### 物理代写|电磁学代写electromagnetism代考|PHYS3040

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|电磁学代写electromagnetism代考|Gauss’s Law for a System of Charges

Consider a system of $N$ discrete point charges, $q_{1}, q_{2}, \ldots, q_{N}$, positive or negative, as shown in Fig. 2.8. Using the superposition principle of electric field, discussed in Chap. 1, we can write
$$\mathbf{E}=\sum_{i=1}^{N} \mathbf{E}{i}$$ Then, the net electric flux through a closed surface is \begin{aligned} \Phi{E} &=\oint_{S} \mathbf{E} \cdot d \mathbf{A} \ &=\oint_{S}\left(\sum_{i=1}^{N} \mathbf{E}_{i}\right) \cdot d \mathbf{A} \end{aligned}

$$=\sum_{i=1}^{N} \oint_{S} \mathbf{E}{i} \cdot d \mathbf{A}$$ where $\mathbf{E}{i}$ is electric field of charge $q_{i}$.
The electric flux for the charge $q_{i}$ through the closed surface is
\begin{aligned} \Phi_{E, i} &=\oint_{S} \mathbf{E}{i} \cdot d \mathbf{A} \ &= \begin{cases}\frac{q{i}}{\epsilon_{0}}, & \text { if } q_{i} \text { inside closed surface } \ 0, & \text { if } q_{i} \text { outside closed surface }\end{cases} \end{aligned}
Therefore, the net electric flux becomes
\begin{aligned} \Phi_{E} &=\sum_{i=1}^{N} \Phi_{E, i} \ &=\sum_{i=1}^{N_{i n}} \frac{q_{i}}{\epsilon_{0}}=\frac{Q_{i n}}{\epsilon_{0}} \end{aligned}
where $Q_{i n}$ is the net charge inside the closed surface and $N_{i n}$ denotes the number of charges inside the closed surface.
$$Q=\sum_{i=1}^{N_{\text {in }}} q_{i}$$
Gauss’s law for a system of charges $q_{1}, q_{1}, \ldots, q_{N}$ says that the net electric flux through any close surface is given by Eq. (2.21). When using this equation, we should note that

1. The charge $Q_{i n}$ is the net charge inside the closed surface.
2. E represents the total electric field, which includes contributions from charges both inside and outside the surface.

## 物理代写|电磁学代写electromagnetism代考|Applications of Gauss’s Law to Insulators

In the following, we are summarizing some tips for solving problems using Gauss’s law. First, Gauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. We should pay attention to ways of choosing the Gaussian surface over which the surface integral given by either Eq. (2.21) or Eq. (2.27) can be simplified and the electric field is determined.
In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove $\mathbf{E}$ from the integral and solve it. Using this calculation, we can determine a surface that satisfies one or more of the following conditions:

1. The value of the electric field can be argued by symmetry to be constant over the surfacc.
2. The dot product can be expressed as a simple algebraic product $E d A$ because $\mathbf{E}$ and $d \mathbf{A}$ are parallel.
3. The dot product is zero because $\mathbf{E}$ and $d \mathbf{A}$ are perpendicular.
4. The field is zero over the surface.

## 物理代写|电磁学代写electromagnetism代考|Electrostatic Potential Energy

Suppose a test charge $q_{0}$ is placed in the electric field $\mathbf{E}$ of some other charged object. Then, electric force on $q_{0}$ is $\mathbf{F}=q_{0} \mathbf{E}$. If $\mathbf{E}$ is created by a system of charges $q_{1}, \ldots, q_{N}$, then from superposition principle
$$\mathbf{E}{R}=\sum{i=1}^{N} \mathbf{E}{i}$$ The electric force acting on $q{0}$ is
$$\mathbf{F}=\sum_{i=1}^{N} q_{0} \mathbf{E}{i}=q{0} \mathbf{E}_{R}$$

Therefore, if the test charge $q_{0}$ moves in the electric field $\mathbf{E}$, the electrostatic forces (see also Eq. (3.2)) do work on $q_{0}$.

Suppose that $q_{0}$ moves in the field $\mathbf{E}$ by some external agent, then the work done by electric field is negative of the work done by external agent. Let $d \mathbf{s}$ be an infinitesimal displacement in the electric field, then work done by the field on test charge $q_{0}$ is calculated as
$$d W_{e}=\mathbf{F} \cdot d \mathbf{s}=q_{0} \mathbf{E} \cdot d \mathbf{s}=-d U$$
In Eq. (3.3), $-d U$ is the decrease of the potential energy of charge-field system. Therefore,
$$d U=-d W_{e}=-q_{0} \mathbf{E} \cdot d \mathbf{s}$$
That is, the electric field’s work $\mathbf{E}$ decreases electrostatic potential energy of the charge moving in the field.

## 物理代写|电磁学代写electromagnetism代考|Gauss’s Law for a System of Charges

=∑一世=1ñ∮小号和一世⋅d一个在哪里和一世是电荷电场q一世.

1. 费用问一世n是封闭表面内的净电荷。
2. E 代表总电场，包括表面内外电荷的贡献。

## 物理代写|电磁学代写electromagnetism代考|Applications of Gauss’s Law to Insulators

1. 电场的值可以通过对称性论证为在表面上是恒定的。
2. 点积可以表示为简单的代数积和d一个因为和和d一个是平行的。
3. 点积为零，因为和和d一个是垂直的。
4. 场在表面上为零。

## 物理代写|电磁学代写electromagnetism代考|Electrostatic Potential Energy

F=∑一世=1ñq0和一世=q0和R

d在和=F⋅ds=q0和⋅ds=−d在

d在=−d在和=−q0和⋅ds

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## MATLAB代写

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