eifueo/docs/1b/ece106.md

ECE 106: Electricity and Magnetism

MATH 117 review

!!! definition A definite integral is composed of:

- the **upper limit**, $b$,
- the **lower limit**, $a$,
- the **integrand**, $f(x)$, and
- the **differential element**, $dx$.

\[\int^b_a f(x)\ dx\]

The original function cannot be recovered from the result of a definite integral unless it is known that \(f(x)\) is a constant.

N-dimensional integrals

Much like how \(dx\) represents an infinitely small line, \(dx\cdot dy\) represents an infinitely small rectangle. This means that the surface area of an object can be expressed as:

\[dS=dx\cdot dy\]

Therefore, the area of a function can be expressed as:

\[S=\int^x_0\int^y_0 dy\ dx\]

where \(y\) is usually equal to \(f(x)\), changing on each iteration.

!!! example The area of a circle can be expressed as \(y=\pm\sqrt{r^2-x^2}\). This can be reduced to \(y=2\sqrt{r^2-x^2}\) because of the symmetry of the equation.

$$
\begin{align*}
A&=\int^r_0\int^{\sqrt{r^2-x^2}}_0 dy\ dx \\
&=\int^r_0\sqrt{r^2-x^2}\ dx
\end{align*}
$$

!!! warning Similar to parentheses, the correct integral squiggly must be paired with the correct differential element.

These rules also apply for a system in three dimensions:

Vector	Length	Area	Volume
\(x\)	\(dx\)	\(dx\cdot dy\)	\(dx\cdot dy\cdot dz\)
\(y\)	\(dy\)	\(dy\cdot dz\)
\(z\)	\(dz\)	\(dx\cdot dz\)

Although differential elements can be blindly used inside and outside an object (e.g., area), the rules break down as the boundary of an object is approached (e.g., perimeter). Applying these rules to determine an object’s perimeter will result in the incorrect deduction that \(\pi=4\).

Therefore, further approximations can be made using the Pythagorean theorem to represent the perimeter.

\[dl=\sqrt{(dx^2) + (dy)^2}\]

Polar coordinates

Please see MATH 115: Linear Algebra#Polar form for more information.

In polar form, the difference in each “rectangle” side length is slightly different.

Vector	Length difference
\(\hat r\)	\(dr\)
\(\hat\phi\)	\(rd\phi\)

Therefore, the change in surface area can be approximated to be a rectangle and is equal to:

\[dS=(dr)(rd\phi)\]

!!! example The area of a circle can be expressed as \(A=\int^{2\pi}_0\int^R_0 r\ dr\ d\phi\).

$$
\begin{align*}
A&=\int^{2\pi}_0\frac{1}{2}R^2\ d\phi \\
&=\pi R^2
\end{align*}
$$

If \(r\) does not depend on \(d\phi\), part of the integral can be pre-evaluated:

\[ \begin{align*} dS&=\int^{2\pi}_{\phi=0} r\ dr\ d\phi \\ dS^\text{ring}&=2\pi r\ dr \end{align*} \]

So long as the variables are independent of each other, their order does not matter. Otherwise, the dependent variable must be calculated first.

!!! tip There is a shortcut for integrals of cosine and sine squared, so long as \(a=0\) and \(b\) is a multiple of \(\frac\pi 2\):

$$
\int^b_a\cos^2\phi=\frac{b-a}{2} \\
\int^b_a\sin^2\phi=\frac{b-a}{2}
$$

The side length of a curve is as follows:

\[dl=\sqrt{(dr^2+(rd\phi)^2}\]

!!! example The side length of the curve \(r=e^\phi\) (Archimedes’ spiral) from \(0\) to \(2\pi\):

\begin{align*}
dl &=d\phi\sqrt{\left(\frac{dr}{d\phi}\right)^2 + r^2} \\
\tag{$\frac{dr}{d\phi}=e^\phi$}&=d\phi\sqrt{e^{2\phi}+r^2} \\
&=????????
\end{align*}

Polar volume is the same as Cartesian volume:

\[dV=A\ dr\]

!!! example For a cylinder of radius \(R\) and height \(h\):

$$
\begin{align*}
dV&=\pi R^2\ dr \\
V&=\int^h_0 \pi R^2\ dr \\
&=\pi R^2 h
\end{align*}
$$

Moment of inertia

The mass distribution of an object varies depending on its surface density \(\rho_s\). In objects with uniformly distributed mass, the surface density is equal to the total mass over the total area.

\[dm=\rho_s\ dS\]

The formula for the moment of inertia of an object is as follows, where \(r_\perp\) is the distance from the axis of rotation:

\[dI=(r_\perp)^2dm\]

If the axis of rotation is perpendicular to the plane of the object, \(r_\perp=r\). If the axis is parallel, \(r_\perp\) is the shortest distance to the axis. Setting an axis along the axis of rotation is easier.

