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Next: Gaussの法則 Up: denjiki Previous: 電磁気で使う数学

Coulombの法則

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item Coulomb力
\begin{center}
\parbo...
...} \frac{\bm{r}}{r^3} \\
\bm{F} &=& Q\bm{E}
\end{eqnarray*}
\end{itemize}
}}

[ 例題1 ] 電気双極子 ($ r \gg d$ )

\includegraphics[width=3cm]{dipole.eps}

$\displaystyle \bm{E}(\bm{r})$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_0} \frac{\bm{r}-\bm{d}}{\vert\bm{r}-\bm{d}\vert^3} - \frac{q}{4\pi\epsilon_0} \frac{\bm{r}+\bm{d}}{\vert\bm{r}+\bm{d}\vert^3}$  
  $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_0} \left\{ \frac{\bm{r}-\bm{d}}{(r^2+d^2-2dr\cos\theta)^{3/2}} - \frac{\bm{r}+\bm{d}}{(r^2+d^2+2dr\cos\theta)^{3/2}} \right\}$  
    $\displaystyle {\color{blue} \bm{d}\cdot\bm{r} = dr\cos\theta}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r^3} \left\{ (\bm{r}-\bm{d}) \left( 1 - \f...
...bm{r}+\bm{d}) \left( 1 + \frac{2\bm{d}\cdot\bm{r}}{r^2} \right)^{-3/2} \right\}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r^3} \left\{ (\bm{r}-\bm{d}) \left( 1 + \f...
...t) - (\bm{r}+\bm{d}) \left( 1 - \frac{3\bm{d}\cdot\bm{r}}{r^2} \right) \right\}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r^3} \left\{ \frac{6(\bm{d}\cdot\bm{r})\bm{r}}{r^2} - 2\bm{d} \right\}$  
    $\displaystyle {\color{blue} \bm{p} = 2q\bm{d} : 電気双極子モーメント}$  
  $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_0r^3} \left\{ \frac{3(\bm{p}\cdot\bm{r})\bm{r}}{r^2} - \bm{p} \right\}$  

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item 電位 (ポテンシャル)
\begin{eqnar...
...laymath}
\phi = \frac{q}{4\pi\epsilon_0r}
\end{displaymath}
\end{itemize}
}}

[ 例題2 ] 電気双極子 ($ r \gg d$ )

\includegraphics[width=3cm]{dipole.eps}

$\displaystyle \phi(\bm{r})$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_0} \frac{1}{\vert\bm{r}-\bm{d}\vert} - \frac{q}{4\pi\epsilon_0} \frac{1}{\vert\bm{r}+\bm{d}\vert}$  
  $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_0} \left\{ \frac{1}{\sqrt{r^2+d^2-2dr\cos\theta}} - \frac{1}{\sqrt{r^2+d^2+2dr\cos\theta}} \right\}$  
    $\displaystyle {\color{blue} \bm{d}\cdot\bm{r} = dr\cos\theta}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r} \left\{ \left( 1 - \frac{2\bm{d}\cdot\b...
...ght)^{-1/2} - \left( 1 + \frac{2\bm{d}\cdot\bm{r}}{r^2} \right)^{-1/2} \right\}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r} \left\{ \left( 1 + \frac{\bm{d}\cdot\bm{r}}{r^2} \right) - \left( 1 - \frac{\bm{d}\cdot\bm{r}}{r^2} \right) \right\}$  
  $\displaystyle \simeq$ $\displaystyle \frac{q}{4\pi\epsilon_0r} \frac{2\bm{d}\cdot\bm{r}}{r^2}$  
    $\displaystyle {\color{blue} \bm{p} = 2q\bm{d} : 電気双極子モーメント}$  
  $\displaystyle =$ $\displaystyle \frac{\bm{p}\cdot\bm{r}}{4\pi\epsilon_0r^3}$  

$\displaystyle \bm{E}(\bm{r}) = -\nabla\phi
= -\frac{1}{4\pi\epsilon_0} \nabla \left( \frac{\bm{p}\cdot\bm{r}}{r^3} \right)
$


