電動(dòng)力學(xué)每日一題 2021/10/10

上大學(xué)以前覺(jué)得自己大概數(shù)理化都能學(xué)得不錯(cuò)，后來(lái)大一有兩門課讓我認(rèn)清了現(xiàn)實(shí)，一門是程序設(shè)計(jì)，另一門是模電。程序設(shè)計(jì)學(xué)C語(yǔ)言，我當(dāng)時(shí)學(xué)得勤奮刻苦，每次上機(jī)課都會(huì)主動(dòng)多待兩個(gè)小時(shí)，但期中居然沒(méi)及格。模電就更不要說(shuō)了，甚至要分?jǐn)?shù)乘二才能及格。經(jīng)過(guò)這兩門課后我接受了人確實(shí)會(huì)有很多不擅長(zhǎng)的而且怎么學(xué)也學(xué)不會(huì)的東西。電磁理論電磁波經(jīng)典電動(dòng)力學(xué)基礎(chǔ)內(nèi)容差別不大，共同點(diǎn)是都帶“電”，于是在模電課上發(fā)生的估計(jì)又要發(fā)生一次了。為了這學(xué)期期末分?jǐn)?shù)不至于太難看，我打算試試每天記錄一道題，日復(fù)一日必有精進(jìn)。

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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(shuō)的是
$$??\text{S}+\frac{?\mathcal{E}}{?t}=0$$

在這個(gè)問(wèn)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(dāng)$\theta \to 0$或者$\theta \to \pi$時(shí)，$\sin \theta \to 0$，$??S$成了$\frac{0}{0}$不定型。所以(c)問(wèn)要討論能量守恒需要用積分形式，在小體積內(nèi)，Poynting vector的曲面積分應(yīng)該等于電線攜帶的能量變化，也就應(yīng)該等于電磁場(chǎng)對(duì)電線做的功。

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