如果你也在 怎样代写电动力学electrodynamics这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。电动力学electrodynamics是物理学的一个分支，涉及到对电磁力的研究，这是一种发生在带电粒子之间的物理作用。电磁力是由电场和磁场组成的电磁场所承载的，它是诸如光这样的电磁辐射的原因。它与强相互作用、弱相互作用和引力一起，是自然界的四种基本相互作用（通常称为力）之一。
电动力学electrodynamics电磁现象是以电磁力来定义的，有时也称为洛伦兹力，它包括电和磁，是同一现象的不同表现形式。电磁力在决定日常生活中遇到的大多数物体的内部属性方面起着重要作用。原子核和其轨道电子之间的电磁吸引力将原子固定在一起。电磁力负责原子之间形成分子的化学键，以及分子间的力量。电磁力支配着所有的化学过程，这些过程是由相邻原子的电子之间的相互作用产生的。电磁学在现代技术中应用非常广泛，电磁理论是电力工程和电子学包括数字技术的基础。
my-assignmentexpert™ 电动力学electrodynamics作业代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。my-assignmentexpert™， 最高质量的电动力学electrodynamics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此电动力学electrodynamics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。
想知道您作业确定的价格吗? 免费下单以相关学科的专家能了解具体的要求之后在1-3个小时就提出价格。专家的 报价比上列的价格能便宜好几倍。
my-assignmentexpert™ 为您的留学生涯保驾护航 在物理physics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的电动力学electrodynamics代写服务。我们的专家在物理physics代写方面经验极为丰富，各种电动力学electrodynamics相关的作业也就用不着 说。
我们提供的电动力学electrodynamics及其相关学科的代写，服务范围广, 其中包括但不限于:
物理代写|电动力学作业代electrodynamics代考|Electric Scalar Potential
The electric potential is a scalar quantity which can be interpreted either as a scalar field or as electrostatic potential energy per unit charge. The negative gradient of the scalar field at a point in space is a measure of the electric field at that point. The electric potential at a point in space is also equal to the work done to bring a unit test charge from infinity to that point. Imagine a source charge $q_{i}$ placed at $\mathbf{x}{\mathbf{i}}^{\prime}$ which leads to an electric field $\mathbf{E}$ around it. The work done to bring a test charge $Q$ from infinity to a point $\mathbf{x}$ in space is given by $$ W=-\int{\infty}^{\mathbf{x}} \frac{q_{i} Q}{\left|\mathbf{x}-\mathbf{x}{\mathbf{i}}^{\prime}\right|^{3}}\left(\mathbf{x}-\mathbf{x}{\mathbf{i}}^{\prime}\right) \cdot \mathbf{d l} .
$$
Here, dl is the infinitesimal displacement along some path from infinity to $\mathbf{x}$. Without loss of generality, we can place the source charge at the origin. Then we have
$$
W=-\int_{\infty}^{\mathbf{r}} \frac{q_{i} Q}{|\mathbf{r}|^{3}} \mathbf{r} \cdot \mathbf{d} \mathbf{l}=q_{i} Q \int_{\infty}^{\mathbf{r}} \frac{d \tilde{r}}{\tilde{r}^{2}}=\frac{q_{i} Q}{r}
$$
Noting that $|\mathbf{r}|$ is simply the distance between the source charge and the test charge, we can write a more general expression for the work done per unit test charge or the electric potential at any point $\mathbf{x}$ due to a charge $q_{i}$ placed at $\mathbf{x}{\mathbf{i}}^{\prime}$, as follows: $$ \Phi=\frac{W}{Q}=\frac{q{i}}{\left|\mathbf{x}-\mathbf{x}_{\mathbf{i}}{ }^{\prime}\right|}
$$
物理代写|电动力学作业代electrodynamics代考|Poisson and Laplace Equations
- Let us quickly summarize the important conclusions drawn so far. The electrostatic field is completely specified by the following two equations: $\nabla \cdot \mathbf{E}(\mathbf{x})=4 \pi \rho(\mathbf{x})$ and
$\nabla \times \mathbf{E}(\mathbf{x})=0$. The latter suggests that electric field can be written as a gradient of scalar potential $\Phi$. When the electric field is expressed in terms of potential, the differential form of Gauss’s law is reduced to the following form.
$$
\nabla^{2} \Phi=-4 \pi \rho
$$
The above equation is named the Poisson equation. We are already familiar with the formal solution of the Poisson equation, which is given below.
