贵州省网站节约化建设通知,网络维护工程师工资多少,怎么做用网站赚钱吗,怎么做网站的导航条1. 偶极子: 相距为 $l$, 带电量分别为 $\pm q$ 的一对电荷组成的系统. 称 $$\bex {\bf m}q{\bf l} \eex$$ 为电偶极矩, 其中 ${\bf l}$ 为 $-q$ 到 $q$ 的向量. 2. 取 ${\bf l}$ 为 $z$ 轴, 考虑偶极子的振动: $$\bex {\bf l}(t)l_0 e^{-i\omega t} {\bf e}_3. \eex$$ 则 (1) $…1. 偶极子: 相距为 $l$, 带电量分别为 $\pm q$ 的一对电荷组成的系统. 称 $$\bex {\bf m}q{\bf l} \eex$$ 为电偶极矩, 其中 ${\bf l}$ 为 $-q$ 到 $q$ 的向量. 2. 取 ${\bf l}$ 为 $z$ 轴, 考虑偶极子的振动: $$\bex {\bf l}(t)l_0 e^{-i\omega t} {\bf e}_3. \eex$$ 则 (1) ${\bf j}\rho {\bf v}\rho\cfrac{\rd {\bf l}}{\rd t}-i\omega l_0\rho e^{-i\omega t}{\bf e}_3$. (2) $$\beex \bea \quad \cfrac{1}{c^2} \cfrac{\p^2{\bf A}}{\p t^2}-\lap{\bf A}\mu{\bf j}\\ \ra {\bf A}(t,P) \cfrac{\mu_0}{4\pi}\int_{r_{PP}\leq ct} \cfrac{{\bf j}\sex{t-\cfrac{r_{PP}}{c},P}}{r_{PP}}\rd V_{P}\\ \quad-\cfrac{i\mu_0\omega l_0}{4\pi r}e^{-i\omega ti\frac{\omega r}{c}}\int_{r_{PP}\leq ct}\rho \rd V_{P} {\bf e}_3\\ \quad-i\cfrac{\mu_0\omega}{4\pi r}e^{ikr}{\bf m}(t)\quad\sex{r\gg l}\\ \quad\sex{{\bf m}(t)q{\bf l}(t),\ k\cfrac{\omega}{c}\cfrac{2\pi }{cT}:\mbox{ 波数}}. \eea \eeex$$ (3) $$\beex \bea \quad \cfrac{\p}{\p z}\sex{-i\cfrac{\mu_0\omega}{4\pi r}e^{ikr}m(t)}\cfrac{1}{c^2}\cfrac{\p \phi}{\p t}0\quad\sex{Lorentz\mbox{ 条件}}\\ \ra \cfrac{\p}{\p t} \sez{ \cfrac{\p}{\p z}\sex{\cfrac{1}{4\pi \ve_0r}e^{ikr}}m(t)\phi }0\\ \ra \phi-\cfrac{\p}{\p z}\sex{\cfrac{1}{4\pi \ve_0r}e^{ikr}}m(t)\phi_0(x,y,z)\\ \ra \phi-\cfrac{\p}{\p z}\sex{\cfrac{1}{4\pi \ve_0r}e^{ikr}}m(t). \eea \eeex$$ (4) $$\beex \bea {\bf B}\rot {\bf A}\\ \cfrac{\mu_0ck^2}{4\pi r} e^{ikr}\sex{1-\cfrac{1}{ikr}}{\bf n}\times {\bf m}\quad\sex{{\bf n}\cfrac{{\bf r}}{r}}. \eea \eeex$$ (5) $$\beex \bea -\cfrac{\p {\bf A}}{\p t} i\cfrac{\mu_0\omega}{4\pi r}e^{ikr}\sez{-i\omega {\bf m}}\\ \cfrac{\mu_0\omega^2}{4\pi r}e^{ikr}{\bf m}\\ \cfrac{k^2}{4\pi \ve_0r}e^{ikr}{\bf m},\\ -\n \phi\cfrac{1}{4\pi \ve_0}\n \cfrac{\p}{\p z}\sex{\cfrac{1}{r}e^{ikr}}m\\ \cfrac{1}{4\pi \ve_0} \n \sex{-\cfrac{z}{r^3}e^{ikr}\cfrac{ikz}{r^2}e^{ikr}}m\\ \cfrac{1}{4\pi \ve_0} \n\sez{ \sex{-\cfrac{1}{r^3}\cfrac{ik}{r^2}}e^{ikr}z }m\\ \cfrac{1}{4\pi \ve_0} \left[ \sex{\cfrac{3}{r^4}{\bf n} -\cfrac{2ik}{r^3}{\bf