### 物理代写|电磁学代写electromagnetism代考|PHYS2213

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## 物理代写|电磁学代写electromagnetism代考|Uniform Electric Field

The electric flux concept describes quantitatively the electric lines. The number of field lines per unit area (also called line density) going through a rectangular surface of area $A$, which is perpendicular to the field, is proportional to the magnitude of electric field, $\mathbf{E}$, as shown in Fig. 2.1. Furthermore, the total number of lines penetrating the surface is proportional to the product $|\mathbf{E}| A$. By definition, the product of the magnitude of electric field $|\mathbf{E}|$ and surface area $A$ perpendicular to the field is called the electric flux:
$$\Phi_{E}=|\mathbf{E}| A$$
Using Eq. (2.1), from the SI units of $E$ and $A$, we derive the SI units of the electric flux:
$$[E]=\left[\frac{\mathrm{N}}{\mathrm{C}}\right],[A]=\left[\mathrm{m}^{2}\right]$$

Note that the electric flux is proportional to the number of electric field lines penetrating some surface.

Moreover, consider the electric flux on any surface with an arbitrary orientation with respect to electric field $\mathbf{E}$, as shown in Fig. 2.2. Electric flux going through the surface (with area $A$ ) not perpendicular to $\mathbf{E}$ is smaller than the product $|\mathbf{E}| A$. That is, the number of lines that cross this area $A$ is equal to the number of lines that cross the area $A^{\prime}=A \cos \theta$, which is a projection of $A$ aligned perpendicular to the field. Mathematically, the electric flux is given by (Fig. 2.2)
$$\Phi_{E}=|\mathbf{E}| A^{\prime}=|\mathbf{E}| A \cos \theta$$
From the definition, Eq. (2.4), we can say that the maximum electric flux is achieved when $\theta=0^{\circ}$; that is, the surface is perpendicular to $\mathbf{E}$ : $\Phi_{E}^{\max }=|\mathbf{E}| A$ (see also Eq. (2.1)). Or, equivalently, when normal vector $\mathbf{n}$ to the surface is parallel

to $\mathbf{E}$. On the other hand, the minimum electric flux is achieved when $\theta=90^{\circ}$, that is, the surface is parallel to $\mathbf{E}^{:} \Phi_{E}^{\min }=0$. In this case, normal vector $\mathbf{n}$ to the surface is perpendicular to $\mathbf{E}$. In general, denoting the vector $\mathbf{A}=A \mathbf{n}$, we can write
$$\Phi_{E}=\mathbf{E} \cdot \mathbf{A}$$

to $\mathbf{E}$. On the other hand, the minimum electric flux is achieved when $\theta=90^{\circ}$, that is, the surface is parallel to $\mathbf{E}: \Phi_{E}^{\min }=0$. In this case, normal vector $\mathbf{n}$ to the surface is perpendicular to $\mathbf{E}$. In general, denoting the vector $\mathbf{A}=A \mathbf{n}$, we can write
$$\Phi_{E}=\mathbf{E} \cdot \mathbf{A}$$

## 物理代写|电磁学代写electromagnetism代考|General Electric Field Flux

Let us consider the case of a general surface with a nonuniform electric field $\mathbf{E}{i}$, as shown in Fig. 2.3. We can partition the surface on small infinitesimal elements $\Delta A{i}$, such that electric field is constant on every point of $\Delta A_{i}$, then the electric flux through $\Delta A_{i}$ is
$$\Delta \Phi_{E, i}=\mathbf{E}{i} \cdot \Delta \mathbf{A}{i}$$
The total flux can be approximated as
$$\Phi_{E} \approx \sum_{i} \mathbf{E}{i}, \Delta \mathbf{A}{i}$$
Taking the limit when $\Delta \mathbf{A}{i} \rightarrow 0$ on both sides of Eq. (2.7), we obtain the exact electric flux through general surface (Fig. 2.3): $$\Phi{E}=\lim {\Delta A{i} \rightarrow 0} \sum_{i} \mathbf{E}{i} \cdot \Delta \mathbf{A}{i}=\int_{\text {surface }} \mathbf{E} \cdot d \mathbf{A}$$

Note that Eq. $(2.8)$ gives the electric flux through any surface. Often, the electric flux is calculated through a closed surface.

A closed surface defines the surface that divides space into the inside and outside regions, and to move from one region to another, one has to cross the surface. For example, the surface of a sphere, ellipsoid, etc., are all closed surfaces.

