\hypertarget{vector-analysis}{%
\subsection{1.1 Vector Analysis}\label{vector-analysis}}

\hypertarget{scalar}{%
\paragraph{Scalar}\label{scalar}}

A measure described by one real number. Examples include temperature,
size, and mass. A scalar is a \(1 \times 1\) matrix.

\hypertarget{vector}{%
\paragraph{Vector}\label{vector}}

A measure described by more than one real number (direction and
magnitude). Examples include force, velocity, and it's derivatives.
Vectors are often described by \(n \times 1\) or \(1 \times n\)
matricies.

\hypertarget{unit-vector}{%
\paragraph{Unit Vector}\label{unit-vector}}

A unit (direction or normalized) vector is signified with a \(\hat{ }\)
symbol. Common unit vectors include \(\hat{x}\), \(\hat{y}\), and
\(\hat{z}\). The normalized version of any vector is defined as:
\[\hat{a} = \frac{\vec{A}}{\lVert\vec{A}\rVert}\]

\hypertarget{dot-product}{%
\paragraph{Dot Product}\label{dot-product}}

The dot product is a measure of how \emph{parallel} two vectors are,
scaled by the magnitudes of the two vectors. To compute it, find the sum
of the products of the like components of two vectors. It is also
defined as the product of the magnitudes of the vectors normalized by
the cosine of the angle between them. It is defined as:
\[W = \vec{F} \cdot \vec{r} =  \lVert\vec{F}\rVert\lVert\vec{r}\rVert \cos{\alpha}\]

\hypertarget{cross-product}{%
\paragraph{Cross Product}\label{cross-product}}

The cross product is a measure of how \emph{perpendicular} two vectors
are. This operation yeilds a vector quantity \emph{orthoganal} to both
original vectors. The direction vector for the cross product is
\(\hat{a}_c\), and its magnitude is the product of the magnitudes and
the sine of the angle between them. It is defined as:
\[\vec{C} = \vec{A} \times \vec{B} = \lVert\vec{A}\rVert \lVert\vec{B}\rVert \sin(\alpha) \hat{a}_c = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}\]

\hypertarget{right-hand-rule}{%
\paragraph{Right Hand Rule}\label{right-hand-rule}}

The right hand rule is a quick way to find \(\hat{a}_c\).
\[\overrightarrow{thumb} = \overrightarrow{pointer} \times \overrightarrow{middle}\]

\hypertarget{cartesian-to-cylindrical}{%
\paragraph{Cartesian to Cylindrical}\label{cartesian-to-cylindrical}}

\[\vec{A} = 4\hat{x} + 4\hat{y} - 2\hat{z}\] \[r = \sqrt{x^2 + y^2}\]
\[\varphi = \arctan\left(\frac{y}{x}\right)\] \[z = z\]
\[\vec{A} = A_\rho \hat{\rho} + A_\varphi \hat{\varphi} + A_z \hat{z}\]
\[A_\rho = A_x \cos(\varphi) + A_y\sin(\varphi) = 4\sqrt{2}\]
\[A_\varphi = -A_x\sin(\varphi) + A_y\cos(\varphi) = 0\]
\[A_z = A_z = -2\]

\hypertarget{vector-calculus}{%
\subsection{1.2 Vector Calculus}\label{vector-calculus}}

The differential along some path, \(d\vec{l}\), is defined as:
\[d\vec{l} = dx\hat{x} + dy\hat{y} + dz\hat{z} = d\vec{x} + d\vec{y} + d\vec{z}\]

\hypertarget{the-del-leftnablaright-operator}{%
\paragraph{\texorpdfstring{The ``Del'' \(\left(\nabla\right)\)
operator}{The ``Del'' \textbackslash left(\textbackslash nabla\textbackslash right) operator}}\label{the-del-leftnablaright-operator}}

The gradient of the scalar field \(\left(\nabla f\right)\).
\[\nabla = \frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y} + \frac{\partial}{\partial z} \hat{z}\]
\[df = \nabla f \cdot d\vec{l} = \left(\frac{\partial f}{\partial x}\right)dx + \left(\frac{\partial f}{\partial y}\right)dy + \left(\frac{\partial f}{\partial z}\right)dz\]

\hypertarget{directional-derivative}{%
\subsubsection{Directional Derivative:}\label{directional-derivative}}

