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\subsection{Conductors in Electrostatic Equilibrium}
This subsection summarizes the standard properties of conductors in electrostatics and shows how to apply them in AP-style reasoning.
\dfn{Electrostatic equilibrium in a conductor}{Let a conductor contain mobile charges, and let $\vec{E}$ denote the electric field inside the conducting material. The conductor is in \emph{electrostatic equilibrium} when the free charges have redistributed so that there is no sustained macroscopic charge motion. In an ideal conductor, this requires
\[
\vec{E}=\vec{0}
\]
everywhere inside the conducting material. Otherwise a nonzero electric field would exert a force on the free charges and they would continue to move.}
\nt{Free charges in a conductor move in response to any interior electric field. Electrons drift until their rearranged surface charges create an induced field that cancels the original field inside the metal. Equilibrium therefore requires no tangential field along the surface and no field anywhere inside the conducting material. If either existed, the charges would keep moving, so the situation would not be electrostatic.}
\ex{Illustrative example}{A neutral metal sphere is placed in an initially uniform external electric field directed to the right. At first the sphere's free electrons move to the left, leaving an induced positive region on the right surface and an induced negative region on the left surface. This redistribution continues until the induced field cancels the external field everywhere inside the metal. In the final equilibrium state,
\[
\vec{E}=\vec{0}
\]
inside the sphere, and the electric field just outside the surface is perpendicular to the surface.}
\mprop{Standard properties of conductors in electrostatic equilibrium}{Let a conductor be in electrostatic equilibrium. Let $A$ and $B$ be any two points in the conducting material. Let $\hat{n}$ denote the outward unit normal just outside the surface, and let $\sigma$ denote the surface charge density at that point on the surface. Then:
\begin{enumerate}[label=(\alph*)]
\item The electric field inside the conducting material is zero:
\[
\vec{E}_{\mathrm{inside}}=\vec{0}.
\]
\item The conductor is an equipotential, so
\[
V_A=V_B.
\]
In particular, the entire surface of a connected conductor is at one potential.
\item Any excess charge placed on an isolated conductor resides on its surface rather than in the interior bulk material.
\item The electric field just outside the surface is perpendicular to the surface. Its tangential component is zero; otherwise surface charges would move.
\item For an ideal conductor, the field just outside satisfies
\[
\vec{E}_{\mathrm{outside}}=\frac{\sigma}{\varepsilon_0}\hat{n},
\]
because the field just inside is zero.
\end{enumerate}}
\qs{Worked AP-style problem}{An isolated solid conducting sphere has radius
\[
R=0.20\,\mathrm{m}
\]
and net charge
\[
Q=+6.0\times 10^{-9}\,\mathrm{C}.
\]
Let point $A$ be at the center of the sphere, let point $B$ be just inside the metal surface, and let point $C$ be just outside the surface. Let $\hat{n}$ denote the outward unit normal at point $C$. Take
\[
k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}
\]
and proton charge
\[
q_p=+1.60\times 10^{-19}\,\mathrm{C}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the electric field at $A$ and at $B$,
\item the magnitude and direction of the electric field at $C$,
\item the potential difference $V_A-V_B$, and
\item the magnitude and direction of the electric force on a proton placed at $C$.
\end{enumerate}}
\sol Because the sphere is a conductor in electrostatic equilibrium, the electric field everywhere inside the conducting material is zero. Therefore,
\[
\vec{E}(A)=\vec{0},
\qquad
\vec{E}(B)=\vec{0}.
\]
For a charged conducting sphere, the external field is the same as that of a point charge $Q$ located at the center. Point $C$ is just outside the surface, so its distance from the center is essentially
\[
r=R=0.20\,\mathrm{m}.
\]
Thus the field magnitude there is
\[
E(C)=k\frac{Q}{R^2}.
\]
Substitute the given values:
\[
E(C)=\left(8.99\times 10^9\right)\frac{6.0\times 10^{-9}}{(0.20)^2}\,\mathrm{N/C}.
\]
Since
\[
(0.20)^2=0.040,
\]
we get
\[
E(C)=\frac{53.94}{0.040}\,\mathrm{N/C}=1.35\times 10^3\,\mathrm{N/C}.
\]
Because $Q$ is positive, the field points radially outward, perpendicular to the surface. So
\[
\vec{E}(C)=(1.35\times 10^3\,\mathrm{N/C})\hat{n},
\]
where $\hat{n}$ is the outward normal.
Next, the entire conductor is an equipotential in electrostatic equilibrium. Since both $A$ and $B$ lie in the conductor,
\[
V_A=V_B.
\]
Therefore,
\[
V_A-V_B=0\,\mathrm{V}.
\]
Finally, the electric force on a proton at $C$ has magnitude
\[
F=q_pE(C).
\]
Substitute the values:
\[
F=\left(1.60\times 10^{-19}\,\mathrm{C}\right)\left(1.35\times 10^3\,\mathrm{N/C}\right).
\]
This gives
\[
F=2.16\times 10^{-16}\,\mathrm{N}.
\]
Because the proton has positive charge, its force is in the same direction as $\vec{E}$, so the force is radially outward:
\[
\vec{F}=(2.16\times 10^{-16}\,\mathrm{N})\hat{n}.
\]
Therefore,
\[
\vec{E}(A)=\vec{0},
\qquad
\vec{E}(B)=\vec{0},
\]
\[
\vec{E}(C)=(1.35\times 10^3\,\mathrm{N/C})\hat{n},
\]
\[
V_A-V_B=0\,\mathrm{V},
\qquad
\vec{F}=(2.16\times 10^{-16}\,\mathrm{N})\hat{n}.
\]

