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docs/physics/electrostatics/2-coulomb.md

@@ -2,32 +2,32 @@
 
 ## 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.
+Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\mathbf{r}$ relative to the origin.
 
 Thus, according to Coulomb's Law,
 
 $$
 \begin{align}
-    F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
-    F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+    F^e_{qQ}(\mathbf{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
+    F^m_{qQ}(\mathbf{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
 \end{align}
 $$
 
-Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\vb{r}$.
+Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\mathbf{r}$.
 
 $$
 \begin{align}
-    E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
-    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+    E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
+    H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
 \end{align}
 $$
 
-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.
+Now, let $Q$ be at point $\mathbf{r'}$. Then, the unit vector becomes $\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}}|$, and we see the following.
 
 $$
 \begin{align}
-    E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
-    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+    E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
+    H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
 \end{align}
 $$
 
@@ -35,97 +35,89 @@ With multiple charges, we can apply the *superposition principal* to see the fol
 
 $$
 \begin{align}
-    E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
-    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+    E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
+    H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
 \end{align}
 $$
 
 We can convert this to an integral as $N$ goes to infinity.
 
-$$
-\begin{align}
-    E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
-    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
-\end{align}
-$$
+$$\begin{align}
+E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \\
+H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V'
+\end{align}$$
 
 ## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
 
 From a lot of advanced math, we know that
 
-$$
-\div{\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}} = 4 \pi \delta(\vb{r}-\vb{r'})
-$$
+$$\nabla \cdot \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = 4 \pi \delta(\mathbf{r}-\mathbf{r'})$$
 
-Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnetostatic fields.
+Now, apply the divergence operator over $\mathbf{r}$ to the electrostatic and magnetostatic fields.
 
-$$
-\begin{align}
-    \div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
-    \div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
-\end{align}
-$$
+$$\begin{align}
+\nabla \cdot E(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \\
+\nabla \cdot H(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V')
+\end{align}$$
 
-As the divergence operator does not operate on $\vb{r'}$, we see that
+As the divergence operator does not operate on $\mathbf{r'}$, we see that
 
-$$
-\begin{align}
-    \div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
-        &= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
-        &= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\
-    \div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
-        &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
-        &= \frac{\rho_m(\vb{r})}{\mu_0}
-\end{align}
-$$
+$$\begin{align}
+\nabla \cdot E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
+    &= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
+    &= \frac{\rho_e(\mathbf{r})}{\varepsilon_0} \\
+\nabla \cdot H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
+    &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
+    &= \frac{\rho_m(\mathbf{r})}{\mu_0}
+\end{align}$$
 
 The curl of an electrostatic or magnetostatic is relatively simple.
 
 $$
 \begin{align}
-    \curl{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
-    \curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+    \nabla \times{E(\mathbf{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
+    \nabla \times{H(\mathbf{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
 \end{align}
 $$
 
-Additionally, we know $\curl{f\vb{A}} = f \curl{\vb{A}} + \grad{f}\cross\vb{A}$. Thus,
+Additionally, we know $\nabla \times{f\mathbf{A}} = f \nabla \times{\mathbf{A}} + \nabla{f}\times\mathbf{A}$. Thus,
 
 $$
 \begin{align}
-    \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} &= \frac{1}{\abs{\vb{r}-\vb{r'}}^3} \curl{(\vb{r}-\vb{r'})} + (\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}}) \cross (\vb{r}-\vb{r'}) \\
+    \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} &= \frac{1}{|\mathbf{r}-\mathbf{r'}|^3} \nabla \times{(\mathbf{r}-\mathbf{r'})} + (\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}}) \times (\mathbf{r}-\mathbf{r'}) \\
 \end{align}
 $$
 
-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.
+We can verify that $\nabla \times{(\mathbf{r}-\mathbf{r'})} = 0$, cancelling the first term. Additionally, $\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}} = -3 \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^5}$, which when crossed with $\mathbf{r}-\mathbf{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
 
 ## Section 2.4 - Electric and Magnetic Flux Densities
 
-The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
+The electric and magnetic flux density vectors are given by $\varepsilon_0 \mathbf{E}$ and $\mu_0 \mathbf{H}$.
 
 Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
 
 $$
-\Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
+\Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d S = Q_m
 $$
 
 Thus, applying divergence theorem,
 
 $$
-Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{E}} \dd V
+Q_e = \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V
 $$
 
 $$
-Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{H}} \dd V
+Q_m = \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{H} d V
 $$
 
-Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
+Since $Q_e = \int_V \rho_e d V$ and $Q_m = \int_V \rho_m d V$, we see that
 
 $$
 \begin{align}
-    \int_V \rho_e \dd V &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
-    \rho_e &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
-    \int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
-    \rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
+    \int_V \rho_e d V &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
+    \rho_e &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
+    \int_V \rho_m d V &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
+    \rho_m &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
 \end{align}
 $$