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Chapter 26: Q. 66 (page 740)

High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE P26.66. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius $R_{1}$ while the outer conductor of radius $R_{2}$ is grounded (i.e., at $V = 0 V$ ). An insulating material fills the space between them, and an insulating plastic coating goes around the outside.
a. Find an expression for the capacitance per meter of a coaxial cable. Assume that the insulating material between the cylinders is air.
b. Evaluate the capacitance per meter of a cable having $R_{1} = 0.50 m m$ and $R_{2} = 3.0 m m .$

Short Answer

Expert verified

a. The expression for the capacitance per meter of a coaxial cable is $\frac{2 蟺蔚_{0}}{\ln (\frac{R_{2}}{R_{1}})} .$

b. The capacitance of a cable is $31 pF / m .$

Step by step solution

01

Given Information (Part a)

The signal is carried on a wire of radius $R_{1} a n d R_{2}$

Voltage $= 0 V$

02

Explanation (Part a)

a. Applying Gauss's law for the cylinder with linear charge density $位$ , to get electric field outside of a cylinder as

$E 路 2 蟺 r L = \frac{Q}{蔚_{0}} 鈬� E (r) = \frac{Q}{2 蟺蔚_{0} L r}$

Potential difference for two cylinders will be

$螖 V = {鈭�}_{R_{1}}^{R_{2}} E (r) d r = \frac{Q}{2 蟺蔚_{0} L} {鈭�}_{R_{1}}^{R_{2}} \frac{d r}{r} = \frac{Q}{2 蟺蔚_{0} L} \ln (\frac{R_{2}}{R_{1}})$

Find the capacitance per unit length as

$\frac{C}{L} = \frac{Q}{L 螖 V} = \frac{2 蟺蔚_{0}}{\ln (\frac{R_{2}}{R_{1}})}$

Hence, the expression for the capacitance per meter of a coaxial cable is $= \frac{2 蟺蔚_{0}}{\ln 鈦� (\frac{R_{2}}{R_{1}})}$

03

Given Information (Part a)

The signal is carried on a wire of radius $R_{1}$ and $R_{2}$

$Voltage = 0 V$

04

Explanation (Part b)

In the given numerical case, we will have

$\frac{C}{L} = \frac{2 蟺蔚_{0}}{\ln (\frac{R_{2}}{R_{1}})}$

Substitute the values,

localid="1648629781694" $\frac{C}{L} = (\frac{2 蟺脳 8.85 脳 10^{- 12}}{\ln (\frac{3 m m}{0.5 m m})})$

Multiply the expression,

$= 3.1 路 10^{- 11} F / m$

Hence, the capacitance of a cable is $31 p F / m$

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Most popular questions from this chapter

The metal spheres are charged to $\begin{matrix} + \\ - \end{matrix} 300$ . Draw this figure on your paper, then draw a plausible contour map of the potential, showing and labeling the equipotential surfaces.

The electric potential in a region of space is V=200/2x2+y2,where x and y are in meters. What are the strength and direction of the electric field at 1x, y2=12.0 m, 1.0 m2? Give the direction as an angle cw or ccw (specify which) from the positive x-axis

The electric potential along the $x - a x i s is V = 100 x^{2}$ V, where $x$ is in meters. What is $E_{x}$ at

(a) $x = 0 m$ and

(b) $x = 1 m$ ?

The two metal spheres in Figure Q26.9 are connected by a metal wire with a switch in the middle. Initially the switch is open. Sphere 1, with the larger radius, is given a positive charge. Sphere 2, with the smaller radius, is neutral. Then the switch is closed. Afterward, sphere 1 has charge $Q_{1}$ , is at potential $V_{1}$ , and the electric field strength at its surface is $E_{1}$ . The values for sphere 2 are $Q_{2}, V_{2}, and E_{2}$ .

a. Is $V_{1}$ larger than, smaller than, or equal to $V_{2}$ ? Explain.

b. Is $Q_{1}$ larger than, smaller than, or equal to $Q_{2}$ ? Explain.

c. Is $E_{1}$ larger than, smaller than, or equal to $E_{2}$ ? Explain.

The parallel-plate capacitor in Figure Q26.11 is connected to a battery having potential difference $鈭� V_{b a t}$ . Without breaking any of the connections, insulating handles are used to increase the plate separation to $2 d$ .

a. Does the potential difference $鈭� V_{C}$ change as the separation increases? If so, by what factor? If not, why not?

b. Does the capacitance change? If so, by what factor? If not, why not?

c. Does the capacitor charge $Q$ change? If so, by what factor? If not, why not?

See all solutions

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