# Charged cylinders

1. Sep 10, 2009

### fluidistic

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

2. Sep 10, 2009

### mukundpa

What is R and b + R ?

3. Sep 10, 2009

### fluidistic

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)

4. Sep 10, 2009

### 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.

5. Sep 10, 2009

### fluidistic

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

6. Sep 10, 2009

### 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

7. Sep 10, 2009

### fluidistic

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!