Solve Gauss's Law for D: Electric Displacement of Sphere with Polarization kr

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SUMMARY

The discussion centers on solving Gauss's Law for the electric displacement field \( D \) in a sphere with polarization \( k_r \). It is established that the electric displacement \( D \) is zero everywhere due to the absence of free charge (\( q_{free} = 0 \)). The proof utilizes spherical symmetry, leading to the conclusion that \( D \) must be zero as the integral of \( D \) over a closed surface results in zero flux. Additionally, the discussion touches on the relationship between \( D \) and the electric field \( E \) in homogeneous linear dielectrics, emphasizing that \( D \) can be derived from free charge distributions as if no dielectric were present.

PREREQUISITES
  • Understanding of Gauss's Law and its mathematical formulation.
  • Familiarity with electric displacement field \( D \) and its relation to free charge density \( \rho_f \).
  • Knowledge of spherical symmetry in electrostatics.
  • Concepts of polarization in dielectrics and their effects on electric fields.
NEXT STEPS
  • Study the derivation of Gauss's Law in electrostatics.
  • Learn about the properties and applications of electric displacement fields in dielectrics.
  • Explore the mathematical implications of spherical symmetry in electrostatic problems.
  • Investigate the relationship between electric field \( E \) and electric displacement \( D \) in various dielectric materials.
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Students of electromagnetism, physicists, and electrical engineers seeking to deepen their understanding of electric displacement fields and their applications in dielectric materials.

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Homework Statement


We have a sphere with a polarization kr. I need to show that the electric displacement D=0 everywhere.

Homework Equations


closed surfaceD.dS=qfree

The Attempt at a Solution


qfree=0 everywhere so the flux of D is zero everywhere. Clearly D=0 everywhere does solve this, but so could possibly many other things - how do I show D=0 is the solution? This is a very niggly and annoying to think about! Thanks for any help.
 
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You only need to consider the spherical symmetry. Because of that, D is radial and can only depend on r and because you're integrating on a sphere, you're not integrating w.r.t. r and so the integrand is a constant.So we have [itex]\int D \hat{r}\cdot dS\hat r=0 \Rightarrow D\int dS=0 \Rightarrow D 4 \pi R^2=0 \Rightarrow D=0[/itex].
 
Shyan said:
You only need to consider the spherical symmetry. Because of that, D is radial and can only depend on r and because you're integrating on a sphere, you're not integrating w.r.t. r and so the integrand is a constant.So we have [itex]\int D \hat{r}\cdot dS\hat r=0 \Rightarrow D\int dS=0 \Rightarrow D 4 \pi R^2=0 \Rightarrow D=0[/itex].

Ah that was a bit silly of me, thanks.

Another question regarding the D field. My book says that in a homogenous linear dielectric, .Df (free charge density) and xD=0 (I'm fine with that). Then it says D can be found from the free charge just as though the dielectric were not there so D0Evac (where Evac is the field the same free charge distribution would produce in the absence of any dielectric). Then it goes on from here to prove that in such a medium, the vacuum field is reduced by a factor of the relative permittivity, which I'm fine with. I really don't get the reasoning behind the jump from the divergence and curl to D being found as though no dielectric were there. It sort of comes after a discussion about the parallel between E and D.
 

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