The \emph{directional derivative} is used to find the change of a
function along some infinatesimal direction and is defined as:
\[\Delta \varphi = {\nabla}_l \varphi \cdot \Delta \vec{l}\]
\[\therefore {\nabla}_l \varphi = \frac{\Delta \varphi}{\delta \vec{l}} = \frac{d \varphi}{d \vec{l}} = \nabla \varphi \cdot \frac{\vec{l}}{\lVert\vec{l}\rVert}\]

\hypertarget{example-1.1}{%
\paragraph{Example 1.1}\label{example-1.1}}

A function, \(f(x,y,z) = x^2 y^2 + xyz\). Find \(\nabla f\).
\[f(x,y,z) = f(r)\] \[r = \sqrt{x^2 + y^2 + z^2}\]
\[\nabla f = (2xy^2)\hat{x} + (2x^2y + xz)\hat{y} + xy\hat{z}\]

\hypertarget{example-1.2}{%
\paragraph{Example 1.2}\label{example-1.2}}

Consider a function, \(w = x^2y^2 + xyz\). Find \(\nabla_l w\) at
\((2, -1, 0)\) in the direction,
\(\vec{l} = 3\hat{x} + 4\hat{y} + 12\hat{z}\).
\[\lVert\vec{l}\rVert = \sqrt{3^2 + 4^2 + 12^2} = 13\]
\[{\nabla}_l w = \nabla w \cdot \frac{\vec{l}}{\lVert\vec{l}\rVert}\]
Solving at \((2, -1, 0)\):
\[\nabla w = 2(2)(-1)^2\hat{x} + 2(2)^2(-1)\hat{y} + 2(-1)\hat{z} = 4\hat{x} - 8\hat{y} - 2\hat{z}\]
\[{\nabla}_l w= (4\hat{x} - 8\hat{y} - 2\hat{z}) \cdot \left(\frac{3}{13}\hat{x} + \frac{4}{13}\hat{y} + \frac{12}{13}\hat{z}\right) = \frac{-44}{13}\]

\hypertarget{divergence-and-curl}{%
\subsubsection{Divergence and Curl}\label{divergence-and-curl}}

\hypertarget{divergence-of-a-vector-field}{%
\paragraph{Divergence of a Vector
Field}\label{divergence-of-a-vector-field}}

The divergence of a vector field is a measure of outward flux. It is
defined as:
\[\nabla \cdot f = \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot (f_x + f_y + f_z) = \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}\]

If \(\nabla \cdot \vec{A} = 0\), there is no divergence.

\hypertarget{curl-of-a-vector-field}{%
\paragraph{Curl of a Vector Field}\label{curl-of-a-vector-field}}

The curl of a vector field is a measure of circulation in each
infinatesimally small region of the field. It is defined as:
\[\nabla \times f = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_x & f_y & f_z \\ \end{vmatrix} = \left< \frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}, \frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}, \frac{\partial f_y}{\partial z} - \frac{\partial f_x}{\partial z} \right>\]

\hypertarget{solenoidal-field}{%
\paragraph{Solenoidal Field}\label{solenoidal-field}}

A solenoidal field is a vector field without divergence, defined as:
\[\nabla \cdot \vec{f} = 0\]

\hypertarget{conservative-rotational-field}{%
\paragraph{Conservative / Rotational
Field}\label{conservative-rotational-field}}

A conservative field is a vector field without curl, defined as:
\[\nabla \times \vec{f} = 0\]

\hypertarget{example-1.3}{%
\paragraph{Example 1.3}\label{example-1.3}}

Consider the vector field, \(\vec{F} = k \hat{x}\), where both the
direction and magnitude are uniform in all space.
\[\nabla \cdot \vec{F} = 0\] \[\nabla \times \vec{F} = 0\]

\hypertarget{example-1.4}{%
\paragraph{Example 1.4}\label{example-1.4}}

Consider the vector field, \(\vec{F} = k \hat{r}\), where magnitude is
constant and direction is away from a central point.
\[\hat{r} = \sqrt{\hat{x}^2 + \hat{y}^2 + \hat{z}^2}\]
\[\nabla \cdot \vec{F} = \sqrt{3} k\] \[\nabla \times \vec{F} = 0\]