!!! example In a uniformly distributed disk rotating about the origin like a CD with mass \(M\) and radius \(R\):

$$
\begin{align*}
\rho_s &= \frac{M}{\pi R^2} \\
dm &= \rho_s\ r\ dr\ d\phi \\
dI &=r^2\ dm \\
&= r^2\rho_s r\ dr\ d\phi \\
&= \rho_s r^3dr\ d\phi \\
I &=\rho_s\int^{2\pi}_{\phi=0}\int^R_{r=0} r^3dr\ d\phi \\
&= \rho_s\int^{2\pi}_{\phi=0}\frac{1}{4}R^4d\phi \\
&= \rho_s\frac{1}{2}\pi R^4 \\
&= \frac 1 2 MR^2
\end{align*}
$$

Electrostatics

!!! definition - The polarity of a particle is whether it is positive or negative.

The law of conservation of charge states that electrons and charges cannot be created nor destroyed, such that the net charge in a closed system stays the same.

The law of charge quantisation states that charge is discrete — electrons have the lowest possible quantity.

Please see SL Physics 1#Charge for more information.

Coulomb’s law states that for point charges \(Q_1, Q_2\) with distance from the first to the second \(\vec R_{12}\):

\[\vec F_{12}=k\frac{Q_1Q_2}{||R_{12}||^2}\hat{R_{12}}\]

!!! warning Because Coulomb’s law is an experimental law, it does not quite cover all of the nuances of electrostatics. Notably:

- $Q_1$ and $Q_2$ must be point charges, making distributed charges inefficient to calculate, and
- the formula breaks down once charges begin to move (e.g., if a charge moves a lightyear away from another, Coulomb's law says the force changes instantly. In reality, it takes a year before the other charge observes a difference.)

Dipoles

An electric dipole is composed of two equal but opposite charges \(Q\) separated by a distance \(d\). The dipole moment is the product of the two, \(Qd\).

The charge experienced by a positive test charge along the dipole line can be reduced to as the ratio between the two charges decreases to the point that they are basically zero: \[\vec F_q=\hat x\frac{2kQdq}{||\vec x||^3}\]

Maxwell’s theorems

Compared to Coulomb’s law, \(Q_1\) creates an electric field around itself — each point in space is assigned a vector that depends on the distance away from the charge. \(Q_2\) interacts with the field. According to Maxwell, as a charge moves, it emits a wave that carries information to other charges.

The electric field strength \(\vec E\) is the force per unit positive charge at a specific point \(p\):

\[\vec E_p=\lim_{q\to 0}\frac{\vec{F}}{q}\]

Please see SL Physics 1#Electric potential for more information.

Electric field calculations

If charge is distributed over a three-dimensional object, integration similar to moment of inertia can be used. Where \(dQ\) is an infinitely small point charge at point \(P\), \(d\vec E\) is the electric field at that point, and \(r\) is the vector representing the distance from any arbitrary point:

\[d\vec E = \frac{kdQ}{r^2}\hat r\]

!!! warning As the arbitrary point moves, both the direction and the magnitude of the distance from the desired point \(P\) change (both \(\hat r\) and \(r\)).

Generally, if a decomposing the vector into Cartesian forms \(d\vec E_x\), \(d\vec E_y\), and \(d\vec E_z\) is helpful even if it is easily calculated in polar form because of the significantly easier ability to detect symmetry in the shape. Symmetry about the axis allows deductions such as \(\int d\vec E_y=0\), which makes calculations easier.

In a one-dimensional charge distribution (a line), the charge density is used in a similar way as moment of inertia’s surface density:

\[dQ=\rho_\ell d\ell\]

Two-dimensional charge distributions are more or less the same, but polar or Cartesian forms of the surface area work depending on the shape.

\[dQ=\rho_s dS\]

!!! example A rod of uniform charge density and length \(L\) has a charge density of \(p_\ell=\frac{Q}{L}\).

1. Determine the formula for the charge density \(\rho\)
2. Choose an origin and coordinate system (along the axes of the object when possible)
3. Choose an arbitrary point \(A\) on the charge
4. Create a right-angle triangle with \(A\), the desired point, and usually the origin
5. Attempt to find symmetry
6. Solve

Gauss’s law

!!! definition - A closed surface is any closed three-dimensional object. - Electric flux represents the number of electric field lines going through a surface.

