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。
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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|Image Problem Involving Dielectrics
Consider a boundary between two dielectric media labeled as 1 and 2 with dielectric constants $\epsilon_{1}$ and $\epsilon_{2}$, respectively, at $z=0$. The medium 1 occupies the half-space $z>0$, while medium 2, the half-space $z<0$ (Fig.3.3). Let there be a point charge $q$ in medium 1 at $\mathbf{x}{0}=\left(0,0, z{0}\right)$. We want to find the solutions of the following equations: $$ \begin{array}{ll} \nabla^{2} \Phi=4 \pi q \delta\left(\mathbf{x}-\mathbf{x}{0}\right), & z>0 \ \nabla^{2} \Phi=0, & z<0 \end{array} $$ To find the potential for $z>0$, we place an image charge $q^{\prime}$ at $z=-z{0}$. The potential anywhere at $z>0$ due to the two charges $q$ and $q^{\prime}$ at $z=\pm z_{0}$ in a medium of dielectric constant $\epsilon_{1}$ is given by
$$
\Phi_{z>0}(\mathbf{x})=\frac{1}{\epsilon_{1}}\left(\frac{q}{\sqrt{x^{2}+y^{2}+\left(z-z_{0}\right)^{2}}}+\frac{q^{\prime}}{\sqrt{x^{2}+y^{2}+\left(z+z_{0}\right)^{2}}}\right)
$$
物理代写|电动力学作业代electrodynamics代考|Dielectric Sphere in Uniform External Field
Consider a sphere of radius $R$ and dielectric constant $\epsilon$, placed in an originally uniform external electric field $E_{0}$ along $z$-axis. There are a few points to be noted at the outset. Since $\nabla \times \mathbf{E}=0$ everywhere, the electric field $\mathbf{E}$ at any point in space can be expressed as the gradient of the scalar potential $\Phi$ at that point. For a homogeneous, isotropic medium, $\mathbf{D}(\mathbf{x})=\epsilon \mathbf{E}(\mathbf{x})$, where $\epsilon$ is the dielectric constant of the medium. Since $\nabla \cdot \mathbf{D}=0$, with the macroscopic charge density $\rho$ being zero inside the sphere, it follows that $\nabla \cdot \mathbf{E}=0$. The electric field outside the sphere has zero divergence anyway. Hence, the scalar potential $\Phi$ satisfies the Laplace equation both inside and outside the sphere, and we want to find out the potential in these two regions.
The azimuthal symmetry in the problem suggests that the solution must be an expansion in terms of the Legendre polynomials. ${ }^{5}$ For $rR$, potential must not blow up at infinity. Plus, at large distances, the potential must be $-E_{0} r \cos \theta$ such that it leads to uniform electric field $E_{0}$ along $z$-axis. It follows that the solution outside the sphere should be
$$
\Phi_{o u t}(r, \theta)=-E_{0} r \cos \theta+\sum_{l=0}^{\infty} B_{l} r^{-(l+1)} P_{l}(\cos \theta)
$$
物理代写|电动力学作业代ELECTRODYNAMICS代考|Dielectric with Spherical Cavity in Presence of Uniform Electric Field
