V$ makes the stored energy four times as large for the same capacitor.}
+
+\pf{Short derivation from charging work}{Let $q$ denote the instantaneous charge on the capacitor during a slow charging process from $q=0$ to $q=Q$. At that instant, the potential difference is
+\[
+\Delta V(q)=\frac{q}{C}.
+\]
+To move an additional small charge $dq$ onto the capacitor, the external work required is
+\[
+dU=\Delta V(q)\,dq=\frac{q}{C}\,dq.
+\]
+Integrate from $0$ to $Q$:
+\[
+U_C=\int_0^Q \frac{q}{C}\,dq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q=\frac{Q^2}{2C}.
+\]
+Using
+\[
+Q=C\Delta V,
+\]
+this becomes
+\[
+U_C=\frac12 Q\Delta V=\frac12 C(\Delta V)^2.
+\]
+For an ideal vacuum parallel-plate capacitor,
+\[
+C=\varepsilon_0\frac{A}{d}
+\qquad \text{and} \qquad
+E=\frac{\Delta V}{d}.
+\]
+Substituting into $U_C=\tfrac12 C(\Delta V)^2$ gives
+\[
+U_C=\frac12 \left(\varepsilon_0\frac{A}{d}\right)(Ed)^2=\frac12 \varepsilon_0 E^2(Ad).
+\]
+Dividing by the volume $Ad$ yields
+\[
+u_E=\frac{U_C}{Ad}=\frac12 \varepsilon_0 E^2.
+\]}
+
+\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
+\[
+d=1.0\times 10^{-3}\,\mathrm{m}.
+\]
+It is connected to a battery that maintains a potential difference
+\[
+\Delta V=20.0\,\mathrm{V}.
+\]
+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 stored energy $U_C$, and
+\item the electric-field energy density $u_E$ between the plates, then verify that $U_C=u_E(Ad)$.
+\end{enumerate}}
+
+\sol For part (a), use the parallel-plate formula
+\[
+C=\varepsilon_0\frac{A}{d}.
+\]
+Substitute the given values:
+\[
+C=(8.85\times 10^{-12})\frac{2.0\times 10^{-2}}{1.0\times 10^{-3}}\,\mathrm{F}.
+\]
+The geometry factor is
+\[
+\frac{2.0\times 10^{-2}}{1.0\times 10^{-3}}=20.0,
+\]
+so
+\[
+C=(8.85\times 10^{-12})(20.0)\,\mathrm{F}=1.77\times 10^{-10}\,\mathrm{F}.
+\]
+
+For part (b), use
+\[
+Q=C\Delta V.
+\]
+Then
+\[
+Q=(1.77\times 10^{-10}\,\mathrm{F})(20.0\,\mathrm{V})=3.54\times 10^{-9}\,\mathrm{C}.
+\]
+
+For part (c), use any equivalent energy formula. Using $U_C=\tfrac12 C(\Delta V)^2$,
+\[
+U_C=\frac12 (1.77\times 10^{-10})(20.0)^2\,\mathrm{J}.
+\]
+Since
+\[
+(20.0)^2=400,
+\]
+we obtain
+\[
+U_C=\frac12 (1.77\times 10^{-10})(400)\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J}.
+\]
+As a check,
+\[
+U_C=\frac12 Q\Delta V=\frac12 (3.54\times 10^{-9})(20.0)\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J},
+\]
+which agrees.
+
+For part (d), first find the field magnitude between the plates:
+\[
+E=\frac{\Delta V}{d}=\frac{20.0\,\mathrm{V}}{1.0\times 10^{-3}\,\mathrm{m}}=2.0\times 10^4\,\mathrm{V/m}.
+\]
+Now use the vacuum field-energy density formula:
+\[
+u_E=\frac12 \varepsilon_0 E^2.
+\]
+Substitute the values:
+\[
+u_E=\frac12 (8.85\times 10^{-12})(2.0\times 10^4)^2\,\mathrm{J/m^3}.
+\]
+Because
+\[
+(2.0\times 10^4)^2=4.0\times 10^8,
+\]
+we get
+\[
+u_E=\frac12 (8.85\times 10^{-12})(4.0\times 10^8)\,\mathrm{J/m^3}=1.77\times 10^{-3}\,\mathrm{J/m^3}.
+\]
+
+To verify the field viewpoint, compute the volume between the plates:
+\[
+Ad=(2.0\times 10^{-2})(1.0\times 10^{-3})\,\mathrm{m^3}=2.0\times 10^{-5}\,\mathrm{m^3}.
+\]
+Then
+\[
+u_E(Ad)=(1.77\times 10^{-3})(2.0\times 10^{-5})\,\mathrm{J}=3.54\times 10^{-8}\,\mathrm{J}.
+\]
+This matches the capacitor-energy result.
+
+Therefore,
+\[
+C=1.77\times 10^{-10}\,\mathrm{F},
+\qquad
+Q=3.54\times 10^{-9}\,\mathrm{C},
+\]
+\[
+U_C=3.54\times 10^{-8}\,\mathrm{J},
+\qquad
+u_E=1.77\times 10^{-3}\,\mathrm{J/m^3}.
+\]