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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.