III.静电场 ii Electrostatic field ii

• 原文: https://zhuanlan.zhihu.com/p/61608610
• 发布日期: 2019-04-07
• 分类: 电动力学

文章收录于:电动力学前情回顾

✦拉普拉斯方程/*Laplace's equation -*分离变量法:

所解方程: ${\nabla }^2\phi =0$ **#**用分段函数标势不同区域,方程源于 {\nabla }^{2}\varphi =\frac{\rho }_{f}{\varepsilon }.

球坐标系下有通解: #知道就行了. #btw这是在解无自由电荷的静电场.

\varphi \left( r,\theta ,\varphi \right)=\sum\limits_{n,m}{\left( {a}_{nm}{r}^{n}+\frac{b}_{nm}{r}^{n+1} \right)P_{n}^{m}\cos m\varphi }+\sum\limits_{n,m}{\left( {c}_{nm}{r}^{n}+\frac{d}_{nm}{r}^{n+1} \right)P_{n}^{m}\sin m\varphi } #$a_{nm},b_{nm},c_{nm},d_{nm}$ 均为待定常数. $P_n^m$为缔合勒让德函数

球坐标系下体系关于极轴对称时简化为:\varphi \left( r,\theta \right)=\sum\limits_{n=0}^{\infty }{\left( {a}_{n}{r}^{n}+\frac{b}_{n}{r}^{n+1} \right)}{P}_{n}

$\theta$为极角(与极轴成角),$\phi$为方位角. 其中 $P_n$分别为\left\{ \begin{align} & {P}_{0}=1\ \ \ \ \ \ \ \ \ \ \ {P}_{1}=\cos \theta \\ & {P}_{2}=\frac{1}{2}\left( 3{\cos }^{2}\theta -1 \right) \\ & {P}_{3}=\frac{1}{2}\left( 5{\cos }^{2}\theta -3\cos \theta \right) \\ & \cdot \cdot \cdot \\ \end{align} \right.

例:将匀强 {\vec{E}_{0}加在介电常数为 $\epsilon$ 的无穷大介质上,现于介质挖去一个球体,求内外电势.

解:

设挖前球心电势为 {\varphi }_{\text{0} ,记球内电势 {\varphi }_{\text{1}, 球外电势为 {\varphi }_{\text{2}.

求解方程: \left\{ \begin{align} & {\nabla }^{\text{2}{\varphi }_{\text{1}=0\ \ \ \ r\in \left[ 0,R \right] \\ & {\nabla }^{\text{2}{\varphi }_{\text{2}=0\ \ \ \ r\in \left( R,+\infty \right) \\ \end{align} \right.

设通解: \left\{ \begin{align} & {\varphi }_{1}=\sum\limits_{n=0}^{\infty }{\left( {a}_{n}{r}^{n}+\frac{b}_{n}{r}^{n+1} \right)}{P}_{n}\ \ \ r\in \left[ 0,R \right] \\ & {\varphi }_{2}=\sum\limits_{n=0}^{\infty }{\left( {c}_{n}{r}^{n}+\frac{d}_{n}{r}^{n+1} \right)}{P}_{n}\ \ \ r\in \left( R,+\infty \right) \\ \end{align} \right.

分析题目条件: \left\{ \begin{align} & {\left. {\varphi }_{1} \right|}_{r=0}={有限值} \\ & {\left. {\varphi }_{1} \right|}_{r=R}={\left. {\varphi }_{2} \right|}_{r=R} \\ & {\varepsilon }_{0}{\left. \frac{\partial {\varphi }_{1}{\partial r} \right|}_{r=R}=\varepsilon {\left. \frac{\partial {\varphi }_{2}{\partial r} \right|}_{r=R} \\ & {\left. {\varphi }_{2} \right|}_{r=\infty }={\varphi }_{0}-{E}_{0}r\cos \theta \\ \end{align} \right.

