\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,
\[
\Delta V=300\,\mathrm{V},
\qquad
\Delta U=-4.80\times 10^{-17}\,\mathrm{J},
\]
\[
\Delta K=+4.80\times 10^{-17}\,\mathrm{J},
\qquad
v=1.03\times 10^7\,\mathrm{m/s}.
\]