n}}e^{ikr}zm ik\sex{-\cfrac{1}{r^3}\cfrac{ik}{r^2}}e^{ikr}{\bf n} z m\right. \\ \quad\quad\left. \sex{-\cfrac{1}{r^3}\cfrac{ik}{r^2}}e^{ikr}\n z m \right]\\ \cfrac{1}{4\pi \ve_0}\sez{ r({\bf n}\cdot{\bf m}) e^{ikr}{\bf n} \sex{\cfrac{3}{r^4}-\cfrac{3ik}{r^3}-\cfrac{k^2}{r^2} }\sex{-\cfrac{1}{r^3}\cfrac{ik}{r^2}}e^{ikr}{\bf m} }\\ \cfrac{1}{4\pi \ve_0} \sez{ ({\bf n}\cdot{\bf m})e^{ikr} {\bf n}\sex{\cfrac{3}{r^3}-\cfrac{3ik}{r^2}-\cfrac{k^2}{r}} \sex{-\cfrac{1}{r^3}\cfrac{ik}{r^2}}e^{ikr}{\bf m} }\\ -\cfrac{k^2}{4\pi \ve_0r}({\bf n}\cdot{\bf m}){\bf n} \cfrac{1}{4\pi\ve_0r} \sex{\cfrac{1}{r^2}-\cfrac{ik}{r}} e^{ikr}\sez{3({\bf n}\cdot{\bf m}){\bf n}-{\bf m}},\\ {\bf E}-\n\phi-\cfrac{\p {\bf A}}{\p t}\\ \cfrac{k^2}{4\pi\ve_0r}e^{ikr}({\bf n}\times{\bf m})\times {\bf n} \cfrac{1}{4\pi\ve_0r} \sex{\cfrac{1}{r^2}-\cfrac{ik}{r}} e^{ikr}\sez{3({\bf n}\cdot{\bf m}){\bf n}-{\bf m}}. \eea \eeex$$ (5) 当 $r\ll \lmcT\cfrac{2\pi c}{\omega}\cfrac{2\pi}{k}$ 时, $$\bex {\bf B}\cfrac{\mu_0ck}{4\pi r^2}{\bf n}\times {\bf m},\quad {\bf E}\cfrac{1}{4\pi r^3}[3({\bf n}\cdot{\bf m}){\bf n}-{\bf m}]. \eex$$ 而 $B\ll E$, 称为静场区 (电场为主). (6) 当 $r\gg \lm$ 时, $$\bex {\bf B}\cfrac{\mu_0ck^2}{4\pi r}e^{ikr}{\bf n}\times {\bf m},\quad{\bf E}c{\bf B}\times {\bf n}. \eex$$ 而为波动场区 (辐射区). 此时, $$\bex EcB\cfrac{\mu_0\omega^2m_0}{4\pi r}\sin \tt\sev{\cos\sez{\omega\sex{t-\cfrac{r}{c}}}}. \eex$$ 于是电磁场是以电偶极振子为中心的球面波, 相位为 $\omega\sex{t-\cfrac{r}{c}}$ 相同的各点的场强不一定相同, 还与 $\tt$ 有关. 赤道上, 场强最大. (7) 电磁能量密度 $$\bex \cfrac{1}{2}\sex{\ve_0E^2\cfrac{1}{\mu_0}B^2} \cfrac{\mu_0k^2\omega^2m_0^2}{16\pi r^2}\sin^2\tt\cos^2\sez{\omega\sex{t-\cfrac{r}{c}}}; \eex$$ 能量流密度向量 ${\bf S}$ 方向与 ${\bf r}$ 相同, 大小为 $$\bex S\cfrac{c\mu_0k^2\omega^2m_0^2}{16\pi^2r^2}\sin^2\tt\cos^2\sez{\omega\sex{t-\cfrac{r}{c}}}; \eex$$ 单位时间内通过以振子为球心, $r$ 为半径的球面的总能量为 $$\bex P(t)\int_0^\pi S\cdot 2\pi r\sin \tt\cdot r\rd \tt \cfrac{1}{6\pi}c\mu_0k^2\omega^2m_0^2\cos^2\sez{\omega\sex{t-\cfrac{r}{c}}}; \eex$$ 单位时间内辐射的平均能量为 $$\bex \bar P\cfrac{1}{T}\int_0^T P(t)\rd t \cfrac{1}{12}\mu_0\omega^4m_0^2, \eex$$ 其中 $T\cfrac{2\pi}{\omega}$. 转载于:https://www.cnblogs.com/zhangzujin/p/3628504.html
