\subsection{Electric Field as Force per Unit Charge}

This subsection defines the electric field from source charges and shows how it determines the force on any charge placed at a point.

\dfn{Electric field, source charges, and test charges}{Let source charges create an electrostatic interaction in space. Let a field point have position vector $\vec{r}$, and let a small positive test charge $q_0$ be placed at that point. If the electric force on the test charge is $\vec{F}$, then the \emph{electric field} at that point is defined by
\[
\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}.
\]
The source charges are the charges that produce the field. The test charge is a charge used only to probe the field at a location. Because the factor of $q_0$ divides out, the field depends on the source-charge configuration and the location $\vec{r}$, not on the particular test charge used to measure it. By convention, the direction of $\vec{E}$ is the direction of the force on a positive test charge. The SI units of electric field are $\mathrm{N/C}$.}

\wc{Electric field fills all space (even if zero)}{The electric field $\vec{E}$ is defined at every point in space, regardless of whether a test charge is present. At points far from all source charges, the field magnitude may be very small or effectively zero, but the field itself extends throughout space. A test charge placed anywhere will experience $\vec{F}=q\vec{E}$ at that location.}

\thm{Point-charge field law and force relation}{Let a point source charge $Q$ be fixed at position vector $\vec{r}_Q$. Let the field point have position vector $\vec{r}$, and define
\[
\vec{R}=\vec{r}-\vec{r}_Q,
\qquad
R=|\vec{R}|,
\qquad
\hat{R}=\frac{\vec{R}}{R},
\]
with $\vec{r}\neq \vec{r}_Q$. Then the electric field due to the point charge $Q$ is
\[
\vec{E}(\vec{r})=k\frac{Q}{R^2}\hat{R}=k\frac{Q}{R^3}\vec{R},
\]
where
\[
k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
\]
If any charge $q$ is placed at that field point, the electric force on that charge is
\[
\vec{F}=q\vec{E}.
\]}

\pf{Derivation from Coulomb's law}{Let a positive test charge $q_0$ be placed at the field point. Coulomb's law gives the force on the test charge due to the source charge $Q$ as
\[
\vec{F}=k\frac{Qq_0}{R^3}\vec{R}.
\]
Now divide by $q_0$ and use the definition of electric field:
\[
\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}=k\frac{Q}{R^3}\vec{R}=k\frac{Q}{R^2}\hat{R}.
\]
This shows that the field is determined entirely by the source charge and geometry. Once $\vec{E}$ is known at a point, the force on any charge $q$ placed there is obtained by multiplying by $q$, so $\vec{F}=q\vec{E}$.}

\cor{Direction and superposition of electric fields}{Let point source charges $Q_1,Q_2,\dots,Q_N$ be fixed at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$. Let the field point have position vector $\vec{r}$, with $\vec{r}\neq \vec{r}_i$ for all $i$. Then the net electric field is the vector sum of the individual fields:
\[
\vec{E}_{\mathrm{net}}(\vec{r})=\sum_{i=1}^N k\frac{Q_i}{|\vec{r}-\vec{r}_i|^3}(\vec{r}-\vec{r}_i).
\]
For a single source charge, the field points radially away from the charge if $Q_i>0$ and radially toward the charge if $Q_i<0$.}

\qs{Worked AP-style problem}{Two point source charges lie on the $x$-axis. Let
\[
q_1=+3.0\,\mu\mathrm{C}
\qquad\text{at}\qquad
\vec{r}_1=(-0.20\,\mathrm{m})\hat{\imath},
\]
and let
\[
q_2=-2.0\,\mu\mathrm{C}
\qquad\text{at}\qquad
\vec{r}_2=(+0.30\,\mathrm{m})\hat{\imath}.
\]
Let point $P$ be at the origin, so $\vec{r}_P=\vec{0}$. Take $k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$.

Find:
\begin{enumerate}[label=(\alph*)]
\item the electric field $\vec{E}_1$ at $P$ due to $q_1$,
\item the electric field $\vec{E}_2$ at $P$ due to $q_2$,
\item the net electric field $\vec{E}_{\mathrm{net}}$ at $P$, and
\item the electric force on a charge $q=-4.0\,\mathrm{nC}$ placed at $P$.
\end{enumerate}}

\sol First find the displacement vectors from each source charge to the field point $P$.

For $q_1$,
\[
\vec{R}_1=\vec{r}_P-\vec{r}_1=(0.20\,\mathrm{m})\hat{\imath},
\qquad
R_1=0.20\,\mathrm{m}.
\]
For $q_2$,
\[
\vec{R}_2=\vec{r}_P-\vec{r}_2=(-0.30\,\mathrm{m})\hat{\imath},
\qquad
R_2=0.30\,\mathrm{m}.
\]

For part (a), use the point-charge field law:
\[
\vec{E}_1=k\frac{q_1}{R_1^3}\vec{R}_1.
\]
Because $q_1$ is positive, the field points away from $q_1$, which at the origin is in the $+\hat{\imath}$ direction. Its magnitude is
\[
E_1=k\frac{|q_1|}{R_1^2}=(8.99\times 10^9)\frac{3.0\times 10^{-6}}{(0.20)^2}\,\mathrm{N/C}.
\]
So
\[
E_1=6.74\times 10^5\,\mathrm{N/C},
\]
and therefore
\[
\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
\]

For part (b),
\[
\vec{E}_2=k\frac{q_2}{R_2^3}\vec{R}_2.
\]
Here $q_2$ is negative, so the field points toward $q_2$. Since $q_2$ is to the right of the origin, the field at the origin is again in the $+\hat{\imath}$ direction. Its magnitude is
\[
E_2=k\frac{|q_2|}{R_2^2}=(8.99\times 10^9)\frac{2.0\times 10^{-6}}{(0.30)^2}\,\mathrm{N/C}.
\]
Thus,
\[
E_2=2.00\times 10^5\,\mathrm{N/C},
\]
so
\[
\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath}.
\]

For part (c), add the fields as vectors:
\[
\vec{E}_{\mathrm{net}}=\vec{E}_1+\vec{E}_2.
\]
Since both fields point in the same direction,
\[
\vec{E}_{\mathrm{net}}=(6.74\times 10^5+2.00\times 10^5)\hat{\imath}\,\mathrm{N/C}.
\]
Therefore,
\[
\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
\]

For part (d), the force on a charge placed at $P$ is
\[
\vec{F}=q\vec{E}_{\mathrm{net}}.
\]
Substitute $q=-4.0\times 10^{-9}\,\mathrm{C}$:
\[
\vec{F}=(-4.0\times 10^{-9})(8.74\times 10^5)\hat{\imath}\,\mathrm{N}.
\]
So
\[
\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
\]
The negative sign means the force points in the $-\hat{\imath}$ direction. Its magnitude is $3.50\times 10^{-3}\,\mathrm{N}$.

Therefore,
\[
\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath},
\qquad
\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath},
\]
\[
\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath},
\qquad
\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
\]
