Catchup 1

docs/physics/electrostatics/2-coulomb.md

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+# Chapter 2 - Coulomb's Laws, Electric and Magnetic Fields
+
+## Section 2.2 - Parallel Treatment of Electric and Magnetic Fields
+
+Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\vb{r}$ relative to the origin.
+
+Thus, according to Coulomb's Law,
+
+$$
+\begin{align}
+    F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
+    F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+\end{align}
+$$
+
+Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\vb{r}$.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
+    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+\end{align}
+$$
+
+Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}}$, and we see the following.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
+    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+\end{align}
+$$
+
+With multiple charges, we can apply the *superposition principal* to see the following:
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
+    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+\end{align}
+$$
+
+We can convert this to an integral as $N$ goes to infinity.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
+    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
+\end{align}
+$$
+
+## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
+
+From a lot of advanced math, we know that
+
+$$
+\div{\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}} = 4 \pi \delta(\vb{r}-\vb{r'})
+$$
+
+Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnetostatic fields.
+
+$$
+\begin{align}
+    \div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
+    \div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
+\end{align}
+$$
+
+As the divergence operator does not operate on $\vb{r'}$, we see that
+
+$$
+\begin{align}
+    \div{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+        &= \frac{1}{4 \pi \epsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
+        &= \frac{\rho_e(\vb{r})}{\epsilon_0} \\
+    \div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+        &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
+        &= \frac{\rho_m(\vb{r})}{\mu_0}
+\end{align}
+$$
+
+The curl of an electrostatic or magnetostatic is relatively simple.
+
+
+$$
+\begin{align}
+    \curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+    \curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+\end{align}
+$$
+
+Additionally, we know $\curl{f\vb{A}} = f \curl{\vb{A}} + \grad{f}\cross\vb{A}$. Thus,
+
+$$
+\begin{align}
+    \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} &= \frac{1}{\abs{\vb{r}-\vb{r'}}^3} \curl{(\vb{r}-\vb{r'})} + (\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}}) \cross (\vb{r}-\vb{r'}) \\
+\end{align}
+$$
+
+We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Additionally, $\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}} = -3 \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^5}$, which when crossed with $\vb{r}-\vb{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
+
+## Section 2.4 - Eletric and Magnetic Flux Densities
+
+The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
+
+Now, given $S$ is a surfance enclosing $Q_e$ or $Q_m$ total charge, we denotate flux as following:
+
+$$
+\Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
+$$
+
+Thus, applying divergence theorem,
+
+$$
+Q_e = \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{E}} \dd V
+$$
+
+$$
+Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{H}} \dd V
+$$
+
+Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
+
+$$
+\begin{align}
+    \int_V \rho_e \dd V &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
+    \rho_e &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
+    \int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
+    \rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
+\end{align}
+$$
+
+**Definition**. This is known as *Gauss' Law*.
+
+With applicable symnetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.