\hypertarget{example-1.5}{%
\paragraph{Example 1.5}\label{example-1.5}}

Consider the vector field, \(\vec{F} = k \times \hat{r}\), where
magnitude is uniform and the direction is perpendicular to the distance
from a central point for all space. \[\nabla \cdot \vec{F} = 0\]
\[\nabla \times \vec{F} = 2k\]

\hypertarget{curl-and-divergence-identities}{%
\subsubsection{Curl and Divergence
Identities}\label{curl-and-divergence-identities}}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  The Laplacian: can operate on a scalar or vector field
  \[\nabla \cdot (\nabla f) = \nabla^2 f\]
  \[\frac{\partial^2 f}{\partial x} + \frac{\partial^2 f}{\partial y} + \frac{\partial^2 f}{\partial z}\]
\item
  The curl of a gradient is \(0\) \[\nabla \times (\nabla f) = 0\]
\item
  The gradient of divergence is a scalar
  \[\nabla (\nabla \cdot \vec{f})\]
\item
  The divergence of curl is \(0\)
  \[\nabla \cdot (\nabla \times \vec{v}) = 0\]
\item
  Curl of curl
  \[\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}\]
\end{enumerate}

\hypertarget{the-line-integral}{%
\subsubsection{The Line Integral}\label{the-line-integral}}

The line integral is the integral of the tangential component of a
vector field along a path.

The line integral is defined as:
\[\int \vec{A} \cdot d\vec{l} = \int \lVert \vec{A} \rVert \cos(\alpha) \lVert d\vec{l} \rVert\]

Where:

\begin{itemize}
\item
  \(\vec{A}\) is some vector field
\item
  \(d\vec{l} = dx\hat{x} + dy\hat{y} + dz\hat{z}\)
\end{itemize}

If the path of the integral is a closed curve, it is said to be the
circulation of \(\vec{A}\) around \(\vec{l}\), defined as:

\[\oint_c \vec{A} \cdot d\vec{l}\]

\hypertarget{example-1.6}{%
\paragraph{Example 1.6}\label{example-1.6}}

Calculate the circulation of \(\vec{F}\) around the path.

\(\vec{F} = x^2 \hat{x} - xy\hat{y} - y^2\hat{z}\)

Circulation of \(\vec{F}\) around the path:
\[\oint \vec{F} \cdot d \vec{l} = \int_1 + \int_2 + \int_3 + \int_4\]

\hypertarget{path-1}{%
\subparagraph{Path 1}\label{path-1}}

Straight line from \((1,0,0)\) to \((0,0,0)\)

\(x\) varies, \(z=0\), \(y=0\).

Plug into \(\vec{F}\):
\[\vec{F}_1 = x^2\hat{x} - x(0)\hat{y} - (0)^2\hat{z} = x^2\hat{x}\]

Since \(x\) varies at some rate, \(dx\) exists, and since \(y\) and
\(z\) are constant, \(dy\) and \(dz\) are both \(0\).

Plug into \(d\vec{l}\):
\[d\vec{l} = dx\hat{x} + (0)\hat{y} + (0)\hat{z} = dx\hat{x}\]
\[\int \vec{F}_1 \cdot d\vec{l}_1 = \int x^2\hat{x} \cdot dx\hat{x} = \int x^2dx\]

For this specific path:
\[\int_{1}^{0} x^2dx = \left. \frac{x^3}{3} \right\vert_1^0 = 0 - \frac{1}{3} = -\frac{1}{3}\]

\hypertarget{path-2}{%
\subparagraph{Path 2}\label{path-2}}

Straight line from \((0,0,0)\) to \((0,1,0)\)

\(y\) varies, \(x=0\), \(z=0\)

Plug into \(\vec{F}\):
\[\vec{F}_2 = (0)^2\hat{x} - (0)y\hat{y} - y^2\hat{z} = -y^2\hat{z}\]

Since \(y\) varies at some rate, \(dy\) exists, but since \(x\) and
\(z\) are constant, \(dx\) and \(dz\) are both \(0\).

Plug into \(d\vec{l}\):
\[d\vec{l}_2 = (0)\hat{x} + dy\hat{y} + (0)\hat{z} = dy\hat{y}\]
\[\int \vec{F}_2 \cdot d\vec{l} = \int -y^2\hat{z} \cdot dy\hat{y} = \int\limits_0^1 0 = 0\]

\hypertarget{path-3}{%
\subparagraph{Path 3}\label{path-3}}

Straight line from \((0,1,0)\) to \((1,1,1)\)

\(x\) and \(z\) vary at the same rate and always have the same value,
\(y=1\)

\hypertarget{surface-integrals}{%
\subsubsection{Surface Integrals}\label{surface-integrals}}

\(\hat{n}\) is the unit normal vector of a surface.

