High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE P26.68. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius R1 while the outer conductor of radius R2 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. Evaluate the capacitance per meter of a cable having R1=0.50 mm and R2=3.0 mm.

Ch 26: Potential and Field

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 26: Potential and Field

Problem 68b

Chapter 26, Problem 68b

High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE P26.68. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius R₁ while the outer conductor of radius R₂ 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. Evaluate the capacitance per meter of a cable having R₁=0.50 mm and R₂=3.0 mm.

Verified step by step guidance

Step 1: Recognize that the capacitance per unit length of a coaxial cable can be calculated using the formula for the capacitance of a cylindrical capacitor: \( C' = \frac{2 \pi \varepsilon}{\ln(R_2 / R_1)} \), where \( \varepsilon \) is the permittivity of the insulating material, \( R_1 \) is the radius of the inner conductor, and \( R_2 \) is the radius of the outer conductor.

Step 2: Convert the given radii \( R_1 = 0.50 \, \text{mm} \) and \( R_2 = 3.0 \, \text{mm} \) into meters for consistency in SI units. This gives \( R_1 = 0.0005 \, \text{m} \) and \( R_2 = 0.003 \, \(\text{m}\).

Step 3: Identify the permittivity \( \varepsilon \) of the insulating material. If the problem does not specify the material, assume \( \varepsilon = \varepsilon_0 \), the permittivity of free space, which is \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \).

Step 4: Substitute the values of \( \varepsilon \), \( R_1 \), and \( R_2 \) into the formula \( C' = \frac{2 \pi \varepsilon}{\ln(R_2 / R_1)} \). Ensure that the logarithmic term \( \ln(R_2 / R_1) \) is calculated using the natural logarithm.

Step 5: Simplify the expression to find the capacitance per meter \( C' \). The result will be in units of farads per meter (F/m).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a system to store electric charge per unit voltage. It is defined as the ratio of the electric charge stored on a conductor to the potential difference across it. In coaxial cables, capacitance is influenced by the geometry of the conductors and the dielectric material between them, which affects how much charge can be stored.

Coaxial Cable Structure

A coaxial cable consists of two concentric conductors: an inner conductor with radius R1 and an outer conductor with radius R2. The space between these conductors is filled with an insulating material, which serves as a dielectric. This structure allows for efficient transmission of high-frequency signals while minimizing interference and signal loss.

Dielectric Material

Dielectric materials are insulators that can be polarized by an electric field, which enhances the capacitance of a capacitor. In coaxial cables, the dielectric material between the inner and outer conductors affects the cable's capacitance and signal transmission characteristics. The choice of dielectric material can influence the cable's performance, including its bandwidth and attenuation.

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