\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,
\[
E_x(x)=40-10x,
\qquad
\vec{E}(2.0\,\mathrm{m})=(20\,\mathrm{N/C})\hat{\imath},
\]
\[
\Delta V=-45\,\mathrm{V},
\qquad
W_{\text{field}}=9.0\times 10^{-5}\,\mathrm{J}.
\]