Consider a nonuniform electric field penetrating a closed surface, for example, a cylinder, as shown in Fig. 2.4. Every electric field line that passes through a surface element $\Delta A_{i}$ is going to leave the closed surface at some other surface element $\Delta A_{j}$. The electric flux going through the surface element $\Delta \mathbf{A}{i}$ is $$\Delta \Phi{E, i}=\mathbf{E}{i} \Delta \mathbf{A}{i}=\left|\mathbf{E}{i}\right| \Delta A{i} \cos \theta_{i}$$
In Eq. (2.9), $\theta_{i}$ is the angle between electric field $\mathbf{E}{i}$ and unit vector normal to surface element $\mathbf{n}{i}$ :
$$\cos \theta_{i}=\left{\begin{array}{l} 0,0 \leq \theta_{i}<90^{\circ} \ =0, \theta_{i}=90^{\circ} \ <0,90^{\circ}<\theta_{i} \leq 180^{\circ} \end{array}\right.$$
Similarly. the electric flux going through he surface element $\Delta \mathbf{A}{j}$ is $$\Delta \Phi{E, j}=\mathbf{E}{j} \Delta \mathbf{A}{j}=\left|\mathbf{E}{j}\right| \Delta A{j} \cos \theta_{j}$$

## 物理代写|电磁学代写electromagnetism代考|Gauss’s Law

By definition, a Gaussian surface is called a closed surface. Gauss’s law relates the electric flux through the closed surface and the charge enclosed by the surface.

For illustration consider a nonconducting sphere of radius $r$ with an elementary positive charge $+q$ at the center of sphere. Note that the sphere’s surface is equal to Gaussian surface, as indicated in Fig. 2.5. Electric field created by the charge $+q$ at any point at distance $r$ from the charge is
$$\mathbf{E}=k_{e} \frac{q}{r^{2}} \hat{\mathbf{r}}$$

The electric flux through the sphere surface is
\begin{aligned} \Phi_{E} &=\oint_{S} \mathbf{E} \cdot \mathbf{n} d A=\oint_{S} k_{e} \frac{q}{r^{2}} \hat{\mathbf{r}} \cdot \mathbf{n} d A \ &=\oint_{S} k_{e} \frac{q}{r^{2}} d A=k_{e} \frac{q}{r^{2}} \oint_{S} d A=k_{e} \frac{q}{r^{2}}\left(4 \pi r^{2}\right) \ &=4 \pi k_{e} q \end{aligned}
The result given by Eq. (2.16) indicates that $\Phi_{E}$ is independent of radius $r$. Furthermore, since both $q$ and $\epsilon_{0}$ are constants, then the electric flux is constant. Therefore, the same electric flux is passing through the surface of any other sphere with radius $R>r$, which has an elementary charge at the center.

Now, consider several closed surfaces surrounding a charge $q$, as shown in Fig. 2.6, $S_{1}$ (spherical), $S_{2}$ and $S_{3}$ (nonspherical). The flux that passes through $S_{1}$ is given by Eq. (2.16). Since flux is proportional to the number of electric field lines passing through a surface, then the flux through $S_{2}$ and $S_{3}$ also is constant (see Eq. (2.16)).
By definition, the net flux through any closed surface is independent of the shape of that surface. The net flux through an arbitrary closed surface surrounding a point charge $q$ is given by Eq. (2.16).

Now, suppose a charge $+q$ is outside a closed surface (any shape), as shown in Fig. 2.7. In that case, an electric field line that enters the surface leaves the surface at another point. The number of electric field lines entering the surface equals the number leaving the surface, thus
$$\Phi_{E}=0$$
We can conclude that the net electric flux through a closed surface that surrounds no charge is zero (Fig. 2.7).

## 物理代写|电磁学代写electromagnetism代考|Uniform Electric Field

[和]=[ñC],[一个]=[米2]

## 物理代写|电磁学代写electromagnetism代考|General Electric Field Flux

Δ披和,一世=和一世⋅Δ一个一世

Δ披和,一世=和一世Δ一个一世=|和一世|Δ一个一世因⁡θ一世

$$\cos \theta_{i}=\left{\begin{array}{l} 0,0 \leq \theta_{i}<90^{\circ} \ =0, \theta_{i}=90^{\circ} \ <0,90^{\circ}<\theta_{i} \ leq 180^{\circ} \end{array}\right. 小号一世米一世l一个rl是.吨H和和l和C吨r一世CFl在XG○一世nG吨Hr○在GHH和s在rF一个C和和l和米和n吨Δ一个j一世s\ Delta \ Phi {E, j} = \ mathbf {E} {j} \ Delta \ mathbf {A} {j} = \ left | \ mathbf {E} {j} \ right | \ Delta A {j} \ cos \ theta_ {j}$$

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