### 物理代写|几何光学代写Geometrical Optics代考|OPTI502

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• Foundations of Data Science 数据科学基础

## 物理代写|几何光学代写Geometrical Optics代考|Set of Maxwell Equations for Electrostatic Field

First, we introduce the set of Maxwell equations for the electrostatic field in free space. Using Gauss’s Law (see Chap. 2), we can write the electric flux of electric field created by continuous charge distribution in a volume $V$ enclosed by the surface $A$ as
$$\oint_A \mathbf{E} \cdot d \mathbf{A}=\frac{Q_{i n}}{\epsilon_0}$$
Note that in Eq. (4.65) $\mathbf{E}$ is the electrostatic field created by all charges in space, and $Q_{i n}$ is the electric charge inside the volume $V$ enclosed by the surface $A$. The left-hand side of Eq. (4.65) can be written in the following form using Gauss formula: $$\oint_A \mathbf{E} \cdot d \mathbf{A}=\int_V \nabla \cdot \mathbf{E} d V$$
where $V$ is the volume enclosed by the surface $A$. In addition, the right-hand side of Eq. (4.65) can be written as
$$\frac{Q_{i n}}{\epsilon_0}=\int_V \frac{\rho(\mathbf{r})}{\epsilon_0} d V$$
Combining Eqs. (4.65), (4.66) and (4.67), we get
$$\int_V \nabla \cdot \mathbf{E} d V=\int_V \frac{\rho(\mathbf{r})}{\epsilon_0} d V$$
where $\nabla \cdot \mathbf{E}$ is the divergence of the vector $\mathbf{E}$, which produces a scalar.
Comparing both sides of Eq. (4.68), we obtain the first Maxwell equation in free space:
$$\nabla \cdot \mathbf{E}(\mathbf{r})=\frac{\rho(\mathbf{r})}{\epsilon_0}$$
where both $\mathbf{E}$ and $\rho$ can be functions of the position $\mathbf{r}$.
Using the expression of the electrostatic potential difference in free space, Eq. (4.10) (Chap.3), we have
$$\Delta \phi=-\int_A^B \mathbf{E} \cdot d \mathbf{s}$$
where $A$ and $B$ are two points in free space, and $d \mathbf{s}$ is an infinitesimal displacement along the curve joining points $A$ and $B$. If we consider a closed path, that is, $A=B$, then $\Delta \phi=\phi_B-\phi_A=\phi_A-\phi_A=0$, and hence
$$\oint_{\mathcal{L}} \mathbf{E} \cdot d \mathbf{s}=0$$

## 物理代写|几何光学代写Geometrical Optics代考|Maxwell Equations for Dielectric Media Electrostatic Field

We mentioned that in the dielectric medium, an average over macroscopically small volumes, which are microscopically large, is necessary to obtain the Maxwell equations of the macroscopic phenomena.
The first observation is that Eq. (4.74) holds microscopically, that is
$$\nabla \times \mathbf{E}_{\text {micro }}=0$$
When averaging is made of the homogeneous Eq. (4.75), we obtain
$$\nabla \times \mathbf{E}=0$$
Equation (4.76) indicates that Eq. (4.74) holds for the averaged macroscopic electric field $\mathbf{E}$.

Using Eq. (4.57) for the effective charge density in the medium, Eq. (4.69) becomes
$$\nabla \cdot \mathbf{E}(\mathbf{r})=\frac{\rho(\mathbf{r})-\nabla \cdot \mathbf{P}(\mathbf{r})}{\epsilon_0}$$
Rearranging Eq. (4.77), we get
$$\nabla \cdot\left(\epsilon_0 \mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})\right)=\rho(\mathbf{r})$$
Using the definition of the electric displacement vector given by Eq. (4.58), we write Eq. (4.78) as
$$\nabla \cdot \mathbf{D}(\mathbf{r})=\rho(\mathbf{r})$$
Note that Eqs. (4.76) and (4.79) are the macroscopic Maxwell equations in the dielectric medium, which are the counterparts of Eqs. (4.69) and (4.74).

# 几何光学代考

## 物理代写|几何光学代写Geometrical Optics代考|Set of Maxwell Equations for Electrostatic Field

$$\oint_A \mathbf{E} \cdot d \mathbf{A}=\frac{Q_{i n}}{\epsilon_0}$$

$$\oint_A \mathbf{E} \cdot d \mathbf{A}=\int_V \nabla \cdot \mathbf{E} d V$$

$$\frac{Q_{i n}}{\epsilon_0}=\int_V \frac{\rho(\mathbf{r})}{\epsilon_0} d V$$

$$\int_V \nabla \cdot \mathbf{E} d V=\int_V \frac{\rho(\mathbf{r})}{\epsilon_0} d V$$

$$\nabla \cdot \mathbf{E}(\mathbf{r})=\frac{\rho(\mathbf{r})}{\epsilon_0}$$

$$\Delta \phi=-\int_A^B \mathbf{E} \cdot d \mathbf{s}$$

$$\oint_{\mathcal{L}} \mathbf{E} \cdot d \mathbf{s}=0$$

## 物理代写|几何光学代写Geometrical Optics代考|Maxwell Equations for Dielectric Media Electrostatic Field

$$\nabla \times \mathbf{E}_{\text {micro }}=0$$

$$\nabla \times \mathbf{E}=0$$

$$\nabla \cdot \mathbf{E}(\mathbf{r})=\frac{\rho(\mathbf{r})-\nabla \cdot \mathbf{P}(\mathbf{r})}{\epsilon_0}$$

$$\nabla \cdot\left(\epsilon_0 \mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})\right)=\rho(\mathbf{r})$$

$$\nabla \cdot \mathbf{D}(\mathbf{r})=\rho(\mathbf{r})$$

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