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concepts/em/u9/e9-3-field-potential.tex

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