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\subsection{Charge Conservation and Charging Processes}
This subsection treats electric charge as a conserved quantity and introduces the AP-level charging processes of friction, contact, and induction as charge-bookkeeping ideas.
\dfn{Charge conservation and basic charging processes}{Let $q$ denote the net charge of an object or subsystem, measured in coulombs, and let
\[
Q_{\mathrm{tot}}=\sum_i q_i
\]
denote the total charge of a chosen system.
Charge conservation states that for an isolated system,
\[
Q_{\mathrm{tot},f}=Q_{\mathrm{tot},i}.
\]
In ordinary AP charging problems, objects become charged because charge is redistributed:
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item \emph{Charging by friction}: rubbing two materials can transfer electrons from one object to the other.
\item \emph{Charging by contact}: when objects touch, charge can transfer between them before they separate.
\item \emph{Charging by induction}: a nearby charged object causes charge separation in another object; with grounding, charge can enter or leave so the object may be left with a net charge after the process.
\end{enumerate}
In each case, the total charge of the full isolated system remains constant.}
\nt{The sign of charge is bookkeeping. A positive net charge means an electron deficit, while a negative net charge means an electron excess. In ordinary friction, contact, and induction processes, charge is transferred from one place to another; it is not created from nothing. If one part of an isolated system gains $+\Delta q$, the rest of the system must change by $-\Delta q$. When grounding is involved, include the Earth in the system because it can supply or receive the transferred charge.}
\mprop{Practical bookkeeping relations for isolated systems and simple sharing}{Consider a chosen system of objects with initial charges $q_{1,i},q_{2,i},\dots$ and final charges $q_{1,f},q_{2,f},\dots$.
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item For any isolated system,
\[
\sum_k q_{k,f}=\sum_k q_{k,i}
\qquad\text{or}\qquad
\Delta Q_{\mathrm{tot}}=0.
\]
\item For charge transfer between two objects $A$ and $B$ within an isolated system,
\[
\Delta q_A+\Delta q_B=0,
\]
so
\[
q_{A,f}-q_{A,i}=-(q_{B,f}-q_{B,i}).
\]
\item If two identical small conducting spheres with initial charges $q_{A,i}$ and $q_{B,i}$ are touched together and then separated, symmetry gives equal final charges:
\[
q_{A,f}=q_{B,f}=\frac{q_{A,i}+q_{B,i}}{2}.
\]
\item If a neutral object is charged by induction while connected to ground, then charge conservation must be applied to the combined object-Earth system. If the object starts neutral and ends with charge $q_f$, then the Earth changes by
\[
\Delta q_{\mathrm{Earth}}=-q_f.
\]
\end{enumerate}}
\qs{Worked AP-style problem}{Three identical small conducting spheres $A$, $B$, and $C$ are far apart initially. Let their initial charges be
\[
q_{A,i}=+8.0\,\mathrm{nC},
\qquad
q_{B,i}=-2.0\,\mathrm{nC},
\qquad
q_{C,i}=0.
\]
First, sphere $A$ is touched to sphere $B$ and then separated. Next, sphere $B$ is touched to sphere $C$ and then separated.
Find:
\begin{enumerate}[label=(\alph*)]
\item the charge on each sphere after the first contact,
\item the final charge on each sphere after the second contact, and
\item the number of electrons transferred during the second contact.
\end{enumerate}
Take the elementary charge magnitude to be $e=1.60\times 10^{-19}\,\mathrm{C}$.}
\sol Because the spheres are identical, whenever two of them touch and then separate, they share the total charge equally.
For the first contact, apply charge conservation to spheres $A$ and $B$:
\[
q_{A,i}+q_{B,i}=(+8.0\,\mathrm{nC})+(-2.0\,\mathrm{nC})=+6.0\,\mathrm{nC}.
\]
Since the spheres are identical, after they separate each has half of this total charge:
\[
q_{A}=q_{B}=\frac{+6.0\,\mathrm{nC}}{2}=+3.0\,\mathrm{nC}.
