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Next: 誘電体 Up: denjiki Previous: Poisson方程式

静電エネルギー

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item 仕事
\begin{center}
\parbox[b]{...
...\phi dV = \frac{1}{2} \int \epsilon_0E^2dV
\end{displaymath}
\end{itemize}
}}

[ 例題1 ] 3つの電荷からなる系

\includegraphics[width=3.5cm]{charges.eps}

$\displaystyle \phi_1$ $\displaystyle =$ $\displaystyle \frac{Q_2}{4\pi\epsilon_0r_{12}} + \frac{Q_3}{4\pi\epsilon_0r_{31}}$  
$\displaystyle \phi_2$ $\displaystyle =$ $\displaystyle \frac{Q_1}{4\pi\epsilon_0r_{12}} + \frac{Q_3}{4\pi\epsilon_0r_{23}}$  
$\displaystyle \phi_3$ $\displaystyle =$ $\displaystyle \frac{Q_1}{4\pi\epsilon_0r_{31}} + \frac{Q_2}{4\pi\epsilon_0r_{23}}$  


$\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{2} \sum_{i=1}^3 Q_i\phi_i$  
  $\displaystyle =$ $\displaystyle \frac{1}{8\pi\epsilon_0} \left( \frac{Q_1Q_2}{r_{12}} + \frac{Q_1...
...+ \frac{Q_2Q_3}{r_{23}} + \frac{Q_1Q_3}{r_{31}} + \frac{Q_2Q_3}{r_{23}} \right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_0} \left( \frac{Q_1Q_2}{r_{12}} + \frac{Q_2Q_3}{r_{23}} + \frac{Q_3Q_1}{r_{31}} \right)$  

[ 例題2 ] 半径$ a$ の球内に電荷が一様に分布

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

$ r<a$

$\displaystyle \phi$ $\displaystyle =$ $\displaystyle \frac{\rho}{6\epsilon_0} (3a^2 - r^2)$  
$\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{2} \int \rho\phi dV$  
  $\displaystyle =$ $\displaystyle \frac{\rho^2}{12\epsilon_0} \int_0^a (3a^2 - r^2) 4\pi r^2dr$  
  $\displaystyle =$ $\displaystyle \frac{\pi\rho^2}{3\epsilon_0} \left[ (a^2r^3 - \frac{r^5}{5} \right]_0^a$  
  $\displaystyle =$ $\displaystyle \frac{4\pi\rho^2}{15\epsilon_0} a^5$  

[ 例題3 ] 同じ問題を静電エネルギーを用いて

$ r<a$

$\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\rho r}{3\epsilon_0}$  
$\displaystyle U_1$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0}{2} \int_0^a \frac{\rho^2r^2}{9\epsilon_0^2} 4\pi r^2dr$  
  $\displaystyle =$ $\displaystyle \frac{2\pi\rho^2}{9\epsilon_0} \int_0^a r^4dr$  
  $\displaystyle =$ $\displaystyle \frac{2\pi\rho^2}{9\epsilon_0} \left[ \frac{r^5}{5} \right]_0^a$  
  $\displaystyle =$ $\displaystyle \frac{2\pi\rho^2}{45\epsilon_0} a^5$  

$ r>a$

$\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\rho a^3}{3\epsilon_0r^2}$  
$\displaystyle U_2$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0}{2} \int_a^{\infty} \frac{\rho^2a^6}{9\epsilon_0^2r^4} 4\pi r^2dr$  
  $\displaystyle =$ $\displaystyle \frac{2\pi\rho^2a^6}{9\epsilon_0} \int_a^{\infty} \frac{dr}{r^2}$  
  $\displaystyle =$ $\displaystyle -\frac{2\pi\rho^2a^6}{9\epsilon_0} \left[ \frac{1}{r} \right]_a^{\infty}$  
  $\displaystyle =$ $\displaystyle \frac{2\pi\rho^2}{9\epsilon_0} a^5$  

全体で

$\displaystyle U = U_1 + U_2 = \frac{4\pi\rho^2}{15\epsilon_0} a^5
$

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item 静電容量
\begin{eqnarray*}
Q &=& CV \\
U &=& \frac{1}{2} CV^2
\end{eqnarray*}
\end{itemize}
}}

[ 例題4 ] 平行平板コンデンサー

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

$\displaystyle 電荷密度 \hspace{5mm} \sigma = \frac{Q}{S}
$


$\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\sigma}{\pi\epsilon_0} = \frac{Q}{\epsilon_0S}$  
$\displaystyle V$ $\displaystyle =$ $\displaystyle Ed = \frac{Qd}{\epsilon_0S}$  
$\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0S}{d}$  

[ 例題5 ] 同じ問題を静電エネルギーを用いて

$\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{Q}{\epsilon_0S}$  
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{Qd}{\epsilon_0S}$  


$\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0}{2} \int E^2dV$  
  $\displaystyle =$ $\displaystyle \frac{\epsilon_0}{2} \frac{Q^2}{\epsilon_0^2S^2} Sd$  
  $\displaystyle =$ $\displaystyle \frac{\epsilon_0S}{2d} V^2$  
$\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0S}{d}$  


next up previous
Next: 誘電体 Up: denjiki Previous: Poisson方程式
Keiichi Takasugi
平成24年1月25日