Let us consider a dielectric medium with dielectric constant $\epsilon$ having a spherical cavity of radius $R$. The uniform external electric field is $E_{0}$ parallel to z-axis. This is an inversion of problem of the dielectric sphere in uniform electric field. The problem is easy to solve once we recognize that the relative dielectric constant of the cavity to that of the surrounding dielectric medium is now $\frac{1}{\epsilon}$. Then we can immediately write down the potential inside and outside the cavity as
$$
\begin{aligned}
\Phi_{\text {in }}(r, \theta) &=\left(\frac{-3 \epsilon}{1+2 \epsilon}\right) E_{0} r \cos \theta \
\Phi_{\text {out }}(r, \theta) &=-E_{0} r \cos \theta+\left(\frac{1-\epsilon}{1+2 \epsilon}\right) \frac{R^{3}}{r^{2}} E_{0} \cos \theta
\end{aligned}
$$
The electric field inside the cavity is uniform, parallel to external electric field and is given by
$$
\mathbf{E}{\text {in }}=-\nabla \Phi{\text {in }}=\left(\frac{3 \epsilon}{1+2 \epsilon}\right) E_{0} \hat{z}
$$
电动力学代写
物理代写|电动力学作业代ELECTRODYNAMICS代考|IMAGE PROBLEM INVOLVING DIELECTRICS
考虑用介电常数标记为 1 和 2 的两个介电介质之间的边界ε1和ε2，分别在和=0. 介质 1 占据半空间和>0, 而介质 2, 半空间和<0 F一世G.3.3. 让有一个点收费q在介质 1 中 $\mathbf{x} {0}=\left(0,0, z {0}\right).在和在一种n吨吨这F一世nd吨H和s这一世你吨一世这ns这F吨H和F这一世一世这在一世nG和q你一种吨一世这ns:$ \begin{array}{ll} \nabla^{2} \Phi=4 \pi q \delta\left(\mathbf{x}-\mathbf{x} {0}\right), & z>0 \ \nabla^{2} \Phi=0, & z<0 \end{array} $$和>0，我们放置一个图像电荷$$
\Phi_{z>0}(\mathbf{x})=\frac{1}{\epsilon_{1}}\left(\frac{q}{\sqrt{x^{2}+y^{2}+\left(z-z_{0}\right)^{2}}}+\frac{q^{\prime}}{\sqrt{x^{2}+y^{2}+\left(z+z_{0}\right)^{2}}}\right)
$$
物理代写|电动力学作业代ELECTRODYNAMICS代考|DIELECTRIC SPHERE IN UNIFORM EXTERNAL FIELD
考虑一个半径球体R和介电常数ε，置于原本均匀的外电场中和0沿着和-轴。一开始有几点需要注意。自从∇×和=0到处都是电场和在空间中的任何一点都可以表示为标量势的梯度披在那时候。对于均匀的各向同性介质，D(X)=ε和(X)， 在哪里ε是介质的介电常数。自从∇⋅D=0, 与宏观电荷密度ρ在球体内为零，因此∇⋅和=0. 无论如何，球体外部的电场散度为零。因此，标量势披满足球内外的拉普拉斯方程，我们想找出这两个区域的势。
问题中的方位角对称性表明解决方案必须是勒让德多项式的展开。5为了rR，潜力绝不能在无穷大时爆发。另外，在很远的距离，潜力必须是−和0r某物θ从而产生均匀的电场和0沿着和-轴。因此，球外的解决方案应该是
披这你吨(r,θ)=−和0r某物θ+∑一世=0∞乙一世r−(一世+1)磷一世(某物θ)
物理代写|电动力学作业代ELECTRODYNAMICS代考|DIELECTRIC WITH SPHERICAL CAVITY IN PRESENCE OF UNIFORM ELECTRIC FIELD
让我们考虑具有介电常数的电介质ε有一个半径为球形的空腔R. 均匀外电场为和0平行于 z 轴。这是均匀电场中电介质球问题的反演。一旦我们认识到空腔与周围介电介质的相对介电常数现在是1ε. 然后我们可以立即将腔内外的电位写为
披在 (r,θ)=(−3ε1+2ε)和0r某物θ 披出去 (r,θ)=−和0r某物θ+(1−ε1+2ε)R3r2和0某物θ
腔内的电场是均匀的，平行于外部电场，由
$$
\begin{aligned}
\Phi_{\text {in }}(r, \theta) &=\left(\frac{-3 \epsilon}{1+2 \epsilon}\right) E_{0} r \cos \theta \
\Phi_{\text {out }}(r, \theta) &=-E_{0} r \cos \theta+\left(\frac{1-\epsilon}{1+2 \epsilon}\right) \frac{R^{3}}{r^{2}} E_{0} \cos \theta
\end{aligned}
$$
The electric field inside the cavity is uniform, parallel to external electric field and is given by
$$
\mathbf{E}{\text {in }}=-\nabla \Phi{\text {in }}=\left(\frac{3 \epsilon}{1+2 \epsilon}\right) E_{0} \hat{z}
$$
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