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Homework Help: Nabla operator

  1. Dec 6, 2008 #1
    Does

    [tex]\vec{\nabla} \cdot \vec{E} = 0[/tex]

    imply


    [tex]\vec{\nabla}^2 \cdot \vec{E} = 0[/tex]

    ?

    Is this true:

    [tex]\vec{\nabla}^2 \cdot \vec{E} = \vec{\nabla}(\vec{\nabla} \cdot \vec{E})[/tex]
     
    Last edited: Dec 6, 2008
  2. jcsd
  3. Dec 6, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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)).
     
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