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\subsection{Electric Potential Energy}
This subsection introduces electric potential energy as an energy of a charge configuration and relates it to work done by electric forces.
\dfn{Electric potential energy of a system}{Let a system contain interacting charges. Let $U$ denote the \emph{electric potential energy} of the system, measured in joules. Electric potential energy is a property of the \emph{configuration of the system}, not of one charge by itself. For any two configurations,
\[
\Delta U=U_f-U_i,
\]
and if the electric force does work $W_{\mathrm{elec}}$ on the system, then
\[
W_{\mathrm{elec}}=-\Delta U.
\]
Thus, when the electric force does positive work, the system loses electric potential energy, and when an external agent slowly assembles a configuration against the electric force, the system gains electric potential energy.}
\thm{Point-charge pair formula and work relations}{Let two point charges $q_1$ and $q_2$ be separated by distance $r$, and choose the reference value
\[
U(\infty)=0.
\]
Then the electric potential energy of the two-charge system is
\[
U(r)=k\frac{q_1q_2}{r},
\]
where
\[
k=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
\]
If the separation changes from $r_i$ to $r_f$, then
\[
\Delta U=U_f-U_i=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right).
\]
The work done by the electric force is
\[
W_{\mathrm{elec}}=-\Delta U,
\]
and for a slow external rearrangement of the charges,
\[
W_{\mathrm{ext}}=\Delta U.
\]}
\nt{Because $U=kq_1q_2/r$, the sign of $U$ depends on the signs of the two charges. If $q_1q_2>0$, then $U>0$ and positive external work is required to bring the like charges closer together. If $q_1q_2<0$, then $U<0$ and the electric force itself tends to pull the unlike charges together. The important viewpoint is that $U$ belongs to the pair of charges as a system. It is not correct to say that a single isolated charge ``has'' electric potential energy by itself.}
\pf{Derivation from quasistatic assembly}{Let charge $q_1$ be fixed, and let charge $q_2$ be brought slowly from infinity to a final separation $r$. Let $x$ denote the instantaneous separation during the motion, with outward radial unit vector $\hat{r}$. The electric force on $q_2$ is
\[
\vec{F}_{\mathrm{elec}}=k\frac{q_1q_2}{x^2}\hat{r}.
\]
For a quasistatic move, the external force balances the electric force, so
\[
\vec{F}_{\mathrm{ext}}=-\vec{F}_{\mathrm{elec}}.
\]
The external work done in assembling the pair is the increase in potential energy:
\[
U(r)-U(\infty)=W_{\mathrm{ext}}=\int_{\infty}^{r}\vec{F}_{\mathrm{ext}}\cdot d\vec{r}.
\]
Since $d\vec{r}=\hat{r}\,dx$,
\[
U(r)-0=-\int_{\infty}^{r}k\frac{q_1q_2}{x^2}\,dx
=-kq_1q_2\left[-\frac{1}{x}\right]_{\infty}^{r}
=k\frac{q_1q_2}{r}.
\]
This also gives
\[
\Delta U=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right),
\]
and because electric potential energy is defined for a conservative force, the electric-force work satisfies $W_{\mathrm{elec}}=-\Delta U$.}
\qs{Worked AP-style problem}{Two point charges form a system. Let
\[
q_1=+2.0\,\mu\mathrm{C},
\qquad
q_2=-3.0\,\mu\mathrm{C}.
\]
Initially the charges are separated by
\[
r_i=0.50\,\mathrm{m},
\]
and they are moved slowly until their final separation is
\[
r_f=0.20\,\mathrm{m}.
\]
Take
\[
k=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the initial electric potential energy $U_i$,
\item the final electric potential energy $U_f$,
\item the change in electric potential energy $\Delta U$, and
\item the work done by the external agent and by the electric force during the slow move.
\end{enumerate}}
\sol Use
\[
U=k\frac{q_1q_2}{r}.
\]
Because the charges have opposite signs, the product $q_1q_2$ is negative:
\[
q_1q_2=(2.0\times10^{-6}\,\mathrm{C})(-3.0\times10^{-6}\,\mathrm{C})=-6.0\times10^{-12}\,\mathrm{C}^2.
