158 lines
4.6 KiB
TeX
158 lines
4.6 KiB
TeX
\subsection{Equipotentials and Energy Conservation for Moving Charges}
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This subsection explains how equipotential curves or surfaces encode the direction of $\vec{E}$ and how potential differences determine changes in kinetic and electric potential energy for moving charges.
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\dfn{Equipotentials and the energy-change relation}{Let $V(\vec{r})$ denote the electric potential at position vector $\vec{r}$. An \emph{equipotential} is a curve in two dimensions or a surface in three dimensions on which the potential has one constant value:
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\[
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V(\vec{r})=\text{constant}.
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\]
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If a charge $q$ moves from point $A$ to point $B$, define
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\[
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\Delta V=V_B-V_A
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\]
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and
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\[
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\Delta U=U_B-U_A.
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\]
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Then the change in electric potential energy is
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\[
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\Delta U=q\Delta V.
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\]
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Thus potential difference tells how the electric potential energy of a chosen charge changes between two points.}
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\thm{Equipotentials, no work, and kinetic-energy change}{Let $d\vec{\ell}$ denote an infinitesimal displacement in an electrostatic field $\vec{E}$. Then
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\[
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dV=-\vec{E}\cdot d\vec{\ell}.
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\]
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If the displacement is along an equipotential, then $dV=0$, so
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\[
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\vec{E}\cdot d\vec{\ell}=0.
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\]
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Therefore the electric field is perpendicular to an equipotential, and the electric force does no work on a charge moved along an equipotential:
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\[
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W_{\mathrm{elec}}=q\int \vec{E}\cdot d\vec{\ell}=0.
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\]
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For any motion of a charge $q$ from $A$ to $B$ in electrostatics,
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\[
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\Delta U=q\Delta V.
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\]
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If only the electric force does work, conservation of mechanical energy gives
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\[
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K_i+U_i=K_f+U_f,
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\]
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so
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\[
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\Delta K=K_f-K_i=-\Delta U=-q\Delta V.
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\]
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Hence a charge speeds up when its electric potential energy decreases.}
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\ex{Illustrative example}{Points $A$ and $B$ lie on the same equipotential,
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\[
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V_A=V_B=120\,\mathrm{V}.
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\]
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Let a proton of charge
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\[
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q=+e=+1.60\times 10^{-19}\,\mathrm{C}
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\]
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move from $A$ to $B$.
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Because the two points are on the same equipotential,
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\[
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\Delta V=V_B-V_A=0.
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\]
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So the change in electric potential energy is
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\[
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\Delta U=q\Delta V=0,
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\]
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and the work done by the electric field is also zero. The field may still be present, but along that displacement it is perpendicular to the motion.}
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\nt{Be careful to distinguish \emph{lower potential} from \emph{lower potential energy}. Since $\Delta U=q\Delta V$, a positive charge has lower potential energy at lower potential, but a negative charge has lower potential energy at \emph{higher} potential. Thus a positive charge released from rest tends to speed up toward lower $V$, whereas an electron released from rest tends to speed up toward higher $V$. In both cases the rule is the same: the charge moves spontaneously in the direction that makes $U$ decrease and $K$ increase.}
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\qs{Worked AP-style problem}{Two large parallel plates create a uniform electrostatic region. Let point $A$ be near the negative plate and point $B$ be near the positive plate. The potentials are
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\[
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V_A=100\,\mathrm{V},
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\qquad
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V_B=400\,\mathrm{V}.
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\]
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An electron is released from rest at point $A$ and moves to point $B$. Let the electron charge be
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\[
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q=-1.60\times 10^{-19}\,\mathrm{C}
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\]
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and the electron mass be
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\[
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m_e=9.11\times 10^{-31}\,\mathrm{kg}.
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\]
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the potential difference $\Delta V=V_B-V_A$,
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\item the change in electric potential energy $\Delta U$,
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\item the change in kinetic energy $\Delta K$, and
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\item the electron's speed at point $B$.
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\end{enumerate}}
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\sol First compute the potential difference:
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\[
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\Delta V=V_B-V_A=400\,\mathrm{V}-100\,\mathrm{V}=300\,\mathrm{V}.
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\]
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For part (b), use the relation
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\[
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\Delta U=q\Delta V.
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\]
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Substitute the electron charge and the potential difference:
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\[
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\Delta U=(-1.60\times 10^{-19}\,\mathrm{C})(300\,\mathrm{V}).
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\]
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Since $1\,\mathrm{V}=1\,\mathrm{J/C}$,
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\[
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\Delta U=-4.80\times 10^{-17}\,\mathrm{J}.
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\]
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For part (c), only the electric force does work, so
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\[
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\Delta K=-\Delta U.
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\]
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Therefore,
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\[
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\Delta K=+4.80\times 10^{-17}\,\mathrm{J}.
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\]
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This sign makes sense. The electron moves toward higher potential, but because its charge is negative, that motion lowers its electric potential energy and increases its kinetic energy.
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For part (d), the electron starts from rest, so
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\[
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K_i=0.
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\]
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Thus
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\[
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K_f=\Delta K=4.80\times 10^{-17}\,\mathrm{J}.
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\]
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Use the kinetic-energy formula
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\[
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K_f=\frac12 m_e v^2.
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\]
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Solve for the speed $v$:
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\[
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v=\sqrt{\frac{2K_f}{m_e}}.
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\]
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Substitute the values:
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\[
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v=\sqrt{\frac{2(4.80\times 10^{-17}\,\mathrm{J})}{9.11\times 10^{-31}\,\mathrm{kg}}}.
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\]
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This gives
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\[
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v=1.03\times 10^7\,\mathrm{m/s}.
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\]
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Therefore,
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\[
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\Delta V=300\,\mathrm{V},
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\qquad
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\Delta U=-4.80\times 10^{-17}\,\mathrm{J},
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\]
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\[
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\Delta K=+4.80\times 10^{-17}\,\mathrm{J},
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\qquad
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v=1.03\times 10^7\,\mathrm{m/s}.
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\]
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