\]
So after the first contact,
\[
q_{A}=+3.0\,\mathrm{nC},
\qquad
q_{B}=+3.0\,\mathrm{nC},
\qquad
q_{C}=0.
\]
Now sphere $B$ touches sphere $C$. Just before this second contact, their total charge is
\[
q_{B}+q_{C}=(+3.0\,\mathrm{nC})+0=+3.0\,\mathrm{nC}.
\]
Again they are identical, so after separation they share this total equally:
\[
q_{B,f}=q_{C,f}=\frac{+3.0\,\mathrm{nC}}{2}=+1.5\,\mathrm{nC}.
\]
Sphere $A$ is not involved in the second contact, so its charge stays
\[
q_{A,f}=+3.0\,\mathrm{nC}.
\]
Therefore the final charges are
\[
q_{A,f}=+3.0\,\mathrm{nC},
\qquad
q_{B,f}=+1.5\,\mathrm{nC},
\qquad
q_{C,f}=+1.5\,\mathrm{nC}.
\]
To find the number of electrons transferred during the second contact, compute the magnitude of the charge change on either sphere. Sphere $B$ changes from $+3.0\,\mathrm{nC}$ to $+1.5\,\mathrm{nC}$, so
\[
\Delta q_B=(+1.5-3.0)\,\mathrm{nC}=-1.5\,\mathrm{nC}.
\]
The negative change means sphere $B$ gained electrons. Equivalently, sphere $C$ changed from $0$ to $+1.5\,\mathrm{nC}$, so sphere $C$ lost electrons. The magnitude of transferred charge is
\[
|\Delta q|=1.5\times 10^{-9}\,\mathrm{C}.
\]
Let $N$ denote the number of electrons transferred. Then
\[
N=\frac{|\Delta q|}{e}=\frac{1.5\times 10^{-9}}{1.60\times 10^{-19}}=9.375\times 10^9.
\]
To two significant figures,
\[
N\approx 9.4\times 10^9\text{ electrons}.
\]
These electrons moved from sphere $C$ to sphere $B$. This direction makes sense because sphere $B$ became less positive while sphere $C$ became more positive.

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\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.

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

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\subsection{Fields of Continuous Charge Distributions}
This subsection extends electric-field superposition from point charges to rods, arcs, rings, surfaces, and volumes by replacing discrete sums with integrals over charge elements.
\dfn{Charge densities and differential field contribution}{Let the field point have position vector $\vec{r}$. Let a small source element at position vector $\vec{r}'$ carry charge $dq$. Define
\[
\vec{R}=\vec{r}-\vec{r}',
\qquad
R=|\vec{R}|,
\qquad
\hat{R}=\frac{\vec{R}}{R}.
\]
For a continuous distribution, the charge element is written as
\[
dq=\lambda\,dl
\qquad\text{(line charge)},
\qquad
dq=\sigma\,dA
\qquad\text{(surface charge)},
\qquad
dq=\rho\,dV
\qquad\text{(volume charge)},
\]
where $\lambda$ is linear charge density in $\mathrm{C/m}$, $\sigma$ is surface charge density in $\mathrm{C/m^2}$, and $\rho$ is volume charge density in $\mathrm{C/m^3}$. The electric-field contribution of the source element at the field point is
\[
d\vec{E}=k\frac{dq}{R^2}\hat{R}=k\frac{dq}{R^3}\vec{R},
\]
where
\[
k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
\]
}
\thm{Continuous superposition integral for electric field}{For a static continuous charge distribution, the net electric field at the field point $\vec{r}$ is the vector integral
\[
\vec{E}(\vec{r})=\int d\vec{E}
=k\int \frac{1}{R^3}\vec{R}\,dq.
\]
Equivalently,
\[
\vec{E}(\vec{r})
=k\int \frac{1}{R^3}\vec{R}\,\lambda\,dl,
\qquad
\vec{E}(\vec{r})
=k\int \frac{1}{R^3}\vec{R}\,\sigma\,dA,
\qquad
\vec{E}(\vec{r})
=k\int \frac{1}{R^3}\vec{R}\,\rho\,dV,
\]
depending on the geometry. In practice, write $d\vec{E}$ for one source element, resolve it into components, use symmetry to identify any canceling components, and integrate only the nonzero component(s).}
\ex{Illustrative example}{A uniformly charged semicircular arc of radius $R$ lies above the $x$-axis and is centered at the origin. Let its total charge be $Q>0$. Find the electric field at the center.