The total \emph{flux} crossing an area, \(\Delta s\), is given by the
function:
\[\Delta s \left[ \lVert \vec{F} \rVert \cos(\alpha)\right] = \vec{F} \cdot \hat{n} \Delta s\]

Total flux is defined as:
\[\sum_{i=1}^N \vec{F}_i \cos(\alpha_i) \Delta s_i = \int_s \vec{F} \cdot d\vec{s}\]

\hypertarget{gradient-theorem}{%
\paragraph{Gradient Theorem}\label{gradient-theorem}}

The line integral through a gradient field is the difference of the
values a the end points.

\[\int\limits_{r_1}^{r_2} \nabla \varphi \cdot d\vec{l} = \varphi(r_2) - \varphi(r_1)\]

\[\oint \nabla \varphi \cdot d\vec{l} = 0\]

\hypertarget{divergence-theorem}{%
\paragraph{Divergence Theorem}\label{divergence-theorem}}

The divergence in some volume is the same as the flux through its
surface.

\[\iiint_{vol} \nabla \cdot \vec{A} d\tau = \oiint \vec{A} \cdot d\vec{s}\]

\hypertarget{stokes-theorem}{%
\paragraph{Stokes Theorem}\label{stokes-theorem}}

The curl in some region is the same as the circulation of the region's
border.
\[\iint_A (\nabla \times \vec{A}) \cdot d\vec{s} = \oint \vec{A} \cdot d\vec{l}\]

\hypertarget{coulombs-law}{%
\subsection{1.3 Coulomb's Law}\label{coulombs-law}}

Initial observation: \[\vec{F} \propto q_1 q_2\]

Vacuum Permittivity: \[\varepsilon_0 = 8.854 \times 10^{-12}\]

Coulomb's Constant:
\[k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^{-9}\]

Coulomb's Law: \[\vec{F} = k \frac{q_1 q_2}{r^2} \hat{a}_{1 2}\]

Superposition of Coulomb's Law:
\[\vec{F}_{Net} = \sum_{i=1}^{N} k_i \frac{Q q_i}{r_i^2} \hat{r}\]

\textbf{Note}: \(k_i\) depends on material properties. When in a vacuum,
\(k_i = k\).

\hypertarget{example-1.7}{%
\paragraph{Example 1.7}\label{example-1.7}}

Two point charges, \(q_1\) and \(q_2\) are spaced 2cm apart on the
x-axis. A third charge, \(q_3\) is placed between the first two with a
distance \(x_1\) between it and \(q_1\) and \(x_2\) between it and
\(q_2\) such that \(q_3\) is in static equilibrium.

Known: \[\vec{F}_{1 3} + \vec{F}_{2 3} = 0\] By Coulomb's Law:
\[\vec{F}_{1 3} = \frac{k_1 q_3 q_1}{r_{1 3}^2} \hat{z}\]
\[\vec{F}_{2 3} = -\frac{k_2 q_3 q_2}{r_{1 3}^2} \hat{z}\]
\[\therefore \vec{F}_{1 3} + \vec{F}_{2 3} = k q_3 \left( \frac{q_1}{r_{1 3}^2} - \frac{q_2}{r_{2 3}^2} \right)\hat{z} = 0\]
\[\therefore \frac{q_1}{r_{1 3}^2} - \frac{q_2}{r_{2 3}^2} = 0\]

Solve for \(r_{1 3}\): \[r_{1 3} = \pm r_{2 3}\sqrt{\frac{q_1}{q_2}}\]

Known: \[r_{1 3} + r_{2 3} = 2\] \[\therefore r_{2 3} = 2 - r_{1 3}\]

Plug into first equation:
\[r_{1 3} = \pm (2 - r_{1 3}) \sqrt{\frac{q_1}{q_2}}\]

Expand:
\[r_{1 3} = \pm \left(2 \sqrt{\frac{q_1}{q_2}} - r_{1 3}\sqrt{\frac{q_1}{q_2}} \right)\]
\[\therefore r_{1 3} \left(1 \pm \sqrt{\frac{q_1}{q_2}} \right) = \pm 2\sqrt{\frac{q_1}{q_2}}\]
\[\therefore r_{1 3} = \pm 2\sqrt{\frac{q_1}{q_2}} \left( 1 \pm \sqrt{\frac{q_1}{q_2}} \right)^{-1}\]

\hypertarget{electric-field-intensity}{%
\subsubsection{Electric Field
Intensity}\label{electric-field-intensity}}

Call a test charge at the point of measurement \(Q_2\).

