\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
\[
W_{\mathrm{ext}}=2.34\times 10^{-4}\,\mathrm{J}.
\]