The arc length is $\pi R$, so the linear charge density is
\[
\lambda=\frac{Q}{\pi R}.
\]
Let $\theta$ denote the polar angle of a source element, measured from the positive $x$-axis, with $0\le \theta \le \pi$. Then
\[
dq=\lambda R\,d\theta.
\]
Each source element is distance $R$ from the center, so
\[
dE=k\frac{dq}{R^2}.
\]
By symmetry, the $x$-components cancel. The $y$-components all point downward, so
\[
dE_y=-dE\sin\theta=-k\frac{dq}{R^2}\sin\theta.
\]
Substitute $dq=\lambda R\,d\theta$:
\[
dE_y=-k\frac{\lambda}{R}\sin\theta\,d\theta.
\]
Integrate from $0$ to $\pi$:
\[
E_y=-k\frac{\lambda}{R}\int_0^\pi \sin\theta\,d\theta
=-k\frac{\lambda}{R}(2).
\]
Therefore,
\[
\vec{E}=-\frac{2kQ}{\pi R^2}\hat{\jmath}.
\]
The field points downward because the positive charges on the upper arc repel a positive test charge at the center.}
\nt{For continuous distributions, the hardest step is usually not the integral but the geometry. Start with $d\vec{E}$ from one source element, then ask which components cancel by symmetry. On a ring, sideways components cancel and only the axial component survives. On a symmetric finite line, horizontal components cancel at the perpendicular bisector and only the perpendicular component survives. Also choose $dq$ to match the object's dimension: use $dq=\lambda\,dl$ for rods and arcs, $dq=\sigma\,dA$ for sheets, and $dq=\rho\,dV$ for three-dimensional charge distributions.}
\qs{Worked AP-style problem}{A thin ring of radius $a$ is centered at the origin and lies in the $yz$-plane. The ring carries total charge $Q$ distributed uniformly around its circumference. Let point $P$ lie on the ring's axis at position
\[
\vec{r}_P=x\hat{\imath},
\]
where $x>0$. Let
\[
k=\frac{1}{4\pi\varepsilon_0}.
\]
Find the electric field $\vec{E}$ at point $P$ in terms of $Q$, $a$, $x$, and $k$.}
\sol Let $\lambda$ denote the ring's linear charge density. Since the ring circumference is $2\pi a$,
\[
\lambda=\frac{Q}{2\pi a}.
\]
Choose a small source element $dq$ on the ring. Let $\vec{R}$ denote the displacement vector from that source element to point $P$. Every source element on the ring is the same distance from $P$, so
\[
R=\sqrt{x^2+a^2}.
\]
The magnitude of the field due to $dq$ is therefore
\[
dE=k\frac{dq}{R^2}=k\frac{dq}{x^2+a^2}.
\]
Now use symmetry. For each source element on the ring, there is an opposite element whose field contribution has the same magnitude. Their components in the $y$- and $z$-directions cancel, while their components along the axis add. Therefore the net field must point along $\hat{\imath}$, so we only need the $x$-component of $d\vec{E}$.
Let $\phi$ denote the angle between $\vec{R}$ and the positive $x$-axis. Then
\[
\cos\phi=\frac{x}{R}=\frac{x}{\sqrt{x^2+a^2}}.
\]
So the axial component of the differential field is
\[
dE_x=dE\cos\phi
=\left(k\frac{dq}{R^2}\right)\left(\frac{x}{R}\right)
=k\frac{x\,dq}{R^3}.
\]
Since $R=\sqrt{x^2+a^2}$ is constant over the ring,
\[
dE_x=k\frac{x\,dq}{(x^2+a^2)^{3/2}}.
\]
Integrate all the way around the ring:
\[
E_x=\int dE_x
=k\frac{x}{(x^2+a^2)^{3/2}}\int dq.