$$
\Phi(\mathbf{x})=\int_{V} \frac{\rho\left(\mathbf{x}{\mathbf{i}}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}{\mathbf{i}}^{\prime}\right|} d^{3} x^{\prime}
$$
Let us now verify that the above solution does indeed satisfy the Poisson equation. For simplicity, we consider a point charge $q$ located at $\mathbf{x}^{\prime}$. The charge density can be written as $\rho(\mathbf{x})=q \delta\left(\mathbf{x}-\mathbf{x}^{\prime}\right)$. We are already familiar with the mathematical form of the potential due to a point charge. It follows that for $\mathbf{r} \neq \mathbf{r}^{\prime}$, one can easily verify using Cartesian coordinates $\left{x_{i}\right}$, where $i=1,2,3$,
$$
\begin{aligned}
\nabla^{2} \frac{1}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} &=\left(\frac{\partial^{2}}{\partial x_{1}^{2}}+\frac{\partial^{2}}{\partial x_{2}^{2}}+\frac{\partial^{2}}{\partial x_{3}^{2}}\right)\left[\left(x_{1}-x_{1}^{\prime}\right)^{2}+\left(x_{2}-x_{2}^{\prime}\right)^{2}+\left(x_{3}-x_{3}^{\prime}\right)^{2}\right]^{-\frac{1}{2}} \
&=0 .
\end{aligned}
$$
物理代写|电动力学作业代ELECTRODYNAMICS代考|Electrostatic Potential Energy for Discrete and Continuous Charge Distribution
Consider a localized charge distribution $q_{j}$ with $j=1,2, \ldots, n-1$ with corresponding position vectors $\mathbf{x}{\mathbf{j}}$. We have already defined the electrostatic potential $\Phi\left(\mathbf{x}{\mathbf{i}}\right)$ as the work done against the electric field (due to the charge distribution $q_{j}$ in that region) to bring a unit charge from infinity to any point $\mathbf{x}{\mathbf{i}}$. The potential $\Phi\left(\mathbf{x}{\mathbf{i}}\right)$ vanishes at infinity. The potential energy of the charge $q_{i}$ at the point $\mathbf{x}{\mathbf{i}}$ is the work done $W{i}$ to bring a charge $q_{i}$ from infinity to $\mathbf{x}{\mathbf{i}}$ and is given by $$ W{i}=q_{i} \Phi\left(\mathbf{x}{\mathbf{i}}\right) $$ We already know that the potential at the point $\mathbf{x}{\mathbf{i}}$ due to the collection of discrete charges $q_{j}$ is given by
$$
\Phi\left(\mathbf{x}{\mathbf{i}}\right)=\sum{j \neq i}^{n-1} \frac{q_{j}}{\left|\mathbf{x}{\mathbf{i}}-\mathbf{x}{\mathbf{j}}\right|}
$$
Therefore, the potential energy $W_{i}$ of the charge $q_{i}$ can be rewritten as
$$
W_{i}=q_{i} \sum_{j \neq i, j=1}^{n-1} \frac{q_{j}}{\left|\mathbf{x}{\mathbf{i}}-\mathbf{x}{\mathbf{j}}\right|}
$$
电动力学代写
物理代写|电动力学作业代ELECTRODYNAMICS代考|ELECTRIC SCALAR POTENTIAL
电势是一个标量，可以解释为标量场或每单位电荷的静电势能。空间中某点的标量场的负梯度是该点电场的量度。空间中某点的电势也等于将单元测试电荷从无穷远带到该点所做的功。想象一个源电荷q一世放置在$q_{i}$ placed at $\mathbf{x}{\mathbf{i}}^{\prime}$ which leads to an electric field $\mathbf{E}$ around it. The work done to bring a test charge $Q$ from infinity to a point $\mathbf{x}$ in space is given by $$ W=-\int{\infty}^{\mathbf{x}} \frac{q_{i} Q}{\left|\mathbf{x}-\mathbf{x}{\mathbf{i}}^{\prime}\right|^{3}}\left(\mathbf{x}-\mathbf{x}{\mathbf{i}}^{\prime}\right) \cdot \mathbf{d l} .
