Linear Charge Density and Divergence: Finding the Variation of D

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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!