\]
But
\[
\int dq=Q,
\]
so
\[
E_x=k\frac{Qx}{(x^2+a^2)^{3/2}}.
\]
Therefore,
\[
\vec{E}=k\frac{Qx}{(x^2+a^2)^{3/2}}\hat{\imath}.
\]
If $Q>0$, the field points in the $+\hat{\imath}$ direction, and if $Q<0$, it points in the $-\hat{\imath}$ direction.
This result also passes two quick checks. If $x=0$, then
\[
\vec{E}=\vec{0},
\]
which matches the symmetry at the ring's center. If $x\gg a$, then $x^2+a^2\approx x^2$, so
\[
\vec{E}\approx k\frac{Q}{x^2}\hat{\imath},
\]
which is the field of a point charge $Q$ far away from the ring.

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\subsection{Electric Flux}
This subsection introduces electric flux as a signed measure of how much electric field passes through an oriented surface.
\dfn{Area vector and electric flux}{Let $S$ be a surface broken into small area elements of scalar area $dA$. Let $\hat{n}$ denote a chosen unit normal to a surface element. The corresponding \emph{area vector element} is
\[
d\vec{A}=\hat{n}\,dA.
\]
For an open surface, either choice of normal may be used, but the choice must be kept consistent across the surface. For a closed surface, the standard choice is the outward normal.
Let $\vec{E}$ denote the electric field at each point of the surface. The \emph{electric flux} through the oriented surface is the scalar
\[
\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
\]
Its SI units are $\mathrm{N\cdot m^2/C}$.}
\thm{Surface-integral form and uniform-field special case}{Let $S$ be an oriented surface with area vector element $d\vec{A}=\hat{n}\,dA$, and let $\vec{E}$ be the electric field on that surface. Then the electric flux through $S$ is
\[
\Phi_E=\iint_S \vec{E}\cdot d\vec{A}.
\]
This integral adds the component of $\vec{E}$ perpendicular to the surface over the entire surface.
If the surface is flat with area $A$, the field is uniform over it, and $\vec{A}=A\hat{n}$ denotes the surface's area vector, then
\[
\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
\]
where $E=|\vec{E}|$, $\theta$ is the angle between $\vec{E}$ and $\vec{A}$, and $A=|\vec{A}|$. For a closed surface, the same formula is applied piece by piece using outward area vectors on all patches of the surface.}
\ex{Illustrative example}{Let a uniform electric field be
\[
\vec{E}=(300\,\mathrm{N/C})\hat{\imath}.
\]
Let a flat surface have area
\[
A=0.20\,\mathrm{m^2},
\]
and let its area vector make an angle $\theta=60^\circ$ with $\vec{E}$.
Then the flux is
\[
\Phi_E=EA\cos\theta=(300)(0.20)\cos 60^\circ\,\mathrm{N\cdot m^2/C}.
\]
So
\[
\Phi_E=30\,\mathrm{N\cdot m^2/C}.
\]
Because the angle is acute, the flux is positive.}
\nt{Electric flux is not the same thing as electric field magnitude. Flux depends on both the field and the oriented surface. Reversing the chosen normal reverses the sign of $\Phi_E$. A positive flux means the field points generally in the same direction as the chosen area vector, while a negative flux means it points generally opposite that direction. If the field is parallel to the surface, then it is perpendicular to $d\vec{A}$ and the flux is zero even if $|\vec{E}|$ is large.}
\qs{Worked AP-style problem}{A cube of side length $L=0.20\,\mathrm{m}$ is placed in a uniform electric field
\[
\vec{E}=(500\,\mathrm{N/C})\hat{\imath}.
\]
Let the cube's faces be aligned with the coordinate axes. Let the outward area vector of the right face be in the $+\hat{\imath}$ direction, and let the outward area vector of the left face be in the $-\hat{\imath}$ direction.
Find:
\begin{enumerate}[label=(\alph*)]
\item the electric flux through the right face,
\item the electric flux through the left face,
\item the electric flux through any one of the four remaining faces, and
\item the net electric flux through the entire closed cube.
