Electric field lines in a cylinder

[jaumzaum]

In the figure below we have 2 hollow cylinder conductors. The red conductor has radius R1 and negative charge (Q1). We then introduce the blue conductor (radius R2), first connected to the Earth by a conductor thread. Then we disconnect the thread making the blue conductor charge be positive. The length of the cylinders are L (L>>R).
a) What will be the relation between the charges in the 2 cylinders
b) Calculate the electric field inside the blue cylinder and draw the field lines

Final Configuration:

http://img803.imageshack.us/img803/5536/31756482.png

Actually I haven't seen this problem anywhere, I only need it to understand a MIT experience I saw on youtube. I would say for sure the electric field lines inside the blue conductor will be all parallel each other, so the electric fieldbecomes uniform. But then I apply Gauss Law to the red conductor.
2πRL E = Q1/ε
E = Q1/2πRLε

And E depends on R, so it's not uniform! If it's so, how would be the field lines configuration? And the relations of charges? I'm pretty stuck in this.
Any help will be appreciated.

Thanks
John

 

[mfb]

The field cannot be uniform. It has to have a cylindrical symmetry: all field lines point in radial direction.

 

[jaumzaum]

You mean in the cylinder axe? So the electric field in the center would be none? Why is that?
And how can I calculate the e field in function of the height h?

 

[mfb]

You mean in the cylinder axe?

No.

So the electric field in the center would be none? Why is that?

The electric field in the interior of the red cylinder? That is zero. Everything else would violate the laws of electromagnetism.

And how can I calculate the e field in function of the height h?

It does not depend on the height. Do you mean r? With Gauß law, as you did.

 

[jaumzaum]

I thought by radial you mean in the axe direction. So I'm right, E = Q1/2πRLε.
How can I draw the lines?

 

[mfb]

How can I draw the lines?

Draw a disk as part of the cylinder, and find the direction of the electric field there. It is the easiest possible field.