Electrostatics

For a static electric field Faraday's law becomes

$$\nabla \times {\bf E} = 0,$$ (5.4)

where

$$\nabla \cdot \left (\epsilon {\bf E}\right) = \rho_{c}.$$ (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

$$\nabla^{2}F = 0.$$ (5.6)

Hence, we must again solve Laplace's equation.