\end{enumerate}}
\sol Let the area of one face be $A$. Since each face is a square of side length $L$,
\[
A=L^2=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}.
\]
For each face of the cube, use
\[
\Phi_E=\vec{E}\cdot \vec{A}=EA\cos\theta,
\]
where $\theta$ is the angle between the electric field and that face's outward area vector.
For part (a), the right face has outward area vector in the $+\hat{\imath}$ direction, the same direction as $\vec{E}$. Thus
\[
\theta=0^\circ.
\]
So
\[
\Phi_{E,\mathrm{right}}=EA\cos 0^\circ=(500)(0.040)(1)\,\mathrm{N\cdot m^2/C}.
\]
Therefore,
\[
\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C}.
\]
For part (b), the left face has outward area vector in the $-\hat{\imath}$ direction, opposite the field. Thus
\[
\theta=180^\circ.
\]
So
\[
\Phi_{E,\mathrm{left}}=EA\cos 180^\circ=(500)(0.040)(-1)\,\mathrm{N\cdot m^2/C}.
\]
Therefore,
\[
\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C}.
\]
For part (c), on any of the other four faces, the outward area vector is perpendicular to $\vec{E}$. Thus
\[
\theta=90^\circ,
\]
so
\[
\Phi_E=EA\cos 90^\circ=0.
\]
Therefore, the flux through each of those four faces is
\[
0\,\mathrm{N\cdot m^2/C}.
\]
For part (d), add the fluxes from all six faces:
\[
\Phi_{E,\mathrm{net}}=\Phi_{E,\mathrm{right}}+\Phi_{E,\mathrm{left}}+4(0).
\]
So
\[
\Phi_{E,\mathrm{net}}=20+(-20)=0\,\mathrm{N\cdot m^2/C}.
\]
Therefore,
\[
\Phi_{E,\mathrm{right}}=20\,\mathrm{N\cdot m^2/C},
\qquad
\Phi_{E,\mathrm{left}}=-20\,\mathrm{N\cdot m^2/C},
\]
\[
\Phi_E=0\,\mathrm{N\cdot m^2/C}\text{ for each of the other four faces},
\]
and the net flux through the closed cube is
\[
\Phi_{E,\mathrm{net}}=0.
\]

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\subsection{Gauss's Law and Symmetry Reduction}
This subsection states Gauss's law and shows how symmetry can reduce a difficult flux integral to simple algebra when the charge distribution is highly symmetric.
\dfn{Gaussian surface and enclosed charge}{Let $S$ be any closed imaginary surface in space, and let $d\vec{A}$ denote an outward-pointing area element on that surface. The surface $S$ is called a \emph{Gaussian surface}. The \emph{enclosed charge} $q_{\mathrm{enc}}$ is the algebraic sum of all charges contained inside $S$. Charges outside $S$ can affect the electric field on the surface, but they do not contribute to $q_{\mathrm{enc}}$.}
\thm{Gauss's law and when symmetry makes it useful}{Let $S$ be any closed surface with outward area element $d\vec{A}$, and let $q_{\mathrm{enc}}$ be the net charge enclosed by $S$. Then Gauss's law states
\[
\oint_S \vec{E}\cdot d\vec{A}=\frac{q_{\mathrm{enc}}}{\varepsilon_0}.
\]
This law is always true. It becomes a practical method for solving for the electric field when the charge distribution has enough symmetry that one can choose a Gaussian surface for which the magnitude $E=|\vec{E}|$ is constant on each flux-contributing part of the surface and the angle between $\vec{E}$ and $d\vec{A}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the flux integral reduces to algebraic terms such as $EA$, $-EA$, or $0$. Common useful cases are spherical, cylindrical, and planar symmetry.}
\nt{Gauss's law is always true, but it is not always useful for finding $\vec{E}$. In a general asymmetric charge distribution, knowing only the total flux through a closed surface does not tell you the field at each point on that surface. Also, zero net enclosed charge implies zero \emph{net flux}, not necessarily zero field everywhere. The main strategy is therefore: first identify strong symmetry, then choose a Gaussian surface matched to that symmetry.}
\pf{Why symmetry reduces the integral}{Let a point charge $Q$ be at the center of a spherical Gaussian surface of radius $r$. By spherical symmetry, the electric field is radial and has the same magnitude $E(r)$ at every point on the sphere. The outward area element $d\vec{A}$ is also radial, so
\[
\vec{E}\cdot d\vec{A}=E(r)\,dA
\]
everywhere on the surface. Therefore,
\[
\oint_S \vec{E}\cdot d\vec{A}=E(r)\oint_S dA=E(r)(4\pi r^2).
