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 \(\mathbf{r}\) relative to the origin.

Thus, according to Coulomb's Law,

\[ \begin{align} F^e_{qQ}(\mathbf{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\ F^m_{qQ}(\mathbf{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \end{align} \]

Divide by the charge \(q\) to obtain the electric or magnetic field at point \(\mathbf{r}\).

\[ \begin{align} E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\ H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \end{align} \]

Now, let \(Q\) be at point \(\mathbf{r'}\). Then, the unit vector becomes \(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}}|\), and we see the following.

\[ \begin{align} E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\ H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \end{align} \]

With multiple charges, we can apply the superposition principal to see the following:

\[ \begin{align} E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\ H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \end{align} \]

We can convert this to an integral as \(N\) goes to infinity.

\[\begin{align} E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \\ H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \end{align}\]

Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field

From a lot of advanced math, we know that

\[\nabla \cdot \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = 4 \pi \delta(\mathbf{r}-\mathbf{r'})\]

Now, apply the divergence operator over \(\mathbf{r}\) to the electrostatic and magnetostatic fields.

\[\begin{align} \nabla \cdot E(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \\ \nabla \cdot H(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \end{align}\]

As the divergence operator does not operate on \(\mathbf{r'}\), we see that

\[\begin{align} \nabla \cdot E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\ &= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\ &= \frac{\rho_e(\mathbf{r})}{\varepsilon_0} \\ \nabla \cdot H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\ &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\ &= \frac{\rho_m(\mathbf{r})}{\mu_0} \end{align}\]

The curl of an electrostatic or magnetostatic is relatively simple.

\[ \begin{align} \nabla \times{E(\mathbf{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\ \nabla \times{H(\mathbf{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\ \end{align} \]

Additionally, we know \(\nabla \times{f\mathbf{A}} = f \nabla \times{\mathbf{A}} + \nabla{f}\times\mathbf{A}\). Thus,

\[ \begin{align} \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} &= \frac{1}{|\mathbf{r}-\mathbf{r'}|^3} \nabla \times{(\mathbf{r}-\mathbf{r'})} + (\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}}) \times (\mathbf{r}-\mathbf{r'}) \\ \end{align} \]

We can verify that \(\nabla \times{(\mathbf{r}-\mathbf{r'})} = 0\), 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 \(\varepsilon_0 \mathbf{E}\) and \(\mu_0 \mathbf{H}\).

Now, given \(S\) is a surface enclosing \(Q_e\) or \(Q_m\) total charge, we denote flux as following:

\[ \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d S = Q_m \]

Thus, applying divergence theorem,

\[ Q_e = \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \]

\[ Q_m = \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{H} d V \]

Since \(Q_e = \int_V \rho_e d V\) and \(Q_m = \int_V \rho_m d V\), we see that

\[ \begin{align} \int_V \rho_e d V &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\ \rho_e &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\ \int_V \rho_m d V &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\ \rho_m &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\ \end{align} \]

Definition. This is known as Gauss' Law.

With applicable symmetry, the integral factor becomes simply \(E(r)*A\), where \(A\) is the area of the surface at \(r\).