電位移

$\mathbf{D}\ \stackrel{\mathrm{def}}{=}\ \epsilon_{0} \mathbf{E} + \mathbf{P}$

概述

$\nabla\cdot\mathbf{E} = \rho_{total}/\epsilon_0$

$\nabla\cdot\mathbf{P} = - \rho_{bound}$

$\rho_{total} =\rho_{free}+\rho_{bound}$

$\nabla\cdot\mathbf{D}=\rho_{free}$

$\nabla \times \mathbf{D} = \varepsilon_{0}(\nabla \times \mathbf{E}) + (\nabla \times \mathbf{P})$

$\nabla \times \mathbf{D} = \nabla \times \mathbf{P}$

線性電介質

「線性電介質」，對於外電場的施加，會產生線性響應。例如，鐵電材料是非線性電介質。假設線性電介質具有各向同性，則其電場與電極化強度的關係式為[2]

$\mathbf{P}=\chi_e \epsilon_{0} \mathbf{E}$

$\mathbf{D}= (1+\chi_e) \epsilon_0\mathbf{E}=\epsilon\mathbf{E}$

$\nabla\cdot(\epsilon\mathbf{E})=\rho_{free}$

$\nabla\cdot\mathbf{E}=\rho_{free}/\epsilon$

$\epsilon_r\ \stackrel{\mathrm{def}}{=}\ \epsilon/\epsilon_0$

$\epsilon_r=1+\chi_e$

應用範例

$\oint_{\mathbb{S}} \mathbf{D}_+ \cdot d\mathbf{a} = Q$

$2D_+ A= Q$ ;

$D_+=Q/2A$

$D_- =Q/2A$

$E=D/\epsilon=Q/\epsilon A$

$V=Ed=Q d/\epsilon A$

$C=Q/V=\epsilon A/d$

參考文獻

1. ^ Griffiths, David J., Introduction to Electrodynamics (3rd ed.), Prentice Hall, pp. 175, 179–184, 1998, ISBN 0-13-805326-X
2. ^ Jackson, John David, Classical Electrodynamic 3rd., USA: John Wiley & Sons, Inc., pp. 151–154, 1999, ISBN 978-0-471-30932-1