\]
For part (a), the initial potential energy is
\[
U_i=k\frac{q_1q_2}{r_i}
=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.50}\,\mathrm{J}.
\]
So
\[
U_i=-1.08\times10^{-1}\,\mathrm{J}=-0.108\,\mathrm{J}.
\]
For part (b), the final potential energy is
\[
U_f=k\frac{q_1q_2}{r_f}
=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.20}\,\mathrm{J}.
\]
Thus,
\[
U_f=-2.70\times10^{-1}\,\mathrm{J}=-0.270\,\mathrm{J}.
\]
For part (c),
\[
\Delta U=U_f-U_i=(-0.270)-(-0.108)\,\mathrm{J}.
\]
Therefore,
\[
\Delta U=-0.162\,\mathrm{J}.
\]
For part (d), because the move is slow,
\[
W_{\mathrm{ext}}=\Delta U=-0.162\,\mathrm{J}.
\]
The negative sign means the external agent removes energy from the system rather than supplying it. The electric force does the opposite amount of work:
\[
W_{\mathrm{elec}}=-\Delta U=+0.162\,\mathrm{J}.
\]
This result makes physical sense. Unlike charges attract, so when they are brought closer together, the system energy becomes more negative.
Therefore,
\[
U_i=-0.108\,\mathrm{J},
\qquad
U_f=-0.270\,\mathrm{J},
\]
\[
\Delta U=-0.162\,\mathrm{J},
\qquad
W_{\mathrm{ext}}=-0.162\,\mathrm{J},
\qquad
W_{\mathrm{elec}}=+0.162\,\mathrm{J}.
\]

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\subsection{Electric Potential and Voltage}
This subsection defines electric potential as electric potential energy per unit charge, interprets voltage as a potential difference, and emphasizes that potentials from multiple source charges add as scalars.
\dfn{Electric potential and voltage}{Let $U$ denote the electric potential energy of a system containing a chosen test charge $q\neq 0$ at a specified point in an electrostatic field. The \emph{electric potential} $V$ at that point is defined by
\[
V=\frac{U}{q}.
\]
If points $A$ and $B$ have potentials $V_A$ and $V_B$, then the \emph{potential difference} from $A$ to $B$ is
\[
\Delta V=V_B-V_A=\frac{\Delta U}{q},
\]
where
\[
\Delta U=U_B-U_A.
\]
In common AP usage, \emph{voltage} usually means this potential difference between two points.}
\nt{Electric potential is a scalar, not a vector, so it has magnitude and sign but no direction. A positive source charge produces positive potential, and a negative source charge produces negative potential, when the reference value is chosen as $V=0$ at infinity. Only potential differences are directly physical, so the zero of potential is a convenient reference choice rather than an absolute requirement.}
\mprop{Point-charge potential and scalar superposition}{Let fixed point charges $Q_1,Q_2,\dots,Q_N$ be located at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$. Let point $P$ have position vector $\vec{r}$, with $\vec{r}\ne \vec{r}_i$ for all $i$. For each source charge, define the separation distance from $Q_i$ to $P$ by
\[
R_i=|\vec{r}-\vec{r}_i|.
\]
Let
\[
k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
\]
Choosing the reference value $V(\infty)=0$, the electric potential at $P$ due to one point charge $Q_i$ is
\[
V_i(P)=k\frac{Q_i}{R_i}.
\]
Because potential is a scalar, the total potential at $P$ due to all the source charges is
\[
V(P)=\sum_{i=1}^N V_i(P)=k\sum_{i=1}^N \frac{Q_i}{R_i}.
\]
If points $A$ and $B$ have potentials $V_A$ and $V_B$, then the voltage between them is
\[
\Delta V_{A\to B}=V_B-V_A.
\]
For a charge $q$ moved quasistatically from $A$ to $B$ by an external agent,
\[
\Delta U=q\Delta V,
\qquad
W_{\mathrm{ext}}=\Delta U=q\Delta V.