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\subsection{Charge Redistribution, Contact, Induction, and Grounding}
This subsection explains how charge moves on conductors, why touching conductors exchange charge until they reach one potential, and how induction plus grounding can leave an object with a net charge without direct contact.
\dfn{Charge redistribution, contact, induction, and grounding}{Let $q_A,q_B,\dots$ denote the net charges on conductors $A,B,\dots$, and let $V_A,V_B,\dots$ denote their electric potentials.
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item \emph{Charge redistribution}: mobile charge in a conductor moves through the material until electrostatic equilibrium is reached.
\item \emph{Charging by contact}: if conductors touch or are connected by a wire, charge can flow between them until the connected conductors reach one common potential.
\item \emph{Charging by induction}: a nearby charged object causes charges in another object to separate without direct contact.
\item \emph{Grounding}: connecting an object to Earth allows charge to flow between the object and Earth, which acts as a very large charge reservoir.
\end{enumerate}}
\nt{Charge redistribution in a conductor continues until electrostatic equilibrium is reached. In electrostatics, that means the electric field inside the conductor is zero and all parts of one connected conductor are at the same potential. Therefore contact problems are usually solved with charge conservation plus an equal-potential idea. For identical small conducting spheres, symmetry makes equal potential equivalent to equal final charge, but for conductors of different size or shape equal potential does not generally mean equal charge.}
\ex{Illustrative example}{Two identical small conducting spheres $A$ and $B$ are far apart initially. Let their initial charges be
\[
q_{A,i}=+6.0\,\mathrm{nC},
\qquad
q_{B,i}=0.
\]
If the spheres are touched together and then separated, charge is conserved and the identical spheres must finish with equal charge. The total charge is
\[
q_{\mathrm{tot}}=q_{A,i}+q_{B,i}=+6.0\,\mathrm{nC},
\]
so each sphere ends with
\[
q_{A,f}=q_{B,f}=\frac{q_{\mathrm{tot}}}{2}=+3.0\,\mathrm{nC}.
\]
This is a contact example: the charge does not disappear or appear; it redistributes until the connected conductors reach electrostatic equilibrium.}
\mprop{Practical relations and qualitative rules}{Let $q_{\mathrm{tot}}$ denote the total charge of a chosen isolated system, let $q_{A,i},q_{B,i}$ and $q_{A,f},q_{B,f}$ denote initial and final charges, let $q_{\mathrm{object},f}$ denote the final charge of a grounded object, and let $\Delta q_{\mathrm{Earth}}$ denote the charge change of Earth.
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item For any isolated system,
\[
q_{\mathrm{tot},f}=q_{\mathrm{tot},i}.
\]
\item If two identical small conducting spheres touch and then separate,
\[
q_{A,f}=q_{B,f}=\frac{q_{A,i}+q_{B,i}}{2}.
\]
\item Induction without grounding redistributes charge but does not change the net charge of the induced object. For an initially neutral conductor, the near side becomes opposite in sign to the external object, the far side becomes the same sign, and the net charge remains zero.
\item Grounding is charge exchange with Earth. If a positive object is nearby, electrons can flow from Earth onto the grounded conductor. If a negative object is nearby, electrons can flow from the conductor to Earth.
\item In the usual induction-charging sequence, the ground connection is removed first and the external charged object is removed second. Then the conductor is left with a net charge opposite in sign to the inducing object. If the object-Earth system is initially neutral, charge conservation for that system gives
\[
q_{\mathrm{object},f}+\Delta q_{\mathrm{Earth}}=0.
\]
\end{enumerate}}
\qs{Worked AP-style problem}{A neutral metal sphere $S$ is mounted on an insulating stand. A negatively charged rod is brought near the left side of the sphere but does not touch it. While the rod remains in place, the sphere is briefly connected to Earth by a wire. The grounding wire is then removed, and finally the rod is taken away.
Find:
\begin{enumerate}[label=(\alph*)]
\item the signs of the induced charges on the left and right sides of the sphere before the grounding wire is attached,
\item the direction of electron flow while the sphere is grounded,
\item the net charge left on the sphere after the full sequence, and
\item whether charge conservation is violated by the sphere ending with a net charge.
\end{enumerate}}
\sol Before the sphere is grounded, the rod is negative, so it repels mobile electrons in the metal sphere. Those electrons shift toward the right side, farther from the rod.
Therefore, before grounding,
\begin{itemize}
\item the left side of the sphere is induced to be positive, and
\item the right side of the sphere is induced to be negative.
\end{itemize}
Even though the charges have separated, the sphere is still overall neutral at this stage because no charge has entered or left the sphere.
Now the sphere is connected to Earth while the negative rod remains nearby. The excess electrons on the sphere are repelled by the negative rod, and the grounding wire gives those electrons a path to leave. Thus electrons flow
\[
\text{from the sphere to Earth}.
\]
Next, the grounding wire is removed while the rod is still present. Because the sphere is no longer connected to Earth, the electrons that left cannot return. The sphere has lost some electrons, so it now has a net positive charge.
Finally, the rod is taken away. With the rod gone, the remaining positive charge is no longer pulled to one side, so it redistributes over the outer surface of the sphere. The final result is that the sphere is left
\[
\text{positively charged}.
\]
Charge conservation is not violated. The correct isolated system is the sphere together with Earth. During grounding, electrons moved from the sphere to Earth, so the sphere became positive and Earth gained an equal amount of negative charge. If the final charge on the sphere is $q_{S,f}>0$, then
\[
q_{S,f}+\Delta q_{\mathrm{Earth}}=0.
\]
So the total charge of the combined system is unchanged; the process is charge transfer, not charge creation.

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

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

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\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
\[
U_0=9.56\times 10^{-9}\,\mathrm{J},
\qquad
U=2.87\times 10^{-8}\,\mathrm{J}.
\]