Electromagnetism (EM)

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The interaction between electric charges at rest is described by Coulomb's law:


\begin{equation} \vec{F_2}= k \frac{q_1 q_2 \hat{r_{21}}}{r_{21}^2} \end{equation} where: \begin{equation} k = \frac{1}{4 \pi \epsilon_0} = 8.988 \times 10^9 \frac{N m^2}{C^2} \end{equation} \begin{equation} \epsilon_0 = \frac{1}{4 \pi k} = 8.854 \times 10^{-12} \frac{C^2}{N m^2} \end{equation} The work done ( electrical potential energy) by bringing two charges together from infinity to a certain distance is: \begin{equation} W= \int (Force)\cdotp (displacement)=\int_{\infty}^{r_{12}} (-\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}) dr = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}}\end{equation}



Force on particle 2: /$F_2$/=

Work: W=

The sum over pairs can be written as: \begin{equation} U= \frac{1}{2} \sum_{j=1}^{N}\sum_{k \neq j} \frac{1}{4 \pi \epsilon_0} \frac{q_j q_k}{r_{jk}}\end{equation}
The electric field of a charge distribution at point (x,y,z): \begin{equation} E(x,y,z)= \frac{1}{4 \pi \epsilon_0} \sum_{j=1}^{N}\frac{q_j \hat{r_{oj}}}{r_{0j}^2}\end{equation} Therefore, \begin{equation} \vec{F}=q \vec{E} \end{equation} And the electric field for a continuous charge distribution at point (x,y,z): \begin{equation} E(x,y,z)= \frac{1}{4 \pi \epsilon_0} \int \frac{\rho (x',y',z') \hat{r} dx' dy' dz'}{r^2}\end{equation}
Flux, the scalar product of the electric field vector ($\vec{E_j}$) and the area in space ($\vec{a_j}$), such that ( $\Phi= \sum_{all\ j} \vec{E_j} \cdotp \vec{a_j}$ ). At the limit it becomes a surface integral: \begin{equation} \Phi= \int_{entire\_surface} \vec{E} \cdotp d\vec{a} \end{equation} Gauss's law. The flux of the electric field $\vec{E}$ through any closed surface equals $1/\epsilon_0$ times the total charge enclosed by the surface: \begin{equation} \int \vec{E} \cdotp d\vec{a} = \frac{1}{\epsilon_0} \sum_i q_i = \frac{1}{\epsilon_0} \int \rho dv\end{equation}