Linear Charge Density and Divergence: Finding the Variation of D

[dcfan]

Homework Statement

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

So we know that for R=0, p_v = 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) = p_v

For a sphere: divergence(D) = 1/R^2 d/dR(R^2*A_R) + 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:

p_v(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.

 

[hikaru1221]

Why don't you try the integral form of Gauss law? Notice the spherical symmetry of the problem.

 

[dcfan]

I don't see how it can help me more to solve the problem...

 

[hikaru1221]

Gauss law: \oint_S\vec{D}d\vec{A} = \int\int\int_V\rho dV

You know the variation rule of \rho 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, |\vec{D}| 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.

 

[dcfan]

thx very much Hikaru!