Linear Charge Density and Divergence: Finding the Variation of D

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Homework Help Overview

The discussion revolves around a problem involving linear charge density and the application of Gauss's law to find the variation of the electric displacement field D. The charge density is defined to increase linearly from zero at the origin to a specified value at a distance R.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between charge density and the electric displacement field, with one participant attempting to express the charge density as a function of distance. Questions arise regarding how to derive the variation of D from the divergence of D.

Discussion Status

Some participants suggest using the integral form of Gauss's law, noting the spherical symmetry of the problem, while others express uncertainty about how this approach may aid in solving the problem. There is no explicit consensus on the best method to proceed.

Contextual Notes

Participants are working under the constraints of the problem's setup, including the linear increase of charge density and the spherical symmetry, which may influence their reasoning and approaches.

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Homework Statement



If the charge density increases linearly with distance from origin such that pv=0 at the origin and pv=10 C/m^3 at R=2, find the corresponding variation of D.

So we know that for R=0, pv = 0 and for R=2, pv=10 C/m^3.
We know that the charge density increases linearly from origin

Homework Equations



Gauss law: divergence(D) = pv

For a sphere: divergence(D) = 1/R^2 d/dR(R^2*AR) + 1/(R*sin(teta)) d/d(teta) (A(teta)*sin(teta)) + 1/(R*sin(teta))*dA(phi)/d(phi)

The Attempt at a Solution



I found the relationship between sigma and R (because it is linear) and it is:

pv(R) = 5R

Then after I don't know how to find the variation of D. How can I get a variation of D from a divergence? I made some research on the internet and found nothing clear enough for me to understand.
 
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Why don't you try the integral form of Gauss law? :wink: Notice the spherical symmetry of the problem.
 
I don't see how it can help me more to solve the problem...
 
Gauss law: [tex]\oint_S\vec{D}d\vec{A} = \int\int\int_V\rho dV[/tex]

You know the variation rule of [tex]\rho[/tex] in space, so the right hand side can be easily evaluated for any chosen volume V enclosed by a closed surface S. Plus that the system has spherical symmetry about the origin, [tex]|\vec{D}|[/tex] is the same at every point which has the same distance to the origin. Therefore you can choose the Gauss surface S as a sphere whose center is the origin.
 
thx very much Hikaru!
 

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