Coulomb's Law:
\[\vec{F}_{1 2} = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 R^2} \hat{a}_{1 2}\]

The electric field intensisty:
\[\vec{E}_{1 2} = \frac{\vec{F}_{1 2}}{Q_1}\]
\[\vec{E} = \frac{Q_1}{4 \pi \varepsilon_0 r^2} \hat{a}_{1 2}\]
\[\vec{E}_{NET} = \sum_{i=0}^{N} \frac{k_i Q_i}{R_i^2} \hat{a}_{R_i}\]

Force due to electric field: \[\vec{F} = q \vec{E}\]

\hypertarget{electric-field}{%
\paragraph{Electric field}\label{electric-field}}

Always in the same direction as the electric field force.

\hypertarget{electric-field-of-a-dipole}{%
\paragraph{Electric field of a
dipole}\label{electric-field-of-a-dipole}}

At a point equidistant to each pole:
\[\lVert \vec{r}_1 \rVert = \lVert \vec{r_2} \rVert = r\]

By Coulomb's Law: \[\vec{E}_1 = \frac{kq}{r^2} \hat{r}_1\]
\[\vec{E}_2 = \frac{kq}{r^2} \hat{r}_2\]
\[\therefore \lVert \vec{E}_1 \rVert = \lVert \vec{E}_2 \rVert\]

In terms of the component distances: \[r^2 = a^2 + y^2\]
\[\therefore E = \frac{kq}{a^2 + y^2}\]

\[\vec{E}_{NET_y} = \vec{E}_{1_y} + \vec{E}_{2_y} = 2E\cos{\theta}\]

By definition: \[\cos(\theta) = \frac{a}{r}\]
\[\therefore \vec{E}_{NET_y} = \frac{2kq}{a^2 + y^2} \frac{a}{\sqrt{a^2 + y^2}}\hat{y}\]
\[\vec{E}_{NET_y} = 2 \frac{kqa}{(x^2 + y^2)^{3/2}} \hat{y}\]

If \(y \gg a\) (far field): \[\vec{E}_{NET} = 2 \frac{kqa}{y^3}\]

\textbf{Takeaway}: \[\vec{E}_{monopole} \propto \frac{1}{r^2}\]
\[\vec{E}_{dipole} \propto \frac{1}{r^3}\]

\hypertarget{charge-densities}{%
\paragraph{Charge Densities}\label{charge-densities}}

\(\lambda\) - Linear charge density

\(\sigma\) - Surface charge density

\(\rho\) - Volume charge density

Electric flux density: \[\vec{D} = \varepsilon_0 \vec{E}\]

Where:
\[\vec{E} = \lim_{\Delta q \to 0} \frac{k \sum_i \Delta q_i}{\lVert \vec{r}_i \rVert ^2} \hat{r}_i = \int \frac{k}{\lVert \vec{r}_i \rVert ^2} dq\]

\hypertarget{example-1.8}{%
\paragraph{Example 1.8}\label{example-1.8}}

A straight line segment of length \(L\) with uniform charge density
\(\lambda\) extends from the origin in the \(\hat{x}\) direction. Find
the strength of the electric field, \(\vec{E}\) at some arbitrary point,
\(p\) along the ray from the origin in the \(\hat{z}\) direction.

The distance along the line segment in the \(\hat{x}\) direction is
denoted as \(x\), and the distance from the origin to point \(p\) is
denoted as \(z\). The vector, \(\vec{r}\) has length
\(\sqrt{x^2 + z^2}\) and makes an angle \(\theta\) with the ray in the
\(-\hat{z}\) direction.

