Coaxial cables are used extensively for cable television and other electronic applications. A coaxial cable consists of two concentric cylindrical conductors. The region between the conductors is completely filled with silicon, as shown in Figure 27.8a, and current leakage through the silicon,

Question

Coaxial cables are used extensively for cable television and other electronic applications. A coaxial cable consists of two concentric cylindrical conductors. The region between the conductors is completely filled with silicon, as shown in Figure 27.8a, and current leakage through the silicon, in the radial direction, is unwanted. (The cable is designed to conduct current along its length—this is not the current we are considering here.) The radius of the inner conductor is a = 0.500 cm, the radius of the outer one is b = 1.75 cm, and the length is L = 15.0 cm.

Calculate the resistance of the silicon between the two conductors

Accepted Answer

Conceptualize by imagining two currents, as suggested in the text of the problem. The desired current is along the cable, carried within the conductors. The undesired current corresponds to charge leakage through the silicon and its direction is radial. Because we know the resistivity and the geometry of the silicon, we categorize this as a problem in which we find the resistance of the silicon from these parameters, using Equation 27.11. Because the area through which the charges pass depends on the radial position, we must use integral calculus to determine the answer.

To analyze the problem, we divide the silicon into concentric elements of infinitesimal thickness dr (Fig. 27.8b). We start by using the differential form of Equation 27.11, replacing [latex] \ell ext { with } r ext { for the distance variable: } d R= ho d r / A [/latex], where dR is the resistance of an element of silicon of thickness dr and surface area A. In this example, we take as our representative concentric element a hollow silicon cylinder of radius r, thickness dr, and length L, as in Figure 27.8.
Any charge that passes from the inner conductor to the outer one must pass radially through this concentric element, and the area through which this charge passes is [latex] A=2 \pi r L [/latex]. (This is the curved surface area—circumference multiplied by length—of our hollow silicon cylinder of thickness dr.) Hence, we can write the resistance of our hollow cylinder of silicon as

[latex] R= ho \frac{\ell}{A} [/latex] (27.11).

[latex] d R=\frac{ ho}{2 \pi r L} d r [/latex].

Because we wish to know the total resistance across the entire thickness of the silicon, we must integrate this expression from r = a to r = b:

(1) [latex] R=\int_{a}^{b} d R=\frac{ ho}{2 \pi L} \int_{a}^{b} \frac{d r}{r}=\frac{ ho}{2 \pi L} \ln \left(\frac{b}{a} ight) [/latex].

Substituting in the values given, and using [latex] ho=640 \Omega [/latex]. m for silicon, we obtain

[latex] R=\frac{640 \Omega \cdot m }{2 \pi(0.150 m )} \ln \left(\frac{1.75 cm }{0.500 cm } ight)=851 \Omega [/latex].

To finalize this problem, let us compare this resistance to that of the inner conductor of the cable along the 15.0-cm length. Assuming that the conductor is made of copper, we have

[latex] R= ho \frac{\ell}{A}=\left(1.7 imes 10^{-8} \Omega \cdot m ight)\left(\frac{0.150 m }{\pi\left(5.00 imes 10^{-3} m ight)^{2}} ight) [/latex].

[latex] =3.2 imes 10^{-5} \Omega [/latex].

This resistance is much smaller than the radial resistance. As a consequence, almost all of the current corresponds to charge moving along the length of the cable, with a very small fraction leaking in the radial direction.

Question 27.4: Coaxial cables are used extensively for cable television and...

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