The electric displacement vector field and Gauss' law?

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SUMMARY

The electric displacement vector field \( D \) is defined by the equation \(\oint D \cdot dS = \sum Q_{c}\), which relates the field to the free charge enclosed by a Gaussian surface. In cases of spherical symmetry, the displacement vector field is radial due to the uniform distribution of charge. This radial nature arises from the symmetry of the charge distribution, confirming that the direction of the displacement field is inherently radial in such scenarios.

PREREQUISITES
  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric displacement vector field concepts
  • Knowledge of spherical symmetry in charge distributions
  • Basic principles of vector calculus
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  • Study the implications of Gauss' Law for different geometries
  • Explore the relationship between electric fields and displacement fields
  • Investigate applications of electric displacement in dielectric materials
  • Learn about the mathematical derivation of the electric displacement vector
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Hi,

I know that for the electric displacement vector field \oint D.dS=\sum Q_{c} 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.

Thanks.
 
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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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