\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,
\[
C=1.77\times 10^{-10}\,\mathrm{F},
\qquad
Q=3.54\times 10^{-9}\,\mathrm{C},
\]
\[
U_C=3.54\times 10^{-8}\,\mathrm{J},
\qquad
u_E=1.77\times 10^{-3}\,\mathrm{J/m^3}.
\]