\subsection{Dielectrics and Polarization}

This subsection explains how dielectric materials polarize in an electric field and how that polarization changes capacitor behavior in the two common AP settings: fixed charge and fixed voltage.

\dfn{Dielectric, polarization, and dielectric constant}{A \emph{dielectric} is an insulating material placed between capacitor plates. Its charges are not free to flow through the material as they do in a conductor, but the positive and negative charges within its atoms or molecules can shift slightly.

This internal charge separation is called \emph{polarization}. In a polarized dielectric, the side nearer the positive plate becomes slightly negative and the side nearer the negative plate becomes slightly positive, so the dielectric produces an induced electric field that opposes the original field between the plates.

Let $C_0$ denote the capacitance of a capacitor when the gap is vacuum or air, and let $C$ denote the capacitance when a dielectric fully fills the gap. The \emph{dielectric constant} $\kappa$ of the material is the factor by which the capacitance increases:
\[
C=\kappa C_0,
\qquad
\kappa>1.
\]
For an ideal parallel-plate capacitor of plate area $A$ and separation $d$,
\[
C_0=\varepsilon_0\frac{A}{d}
\qquad \text{and therefore} \qquad
C=\kappa\varepsilon_0\frac{A}{d}.
\]}

\nt{The dielectric does not cancel the capacitor's field completely. Instead, polarization creates \emph{bound} charges on the dielectric surfaces, and those bound charges produce a field opposite the field from the capacitor plates. That opposition reduces the net interior field.

The key AP distinction is what stays fixed during insertion.
\begin{itemize}
\item If the capacitor is isolated after charging, then the free charge on the plates cannot change, so this is a fixed-$Q$ situation. The dielectric lowers the field and voltage.
\item If the capacitor remains connected to a battery, then the battery keeps the potential difference fixed, so this is a fixed-$V$ situation. The dielectric still tends to reduce the field, but the battery pushes additional charge onto the plates until the original voltage is restored.
\end{itemize}}

\ex{Illustrative example}{An isolated capacitor has initial capacitance
\[
C_0=10.0\,\mathrm{pF}
\]
and is charged to
\[
V_0=30.0\,\mathrm{V}.
\]
The battery is disconnected, and a dielectric with
\[
\kappa=2.5
\]
is inserted so that it fully fills the gap.

The new capacitance is
\[
C=\kappa C_0=(2.5)(10.0\,\mathrm{pF})=25.0\,\mathrm{pF}.
\]
Because the capacitor is isolated, the charge stays constant:
\[
Q=Q_0=C_0V_0=(10.0\,\mathrm{pF})(30.0\,\mathrm{V})=300\,\mathrm{pC}.
\]
The new voltage is therefore
\[
V=\frac{Q}{C}=\frac{300\,\mathrm{pC}}{25.0\,\mathrm{pF}}=12.0\,\mathrm{V}.
\]
So inserting the dielectric increases the capacitance while reducing the voltage for the same stored charge.}

\mprop{Main capacitor relations for a dielectric that completely fills the gap}{Let $C_0$, $Q_0$, $V_0$, $E_0$, and $U_0$ denote the capacitance, plate-charge magnitude, potential difference, electric-field magnitude, and stored energy before insertion. Let $C$, $Q$, $V$, $E$, and $U$ denote the corresponding quantities after a dielectric of constant $\kappa$ fully fills the gap.

For an ideal parallel-plate capacitor,
\[
C=\kappa C_0=\kappa\varepsilon_0\frac{A}{d}.
\]

If the capacitor is \emph{isolated} after charging (fixed $Q$), then
\[
Q=Q_0,
\qquad
V=\frac{V_0}{\kappa},
\qquad
E=\frac{E_0}{\kappa},
\qquad
U=\frac{U_0}{\kappa}.
\]

If the capacitor remains \emph{connected to a battery} (fixed $V$), then
\[
V=V_0,
\qquad
Q=\kappa Q_0,
\qquad
E=E_0,
\qquad
U=\kappa U_0.
\]

In both cases, the dielectric increases capacitance by the factor $\kappa$. What changes is which other quantity the circuit constraint forces to stay constant.}

\qs{Worked AP-style problem}{A vacuum parallel-plate capacitor has plate area
\[
A=1.50\times 10^{-2}\,\mathrm{m^2}
\]
and plate separation
\[
d=1.00\times 10^{-3}\,\mathrm{m}.
\]
It remains connected to a battery that maintains a potential difference
\[
V_0=12.0\,\mathrm{V}.
\]
A dielectric with dielectric constant
\[
\kappa=3.00
\]
is inserted so that it completely fills the gap. Take
\[
\varepsilon_0=8.85\times 10^{-12}\,\mathrm{F/m}.
\]

