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

In summary, the book says that ∇.D=ρf (free charge density) and ∇xD=0 (I'm fine with that). It goes on to say that D can be found from the free charge just as though the dielectric were not there so D=ε0Evac (where Evac is the field the same free charge distribution would produce in the absence of any dielectric). Then it goes on to say that in such a medium, the vacuum field is reduced by a factor of the relative permittivity, which I'm fine with.
  • #1
fayled
177
0

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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  • #2
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].
 
  • #3
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.
 

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

1. What is Gauss's Law for electric displacement?

Gauss's Law for electric displacement states that the electric displacement (D) through a closed surface is equal to the total charge (Q) enclosed by that surface.

2. How do you solve Gauss's Law for electric displacement?

To solve Gauss's Law for electric displacement, you need to use the formula D = Q/ε0, where ε0 is the permittivity of free space. This formula can be used to calculate the electric displacement for different shapes, such as a sphere with polarization kr.

3. What is the equation for electric displacement for a sphere with polarization kr?

The equation for electric displacement for a sphere with polarization kr is D = (3Q/4πr3)(1 + kr3/3ε0), where Q is the total charge on the sphere, r is the radius of the sphere, and k is the polarization constant.

4. How does polarization affect the electric displacement of a sphere?

Polarization affects the electric displacement of a sphere by increasing it. This is because polarization creates an electric dipole moment within the sphere, which contributes to the total electric displacement.

5. How is electric displacement different from electric field?

Electric displacement (D) and electric field (E) are related but different concepts. While electric field represents the strength and direction of the electric force on a unit charge, electric displacement includes the contribution of both free and bound charges in a medium. Additionally, electric displacement is a vector quantity, while electric field is a scalar quantity.

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