Suppose the coaxial cable is enlarged to twice the overall diameter with two possibilities: (1) the ratio b/a is held fixed, or (2) the difference b - a is held fixed. For which possibility does the leakage current between the inner and outer conductors increase when the voltage is applied between

Question

Suppose the coaxial cable is enlarged to twice the overall diameter with two possibilities: (1) the ratio b/a is held fixed, or (2) the difference b - a is held fixed. For which possibility does the leakage current between the inner and outer conductors increase when the voltage is applied between the two conductors?

Accepted Answer

In order for the current to increase, the resistance must decrease. For possibility (1), in which b/a is held fixed, Equation (1) tells us that the resistance is unaffected. For possibility (2), we do not have an equation involving the difference b - a to inspect. Looking at Figure 27.8b, however, we see that increasing b and a while holding the voltage constant results in charge flowing through the same thickness of silicon but through a larger overall area perpendicular to the flow. This larger area will result in lower resistance and a higher current.

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

Question 27.WI.4: Suppose the coaxial cable is enlarged to twice the overall d...

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