$\displaystyle \frac{\partial}{\partial x} \left( \frac{\bm{p}\cdot\bm{r}}{r^3} \right)$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x} \frac{p_xx+p_yy+p_zz}{(x^2+y^2+z^2)^{3/2}}$  
  $\displaystyle =$ $\displaystyle \frac{p_x}{(\ \cdots\ )^{3/2}} - \frac{3}{2} \frac{p_xx+p_yy+p_zz}{(\ \cdots\ )^{5/2}} 2x$  
  $\displaystyle =$ $\displaystyle \frac{p_x}{r^3} - \frac{3(\bm{p}\cdot\bm{r})x}{r^5}$  
$\displaystyle \nabla \left( \frac{\bm{p}\cdot\bm{r}}{r^3} \right)$ $\displaystyle =$ $\displaystyle \frac{\bm{p}}{r^3} - \frac{3(\bm{p}\cdot\bm{r})\bm{r}}{r^5}$  

$\displaystyle \bm{E}(\bm{r})
= -\frac{1}{4\pi\epsilon_0r^3} \left\{ \frac{3(\bm{p}\cdot\bm{r})\bm{r}}{r^2} - \bm{p} \right\}
$

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item 電荷が分布するとき
\begin{eqnarr...
...ho(\bm{r}')}{\vert\bm{r} - \bm{r}'\vert} dV'
\end{eqnarray*}
\end{itemize}
}}

[ 例題3 ] 半径$ a$ の球内に電荷が一様に分布

\includegraphics[width=5cm]{sphere2.eps}

$ r>a$ のとき

$\displaystyle \phi(\bm{r})$ $\displaystyle =$ $\displaystyle \frac{\rho}{4\pi\epsilon_0} \int \frac{dV'}{\vert\bm{r} - \bm{r}'\vert}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{4\pi\epsilon_0} \iint \frac{2\pi r'^2\sin\theta dr'd\theta}{\sqrt{r^2+r'^2-2rr'\cos\theta}}$  
    $\displaystyle {\color{blue} r^2+r'^2-2rr'\cos\theta = tとおくと}$  
    $\displaystyle {\color{blue} 2rr'\sin\theta d\theta = dt}$  
    $\displaystyle {\color{blue} \sin\theta d\theta = \frac{dt}{2rr'}}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2\epsilon_0} \int_0^a dr' \int_{(r-r')^2}^{(r+r')^2} \frac{r'dt}{2r\sqrt{t}}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2\epsilon_0} \int_0^a dr' \left[ \frac{r'}{r} \sqrt{t} \right]_{(r-r')^2}^{(r+r')^2}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2\epsilon_0r} \int_0^a r'dr' \left( \vert r+r'\vert - \vert r-r'\vert \right)$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2\epsilon_0r} \int_0^a 2r'^2dr' \hspace{1cm} {\color{blue} (r>r')}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{\epsilon_0r} \left[ \frac{r'^3}{3} \right]_0^a$  
  $\displaystyle =$ $\displaystyle \frac{\rho a^3}{3\epsilon_0r}$  

$ r<a$ のとき

$\displaystyle \phi(\bm{r})$ $\displaystyle =$ $\displaystyle \frac{\rho}{2\epsilon_0r} \int_0^r 2r'^2dr' + \frac{\rho}{2\epsilon_0r} \int_r^a 2rr'dr'$  
    $\displaystyle {\color{blue} \hspace{.7cm} (r>r') \hspace{1.8cm} (r<r')}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{\epsilon_0r} \left[ \frac{r'^3}{3} \right]_0^r + \frac{\rho}{\epsilon_0} \left[ \frac{r'^2}{2} \right]_r^a$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{6\epsilon_0r} (3a^2-r^2)$  


next up previous
Next: Gaussの法則 Up: denjiki Previous: 電磁気で使う数学
Keiichi Takasugi
平成24年1月25日