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Next: 静電エネルギー Up: denjiki Previous: Gaussの法則

Poisson方程式

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item Gaussの法則 (微分形)
\begin{eqna...
...r \frac{\partial \phi}{\partial r} \right)
\end{displaymath}
\end{itemize}
}}

[ 例題1 ] 半径$ a$ の球内に電荷が一様に分布

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

$ r>a$ のとき

$\displaystyle \nabla^2\phi = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial\phi}{\partial r} \right) = 0
$


$\displaystyle r^2 \frac{\partial\phi}{\partial r}$ $\displaystyle =$ $\displaystyle C_1$  
$\displaystyle \frac{\partial\phi}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{C_1}{r^2}$  
$\displaystyle \phi$ $\displaystyle =$ $\displaystyle -\frac{C_1}{r} + C_2$  

$\displaystyle r \rightarrow \infty で \phi = 0 より C_2 = 0
$

$ r<a$ のとき

$\displaystyle \nabla^2\phi = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial\phi}{\partial r} \right) = -\frac{\rho}{\epsilon_0}
$


$\displaystyle \frac{\partial}{\partial r} \left( r^2 \frac{\partial\phi}{\partial r} \right)$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_0} r^2$  
$\displaystyle r^2 \frac{\partial\phi}{\partial r}$ $\displaystyle =$ $\displaystyle -\frac{\rho}{3\epsilon_0} r^3 + C_3$  
$\displaystyle \frac{\partial\phi}{\partial r}$ $\displaystyle =$ $\displaystyle -\frac{\rho}{3\epsilon_0} r + \frac{C_3}{r^2}$  
$\displaystyle \phi$ $\displaystyle =$ $\displaystyle -\frac{\rho}{6\epsilon_0} r^2 - \frac{C_3}{r} + C_4$  

$\displaystyle r = 0 で \phi 有限より C_3 = 0
$

$ r=a$ $\displaystyle \frac{\partial\phi}{\partial r}$ の連続

$\displaystyle \left. \frac{\partial\phi}{\partial r} \right\vert _a = \frac{C_1}{a^2} = -\frac{\rho a}{3\epsilon_0} より
C_1 = -\frac{\rho a^3}{3\epsilon_0}
$

$ r=a$$ \phi$ の連続

$\displaystyle \phi(a) = \frac{\rho a^2}{3\epsilon_0} = \frac{\rho a^2}{6\epsilon_0} + C_4 より
C_4 = \frac{\rho a^2}{2\epsilon_0}
$

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item 鏡像法
\begin{center}
\includeg...
...= 0$となることから \\
鏡像電荷$Q'$を考える
}
\end{center}
\end{itemize}
}}

[ 例題2 ] 曲率$ R$ の導体表面に対応する鏡像電荷

\includegraphics[width=4cm]{concave.eps}

$\displaystyle \phi = \frac{Q}{4\pi\epsilon_0} \frac{1}{\sqrt{R^2+(R-a)^2-2R(R-a...
...frac{Q'}{4\pi\epsilon_0} \frac{1}{\sqrt{R^2+(R+a')^2-2R(R+a')\cos\theta}} = 0
$

$ \theta = 0$ のとき

$\displaystyle \frac{Q}{a} = -\frac{Q'}{a'}
$

$ \theta = \pi$ のとき

$\displaystyle \frac{Q}{2R-a} = -\frac{Q'}{2R+a'} = \frac{a'Q}{2R+a'}
$

これから

$\displaystyle 2Ra + aa' = 2Ra' - aa'
$

\begin{displaymath}
\left\{
\begin{array}{ccc}
a' = \displaystyle \frac{Ra}{R...
...yle \frac{RQ}{R-a} & & Q' \rightarrow Q
\end{array}
\right.
\end{displaymath}


next up previous
Next: 静電エネルギー Up: denjiki Previous: Gaussの法則
Keiichi Takasugi
平成24年1月25日