### 物理代写|电磁学代写electromagnetism代考|PHYSICS 2534

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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
• Statistical Machine Learning 统计机器学习
• 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|Differential form of Electric Potential

For an elementary displacement vector $d \mathbf{s}$, the potential difference is given as the following:
$$d \phi=-\mathbf{E} \cdot d \mathbf{s}$$
or
$$\mathbf{E}=-\frac{d \phi}{d \mathbf{s}}$$
It can also be written as
$$\mathbf{E}=-\nabla \phi$$
where $\nabla$ is the gradient. Furthermore, we can write in Cartesian coordinates as
$$\mathbf{E}=-\left(\frac{\partial \phi(x, y, z)}{\partial x} \mathbf{i}+\frac{\partial \phi(x, y, z)}{\partial y} \mathbf{j}+\frac{\partial \phi(x, y, z)}{\partial z} \mathbf{k}\right)$$
where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit vectors along $x, y$, and $z$ axes, respectively. Therefore,
\begin{aligned} &E_{x}=-\frac{\partial \phi(x, y, z)}{\partial x} \ &E_{y}=-\frac{\partial \phi(x, y, z)}{\partial y} \ &E_{z}=-\frac{\partial \phi(x, y, z)}{\partial z} \end{aligned}
In spherical coordinates, Eq. (3.44) can be written as
$$\mathbf{E}=-\left(\hat{\mathbf{r}} \frac{\partial \phi(r, \theta, \psi)}{\partial r}+\hat{\boldsymbol{\theta}} \frac{1}{r} \frac{\partial \phi(r, \theta, \psi)}{\partial \theta}+\hat{\psi} \frac{1}{r \sin \theta} \frac{\partial \phi(r, \theta, \psi)}{\partial \psi}\right)$$
where $\hat{\mathbf{r}}, \hat{\boldsymbol{\theta}}, \hat{\boldsymbol{\psi}}$ are unit vectors along $r, \theta$, and $\psi$ directions, respectively. Therefore,
\begin{aligned} E_{r} &=-\frac{\partial \phi(r, \theta, \psi)}{\partial r} \ E_{\theta} &=-\frac{1}{r} \frac{\partial \phi(r, \theta, \psi)}{\partial \theta} \ E_{\psi} &=-\frac{1}{r \sin \theta} \frac{\partial \phi(r, \theta, \psi)}{\partial \psi} \end{aligned}

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Charged Conductor

We can show that the surface of a charged conductor in equilibrium is an equipotential surface. For that, consider two points $A$ and $B$ on the surface of a charged conductor, as shown in Fig.3.7. Along a surface path connecting these points, $\mathbf{E}$ is always perpendicular to the displacement $d \mathbf{s}$; that is, $\mathbf{E} \cdot d \mathbf{s}=0$. Therefore,
$$\phi_{B}-\phi_{A}=-\int_{A}^{B} \mathbf{E} \cdot d \mathbf{s}=0$$
which implies that $\phi_{B}=\phi_{A}$.
Therefore, $\phi$ is constant everywhere on the surface of a charged conductor in equilibrium. That is, the surface of any charged conductor in electrostatic equilibrium is an equipotential surface. Besides, the electric field is zero inside the conductor, that is
$$E_{r}=-d \phi / d r=0$$

which indicates that the electric potential $\phi$ is constant everywhere inside the conductor, including the surface. Therefore, the potential $\phi$ inside the conductor in equilibrium is equal to its value at the surface.

As an example, let us consider a solid metal conducting sphere of radius $R$ and total positive charge $Q$ (see also Fig. 3.8). Using Gauss’s law with Gaussian surfaces as indicated in Fig. 3.8, we can calculate the electric flux as
$$\Phi_{F}=\oint_{S} \mathbf{E} \cdot d \mathbf{A}=E 4 \pi r^{2}=\frac{Q}{\epsilon_{0}}$$
Equation (3.68) implies that the magnitude of electric field is
$$E= \begin{cases}k_{e} \frac{Q}{r^{2}}, & r \geq R \ 0, & r<R\end{cases}$$
In addition, electric field vector is given as
$$\mathbf{E}= \begin{cases}k_{e} \frac{Q}{r^{2}} \hat{\mathbf{r}}, & r \geq R \ 0, & r<R\end{cases}$$

## 物理代写|电磁学代写electromagnetism代考|Cavity Within a Conductor

Let us consider a cavity of no charge within a conductor of any shape, as shown in Fig.3.9. Based on Gauss’s law ( $Q_{i n}=0$ !), the electric field inside the cavity is zero independent of the charge distribution on the outside surface of the conductor. Furthermore, the field within the cavity is zero, even if the conductor is in an external electric field because every point inside the conductor is at the same electric potential. For instance, consider the points $A$ and $B$ inside the conductor and on the surface of the cavity. Then, for those points:
$$\phi_{A}=\phi_{B}$$
On the other hand, using Eq. $(3.14)$, we have
$$\phi_{D}-\phi_{A}=-\int \mathbf{E} \cdot d \mathbf{s}$$ where $\mathbf{E}$ is electric field inside cavity. Since $\phi_{A}=\phi_{B}, \int \mathbf{E} \cdot d \mathbf{s}=0$, or $\mathbf{E}=0$.

dφ=−和⋅ds

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Charged Conductor

φ乙−φ一个=−∫一个乙和⋅ds=0

## 物理代写|电磁学代写electromagnetism代考|Cavity Within a Conductor

φ一个=φ乙

φD−φ一个=−∫和⋅ds在哪里和是腔内的电场。自从φ一个=φ乙,∫和⋅ds=0， 或者和=0.

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## MATLAB代写

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