Linear Charge Density and Divergence: Finding the Variation of D

In summary, the charge density increases linearly with distance from the origin and can be represented as pv=5R. To find the corresponding variation of D, we can use the integral form of Gauss law and take advantage of the spherical symmetry of the problem. By choosing the Gauss surface as a sphere with the center at the origin, we can easily evaluate the right hand side of the equation and determine the variation of D.
  • #1
dcfan
5
0

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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  • #2
Why don't you try the integral form of Gauss law? :wink: Notice the spherical symmetry of the problem.
 
  • #3
I don't see how it can help me more to solve the problem...
 
  • #4
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.
 
  • #5
thx very much Hikaru!
 

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