Understanding the Nabla Operator and Electric Field: A Comprehensive Guide

[leopard]

Does

$$\vec{\nabla} \cdot \vec{E} = 0$$

imply $$\vec{\nabla}^2 \cdot \vec{E} = 0$$

?

Is this true:

$$\vec{\nabla}^2 \cdot \vec{E} = \vec{\nabla}(\vec{\nabla} \cdot \vec{E})$$

 

[Dick]

No. If you work out grad(div(E)) you are going to get other terms as well. The correct relation is laplacian(E)=grad(div(E))-curl(curl(E)).