Gaussian cylinder in the finite case

[hatsoff]

Homework Statement

Consider two long coaxial metal cylindrical tubes, with radii a and b and length L. (You may assume a,b<<L. Also a<b.) Suppose the inner cylinder is given a charge +Q and the outer cylinder a charge -Q.

Using Gauss' Law, compute the electric field for all r between a and b.

Homework Equations

Gauss' Law, the usual Gaussian surface stuff.

The Attempt at a Solution

As the textbook recommends, I put my Gaussian cylinder far from the ends of the tube. So for a<r<b the enclosed charge is

$$Q_{\text{enc}}=\frac{Q\lambda}{L},$$

where lambda << L is the length of the Gaussian cylinder. So by Gauss' law,

$$2\pi r\lambda|\textbf{E}|\approx\oint_{\text{surface}} \textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}= \frac{Q\lambda}{L\epsilon_0},$$

(What about the ends?) Therefore in cylindrical coordinates

$$\textbf{E}=\frac{Q}{2\pi L\epsilon_0 r}\boldsymbol{\hat r}$$

My questions:

(1) Why do we get to ignore the ends? Or do we get to ignore the ends?

(2) I'm not really sure what the restrictions are for the Gaussian cylinder when the case is finite as opposed to infinite. If we don't put it at the very center of the wire(s)/tube(s)/whatever, then doesn't that ruin symmetry? And even if we just put it far from the edges, then it seems like we can't say anything about those edges, i.e. we can only calculate the electric field at points far from the edges of the wire(s)/tube(s).

I guess I just don't understand the physical interpretation behind the math, and it's giving me trouble. Any help would be much appreciated!

 

[tiny-tim]

hi hatsoff!

Consider two long …
…
(1) Why do we get to ignore the ends? Or do we get to ignore the ends?

in an exam question, "long" always means that you can treat it as infinite

(2) I'm not really sure what the restrictions are for the Gaussian cylinder when the case is finite as opposed to infinite. If we don't put it at the very center of the wire(s)/tube(s)/whatever, then doesn't that ruin symmetry?

you won't be asked about that

but yes, you're right … the field lines will be bent slightly out of shape, and difficult to calculate ​