Cylindrical Charge distribution with dielectric shell

[nich]

Homework Statement

A cylindrical distribution of charge ρ = α/sqrt(r) where α = 2 µC/m^(5/2) extends from 0 cm to 9.3 cm (has radius 9.3 cm). Concentric with this is a dielectric shell with k = 5.44 of inner radius 16.6 cm and outer radius 24.9 cm. What is the electric field at 3.53 cm, 12.6 cm, 21.4 cm, and 33 cm? Answer in units of V/m.

What is the surface charge density on the inner surface of the dielectric?

Homework Equations

p = α/sqrt(r)
E dA = Integral( q(enclosed)/Eps )

The Attempt at a Solution

I already got the Electric Field at 3.53 cm by integrating through the square root of the radius and multiplying by the charge density (to get 28293 V/m), but at these other points you have to account for the dielectric and I don't know how to treat the cylindrical charge distribution outside the cylinder... anyone have pointers?

 

[ehild]

The charge is confined in the cylinder of radius 9.3 cm. If r > 9.3 cm you have to integrate up that radius, to get the charge enclosed by the Gaussian surface.

 

[rude man]

Recall that Gauss says ∫D⋅ds over a closed surface s = free q inside that surface, and that D = εE.