# Cylindrical Charge distribution with dielectric shell

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1. Feb 27, 2015

### nich

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
p = α/sqrt(r)
E dA = Integral( q(enclosed)/Eps )

3. 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?

2. Mar 4, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Mar 4, 2015

### 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.

4. Mar 5, 2015

### rude man

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