Surface Charge density of glass cyclinder

[Taylor_1989]

Homework Statement

Consider an infinitely long one dimensional conducting wire with a homeogenous charge density ##\lambda##, running along the central axis of an infinitely long cyclindrical glass casing of radius b (glass is a dielectric material). Calculate:

a) The displacement vector inside the glass

b) the surface bound charges on the outer surface of the glass

Homework Equations

a) ##\oint D . dA=q##

b) ##\sigma_b=P.\hat n##
##P=\epsilon_0\chi_e E##

The Attempt at a Solution

[/B]
So my attempt for a) I drew a Gaussian surface in terms of a cylinder and calculate all three surfaces like so:

##\oint D . dA=\int_1 D dAcos\theta+\int_2 D dAcos\theta+\int_3 D dAcos\theta##

By calulating ##\theta ## for the respectable surfaces ##1=90, 2=0, 3=90##

Thus leaving me with ##\int_2 D dA=D\int_2 dA=D\times A=D(L)(2\pi r)=q##
##\lambda=\frac{q}{L}##

##D=\frac{\lambda}{2\pi r}## (electric displacement vector)

b) But I am not sure where to even start I have been reading a one book say that the surface charge cannot be calculate, but dosent really give an explanation and the other basically uses the two formulas which I have mentioned above. So I am a bit confused about the whole thing at the moment a would appreciate any help that anyone could give. I understand the equation I have given for part b are for a linear dielectric but I am just not 100% sure if the question I have been asked is a linear dielectric

 

[TSny]

Part (a) looks good.

For part (b) I think you can assume the glass is a linear dielectric. Otherwise, I don't think you could solve the problem.
Besides the two equations you listed for part (b) you will need one more. There is a fundamental equation that relates D, E, and P.

 

[Taylor_1989]

Are you talking with respect to the diplacment vector in terms of ##D= \epsilon_0 E##

 

[TSny]

Are you talking with respect to the diplacment vector in terms of ##D= \epsilon_0 E##

That's not the particular equation I was thinking of. However, your equation ##D= \epsilon_0 E## could be used instead, if you modify it so that it is valid inside the glass.

 

[Taylor_1989]

@TSny Ok I am slightly confuse how would I modify the equation?

 

[Taylor_1989]

Ok I think I have it the correction that needs to be make is following:

I assume the equation you where talking of is this:

##D=\epsilon_0 E+P## so from this then become ##D=\epsilon_0 E+\epsilon_0 \chi_e E##
By factoring out the ##\epsilon_0## I get ##D=\epsilon_0(1+\chi_e)##

Now ##\epsilon_0(1+\chi_e)=\epsilon## So then ##D=\epsilon E##

 

[TSny]

Ok I think I have it the correction that needs to be make is following:

I assume the equation you where talking of is this:

##D=\epsilon_0 E+P## so from this then become ##D=\epsilon_0 E+\epsilon_0 \chi_e E##
By factoring out the ##\epsilon_0## I get ##D=\epsilon_0(1+\chi_e)##

Now ##\epsilon_0(1+\chi_e)=\epsilon## So then ##D=\epsilon E##

Yes. That looks very good.