At an arbitrary surface, the normal to the plane is its vector form:

\[\vec{dS}=\vec n\cdot dS\]

The electric flux density \(\vec D\) is an alternate representation of electric field strength. In a vacuum:

\[\vec D = \epsilon_0\vec E\]

Electric flux is the electric flux density multiplied by the surface area at every point of an object.

\[\phi_e=\epsilon_0\int_s\vec E\bullet\vec{dS}\]

The flux from charges outside a closed surface will always be zero at the surface. A point charge in the centre of a closed space has a flux equal to its charge. Regardless of the charge distribution or shape, the total flux through a closed surface is equal to the total charge within the closed surface.

\[\oint \vec D\bullet\vec{dS}=Q_\text{enclosed}\]

This implies \(\phi_e>0\) is a net positive charge enclosed.

!!! warning Gauss’s law only applies when \(\vec E\) is from all charges in the system

Charge distributed over a line/cylinder

!!! warning “Limitations” To apply this strategy, the following conditions must hold:

- $Q$ must not vary with the length of the cylinder or $\phi$
- The charge must be distributed over either a cylindrical surface or the volume of the cylinder.
- In the real world, $r$ must be significantly smaller than $L$ as an approximation.
- The strategy is more accurate for points closer to the centre of the wire.

Please see Maxwell’s integral equations#Gauss’s law for more information.

Outside the radius \(R\) of the cylinder of the Gaussian surface, the enclosed charge is, where \(L\) is the length of the cylinder:

$\(Q_{enc}=\pi R^2\rho_0L\)

such that the field at any radius \(r>R\) is equal to:

\[\vec E(r)=\frac{\rho_0\pi R^2}{2\pi\epsilon_0r}\hat r\]

Inside the radius \(R\) of the cylinder, the enclosed charge depends on \(r\). For a uniform charge density:

\[Q_{enc}=\pi r^2\rho_0L\]

such that the field at any radius \(r< R\) is equal to:

\[\vec E(r)=\frac{\rho_0}{2\epsilon_0}r\hat r\]

The direction of \(\vec E\) should always be equal to that of \(\vec r\). Generally, where \(lim\) is \(r\) if \(r\) is inside the cylinder or \(R\) otherwise, \(\rho_v\) is the function for charge density based on radius, and \(r_1\) is hell if I know:

\[\epsilon_0 E2\pi rL=\int^{lim}_0\rho_v(r_1)2\pi r_1L\ dr_1\]

Charge distributed over a plane

!!! warning To apply this strategy, the following conditions must hold:

- $Q$ must not vary with the lengths of the plane
- The charge must be distributed over a plane or slab
- In the real world, the thickness $z$ must be significantly smaller than the lengths as an approximation

Where \(\rho_v\) is an even surface density function and \(lim\) is from \(0\) to \(z\) if the desired field is outside of the charge, or \(0\) to field height \(h\) if it is inside the charge:

\[\epsilon_0 E=\int_{lim}\rho_v\ dh_1\]

Any two points have equal electric fields regardless of distance due to the construction of a uniform electric field.

Where \(\rho_v\) is not an even surface density function, \(d\) is the thickness of the slab, and \(E\) is the electric field outside the slab:

\[2\epsilon_0E = \int^d_0\rho_v(A)dh_1\]

Where \(E\) is the electric field inside the slab at some height \(z\):

\[E=\frac{\rho_0}{4\epsilon_0}(2z^2-d^2),0\leq z\leq d\]

If \(E\) is negative, it must point opposite the original direction (\(\hat z\)).

Generally:

1. Determine \(\vec E\) outside the slab.
2. Set one outside surface and one inside surface as a pillbox and apply rules.

Electrostatic potential

At a point \(P\), the electrostatic potential \(V_p\) or voltage is the work done per unit positive test charge from infinity to bring it to point \(P\) by an external agent.

\[ V_p=\lim_{q\to 0^+}\frac{W_i}{q} \\ W_I=\int^p_\infty\vec F_I\bullet \vec{dl} \]

Because the desired force acts opposite to the force from the electric field, as long as \(\vec E\) is known at each point:

\[ V_p=-\int^p_\infty\vec E\bullet\vec{dl} V_p=-\int^p_\infty E\ dr \]

The work done only depends on initial and final positions — it is conservative, thus implying Kirchoff’s voltage law.

Where \(\vec dl\) is the path of the test charge from infinity to the point, and \(\vec dr\) is the direct path from the origin through the point to the charge, because \(dr=-dl\):

\[\vec E\bullet\vec{dl}=Edr\]

Therefore, the potential due to a point charge is equal to:

\[V_p=-\int^p_\infty\frac{kQ}{r^2}dr=\frac{kQ}{r}\]

Positive charges naturally move to lower potentials (\(V\) decreases) while negative charges do the opposite.

In order to calculate the voltage for charge distributions:

• If \(\vec E\) is easy to find via Gauss law:

\[V_p=-\int^p_\infty\vec E\bullet\vec{dl}\]

• If the charge is asymmetric:

\[V_p=\int_\text{charge dist}\frac{kdQ}{r}\]