Question

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 $V=0\mathrm{V}$). 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 $R_1=0.50mm$and $R_2=3.0mm.$

Short answer

a. The expression for the capacitance per meter of a coaxial cable is $\frac{2\pi {\epsilon }_0}{\ln \left(\frac{R_2}{R_1}\right)}.$

b. The capacitance of a cable is $31\mathrm{pF}/\mathrm{m}.$

Step-by-step solution

Given Information (Part a)

The signal is carried on a wire of radius $R_{1}and{R}_2$

Voltage$=0V$

Explanation (Part a)

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

$E·2\pi rL=\frac{Q}{{\epsilon }_0}\Rightarrow E\left(r\right)=\frac{Q}{2\pi {\epsilon }_{0}Lr}$

Potential difference for two cylinders will be

$\Delta V={\int }_{R_1}^{R_2}E\left(r\right)dr=\frac{Q}{2\pi {\epsilon }_{0}L}{\int }_{R_1}^{R_2}\frac{dr}{r}=\frac{Q}{2\pi {\epsilon }_{0}L}\ln \left(\frac{R_2}{R_1}\right)$

Find the capacitance per unit length as

$\frac{C}{L}=\frac{Q}{L\Delta V}=\frac{2\pi {\epsilon }_0}{\ln \left(\frac{R_2}{R_1}\right)}$

Hence, the expression for the capacitance per meter of a coaxial cable is$=\frac{2\pi {\epsilon }_0}{\ln \left(\frac{R_2}{R_1}\right)}$

Given Information (Part a)

The signal is carried on a wire of radius $R_1$and $R_2$

$\text{Voltage}=0\mathrm{V}$

Explanation (Part b)

In the given numerical case, we will have

$\frac{C}{L}=\frac{2\pi {\epsilon }_0}{\ln \left(\frac{R_2}{R_1}\right)}$

Substitute the values,

localid="1648629781694" $\frac{C}{L}=\left(\frac{2\pi \times 8.85\times {10}^{-12}}{\ln \left(\frac{3mm}{0.5mm}\right)}\right)$

Multiply the expression,

$=3.1·{10}^{-11}\mathrm{F}/\mathrm{m}$

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