\]}
\qs{Worked AP-style problem}{Three fixed point charges lie in an $xy$-plane. Let
\[
Q_1=+4.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_1=\vec{0},
\]
\[
Q_2=-2.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_2=(0.30\,\mathrm{m})\hat{\imath},
\]
and
\[
Q_3=+3.0\,\mu\mathrm{C}\quad\text{at}\quad \vec{r}_3=(0.40\,\mathrm{m})\hat{\jmath}.
\]
Let point $P$ be at
\[
\vec{r}_P=(0.30\hat{\imath}+0.40\hat{\jmath})\,\mathrm{m}.
\]
Take
\[
k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the electric potential $V(P)$ relative to infinity, and
\item the external work required to bring a test charge
\[
q_0=+2.0\,\mathrm{nC}
\]
from infinity to point $P$.
\end{enumerate}}
\sol First find the distances from each source charge to point $P$.
From $Q_1$ at the origin to $P$,
\[
R_1=|\vec{r}_P-\vec{r}_1|=\sqrt{(0.30)^2+(0.40)^2}\,\mathrm{m}=0.50\,\mathrm{m}.
\]
From $Q_2$ at $(0.30\,\mathrm{m})\hat{\imath}$ to $P$,
\[
R_2=0.40\,\mathrm{m}.
\]
From $Q_3$ at $(0.40\,\mathrm{m})\hat{\jmath}$ to $P$,
\[
R_3=0.30\,\mathrm{m}.
\]
Because electric potential is a scalar, add the contributions algebraically:
\[
V(P)=k\left(\frac{Q_1}{R_1}+\frac{Q_2}{R_2}+\frac{Q_3}{R_3}\right).
\]
Substitute the given values:
\[
V(P)=8.99\times 10^9\left(\frac{4.0\times 10^{-6}}{0.50}+\frac{-2.0\times 10^{-6}}{0.40}+\frac{3.0\times 10^{-6}}{0.30}\right)\mathrm{V}.
\]
Evaluate each term inside the parentheses:
\[
\frac{4.0\times 10^{-6}}{0.50}=8.0\times 10^{-6},
\qquad
\frac{-2.0\times 10^{-6}}{0.40}=-5.0\times 10^{-6},
\qquad
\frac{3.0\times 10^{-6}}{0.30}=1.0\times 10^{-5}.
\]
So
\[
V(P)=8.99\times 10^9\left(1.3\times 10^{-5}\right)\mathrm{V}.
\]
Therefore,
\[
V(P)=1.17\times 10^5\,\mathrm{V}.
\]
For part (b), the potential at infinity is the chosen reference value,
\[
V(\infty)=0,
\]
so the potential difference from infinity to $P$ is
\[
\Delta V=V(P)-V(\infty)=1.17\times 10^5\,\mathrm{V}.
\]
The external work required to bring the test charge slowly from infinity to $P$ is the change in electric potential energy:
\[
W_{\mathrm{ext}}=\Delta U=q_0\Delta V.
\]
Substitute $q_0=2.0\times 10^{-9}\,\mathrm{C}$:
\[
W_{\mathrm{ext}}=(2.0\times 10^{-9})(1.17\times 10^5)\,\mathrm{J}.
\]
Thus,
\[
W_{\mathrm{ext}}=2.34\times 10^{-4}\,\mathrm{J}.
\]
So the electric potential at point $P$ is
\[
V(P)=1.17\times 10^5\,\mathrm{V},
\]
and the required external work to bring the positive test charge from infinity to $P$ is
\[
W_{\mathrm{ext}}=2.34\times 10^{-4}\,\mathrm{J}.
\]

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

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\subsection{Equipotentials and Energy Conservation for Moving Charges}
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.
\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:
\[
V(\vec{r})=\text{constant}.
\]
If a charge $q$ moves from point $A$ to point $B$, define
\[
\Delta V=V_B-V_A
\]
and
\[
\Delta U=U_B-U_A.
\]
Then the change in electric potential energy is
\[
\Delta U=q\Delta V.