Find:
\begin{enumerate}[label=(\alph*)]
\item the initial capacitance $C_0$,
\item the final capacitance $C$,
\item the initial and final plate-charge magnitudes $Q_0$ and $Q$,
\item the electric-field magnitude between the plates before and after insertion, and
\item the initial and final stored energies $U_0$ and $U$.
\end{enumerate}}

\sol Because the dielectric fully fills the gap, the capacitance increases by the factor $\kappa$:
\[
C=\kappa C_0.
\]
Because the battery remains connected, the voltage stays fixed at
\[
V=V_0=12.0\,\mathrm{V}.
\]

For part (a), first find the initial vacuum capacitance:
\[
C_0=\varepsilon_0\frac{A}{d}.
\]
Substitute the given values:
\[
C_0=(8.85\times 10^{-12})\frac{1.50\times 10^{-2}}{1.00\times 10^{-3}}\,\mathrm{F}.
\]
The geometry factor is
\[
\frac{1.50\times 10^{-2}}{1.00\times 10^{-3}}=15.0,
\]
so
\[
C_0=(8.85\times 10^{-12})(15.0)\,\mathrm{F}=1.33\times 10^{-10}\,\mathrm{F}.
\]

For part (b), multiply by $\kappa=3.00$:
\[
C=\kappa C_0=(3.00)(1.33\times 10^{-10}\,\mathrm{F})=3.98\times 10^{-10}\,\mathrm{F}.
\]

For part (c), before insertion the plate-charge magnitude is
\[
Q_0=C_0V_0.
\]
Thus,
\[
Q_0=(1.33\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V})=1.59\times 10^{-9}\,\mathrm{C}.
\]
After insertion, the voltage is unchanged but the capacitance is larger, so
\[
Q=CV=(3.98\times 10^{-10}\,\mathrm{F})(12.0\,\mathrm{V})=4.78\times 10^{-9}\,\mathrm{C}.
\]
This is also consistent with
\[
Q=\kappa Q_0=(3.00)(1.59\times 10^{-9}\,\mathrm{C})=4.78\times 10^{-9}\,\mathrm{C}.
\]

For part (d), the electric-field magnitude in an ideal parallel-plate capacitor is related to the maintained potential difference by
\[
E=\frac{V}{d}.
\]
Before insertion,
\[
E_0=\frac{V_0}{d}=\frac{12.0\,\mathrm{V}}{1.00\times 10^{-3}\,\mathrm{m}}=1.20\times 10^4\,\mathrm{V/m}.
\]
Because the battery keeps the same voltage across the same plate spacing, the final field magnitude is the same:
\[
E=\frac{V}{d}=\frac{12.0\,\mathrm{V}}{1.00\times 10^{-3}\,\mathrm{m}}=1.20\times 10^4\,\mathrm{V/m}.
\]

For part (e), the initial stored energy is
\[
U_0=\frac12 C_0V_0^2.
\]
Substitute the values:
\[
U_0=\frac12 (1.33\times 10^{-10})(12.0)^2\,\mathrm{J}.
\]
Since
\[
(12.0)^2=144,
\]
we get
\[
U_0=\frac12 (1.33\times 10^{-10})(144)\,\mathrm{J}=9.56\times 10^{-9}\,\mathrm{J}.
\]
After insertion,
\[
U=\frac12 CV^2=\frac12 (3.98\times 10^{-10})(12.0)^2\,\mathrm{J}=2.87\times 10^{-8}\,\mathrm{J}.
\]
Equivalently,
\[
U=\kappa U_0=(3.00)(9.56\times 10^{-9}\,\mathrm{J})=2.87\times 10^{-8}\,\mathrm{J}.
\]

Therefore,
\[
C_0=1.33\times 10^{-10}\,\mathrm{F},
\qquad
C=3.98\times 10^{-10}\,\mathrm{F},
\]
\[
Q_0=1.59\times 10^{-9}\,\mathrm{C},
\qquad
Q=4.78\times 10^{-9}\,\mathrm{C},
\]
\[
E_0=E=1.20\times 10^4\,\mathrm{V/m},
\]
and
\[
U_0=9.56\times 10^{-9}\,\mathrm{J},
\qquad
U=2.87\times 10^{-8}\,\mathrm{J}.
\]
