\subsection{Coulomb's Law and Superposition}

This subsection gives the electrostatic force between point charges and shows how forces from multiple source charges combine by vector addition.

\dfn{Point charges, separation vector, and superposition}{Let point charges $q_1,q_2,\dots,q_N$ be located at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$ in an inertial frame. For two distinct charges $q_i$ and $q_j$, define the separation vector from $q_i$ to $q_j$ by
\[
\vec{r}_{ij}=\vec{r}_j-\vec{r}_i,
\]
let $r_{ij}=|\vec{r}_{ij}|$ be the separation distance, and let $\hat{r}_{ij}=\vec{r}_{ij}/r_{ij}$ be the corresponding unit vector.

A \emph{point charge} is an idealized charged object whose size is negligible compared with the distances of interest. The \emph{superposition principle} states that when several source charges act on a chosen charge, the net electric force is the vector sum of the individual forces exerted by each source charge separately.}

\thm{Coulomb's law in vector form and force superposition}{Let $k=\dfrac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$. For two point charges $q_i$ and $q_j$ with separation vector $\vec{r}_{ij}\neq \vec{0}$, the electric force on $q_j$ due to $q_i$ is
\[
\vec{F}_{i\to j}=k\frac{q_i q_j}{r_{ij}^2}\hat{r}_{ij}
=k\frac{q_i q_j}{r_{ij}^3}\vec{r}_{ij}.
\]
Its magnitude is
\[
F_{i\to j}=k\frac{|q_i q_j|}{r_{ij}^2}.
\]
Thus the force is proportional to the product of the charges, inversely proportional to the square of the separation distance, and directed along the line joining the charges. If $N$ source charges act on $q_j$, then
\[
\vec{F}_{\mathrm{net},j}=\sum_{i\ne j}\vec{F}_{i\to j}
=\sum_{i\ne j} k\frac{q_i q_j}{|\vec{r}_j-\vec{r}_i|^3}(\vec{r}_j-\vec{r}_i).
\]}

\pf{Why the vector law has this form}{For two point charges separated by distance $r_{ij}$, Coulomb's law gives the force magnitude
\[
F_{i\to j}=k\frac{|q_i q_j|}{r_{ij}^2}.
\]
The force must lie along the line connecting the charges, so its direction is either $+\hat{r}_{ij}$ or $-\hat{r}_{ij}$. If $q_i q_j>0$, the charges have the same sign and repel, so the force on $q_j$ points away from $q_i$, which is $+\hat{r}_{ij}$. If $q_i q_j<0$, the charges have opposite signs and attract, so the force on $q_j$ points toward $q_i$, which is $-\hat{r}_{ij}$. Writing the force as
\[
\vec{F}_{i\to j}=k\frac{q_i q_j}{r_{ij}^2}\hat{r}_{ij}
\]
captures both cases automatically through the sign of $q_i q_j$. Because force is a vector, multiple electric forces combine by ordinary vector addition, giving the superposition formula.}

\cor{Collinear charges on the $x$-axis}{Let fixed source charges $q_1,\dots,q_N$ lie on the $x$-axis at coordinates $x_1,\dots,x_N$. Let a test charge $q$ be at coordinate $x$, with $x\neq x_i$ for all $i$. Then the net force on the test charge is purely along the $x$-axis:
\[
\vec{F}_{\mathrm{net}}=kq\left(\sum_{i=1}^N q_i\frac{x-x_i}{|x-x_i|^3}\right)\hat{\imath}.
\]
So in one dimension, Coulomb superposition reduces to an algebraic sum of signed $x$-components. In particular, if two equal source charges $+Q$ are placed at $x=-a$ and $x=+a$, then at the midpoint $x=0$ their forces cancel, so $\vec{F}_{\mathrm{net}}=\vec{0}$ on any test charge placed there.}

\qs{Worked AP-style problem}{In an $xy$-plane, let $q_1=+4.0\,\mu\mathrm{C}$ be fixed at $\vec{r}_1=\vec{0}$, let $q_2=-2.0\,\mu\mathrm{C}$ be fixed at $\vec{r}_2=(0.30\,\mathrm{m})\hat{\imath}$, and let $q_3=+1.5\,\mu\mathrm{C}$ be located at $\vec{r}_3=(0.40\,\mathrm{m})\hat{\jmath}$. Let $k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$.