\]
Thus potential difference tells how the electric potential energy of a chosen charge changes between two points.}
\thm{Equipotentials, no work, and kinetic-energy change}{Let $d\vec{\ell}$ denote an infinitesimal displacement in an electrostatic field $\vec{E}$. Then
\[
dV=-\vec{E}\cdot d\vec{\ell}.
\]
If the displacement is along an equipotential, then $dV=0$, so
\[
\vec{E}\cdot d\vec{\ell}=0.
\]
Therefore the electric field is perpendicular to an equipotential, and the electric force does no work on a charge moved along an equipotential:
\[
W_{\mathrm{elec}}=q\int \vec{E}\cdot d\vec{\ell}=0.
\]
For any motion of a charge $q$ from $A$ to $B$ in electrostatics,
\[
\Delta U=q\Delta V.
\]
If only the electric force does work, conservation of mechanical energy gives
\[
K_i+U_i=K_f+U_f,
\]
so
\[
\Delta K=K_f-K_i=-\Delta U=-q\Delta V.
\]
Hence a charge speeds up when its electric potential energy decreases.}
\ex{Illustrative example}{Points $A$ and $B$ lie on the same equipotential,
\[
V_A=V_B=120\,\mathrm{V}.
\]
Let a proton of charge
\[
q=+e=+1.60\times 10^{-19}\,\mathrm{C}
\]
move from $A$ to $B$.
Because the two points are on the same equipotential,
\[
\Delta V=V_B-V_A=0.
\]
So the change in electric potential energy is
\[
\Delta U=q\Delta V=0,
\]
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.}
\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.}
\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
\[
V_A=100\,\mathrm{V},
\qquad
V_B=400\,\mathrm{V}.
\]
An electron is released from rest at point $A$ and moves to point $B$. Let the electron charge be
\[
q=-1.60\times 10^{-19}\,\mathrm{C}
\]
and the electron mass be
\[
m_e=9.11\times 10^{-31}\,\mathrm{kg}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the potential difference $\Delta V=V_B-V_A$,
\item the change in electric potential energy $\Delta U$,
\item the change in kinetic energy $\Delta K$, and
\item the electron's speed at point $B$.
\end{enumerate}}
\sol First compute the potential difference:
\[
\Delta V=V_B-V_A=400\,\mathrm{V}-100\,\mathrm{V}=300\,\mathrm{V}.
\]
For part (b), use the relation
\[
\Delta U=q\Delta V.
\]
Substitute the electron charge and the potential difference:
\[
\Delta U=(-1.60\times 10^{-19}\,\mathrm{C})(300\,\mathrm{V}).
\]
Since $1\,\mathrm{V}=1\,\mathrm{J/C}$,
\[
\Delta U=-4.80\times 10^{-17}\,\mathrm{J}.
\]
For part (c), only the electric force does work, so
\[
\Delta K=-\Delta U.
\]
Therefore,
\[
\Delta K=+4.80\times 10^{-17}\,\mathrm{J}.
\]
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.
For part (d), the electron starts from rest, so
\[
K_i=0.
\]
Thus
\[
K_f=\Delta K=4.80\times 10^{-17}\,\mathrm{J}.
\]
Use the kinetic-energy formula
\[
K_f=\frac12 m_e v^2.
\]
Solve for the speed $v$:
\[
v=\sqrt{\frac{2K_f}{m_e}}.
\]
Substitute the values:
\[
v=\sqrt{\frac{2(4.80\times 10^{-17}\,\mathrm{J})}{9.11\times 10^{-31}\,\mathrm{kg}}}.
\]
This gives
\[
v=1.03\times 10^7\,\mathrm{m/s}.
\]
Therefore,
\[
\Delta V=300\,\mathrm{V},
\qquad
\Delta U=-4.80\times 10^{-17}\,\mathrm{J},
\]
\[
\Delta K=+4.80\times 10^{-17}\,\mathrm{J},
\qquad
v=1.03\times 10^7\,\mathrm{m/s}.
\]