Capacitance with a Geiger Tube

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Homework Statement


The radius and the length of the central wire in a Geiger tube are 0.180 mm and 10.0 cm, respectively. The outer surface of the tube is a conducting cylindrical shell that has an inner radius of 1.50 cm. The shell is coaxial with the wire and has the same length (10.0 cm). The tube is filled with a gas that has a dielectric constant of 1.08 and a dielectric strength of 1.00*10^6 V/m.

(a) What is the maximum potential difference that can be maintained between the wire and shell?


(b) What is the maximum charge per unit length on the wire?



Homework Equations


C/L=2[tex]\pi\epsilon[/tex]/ln[tex]a/b[/tex]

The Attempt at a Solution


I am completely befuddled by this. I'm not sure where to start even, I'd just like some one to please point me towards the start. I have no trouble working with single cylinders or with parallel plates but this is just beyond confusing to me. Advice? Thank you in advanced.

edit:whoa...that equation came out ugly...
0220.png
hopefully that works
 
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ideasrule said:
Yup, that's the right answer.

Sarcasm is not appreciated, I'm not asking for someone to do it for me...just some guidance.
 
I really wasn't trying to be sarcastic. You said:

1st2fall said:
edit:whoa...that equation came out ugly...
0220.png
hopefully that works

That IS the right answer for the capacitance. You couldn't have gotten that without answering at least part of (b) correctly, so I assumed you didn't need any more help. Now I know that you probably found that equation in the textbook and didn't derive it, but that wasn't clear from your first post.

For (b), if you take a cylindrical Gaussian surface surrounding the line of charge, you can calculate E. If the maximum E equals the dielectric breakdown value, that's when the charge held by the capacitor is the maximum possible.

For (a), you have to first derive a formula for potential difference. You already have an equation relating E and r from part (b), so integrate E*dr from a to b and you'll get the equation.