127 lines
5.8 KiB
Markdown
127 lines
5.8 KiB
Markdown
# Chapter 2 - Coulomb's Laws, Electric and Magnetic Fields
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## Section 2.2 - Parallel Treatment of Electric and Magnetic Fields
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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.
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Thus, according to Coulomb's Law,
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$$
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\begin{align}
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F^e_{qQ}(\mathbf{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
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F^m_{qQ}(\mathbf{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
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\end{align}
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$$
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Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\mathbf{r}$.
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$$
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\begin{align}
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E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
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H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
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\end{align}
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$$
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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.
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$$
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\begin{align}
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E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
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H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
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\end{align}
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$$
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With multiple charges, we can apply the *superposition principal* to see the following:
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$$
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\begin{align}
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E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
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H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
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\end{align}
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$$
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We can convert this to an integral as $N$ goes to infinity.
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$$\begin{align}
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E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \\
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H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V'
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\end{align}$$
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## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
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From a lot of advanced math, we know that
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$$\nabla \cdot \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = 4 \pi \delta(\mathbf{r}-\mathbf{r'})$$
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Now, apply the divergence operator over $\mathbf{r}$ to the electrostatic and magnetostatic fields.
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$$\begin{align}
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\nabla \cdot E(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \\
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\nabla \cdot H(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V')
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\end{align}$$
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As the divergence operator does not operate on $\mathbf{r'}$, we see that
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$$\begin{align}
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\nabla \cdot E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
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&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
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&= \frac{\rho_e(\mathbf{r})}{\varepsilon_0} \\
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\nabla \cdot H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
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&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
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&= \frac{\rho_m(\mathbf{r})}{\mu_0}
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\end{align}$$
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The curl of an electrostatic or magnetostatic is relatively simple.
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$$
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\begin{align}
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\nabla \times{E(\mathbf{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
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\nabla \times{H(\mathbf{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
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\end{align}
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$$
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Additionally, we know $\nabla \times{f\mathbf{A}} = f \nabla \times{\mathbf{A}} + \nabla{f}\times\mathbf{A}$. Thus,
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$$
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\begin{align}
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\nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} &= \frac{1}{|\mathbf{r}-\mathbf{r'}|^3} \nabla \times{(\mathbf{r}-\mathbf{r'})} + (\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}}) \times (\mathbf{r}-\mathbf{r'}) \\
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\end{align}
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$$
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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.
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## Section 2.4 - Electric and Magnetic Flux Densities
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The electric and magnetic flux density vectors are given by $\varepsilon_0 \mathbf{E}$ and $\mu_0 \mathbf{H}$.
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Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
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$$
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\Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d S = Q_m
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$$
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Thus, applying divergence theorem,
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$$
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Q_e = \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V
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$$
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$$
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Q_m = \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{H} d V
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$$
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Since $Q_e = \int_V \rho_e d V$ and $Q_m = \int_V \rho_m d V$, we see that
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$$
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\begin{align}
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\int_V \rho_e d V &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
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\rho_e &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
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\int_V \rho_m d V &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
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\rho_m &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
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\end{align}
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$$
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**Definition**. This is known as *Gauss' Law*.
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With applicable symmetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.
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