# 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](/1a/math115/#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](/sph3u7/#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](/sph3u7/#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](https://en.wikiversity.org/wiki/MyOpenMath/Solutions/Maxwell%27s_integral_equations) 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$$