\]
Since the enclosed charge is $q_{\mathrm{enc}}=Q$, Gauss's law gives
\[
E(r)(4\pi r^2)=\frac{Q}{\varepsilon_0},
\qquad
E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.
\]
The law itself is general, but the symmetry is what allowed $E(r)$ to be pulled outside the integral.}
\qs{Worked AP-style problem}{A very long straight wire carries a uniform positive linear charge density
\[
\lambda=3.0\times 10^{-6}\,\mathrm{C/m}.
\]
Let point $P$ be at perpendicular distance
\[
r=0.20\,\mathrm{m}
\]
from the wire. Choose a cylindrical Gaussian surface of radius $r$ and length $L$ coaxial with the wire.
Find:
\begin{enumerate}[label=(\alph*)]
\item the enclosed charge $q_{\mathrm{enc}}$ for that Gaussian surface,
\item the electric flux through the curved side and through the two flat end caps, and
\item the magnitude and direction of the electric field at $P$.
\end{enumerate}}
\sol Let the cylinder have radius $r$ and length $L$. Because the wire has uniform linear charge density $\lambda$, the charge enclosed by the Gaussian surface is
\[
q_{\mathrm{enc}}=\lambda L.
\]
By cylindrical symmetry, the electric field due to the long wire points radially outward from the wire and has the same magnitude $E(r)$ everywhere on the curved side of the Gaussian cylinder. Let $\hat{s}$ denote the outward radial unit vector from the wire.
On the curved side, the area element $d\vec{A}$ also points radially outward, so $\vec{E}$ is parallel to $d\vec{A}$. Thus,
\[
\vec{E}\cdot d\vec{A}=E(r)\,dA
\]
on the curved surface.
On each flat end cap, the area element $d\vec{A}$ points along the axis of the wire, while $\vec{E}$ points perpendicular to that axis. Therefore,
\[
\vec{E}\cdot d\vec{A}=0
\]
on both end caps, so the flux through each end cap is zero.
The total flux is therefore entirely through the curved side:
\[
\oint_S \vec{E}\cdot d\vec{A}=E(r)\bigl(2\pi rL\bigr).
\]
Apply Gauss's law:
\[
E(r)(2\pi rL)=\frac{\lambda L}{\varepsilon_0}.
\]
Cancel $L$:
\[
E(r)=\frac{\lambda}{2\pi \varepsilon_0 r}.
\]
Now substitute $\lambda=3.0\times 10^{-6}\,\mathrm{C/m}$, $r=0.20\,\mathrm{m}$, and $\varepsilon_0=8.85\times 10^{-12}\,\mathrm{C^2/(N\,m^2)}$:
\[
E(r)=\frac{3.0\times 10^{-6}}{2\pi(8.85\times 10^{-12})(0.20)}\,\mathrm{N/C}.
\]
This gives
\[
E(r)=2.7\times 10^5\,\mathrm{N/C}.
\]
So the fluxes are
\[
\Phi_{\mathrm{curved}}=E(r)(2\pi rL)=\frac{\lambda L}{\varepsilon_0},
\qquad
\Phi_{\mathrm{cap\ 1}}=0,
\qquad
\Phi_{\mathrm{cap\ 2}}=0,
\]
and the electric field at $P$ is
\[
\vec{E}(P)=(2.7\times 10^5\,\mathrm{N/C})\hat{s},
\]
where $\hat{s}$ points radially away from the positively charged wire.