$$
Here, dl is the infinitesimal displacement along some path from infinity to $\mathbf{x}$. Without loss of generality, we can place the source charge at the origin. Then we have
$$
W=-\int_{\infty}^{\mathbf{r}} \frac{q_{i} Q}{|\mathbf{r}|^{3}} \mathbf{r} \cdot \mathbf{d} \mathbf{l}=q_{i} Q \int_{\infty}^{\mathbf{r}} \frac{d \tilde{r}}{\tilde{r}^{2}}=\frac{q_{i} Q}{r}
$$
注意|r|只是源电荷和测试电荷之间的距离，我们可以将单位测试电荷所做的功或任何点的电势写成更一般的表达式X由于收费q一世放置在$\mathbf{x}$ due to a charge $q_{i}$ placed at $\mathbf{x}{\mathbf{i}}^{\prime}$, as follows: $$ \Phi=\frac{W}{Q}=\frac{q{i}}{\left|\mathbf{x}-\mathbf{x}_{\mathbf{i}}{ }^{\prime}\right|}
$$
物理代写|电动力学作业代ELECTRODYNAMICS代考|POISSON AND LAPLACE EQUATIONS
- 让我们快速总结一下迄今为止得出的重要结论。静电场完全由以下两个等式指定：∇⋅和(X)=4圆周率ρ(X)和
∇×和(X)=0. 后者表明电场可以写成标量势的梯度披. 当电场以电势表示时，高斯定律的微分形式简化为以下形式。
∇2披=−4圆周率ρ
上述方程称为泊松方程。我们已经熟悉了 Poisson 方程的形式解，如下所示。
$$
\披X=\int_{V} \frac{\rho\left(\mathbf{x} {\mathbf{i}}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x} {\mathbf{i}}^{\prime}\right|} d^{3} x^{\prime}
现在让我们验证上述解确实满足泊松方程。为简单起见，我们考虑位于 $\mathbf{x}^{\prime}$ 的点费 $q$。电荷密度可以写成$\rho(\mathbf{x})=q \delta\left(\mathbf{x}-\mathbf{x}^{\prime}\right)$。我们已经熟悉了由点电荷引起的电势的数学形式。由此可见，对于 $\mathbf{r} \neq \mathbf{r}^{\prime}$，可以使用笛卡尔坐标 $\left{x_{i}\right}$ 轻松验证，其中 $i=1， 2,3$,现在让我们验证上述解确实满足泊松方程。为简单起见，我们考虑位于 $\mathbf{x}^{\prime}$ 的点费 $q$。电荷密度可以写成$\rho(\mathbf{x})=q \delta\left(\mathbf{x}-\mathbf{x}^{\prime}\right)$。我们已经熟悉了由点电荷引起的电势的数学形式。由此可见，对于 $\mathbf{r} \neq \mathbf{r}^{\prime}$，可以使用笛卡尔坐标 $\left{x_{i}\right}$ 轻松验证，其中 $i=1， 2,3$,
∇21|X−X′|=(∂2∂X12+∂2∂X22+∂2∂X32)[(X1−X1′)2+(X2−X2′)2+(X3−X3′)2]−12 =0.
$$
物理代写|电动力学作业代ELECTRODYNAMICS代考|ELECTROSTATIC POTENTIAL ENERGY FOR DISCRETE AND CONTINUOUS CHARGE DISTRIBUTION
考虑局部电荷分布qj和j=1,2,…,n−1对应的位置向量 $q_{j}$ with $j=1,2, \ldots, n-1$ with corresponding position vectors $\mathbf{x}{\mathbf{j}}$. We have already defined the electrostatic potential $\Phi\left(\mathbf{x}{\mathbf{i}}\right)$ as the work done against the electric field (due to the charge distribution $q_{j}$ in that region) to bring a unit charge from infinity to any point $\mathbf{x}{\mathbf{i}}$. The potential $\Phi\left(\mathbf{x}{\mathbf{i}}\right)$ vanishes at infinity. The potential energy of the charge $q_{i}$ at the point $\mathbf{x}{\mathbf{i}}$ is the work done $W{i}$ to bring a charge $q_{i}$ from infinity to $\mathbf{x}{\mathbf{i}}$ and is given by $$ W{i}=q_{i} \Phi\left(\mathbf{x}{\mathbf{i}}\right) $$ We already know that the potential at the point $\mathbf{x}{\mathbf{i}}$ due to the collection of discrete charges $q_{j}$ is given by
$$
\Phi\left(\mathbf{x}{\mathbf{i}}\right)=\sum{j \neq i}^{n-1} \frac{q_{j}}{\left|\mathbf{x}{\mathbf{i}}-\mathbf{x}{\mathbf{j}}\right|}
$$
Therefore, the potential energy $W_{i}$ of the charge $q_{i}$ can be rewritten as
$$
W_{i}=q_{
物理代写|电动力学作业代写electrodynamics代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
