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Electrostatics

For a static electric field Faraday's law becomes
\begin{displaymath}
\nabla \times {\bf E} = 0,
\end{displaymath} (5.4)

where
\begin{displaymath}
\nabla \cdot \left (\epsilon {\bf E}\right) = \rho_{c}.
\end{displaymath} (5.5)

(5.5) reduces to (3.4) when $\epsilon$ is a constant. (5.4) is satisfied identically when ${\bf E} = - \nabla F$. Then, when there is no charge density ($\rho_{c}=0$) and $\epsilon$ is constant, (5.5) becomes
\begin{displaymath}
\nabla^{2}F = 0.
\end{displaymath} (5.6)

Hence, we must again solve Laplace's equation.



Prof. Alan Hood
2000-11-06