Question

long thin metal wire with radius $\mathit{r}$and length$\mathit{L}$is surrounded by a concentric long narrow metal tube of radius $\mathit{R}$, where$\mathit{R}\mathbf{>}\mathbf{>}\mathit{L}$, as shown in Figure 16.86. 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 as shown. There is a charge$\mathbf{+}\mathit{Q}$on the inner wire and a charge$\mathbf{-}\mathbf{Q}$on the outer tube. As we will see when we study Gauss’s law in a later chapter, the electric field inside the tube is contributed solely by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. (a) In terms of the charge$\mathbf{Q}$, length$\mathit{L}$, inner radius$\mathit{r}$, and outer radius$\mathit{R}$ , what is the potential difference${\mathbf{V}}_{\mathbf{tube}}\mathbf{-}{\mathbf{V}}_{\mathbf{wire}}$ between the inner wire and the outer tube? Explain, and include checks on your answer. (b) The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. Calculate this power-supply voltage (give a numerical value), and explain your calculation. The length$\mathit{L}\mathbf{=}\mathbf{80}\mathbf{\text{\hspace{0.17em}cm}}$ , the inner radius$\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.7}\mathbf{\text{\hspace{0.17em}mm}}$, and the outer radius$\mathit{R}\mathbf{=}\mathbf{3}\mathbf{\text{\hspace{0.17em}cm}}$. This device is called a “Geiger–Müller tube” and was one of the first electronic particle detectors. The voltage is set just below the threshold for making the air glow near the wire. A charged particle that passes near the center wire can trigger breakdown in the air, leading to a large current that can be easily measured.

Short answer

(a) The potential difference${\mathrm{V}}_{\mathrm{tube}}-{\mathrm{V}}_{\mathrm{wire}}$ is $-\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)$.

(b) The power supplied voltage is $-7891.53\text{\hspace{0.17em}V}$.

Step-by-step solution

Identification of given data

The given data is listed below,

• The charge on the inner wire is,$+Q$
• The charge on the outer wire is,$-Q$
• The length of the wire is,$L=80\text{\hspace{0.17em}cm}\times \frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}}=0.8\text{\hspace{0.17em}m}$
• The inner radius of the wire is,$r=0.7\text{\hspace{0.17em}mm}\times \frac{1\text{\hspace{0.17em}m}}{1000\text{\hspace{0.17em}mm}}=0.7\times {10}^{-3}\text{\hspace{0.17em}m}$
• The outer radius of the wire is,$R=3\text{\hspace{0.17em}cm}\times \frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}}=0.03\text{\hspace{0.17em}m}$

Significance of the electric potential. 

When two charged bodies collide, the charge flows from one conductor to the another. The electric potential is the electric state that governs the transfer of charge from one conductor to another that is in contact.

(a) Determination of the potential difference Vtube−Vwire between the inner wire and the outer tube.

The potential difference should be negative because the field of the inner wire, which is approximately given by,

${\mathrm{E}}_{\mathrm{wire}}=\frac{2\mathrm{kQ}}{\mathrm{Lr}}$

Here, kis the coulomb constant with the value is$9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2$ , Q is the charge on the wire, L is the length of the wire, and r is the inner radius of the wire.

The filed inside with tube is zero so it will not contribute in the potential difference so the final potential difference is given by,

$\begin{array}{c}\Delta V=V_{tube}-V_{wire}\\ V_{tube}-V_{wire}=-\underset{r_1}{\overset{r_2}{\int }}Edr\\ =-{\int }_r^R\frac{2kQ}{Lr}dr\\ =-\frac{2kQ}{L}{\int }_r^R\frac{1}{r}dr\\ =-\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)\end{array}$

Thus, the potential difference${\mathrm{V}}_{\mathrm{tube}}-{\mathrm{V}}_{\mathrm{wire}}$ is $-\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)$.

(b) Determination of the power-supply voltage.

The electric charge on the wire is given by,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{wire}}=\frac{2\mathrm{kQ}}{\mathrm{Lr}}\\ \mathrm{Q}=\frac{\mathrm{rLE}}{2\mathrm{k}}\end{array}$

Here, L is the length of the wire, E is the electric field for air breakdown in inner wire with an assumed value,$3\times {10}^6\text{\hspace{0.17em}N/C}$ , and r is the inner radius of the wire.

Substitute all the values in the above equation.

$\begin{array}{c}Q=\frac{(0.7\times {10}^{-3}\text{\hspace{0.17em}m})\left(0.8\text{\hspace{0.17em}m}\right)(3\times {10}^6\text{\hspace{0.17em}N/C})}{2(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)}\\ =9.35\times {10}^{-8}\text{\hspace{0.17em}C}\end{array}$

The voltage supplied is given by,

$\Delta V=-\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)$

Substitute all the values in the above,

$\begin{array}{c}\Delta V=-\frac{2(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(9.35\times {10}^{-8}\text{\hspace{0.17em}C})}{\left(0.8\text{\hspace{0.17em}m}\right)}\ln \left(\frac{0.03\text{\hspace{0.17em}m}}{0.7\times {10}^{-3}\text{\hspace{0.17em}m}}\right)\\ =-7891.53\text{\hspace{0.17em}N}\cdot \text{m/C}\left(\frac{1\text{\hspace{0.17em}V}}{1\text{\hspace{0.17em}N}\cdot \text{m/C}}\right)\\ =-7891.53\text{\hspace{0.17em}V}\end{array}$

Thus, the power supplied voltage is $-7891.53\text{\hspace{0.17em}V}$.