Question

A Geiger counter detects radiation such as alpha particles by using the fact That the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it (Fig. P23.62). A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible “click.” Suppose the radius of the central wire is 145 $\mathit{\mu }\mathit{m}$and the radius of the hollow cylinder is 1.80 cm. What potential difference between the wire and the cylinder produces an electric field of at a distance of 1.20 cm from the axis of the wire? (The wire and cylinder are both very long in comparison to their radii, so the results of Problem 23.61 apply.)

Short answer

The potential difference is$v_{ab}=1.157\times {10}^{3}V$

Step-by-step solution

Step 1:

Given data:

$$\begin{gathered} b=1.8cm \\ a=145\mu C \\ E=2.00\times {10}^{4}V/m \\ r=1.2cm \end{gathered}$$

Step 2:

Therefore, the electric field at any point is;

$E\left(r\right)=\frac{v_{ab}}{r\ln \left(b/a\right)}$

Putting all the values,

$$\begin{gathered} 2.00\times {10}^4=\frac{v_{ab}}{\left(1.2\times {10}^{-2}\right)\ln \left({\displaystyle \frac{1.8\times {10}^{-2}}{145\times {10}^{-6}}}\right)} \\ {v}_{ab}=1.157\times {10}^{3}V \end{gathered}$$

Hence, The potential difference is $v_{ab}=1.157\times {10}^{3}V$.