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The electric displacement vector field and Gauss' law?

  1. Feb 24, 2014 #1


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    I know that for the electric displacement vector field [itex]\oint D.dS[/itex]=[itex]\sum Q_{c}[/itex] does this mean that I can just use a Gaussian surface to explain why the displacement vector field for a sphere is radial or not without having to talk about the electric field. If not what is the reasoning to explain that the displacement vector field is radial? or is there just a definition for which direction the displacement field always points in.

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  3. Feb 24, 2014 #2

    Meir Achuz

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    There has to be spherical symmetry for the charge distribution. Then symmetry shows that the only direction the D field can have is radial.
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