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# 物理代写|电磁学代写Electromagnetism代考|ENGR2861 Metallic Sphere in a Uniform Electric Field

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## 物理代写|电磁学代写Electromagnetism代考|Metallic Sphere in a Uniform Electric Field

Based on the just mentioned assumption and using the quantities from Fig. 2.36, one makes the following Ansatz:
\begin{aligned} \varphi &=\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}-E_{a, \infty} z \ &=\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}-E_{a, \infty} r \cos \theta . \end{aligned}
$E_{a, \infty}$ is the externally applied field, which at a sufficiently large distance, is not distorted by the metallic sphere. The potential is generated by the dipole according to eq. (2.60) and by a part that belongs to the uniform outside field. This assumption is confirmed if we can choose $p$ such that $\varphi$ is constant for all $r=r_s$ :
$$\varphi=\varphi_0=\frac{p \cos \theta}{4 \pi \varepsilon_0 r_s^2}-E_{a, \infty} r_s \cos \theta .$$
$\varphi$ will, in fact, be constant for $r=r_s$, provided one chooses
$$p=4 \pi \varepsilon_0 r_s^3 E_{a, \infty} .$$
Thus
$$\varphi=E_{a, \infty}\left(\frac{r_s^3}{r^2}-r\right) \cos \theta .$$

## 物理代写|电磁学代写Electromagnetism代考|Metallic Cylinder in the Field of a Line Charge

Consider a metallic cylinder to be located within the field of a uniform line charge with its axis oriented parallel to the line charge (see Fig. 2.41). One can think of the overall field outside the cylinder as being created by the given line charge $q$ (outside the cylinder) and its image charge, also a line charge -q (inside). The product of the distances of the two line charges from the cylinder axis equals the square of the cylinder radius, i.e. the piercing points of the two line charges emerge by reflection at the circle $r=r_C$, (where $r_C$ is the radius of the cylinder). Thus
$$x_1 \cdot x_2=r_C^2$$
The proof is easy. Based on eq. (2.46), one first calculates the potential of the two line charges at a field point $(x, y, z)$

$$\varphi=-\frac{q}{2 \pi \varepsilon_0} \ln \frac{1}{r_B}+\frac{q}{2 \pi \varepsilon_0} \ln \frac{2}{r_B}=\frac{q}{2 \pi \varepsilon_0} \ln \frac{2}{r_1} .$$
where
$$r_1^2=\left(x-x_1\right)^2+y^2$$
and
$$r_2^2=\left(x-x_2\right)^2+y^2=\left(x-\frac{r_C^2}{x_1}\right)^2+y^2$$
On the cylinder wall we have
$$x^2+y^2=r_C^2$$
and thus
\begin{aligned} \frac{r_2^2}{r_1^2} &=\frac{x^2-\frac{2 x r_C^2}{x_1}+\frac{r_C^4}{x_1^2}+y^2}{x^2-2 x x_1+x_1^2+y^2} \ &=\frac{r_C^2-\frac{2 x r_C^2}{x_1}+\frac{r_C^4}{x_1^2}}{r_C^2-2 x x_1+x_1^2}=\frac{r_C^2}{x_1^2}=\text { const } \end{aligned}
Therefore $r_2 / r_1$, and thereby also $\varphi$ are constant on the cylinder wall. The location of all the geometrical points for which the distance ratios $r_2 / r_1$ from the two fixed points is constant, as shown in Fig. $2.41$ These are known in geometry as the circles of Apollonius. The circular cross section of the cylinder constitutes one of those circles.

## 物理代写电磁学代写Electromagnetism代考|Metallic Sphere in a Uniform Electric Field

$$\varphi=\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}-E_{a, \infty} z \quad=\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}-E_{a, \infty} r \cos \theta .$$
$E_{a, \infty}$ 是外部施加的场，在足够大的距离处，不会被金属球扭曲。电势由偶极子根据方程式产生。 $(2.60)$ 和属于统一外场的部分。 如果我们可以选择，这个假设得到证实 $p$ 这样 $\varphi$ 对所有人都是但定的 $r=r_s$ :
$$\varphi=\varphi_0=\frac{p \cos \theta}{4 \pi \varepsilon_0 r_s^2}-E_{a, \infty} r_s \cos \theta .$$
$\varphi$ 实际上，将是恒定的 $r=r_s$ 提供一个选择
$$p=4 \pi \varepsilon_0 r_s^3 E_{a, \infty} .$$

$$\varphi=E_{a, \infty}\left(\frac{r_s^3}{r^2}-r\right) \cos \theta .$$

## 物理代与目磁学代写Electromagnetism代考|Metallic Cylinder in the Field of a Line Charge

$$x_1 \cdot x_2=r_C^2$$

$$\varphi=-\frac{q}{2 \pi \varepsilon_0} \ln \frac{1}{r_B}+\frac{q}{2 \pi \varepsilon_0} \ln \frac{2}{r_B}=\frac{q}{2 \pi \varepsilon_0} \ln \frac{2}{r_1} .$$

$$r_1^2=\left(x-x_1\right)^2+y^2$$

$$r_2^2=\left(x-x_2\right)^2+y^2=\left(x-\frac{r_C^2}{x_1}\right)^2+y^2$$

$$x^2+y^2=r_C^2$$

$$\frac{r_2^2}{r_1^2}=\frac{x^2-\frac{2 x r_C^2}{x_1}+\frac{r_{\mathcal{C}}^4}{x_1^2}+y^2}{x^2-2 x x_1+x_1^2+y^2} \quad=\frac{r_C^2-\frac{2 x x_C^2}{x_1}+\frac{r_{\mathcal{4}}^{\mathcal{C}}}{x_1^2}}{r_C^2-2 x x_1+x_1^2}=\frac{r_C^2}{x_1^2}=\text { const }$$

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