The contribution \(d\vec{E}\) to the total electric field, \(\vec{E}\),
at point \(p\) is defined as:
\[d\vec{E} = \frac{k dq}{\lVert \vec{r} \rVert ^2} \hat{r}\]

For a linear charge distribution: \[dq = \lambda dl\] The distance,
\(r\), to each \(x\) alond the line:
\[\lVert \vec{r} \rVert ^2 = x^2 + z^2\] The components of the
\(\vec{E}\)-field at point \(p\):
\[d\vec{E}_x = \lVert d\vec{E} \rVert \sin(\theta) \hat{x}\]
\[d\vec{E}_z = \lVert d\vec{E} \rVert \cos(\theta) \hat{z}\]

\hypertarget{gausss-law}{%
\subsection{Gauss's Law}\label{gausss-law}}

\hypertarget{e-flux-density}{%
\subsection{E-Flux Density}\label{e-flux-density}}

\hypertarget{emf}{%
\subsection{EMF}\label{emf}}

Measured in \emph{volts}, electromotive force (EMF), is denoted by
\(\mathcal{E}\). The value for EMF is defined as:
\[\mathcal{E} = \oint \vec{E} \cdot d \vec{l} = -\frac{d}{dt} \int_s B_z(t) \cdot ds = -\frac{d}{dt}\psi_m\]

Where:

\[\psi_m = \int_s \vec{B} \cdot d\vec{s}\]

A perfectly conducting ring with radius, \(\rho_0\), centered on the
origin in the x-y plane.

The charge distribution: \[\rho = \rho_0 + \rho_0 \sin(\omega t)\]
\[\vec{B}(t) = B_0 \cos(\omega t)\hat{z}\]
\[\mathcal{E} = \oint \vec{E} \cdot d\vec{l} = \iint\limits_{\phi R} B_0 \cos(\omega t) dr d\phi\]

Where:

\(\varphi: [0, 2\pi]\)

\(R: [0, \rho(t)]\)

Therefore: \[\psi_m = (B_0 \cos(\omega t) \hat{z})(2\pi \rho(t))\]

\[\mathcal{E} = -\frac{d}{dt} B_0 2\pi \cos(\omega t)(\rho_0 + \rho_0\sin(\omega t))\]

\hypertarget{filling-in-some-gaps}{%
\subsubsection{Filling in some Gaps}\label{filling-in-some-gaps}}

\[\vec{F} = -\nabla \vec{u}\] Where: \(\vec{u}\) is potential.

\[\vec{E} = -\nabla \vec{v}\] Where: \(\vec{v}\) is electric potential.

\[W = \vec{F} \cdot \vec{d} = q\vec{E} \cdot \vec{d}\]

Work done by the \(\vec{E}\)-field on a charge will reduce the electric
potential:

\[-\Delta u = u_B - u_A = -\Delta W = -qEd\]

The total change is:

\[\Delta u = -q \int\limits_A^B \vec{E} \cdot d\vec{l}\]

\[\therefore \frac{\Delta u}{q} = \int\limits_A^B \vec{E} \cdot d\vec{l} = \Delta V\]

\[W = q \int \vec{E} \cdot d\vec{l}\]

\[V(r) = \frac{kq}{r}\]

For \(N\) discrete charges,
\[V = \sum\limits_{i=1}^{N} \frac{k q_i}{r_i}\]

In cartesian coordinates:

\[\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{z}\]

Where:

\(F_x = \frac{dw}{dx}\)

\(F_y = \frac{dw}{dy}\)

\(F_z = \frac{dw}{dz}\)

Therefore:
\[\vec{F} = \frac{\partial w}{\partial x}\hat{i} + \frac{\partial w}{\partial y}\hat{j} + \frac{\partial w}{\partial z}\hat{k} = \nabla \vec{w}\]

\[q\vec{E} = \nabla (-\vec{u})\]

\[\vec{F} = -\nabla \vec{u}\]

\[\vec{E} = -\nabla \left( \frac{\vec{u}}{q} \right) = -\nabla \vec{V}\]

\hypertarget{electric-flux-density}{%
\paragraph{Electric Flux Density}\label{electric-flux-density}}

\[\vec{D} = \varepsilon \vec{E}\] Where
\(\varepsilon = \varepsilon_r \varepsilon_0\) (permittivity).

\hypertarget{magnetic-flux-density}{%
\paragraph{Magnetic Flux Density}\label{magnetic-flux-density}}

\[\vec{B} = \mu\vec{H}\] Where \(\mu = \mu_r \mu_0\) (permeability).

