notes-archive/docs/physics/electrostatics/2-coulomb.md

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.