# Charge needed to ionize the air

1. Feb 8, 2014

### Lee33

1. The problem statement, all variables and given/known data
A Geiger tube contains a long thin metal wire, surrounded by a concentric long narrow metal tube. Insulating spokes hold the wire in the center of the tube and prevent electrical contact between the wire and the tube. A variable power supply is connected to the device. The power supply maintains opposite charges on the wire and the tube.

a. The electric field in the space between the wire and the outer tube is due only to the wire. When the electric field at any location in air reaches 3e6 volts/meter, the air ionizes and becomes a conductor. For the tube, the length L = 80 cm, the inner radius r = 0.7 mm, and the outer radius R = 2.5 cm. How much charge would there need to be on the inner wire in order to ionize the air near the wire (where the field is largest)?

2. Relevant equations

$E = \frac{2K(Q/L)}{r}, \ r<< L$

3. The attempt at a solution

Do I treat the tube as a rod? I am not sure how I can go about doing this problem. If I treat the tube as a rod then I can use the equation given above to find the charge necessary to ionize the air. Is that correct?

2. Feb 8, 2014

### SammyS

Staff Emeritus
Treat the wire as a rod.

3. Feb 8, 2014

### Lee33

Ah, yes, that is what I meant. That was a mistake by me, I meant wire instead of tube. But if I do treat it as a rod then

$E = \frac{2K(Q/L)}{r}$

$Q = \frac{rLE}{2K} = \frac{(0.8m)(3e6v/m)r}{2(9e9)}$, where $v/m$ is volts per meter.

Is that correct? If so, what will be my $r$?

4. Feb 9, 2014

### SammyS

Staff Emeritus
What value of r will give the greatest value for the electric field?

5. Feb 9, 2014

### Lee33

Since the wire has a radius of 0.7 mm then will my r be 0.0007 m?

6. Feb 9, 2014

### SammyS

Staff Emeritus
Yes.

7. Feb 10, 2014

Thank you!