\hypertarget{amperes-law}{%
\subsubsection{Ampere's Law}\label{amperes-law}}

The total current crossing an area, \(s\), that is enclosed by the
contour \(C\): \[\oint_C \frac{\vec{B}}{\mu_0} \cdot d\vec{l}\]

The total current is the sum of the current due to charge flow and the
current due to the time rate of change of the electric flux crossing an
area, \(s\). Maxwell was able to unify electricity and magnetism by
adding the current due to the time rate of change of electric flux.
\[\oint_C \vec{H} \cdot d\vec{l} = \int_S \vec{J} \cdot d\vec{s} + \frac{d}{dt}\int \varepsilon_0\vec{E} \cdot d\vec{s}\]

\hypertarget{simplified-amperes-law}{%
\paragraph{Simplified Ampere's Law}\label{simplified-amperes-law}}

\[\oint_C \vec{H} \cdot d\vec{l} = I = \int_s \vec{J} \cdot d\vec{s}\]

The charge density, \(J = \rho v\), has units
\(\left[ \frac{A}{m^2} \right]\).

\hypertarget{example}{%
\paragraph{Example}\label{example}}

A current, \(I\), in an infinitely long cylindrical wire with radius,
\(R\). \[\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\]

Measuring the magnetic field at some distance, \(r\), from the center of
the conductor: \[B\int dl = \mu_0 I\] Where \(\int dl\) is the
circumfrece of measurement. \[B(2\pi r) = \mu_0 I\]
\[B_{out} = \frac{\mu_0 I}{2\pi r} \hat{\varphi}\] Inside the wire:
\[\int \vec{B} \cdot d\vec{l} = \mu_0 I\]
\[B\int dl = \frac{\mu_0 I r^2}{R^2} = 2\pi rB\]
\[B = \frac{\mu_0 I}{2\pi R^2} r\]
\[\vec{B} = \frac{\mu_0 I}{2\pi R^2}r \hat{\varphi}\]

\hypertarget{coulombs-law-1}{%
\subsubsection{Coulomb's Law}\label{coulombs-law-1}}

The total displacement flux of charge:
\[\int_s \varepsilon_0 \vec{E} \cdot d\vec{s}\]

The total current (charge with respect to time):
\[I = \frac{d}{dt} \int_s \varepsilon_0 \vec{E} \cdot d\vec{s}\]

\hypertarget{faradays-law}{%
\subsubsection{Faraday's Law}\label{faradays-law}}

Work done in moving a unit positive test charge around a closed path,
\(C\): \[\oint_C \vec{E} \cdot d\vec{l}\]

Magnetic force on a poving charge and is directed perpendicular to both
the direction of the motion of the charge and the magnetic field.
\[\oint_C \vec{B} \cdot d\vec{l}\]

\hypertarget{solenoid-ideal}{%
\subsubsection{Solenoid (Ideal)}\label{solenoid-ideal}}

For an ideal solenoid with constant current, \(I\), assume uniform
\(\vec{B}\) inside, \(\vec{B} = 0\) outside, and infinite length.

By Ampere's Law: \[\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\]

For a square loop with one side in the solenoid, and it's parallel side
outside the loop:
\[\oint \vec{B} \cdot d\vec{l} = \int_1 + \int_2 + \int_3 + \int_4\]

Sides 2 and 4 are parallel, and side 3 is outside the solenoid, so:
\[\oint \vec{B} \cdot d\vec{l} = \int_1 = Bl\]

Back to Ampere's Law: \[Bl = \mu_0 I N\]
\[\therefore B = \frac{\mu_0 I N}{l} = \mu_0 I n\] Where:

\(N\) is the total number of windings,

\(l\) is the sidelength of the Amperian loop,

\(n\) is the number of windings per unit length \(\frac{N}{l}\)

\hypertarget{toroid-ideal}{%
\subsubsection{Toroid (Ideal)}\label{toroid-ideal}}

From a symmetric \(\vec{B}\)-field, lines form concentric circles inside
the toroid. For an ideal toroid, assume \(\vec{B} = 0\) outside, and
Ampere's law inside.

\[\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\] For some circular
Amperian loop inside the toroid: \[\vec{B} = B\hat{\varphi}\]
\[d\vec{l} = 2\pi r \vec{\varphi}\] \[B(2\pi r) = \mu_0 N I\]
\[\therefore B = \frac{\mu_0 N I}{2\pi r}\]
