1. Coulomb’s Law

First introduced in Chapter 3.

Coulomb’s Law

Electric charges always exert a force on each other.

Magnitude of the force is proportional to the magnitude of each charge and inversely proportional to the square of the distance between them.

\[\mathbf{\vec{F}} = \frac{Q_1 Q_2}{4 \pi \epsilon_0 R^2} \, \mathbf{\vec{a_R}}\label{eq-coulomb}\]

While fundamental, this force is difficult to deal with because it depends on the specifics of the charges. What is better is to define a vector field that describes the force a “test charge” q would experience due to another charge.

\[\mathbf{\vec{E}} \equiv \frac{\mathbf{\vec{F}}}{q} \label{eq-e-definition}\]

Combining with Coulomb’s law yields the electric field (E-field) surrounding a point charge

\[\mathbf{\vec{E}} = \frac{Q}{4 \pi \epsilon_0 R^2} \, \mathbf{\vec{a_R}} \label{eq-e-field}\]

The Coulombic force is in newtons, \(\mathrm{N = \frac{kg \cdot m}{s^2}}\). Dividing by charge gives units of newtons per coulomb \(\mathrm{\frac{N}{C} = \frac{kg \cdot m}{s^2 \cdot C}}\) for the electric field.

  • a volt has units \(\mathrm{\frac{J}{C}}\) or \(\mathrm{\frac{W}{A}}\) or \(\mathrm{\frac{kg \cdot m^2}{s^3 \cdot A}}\)

  • charge is \(\mathrm{s\!\cdot\!A}\) in SI base units.

Substituting, electric field can thus be expressed in units of \(\mathrm{\frac{V}{m}}\).

2. Gauss’s Law

\[\nabla \bullet \mathbf{\vec{E}} = \frac{\rho_V}{\epsilon_0} \quad \text{Gauss's Law (point)}\label{eq-gauss-point}\]
\[\oint \mathbf{\vec{E}}\bullet\mathbf{\vec{dS}} = \frac{Q_{enc}}{\epsilon_0} \quad \text{Gauss's Law (integral)}\label{eq-gauss-integral}\]
A “Gaussian surface” is simply a closed surface.

3. Ampere’s Law

\[\oint \vec{B} \bullet \mathrm{d}\vec{l} = \mu_0 \cdot I_{\mathrm{enc}}\]
\[\vec{\nabla} \times \vec{B} = \mu_0 \cdot \vec{J}\]

4. Faraday’s Law of Induction

5. Vector math

5.1. Divergence

Chapter 5.2

\[\nabla \bullet \mathbf{\vec{A}} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}\]

5.2. Gradient

Chapter 7

vector derivative

5.3. Curl

\[\mathop{curl} \vec{A} = \vec{\nabla} \times \vec{A} = \lim_{\Delta s \rightarrow 0} \frac{\hat{a}_n \oint \vec{A} \bullet \mathrm{d}\vec{l}}{\Delta s}\]

The determinant form is most useful

Cartesian
\[\vec{\nabla} \times \vec{A} = \begin{vmatrix} \hat{a}_x & \hat{a}_y & \hat{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}\]
Cylindrical
\[\vec{\nabla} \times \vec{A} = \frac{1}{\rho} \cdot \begin{vmatrix} \hat{a}_\rho & \rho \hat{a}_\phi & \hat{a}_z \\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ A_\rho & \rho A_\phi & A_z \end{vmatrix}\]
Spherical
\[\vec{\nabla} \times \vec{A} = \frac{1}{\rho^2 \sin{\theta}} \cdot \begin{vmatrix} \hat{a}_r & r \, \hat{a}_\theta & r \sin{\theta} \, \hat{a}_\phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ A_r & r A_\theta & r \sin{\theta} \, A_\phi \end{vmatrix}\]

6. Other

Think first, then click for an explanation

Melting wires has to do with heat transfer.

Power absorbed in the wire heats it, increasing the wire’s temperature. Eventually, the temperature will reach the melting point of copper (1084 °C) and …​ game over.

However at the same time, heat is lost to the environment via radiation and convection, counteracting the temperature rise. When the system reaches steady-state, the electrical power absorbed will equal the heat power lost at some elevated temperature.

Both convection and radiation depend on (the wire’s) surface area, but, as we learned in Lesson 10, power is absorbed per volume.

The ratio of volume to surface area (or cross-section area to circumference in 2-D) is proportional to diameter, meaning that the power absorbed increases faster than the “cooling power” as the wire diameter increases. This effect makes for higher temperatures for larger wires at the same current density.

A similar volume to surface area effect was involved in the construction technique for the Hoover Dam. It would have taken about 125 years to cool off from the heat generated from the concrete curing process without active cooling. See The Story of Hoover Dam - Concrete.