160 lines
4.6 KiB
TeX
160 lines
4.6 KiB
TeX
\subsection{The Field-Potential Relation}
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This subsection connects electric field and electric potential, both globally through a line integral and locally through the slope or gradient of the potential.
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\dfn{Potential difference from the electric field}{Let $A$ and $B$ be two points in an electrostatic region, let $C$ be any path from $A$ to $B$, and let $d\vec{\ell}$ denote an infinitesimal displacement along that path. If the electric field is $\vec{E}$, then the infinitesimal potential change is
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\[
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dV=-\vec{E}\cdot d\vec{\ell}.
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\]
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Integrating from $A$ to $B$ gives the potential difference
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\[
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\Delta V=V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{\ell}.
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\]
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For electrostatics, this value is independent of the path because the electric field is conservative.}
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\thm{Global and local field-potential relations}{Let $V(\vec{r})$ denote the electric potential at position vector $\vec{r}$, and let $\vec{E}(\vec{r})$ denote the electric field there. Then in electrostatics,
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\[
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\Delta V=V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{\ell}.
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\]
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Locally,
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\[
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dV=-\vec{E}\cdot d\vec{\ell}.
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\]
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Since also
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\[
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dV=\nabla V\cdot d\vec{\ell},
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\]
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comparison gives the vector relation
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\[
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\vec{E}=-\nabla V.
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\]
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For one-dimensional motion along the $x$-axis,
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\[
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E_x=-\frac{dV}{dx}.
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\]
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Thus the $x$-component of the electric field is the negative slope of the potential graph.}
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\pf{Derivation from work per unit charge}{Let a charge $q$ move through an infinitesimal displacement $d\vec{\ell}$ in an electric field $\vec{E}$. The electric force is
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\[
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\vec{F}=q\vec{E},
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\]
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so the infinitesimal work done by the field is
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\[
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dW=\vec{F}\cdot d\vec{\ell}=q\vec{E}\cdot d\vec{\ell}.
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\]
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Electric potential difference is potential-energy change per unit charge, so
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\[
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dV=\frac{dU}{q}.
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\]
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Because the work done by the electric field decreases electric potential energy,
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\[
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dU=-dW.
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\]
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Therefore,
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\[
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dV=\frac{-dW}{q}=-\vec{E}\cdot d\vec{\ell}.
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\]
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Integrating between two points gives
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\[
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\Delta V=-\int \vec{E}\cdot d\vec{\ell}.
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\]
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Comparing this with the differential identity $dV=\nabla V\cdot d\vec{\ell}$ yields $\vec{E}=-\nabla V$.}
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\cor{Useful special cases}{Let $x$ denote position along the $x$-axis.
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\begin{enumerate}[label=(\alph*)]
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\item In one dimension,
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\[
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E_x=-\frac{dV}{dx}.
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\]
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So where a graph of $V$ versus $x$ slopes downward, $E_x$ is positive, and where it slopes upward, $E_x$ is negative.
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\item If the electric field is uniform and parallel to the displacement, so $\vec{E}=E\hat{u}$ and $\Delta \vec{\ell}=\Delta s\hat{u}$, then
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\[
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\Delta V=-E\Delta s.
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\]
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In particular, between parallel plates with nearly uniform field magnitude $E$ and separation $d$ measured in the field direction,
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\[
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|\Delta V|=Ed.
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\]
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\end{enumerate}}
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\qs{Worked AP-style problem}{Along the $x$-axis, the electric potential is
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\[
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V(x)=120-40x+5x^2,
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\]
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where $V$ is in volts and $x$ is in meters.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the electric-field component $E_x(x)$,
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\item the electric field at $x=2.0\,\mathrm{m}$,
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\item the potential difference $\Delta V=V(4.0\,\mathrm{m})-V(1.0\,\mathrm{m})$, and
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\item the work done by the electric field on a charge $q=+2.0\,\mu\mathrm{C}$ moving from $x=1.0\,\mathrm{m}$ to $x=4.0\,\mathrm{m}$.
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\end{enumerate}}
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\sol For part (a), use the one-dimensional field-potential relation:
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\[
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E_x=-\frac{dV}{dx}.
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\]
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Differentiate the given potential function:
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\[
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\frac{dV}{dx}=\frac{d}{dx}(120-40x+5x^2)=-40+10x.
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\]
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Therefore,
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\[
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E_x=-( -40+10x)=40-10x.
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\]
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So the field as a function of position is
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\[
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E_x(x)=40-10x
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\]
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in units of $\mathrm{N/C}$.
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For part (b), substitute $x=2.0\,\mathrm{m}$:
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\[
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E_x(2.0)=40-10(2.0)=20\,\mathrm{N/C}.
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\]
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Since this value is positive, the electric field points in the $+x$ direction:
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\[
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\vec{E}(2.0\,\mathrm{m})=(20\,\mathrm{N/C})\hat{\imath}.
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\]
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For part (c), first evaluate the potential at each position:
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\[
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V(4.0)=120-40(4.0)+5(4.0)^2=120-160+80=40\,\mathrm{V},
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\]
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and
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\[
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V(1.0)=120-40(1.0)+5(1.0)^2=120-40+5=85\,\mathrm{V}.
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\]
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Thus,
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\[
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\Delta V=V(4.0)-V(1.0)=40-85=-45\,\mathrm{V}.
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\]
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For part (d), the work done by the electric field is related to potential difference by
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\[
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W_{\text{field}}=-q\Delta V.
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\]
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Substitute $q=+2.0\times 10^{-6}\,\mathrm{C}$ and $\Delta V=-45\,\mathrm{V}$:
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\[
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W_{\text{field}}=-(2.0\times 10^{-6})(-45)\,\mathrm{J}=9.0\times 10^{-5}\,\mathrm{J}.
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\]
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So the field does positive work:
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\[
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W_{\text{field}}=9.0\times 10^{-5}\,\mathrm{J}=90\,\mu\mathrm{J}.
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\]
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Therefore,
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\[
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E_x(x)=40-10x,
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\qquad
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\vec{E}(2.0\,\mathrm{m})=(20\,\mathrm{N/C})\hat{\imath},
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\]
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\[
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\Delta V=-45\,\mathrm{V},
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\qquad
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W_{\text{field}}=9.0\times 10^{-5}\,\mathrm{J}.
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\]
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