Riemann surface
matlab

无需担心!我们的电动力学代写专家团队将专业地解决您在电动力学学习中遇到的各种挑战。我们拥有广泛的专业知识和丰富的经验,可以协助您完成高水平的作业和论文,确保您在学习道路上顺利前行!

以下是一些我们可以帮助您解决的问题:

电场与电势:涵盖电场、电势的概念、性质和计算方法,如库仑定律、电势差等。

电荷分布与电场:研究电荷分布对电场产生的影响,如连续分布、点电荷等。

电磁感应与法拉第定律:常见的电磁感应现象和法拉第电磁感应定律的应用,如感应电动势、电磁感应中的能量转换等。

电场与介质:介绍电场在不同介质中的传播和相互作用,如电介质极化、介质中的电场能量储存等。

安培定律与电流:研究电流和安培定律的应用,如电流分布、电流在电路中的传输等。

电磁场的辐射和辐射场:介绍电磁场的辐射和辐射场的产生、传播和特性,如辐射强度、辐射模式等。

麦克斯韦方程组:研究和应用麦克斯韦方程组,如电磁波的传播和解析解等。

无论您面临的电动力学问题是什么,我们都会竭尽全力提供专业的帮助,确保您的学习之旅顺利无阻!

问题 1.


(a) Show that the work done to remove the charge $q$ from a distance $r>a$ to infinity against the force, Eq. (2.6), of a grounded conducting sphere is
$$
W=\frac{q^2 a}{8 \pi \epsilon_0\left(r^2-a^2\right)}
$$
Relate this result to the electrostatic potential, Eq. (2.3), and the energy discussion of Section 1.11.


(a) The force is
$$
|F|=\frac{q^2 a}{4 \pi \epsilon_0} \frac{1}{y^3\left(1-a^2 / y^2\right)^2}
$$
directed radially inward. The work is
$$
\begin{aligned}
W & =-\int_r^{\infty} F d y \
& =\frac{q^2 a}{4 \pi \epsilon_0} \int_r^{\infty} \frac{d y}{y^3\left(1-a^2 / y^2\right)^2} \
& =\frac{q^2 a}{4 \pi \epsilon_0} \int_r^{\infty} \frac{y d y}{\left(y^2-a^2\right)^2} \
& =\frac{q^2 a}{4 \pi \epsilon_0} \int_{r^2-a^2}^{\infty} \frac{d u}{2 u^2} \
& =\frac{q^2 a}{4 \pi \epsilon_0}\left|-\frac{1}{2 u}\right|_{r^2-a^2}^{\infty} \
& =\frac{q^2 a}{8 \pi \epsilon_0\left(r^2-a^2\right)}
\end{aligned}
$$
To relate this to earlier results, note that the image charge $q^{\prime}=-(a / r) q$ is located at radius $r^{\prime}=a^2 / r$. The potential energy between the point charge and

its image is
$$
\begin{aligned}
P E & =\frac{1}{4 \pi \epsilon_0}\left(\frac{q q^{\prime}}{\left|r-r^{\prime}\right|}\right) \
& =\frac{1}{4 \pi \epsilon_0}\left(\frac{-q^2 a}{r\left(r-a^2 / r\right)}\right) \
& =\frac{1}{4 \pi \epsilon_0}\left(\frac{-q^2 a}{r^2-a^2}\right)
\end{aligned}
$$
Result (7) is only half of (8). This would seem to violate energy conservation. It would seem that we could start with the point charge at infinity and allow it to fall in to a distance $r$ from the sphere, liberating a quantity of energy (8), which we could store in a battery or something. Then we could expend an energy equal to (7) to remove the charge back to infinity, at which point we would be back where we started, but we would still have half of the energy saved in the battery. It would seem that we could keep doing this over and over again, storing up as much energy in the battery as we pleased.

I think the problem is with equation (8). The traditional expression $q_1 q_2 / 4 \pi \epsilon_0 r$ for the potential energy of two charges comes from calculating the work needed to bring one charge from infinity to a distance $r$ from the other charge, and it is assumed that the other charge does not move and keeps a constant charge during the process. But in this case one of the charges is a fictitious image charge, and as the point charge $q$ is brought in from infinity the image charge moves out from the center of the sphere, and its charge increases. So the simple expression doesn’t work to calculate the potential energy of the configuration, and we should take (7) to be the correct result.

问题 2.


(b) Repeat the calculation of the work done to remove the charge $q$ against the force, Eq. (2.9), of an isolated charged conducting sphere. Show that the work done is
$$
W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q^2 a}{2\left(r^2-a^2\right)}-\frac{q^2 a}{2 r^2}-\frac{q Q}{r}\right] .
$$
Relate the work to the electrostatic potential, Eq. (2.8), and the energy discussion of Section 1.11.

(b) In this case there are two image charges: one of the same charge and location as in part a, and another of charge $Q-q^{\prime}$ at the origin. The work needed to remove the point charge $q$ to infinity is the work needed to remove the point charge from its image charge, plus the work needed to remove the point charge from the extra charge at the origin. We calculated the first contribution above. The second contribution is
$$
\begin{aligned}
-\int_r^{\infty} \frac{q\left(Q-q^{\prime}\right) d y}{4 \pi \epsilon_0 y^2} & =-\frac{1}{4 \pi \epsilon_0} \int_r^{\infty}\left[\frac{q Q}{y^2}+\frac{q^2 a}{y^3}\right] d y \
& =-\frac{1}{4 \pi \epsilon_0}\left|-\frac{q Q}{y}-\frac{q^2 a}{2 y^2}\right|_r^{\infty} \
& =-\frac{1}{4 \pi \epsilon_0}\left[\frac{q Q}{r}+\frac{q^2 a}{2 r^2}\right]
\end{aligned}
$$
so the total work done is
$$
W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q^2 a}{2\left(r^2-a^2\right)}-\frac{q^2 a}{2 r^2}-\frac{q Q}{r}\right] .
$$

E-mail: help-assignment@gmail.com  微信:shuxuejun

help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学,统计,金融,经济,计算机科学,物理的作业代写服务

发表回复

您的电子邮箱地址不会被公开。 必填项已用 * 标注