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concepts/em/u10/e10-1-conductors.tex

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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,
+\[
+\vec{E}(A)=\vec{0},
+\qquad
+\vec{E}(B)=\vec{0},
+\]
+\[
+\vec{E}(C)=(1.35\times 10^3\,\mathrm{N/C})\hat{n},
+\]
+\[
+V_A-V_B=0\,\mathrm{V},
+\qquad
+\vec{F}=(2.16\times 10^{-16}\,\mathrm{N})\hat{n}.
+\]