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Figure 23-61 shows a Geiger counter, a device used to detect ionizing radiation, which causes ionization of atoms. A thin, positively charged central wire is surrounded by a concentric, circular, conducting cylindrical shell with an equal negative charge, creating a strong radial electric field. The shell contains a low-pressure inert gas. A particle of radiation entering the device through the shell wall ionizes a few of the gas atoms. The resulting free electrons (e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting 鈥渁valanche鈥� of electrons is collected by the wire, generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is 25 mm, the inner radius of the shell 1.4 cm, and the length of the shell 16 cm. If the electric field at the shell鈥檚 inner wall is,$\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{脳}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$ what is the total positive charge on the central wire?

Step by step solution

01

The given data

1. The radius of the central wire is $R=25\mathrm{渭尘}$, the inner radius of the shell, ${\mathrm{r}}_{\mathrm{in}}=1.4\mathrm{cm}$and the length of the shell t = 16 cm.
2. The electric field at the shell鈥檚 inner wall is${\mathrm{E}}_{\mathrm{in}}=2.9脳{10}^4\mathrm{N}/\mathrm{C}$

02

Understanding the concept of the electric flux

Using the concept of the Gauss flux theorem, we can get the required value of the positive charge enclosed q within the given volume of the Gaussian cylindrical surface.

Formula:

The electric flux distribution within an enclosed surface, $蠒=EA=q/{蔚}_0$ (i)

03

Calculation of the total positive charge on the central wire

The electric field is radially outward from the central wire.

Since the magnitude of the field at the cylinder wall is known, we take the Gaussian surface to coincide with the wall. Thus, the Gaussian surface is a cylinder with radius R and length L, coaxial with the wire.

Only the charge on the wire is actually enclosed by the Gaussian surface; we denote it by q. The area of the Gaussian cylindrical surface is given as:

$A=2蟺RL$,

Now, the flux through it is given using equation (i) as:

$蠒=2\mathrm{蟺搁尝贰}$

We assume there is no flux through the ends of the cylinder, so this is the total flux. Thus, substituting the given values in equation (i) is given as:

$$\begin{gathered} q=2\mathrm{蟺}(8.85脳{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)(0.014\mathrm{m})(0.16\mathrm{m})(2.9脳{10}^4\mathrm{N}/\mathrm{C}) \\ =3.6脳{10}^{-9}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $3.6脳{10}^{-9}\mathrm{C}$.

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