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Electostatic + Electrodynamics Use varepsilon

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Indigo5684
2025-09-30 13:19:33 -05:00
parent 3a9785a406
commit 6e16e4af8f
5 changed files with 72 additions and 52 deletions
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32
docs/physics/electrostatics/2-coulomb.md
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@@ -8,7 +8,7 @@ Thus, according to Coulomb's Law,
$$
\begin{align}
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
\end{align}
$$
@@ -17,7 +17,7 @@ Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\v
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
\end{align}
$$
@@ -26,7 +26,7 @@ Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}-
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
\end{align}
$$
@@ -35,7 +35,7 @@ With multiple charges, we can apply the *superposition principal* to see the fol
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
\end{align}
$$
@@ -44,7 +44,7 @@ We can convert this to an integral as $N$ goes to infinity.
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
\end{align}
$$
@@ -61,7 +61,7 @@ Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnet
$$
\begin{align}
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
\end{align}
$$
@@ -70,9 +70,9 @@ As the divergence operator does not operate on $\vb{r'}$, we see that
$$
\begin{align}
\div{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \epsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\epsilon_0} \\
\div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\
\div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_m(\vb{r})}{\mu_0}
@@ -83,7 +83,7 @@ The curl of an electrostatic or magnetostatic is relatively simple.
$$
\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{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' \\
\end{align}
$$
@@ -100,30 +100,30 @@ We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Add
## Section 2.4 - Electric and Magnetic Flux Densities
The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
$$
\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
\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
$$
Thus, applying divergence theorem,
$$
Q_e = \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{E}} \dd V
Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_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
Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_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_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 \\
\end{align}