# Cylindrical capacitor

1. Dec 4, 2006

### phalanx123

A capacitor consists of two thin infinite concentric cylinderss with inner and outer radii a and b. If the outer cylinder has radius 10mm and the breakdown electric field of air is 3V per micrometre, what radius should be choosen for the inner cylinder in order to maximum the potential diffence V and what is that maximum.

I am really stuck on getting the radius of the inner cylinder. I know that the maximum electric field strength on the surface of the inner cylinder is 3V per micrometre. And I calculated the electric field strength on the surface of the inner cylinder E is E=λ /(2πεa)=3V per micrometre. where λ is the charge density per unit length. but I don't know this value, so I couldn't get the value a .

could somebody help me please? many thanks

2. Dec 4, 2006

### Staff: Mentor

What is the equation for the electric field between the inner and outer plates of a cylindrical capacitor? The E field is not independent of radius r.

3. Dec 4, 2006

### phalanx123

The E field between the two cylinder would be

E=λ /(2πεr)

4. Dec 4, 2006

### Staff: Mentor

So where in the capacitor is the E field maximum? (and I don't know the answer to this last part...) Will the capacitor break down when the 3V/um happens at the highest E field area or at the lowest E field area. I think I know the answer, but I'm not sure.

5. Dec 4, 2006

### phalanx123

E field maximum will be on the surface of the inner cylinder. And I think the capacitor will break down when this happens on the lowest field area which is the surface of the outer cylinder I guess. But I don't know what this is leading to ?

6. Dec 4, 2006

### Staff: Mentor

I think you may be able to cancel out the lamda in the optimization part of this. Consider if you make a almost as big as b, then the separation is very small, and arcover is more likely. And if you make a very small, there is a very high E field close to the tiny inner shell, which would make the initial breakdown more likely. (Based on how the problem is stated, I'm guessing that as soon as the breakdown starts near the inner conductor, the full arc-over will occur.)

So there is some radius b<a<0 that gives the highest overall arcover voltage. See if you can set up the right thing to differentiate and set it equal to zero. Hopefully you can get the charge density out of the equation....

7. Dec 5, 2006

### phalanx123

thank you very much for the help, I'll try to solve it that way. But you said for some radius b<a<0, that's a negative value, it is suppose to mean b>r>a where r is the optimized radius?

Last edited: Dec 5, 2006
8. Dec 5, 2006

### phalanx123

Sorry one more question what is a arcover? I've never heard it before.

9. Dec 5, 2006

### Staff: Mentor

Oopsies. Sorry for my typo. I meant 0<a<b. a is the inner cylinder radius and b is the outer cylinder radius.

Arcover is a term for when you get the electric breakdown in the air/gas due to ionization happening. That's what this problem is asking about. BTW, there is also a strong effect of the gas pressure on the arcover voltage. There is a set of "Paschen" curves that show the arcover voltage versus pressure.