電動力學每日一題 2021/10/10

上大學以前覺得自己大概數理化都能學得不錯，后來大一有兩門課讓我認清了現實，一門是程序設計，另一門是模電。程序設計學C語言，我當時學得勤奮刻苦，每次上機課都會主動多待兩個小時，但期中居然沒及格。模電就更不要說了，甚至要分數乘二才能及格。經過這兩門課后我接受了人確實會有很多不擅長的而且怎么學也學不會的東西。電磁理論電磁波經典電動力學基礎內容差別不大，共同點是都帶“電”，于是在模電課上發生的估計又要發生一次了。為了這學期期末分數不至于太難看，我打算試試每天記錄一道題，日復一日必有精進。

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(a) There’s no electric field inside the shell, so
$$\text{E}(\text{r})=\{\begin{array}{l}0,r<R\\ \frac{Q\hat{r}}{4\pi {?}_0{r}^2},r>R\end{array}$$

The energy density is
$$\mathcal{E}(\text{r})=\frac{1}{2}{?}_0|\text{E}{|}^2+\frac{1}{2}{\mu }_0|\text{H}{|}^2=\{\begin{array}{l}\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2{\sin }^2\theta },r<R\\ \frac{Q^2}{32{\pi }^2{?}_0{r}^4}+\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2{\sin }^2\theta },r>R\end{array}$$

(b) The Poynting vector is
$$\text{S}(\text{r})=\text{E}\times \text{H}=\{\begin{array}{l}0,r<R\\ ?\frac{QI}{8{\pi }^2{?}_0{r}^3\sin \theta }\hat{\theta },r>R\end{array}$$

Evaluate divergence of Poynting vector under spherical coordinate system, if $\theta \in (0,\pi )$, $r>R$,
$$\begin{array}{rl}??\text{S}& =\frac{1}{r\sin \theta }\frac{?}{?\theta }(\sin \theta S_{\theta })\\ & =\frac{1}{r\sin \theta }\frac{?}{?\theta }\left(\sin \theta \frac{?QI}{8{\pi }^2{?}_0{r}^3\sin \theta }\right)=0\end{array}$$

(c) Along the z-axis, the Poynting vector is directed toward the wire where $z>R,\theta \to 0$ and away from the wire where $z<?R,\theta \to \pi$. Taking a small cylinder of radius $?$ and height $\delta$ whose center on the bottom locates on $(0,0,R)$, the surface integral of $\text{S}(\text{r})$ over the cylinder is
$${\oint }_{Cylinder}\text{S}?d\text{a}={\int }_{SideofCylinder}\text{S}?d\text{a}=?(2\pi ?\delta )\left(\frac{QI}{8{\pi }^2{?}_0{r}^3\sin \theta }\right)$$

Now $\theta$ is so small that $?\approx r\sin \theta$, $|z|=|r\cos (\theta )|\approx r$,
$$?(2\pi ?\delta )\left(\frac{QI}{8{\pi }^2{?}_0{r}^3\sin \theta }\right)\approx ?\frac{QI\delta }{4\pi {?}_0{z}^2}$$

Consider the electric field acting on the wire at $z$ over a short segment of length $\delta$. The work is
$${\int }_{Segment}\text{E}?{\text{J}}_{free}dl=\frac{QI\delta }{4\pi {?}_0{z}^2}$$

Thus, the energy re-enters the wire in the region above the shell. Similarly, the energy emanates from the wire in the region below the shell.

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能量守恒說的是
$$??\text{S}+\frac{?\mathcal{E}}{?t}=0$$

在這個問題中
$$??\text{S}=\frac{1}{r\sin \theta }\frac{?}{?\theta }(\sin \theta S_{\theta })=\frac{1}{r\sin \theta }\frac{?}{?\theta }\left(\frac{?QI}{8{\pi }^2{?}_0{r}^3}\right)$$

當$\theta \to 0$或者$\theta \to \pi$時，$\sin \theta \to 0$，$??S$成了$\frac{0}{0}$不定型。所以(c)問要討論能量守恒需要用積分形式，在小體積內，Poynting vector的曲面積分應該等于電線攜帶的能量變化，也就應該等于電磁場對電線做的功。

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