Solving Electric Field & Potential for Charged Cylinders

[fluidistic]

Homework Statement

2 long cylinders with radius a and b (a<b) have a charge density by unit of length worth - \lambda and \lambda respectively.
Use Gauss's law to find the electric field in every point of the space.
Find the potential in all points of the space, assuming that the potential is worth V_b over the cylinder with radius b.



2. The attempt at a solution
I believe that the electric field inside both cylinders is null.
For any point r such that a<r<b, I believe that only the cylinder with radius a contributes to the electric field. I get that E=-\frac{2 \lambda k}{r} using Gauss's law.

And outside both cylinder, E is determined by both cylinders, so E=-\frac{2 \lambda k}{b+R}+\frac{2 \lambda k}{R}

I know I'm completely wrong. I remember a helper at university saying that outside both cylinders, the electric field is null. He also said that the situation is not the one of a capacitor... I don't know why.
I really need help... Thanks in advance!

 

[mukundpa]

What is R and b + R ?

 

[fluidistic]

What is R and b + R ?

First : thanks for helping me!
Second : Sorry, I should have mentioned it : R would be the distance from the surface of the biggest cylinder to any point outside both cylinders.
b is the radius of the biggest cylinder, hence b+R is the distance from the center of both cylinders to any point outside them.

Do you understand what I'm saying? (Sorry for my English)

 

[mukundpa]

But in the formula r is the distance from axis of the charge system. Actually this is the radius of Gaussian cylinder (virtual closed surface) considered coaxially to find field at distance r from axis.

 

[fluidistic]

But in the formula r is the distance from axis of the charge system. Actually this is the radius of Gaussian cylinder (virtual closed surface) considered coaxially to find field at distance r from axis.

Right.
I don't see the inconstancy. R is not r.

 

[mukundpa]

As the distance r of the point (at which we find field outside both cylinders) is taken from the axis of the system it is same for both cylinders

 

[fluidistic]

As the distance r of the point (at which we find field outside both cylinders) is taken from the axis of the system it is same for both cylinders

Ah you're right. I see my error! So indeed the electric field outside both cylinders is 0.
Was I right when I wrote E=-\frac{2 \lambda k}{r} inside the biggest cylinder but outside the smaller one?

Thank you very much for all!