Find:
\begin{enumerate}[label=(\alph*)]
\item the force $\vec{F}_{1\to 3}$ on $q_3$ due to $q_1$,
\item the force $\vec{F}_{2\to 3}$ on $q_3$ due to $q_2$, and
\item the net force $\vec{F}_{\mathrm{net},3}$ on $q_3$, including its magnitude and direction measured counterclockwise from the positive $x$-axis.
\end{enumerate}}

\sol First find the separation vectors to the charge $q_3$.

For the force due to $q_1$,
\[
\vec{r}_{13}=\vec{r}_3-\vec{r}_1=(0.40\,\mathrm{m})\hat{\jmath},
\qquad
r_{13}=0.40\,\mathrm{m}.
\]
For the force due to $q_2$,
\[
\vec{r}_{23}=\vec{r}_3-\vec{r}_2=(-0.30\hat{\imath}+0.40\hat{\jmath})\,\mathrm{m},
\qquad
r_{23}=\sqrt{(0.30)^2+(0.40)^2}\,\mathrm{m}=0.50\,\mathrm{m}.
\]

For part (a), use Coulomb's law. Since $q_1$ and $q_3$ are both positive, the force on $q_3$ is repulsive and points away from $q_1$, which is in the $+\hat{\jmath}$ direction:
\[
\vec{F}_{1\to 3}=k\frac{q_1 q_3}{r_{13}^3}\vec{r}_{13}.
\]
Its magnitude is
\[
F_{1\to 3}=k\frac{|q_1 q_3|}{r_{13}^2}
=(8.99\times 10^9)\frac{(4.0\times 10^{-6})(1.5\times 10^{-6})}{(0.40)^2}\,\mathrm{N}.
\]
So
\[
F_{1\to 3}=0.337\,\mathrm{N},
\]
and therefore
\[
\vec{F}_{1\to 3}=(0.337\,\mathrm{N})\hat{\jmath}.
\]

For part (b), $q_2$ is negative and $q_3$ is positive, so the force on $q_3$ is attractive and points from $q_3$ toward $q_2$. Let $\hat{u}_{3\to 2}$ denote the unit vector from $q_3$ to $q_2$. Then
\[
\hat{u}_{3\to 2}=\frac{(0.30\hat{\imath}-0.40\hat{\jmath})\,\mathrm{m}}{0.50\,\mathrm{m}}
=0.60\hat{\imath}-0.80\hat{\jmath}.
\]
The magnitude is
\[
F_{2\to 3}=k\frac{|q_2 q_3|}{r_{23}^2}
=(8.99\times 10^9)\frac{(2.0\times 10^{-6})(1.5\times 10^{-6})}{(0.50)^2}\,\mathrm{N}.
\]
Thus
\[
F_{2\to 3}=0.108\,\mathrm{N}.
\]
So the vector force is
\[
\vec{F}_{2\to 3}=F_{2\to 3}\hat{u}_{3\to 2}
=(0.108)(0.60\hat{\imath}-0.80\hat{\jmath})\,\mathrm{N}.
\]
Therefore,
\[
\vec{F}_{2\to 3}=(0.0647\hat{\imath}-0.0863\hat{\jmath})\,\mathrm{N}.
\]

For part (c), add the forces componentwise:
\[
\vec{F}_{\mathrm{net},3}=\vec{F}_{1\to 3}+\vec{F}_{2\to 3}.
\]
So
\[
\vec{F}_{\mathrm{net},3}=(0.0647\hat{\imath}+0.2507\hat{\jmath})\,\mathrm{N}.
\]
Its magnitude is
\[
|\vec{F}_{\mathrm{net},3}|=\sqrt{(0.0647)^2+(0.2507)^2}\,\mathrm{N}=0.259\,\mathrm{N}.
\]
Let $\theta$ denote the direction measured counterclockwise from the positive $x$-axis. Then
\[
\tan\theta=\frac{0.2507}{0.0647}=3.88,
\]
so
\[
\theta=\tan^{-1}(3.88)=75.5^\circ.
\]

Therefore,
\[
\vec{F}_{1\to 3}=(0.337\,\mathrm{N})\hat{\jmath},
\qquad
\vec{F}_{2\to 3}=(0.0647\hat{\imath}-0.0863\hat{\jmath})\,\mathrm{N},
\]
and the net force on $q_3$ is
\[
\vec{F}_{\mathrm{net},3}=(0.0647\hat{\imath}+0.2507\hat{\jmath})\,\mathrm{N},
\]
with magnitude
\[
|\vec{F}_{\mathrm{net},3}|=0.259\,\mathrm{N}
\]
at angle
\[
\theta=75.5^\circ
\]
above the positive $x$-axis.