${{\phi }_1|}_{r=0}=有限值\Rightarrow b_n=0\Rightarrow {\phi }_1=\sum _{n=0}^{\infty }{a}_n{r}^n{P}_n$

\begin{align} & {\left. {\varphi }_{2} \right|}_{r=\infty }=\sum\limits_{n=0}^{\infty }{c}_{n}{r}^{n}{P}_{n}={\varphi }_{0}-{E}_{0}r\cos \theta \\ & \ \ \ \Rightarrow {\varphi }_{2}={\varphi }_{0}-{E}_{0}r\cos \theta +\sum\limits_{n=0}^{\infty }{\frac{d}_{n}{r}^{n+1}{P}_{n} \\ \end{align}

{\left. {\varphi }_{1} \right|}_{r=R}=\sum\limits_{n=0}^{\infty }{a}_{n}{R}^{n}{P}_{n}={\left. {\varphi }_{2} \right|}_{r=R}={\varphi }_{0}-{E}_{0}R\cos \theta +\sum\limits_{n=0}^{\infty }{\frac{d}_{n}{R}^{n+1}{P}_{n} \Rightarrow [1]\left\{ \begin{align} & {a}_{0}={\varphi }_{0}+\frac{d}_{0}{R} \\ & {a}_{1}=-{E}_{0}+\frac{d}_{1}{R}^{3} \\ & {a}_{n}=\frac{d}_{n}{R}^{2n+1}\ \ \ \left( n\ge 2 \right) \\ \end{align} \right. #就是对照等式两边 $P_n$的系数得到

{\varepsilon }_{0}{\left. \frac{\partial {\varphi }_{1}{\partial r} \right|}_{r=R}=\sum\limits_{n=0}^{\infty }{n{\varepsilon }_{0}{a}_{n}{R}^{n-1}{P}_{n}=\varepsilon {\left. \frac{\partial {\varphi }_{2}{\partial r} \right|}_{r=R}=-\varepsilon E\cos \theta -\sum\limits_{n=0}^{\infty }{\frac{\left( n+1 \right)\varepsilon {d}_{n}{R}^{n+2}{P}_{n} \Rightarrow [2]\left\{ \begin{align} & {d}_{0}=0 \\ & {\varepsilon }_{0}{a}_{1}=-\varepsilon {E}_{0}-\frac{2\varepsilon }{R}^{3}{d}_{1} \\ & {a}_{n}=\frac{-\varepsilon \left( n+1 \right)}{\varepsilon }_{0}n{R}^{2n+1}{d}_{n}\ \left( n\ge 2 \right) \\ \end{align} \right. #就是对照等式两边 $P_n$的系数得到

结合[1]与[2]得到 \left\{ \begin{align} & {a}_{0}={\varphi }_{0}\ \ \ {a}_{1}=-\frac{3\varepsilon }{2\varepsilon +{\varepsilon }_{0}{E}_{0}\ \ \ \ {a}_{0}=0\ \left( n\ge 2 \right) \\ & {d}_{0}=0\ \ \ \ {d}_{1}=\frac{\varepsilon }_{0}-\varepsilon }{2\varepsilon +{\varepsilon }_{0}{E}_{0}{R}^{3}\ \ \ {d}_{n}=0\ \left( n\ge 2 \right) \\ \end{align} \right.

综上所述: \left\{ \begin{align} & {\varphi }_{1}={\varphi }_{0}-\frac{3\varepsilon }{2\varepsilon +{\varepsilon }_{0}r\cos \theta \ \ \ r\in \left[ 0,R \right] \\ & {\varphi }_{2}={\varphi }_{0}-Er\cos \theta +\frac{\varepsilon }_{0}-\varepsilon }{2\varepsilon +{\varepsilon }_{0}{E}_{0}{R}^{3}{r}^{-2}\cos \theta \ \ \ \ r\in \left( R,+\infty \right) \\ \end{align} \right.