A coaxial cable used in a transmission line has an inner radius of 0.14 mm and an outer radius of 0.54 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with a material with a dielectric constant of 3.1. Number i 1.2E-10 Units pF/m
A coaxial cable used in a transmission line has an inner radius of 0.14 mm and an outer radius of 0.54 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with a material with a dielectric constant of 3.1. Number i 1.2E-10 Units pF/m
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Publisher:Raymond A. Serway, Chris Vuille
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![### Example Problem: Calculating Capacitance of a Coaxial Cable
**Problem Statement:**
A coaxial cable used in a transmission line has an inner radius of 0.14 mm and an outer radius of 0.54 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with a material with a dielectric constant of 3.1.
**Solution:**
To solve for the capacitance per meter for a coaxial cable, we can use the following formula for the capacitance \(C'\) per unit length of a coaxial cable:
\[ C' = \frac{2 \pi \epsilon_0 \epsilon_r}{\ln(b/a)} \]
where:
- \( \epsilon_0 \) is the permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \)
- \( \epsilon_r \) is the relative permittivity (dielectric constant) of the material between the conductors
- \( a \) is the inner radius
- \( b \) is the outer radius
- \( \ln \) denotes the natural logarithm
Given:
- \( a = 0.14 \, \text{mm} = 0.14 \times 10^{-3} \, \text{m} \)
- \( b = 0.54 \, \text{mm} = 0.54 \times 10^{-3} \, \text{m} \)
- \( \epsilon_r = 3.1 \)
First, we need to convert the given dimensions to meters and substitute them into the formula.
\[ C' = \frac{2 \pi \epsilon_0 \epsilon_r}{\ln(0.54 \times 10^{-3} / 0.14 \times 10^{-3})} \]
Next, we calculate the natural logarithm:
\[ \ln(0.54 \times 10^{-3} / 0.14 \times 10^{-3}) = \ln(0.54 / 0.14) \approx \ln(3.857) \approx 1.35 \]
Now we substitute the values:
\[ C' = \frac{2 \pi \times (8.854 \times 10^{-12} \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba35cc5f-959d-4f80-acdd-e9cc0bd31dfa%2F85935b56-98c4-4398-b5b3-4ce2cb786fc9%2Fm8cpmfj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem: Calculating Capacitance of a Coaxial Cable
**Problem Statement:**
A coaxial cable used in a transmission line has an inner radius of 0.14 mm and an outer radius of 0.54 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with a material with a dielectric constant of 3.1.
**Solution:**
To solve for the capacitance per meter for a coaxial cable, we can use the following formula for the capacitance \(C'\) per unit length of a coaxial cable:
\[ C' = \frac{2 \pi \epsilon_0 \epsilon_r}{\ln(b/a)} \]
where:
- \( \epsilon_0 \) is the permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \)
- \( \epsilon_r \) is the relative permittivity (dielectric constant) of the material between the conductors
- \( a \) is the inner radius
- \( b \) is the outer radius
- \( \ln \) denotes the natural logarithm
Given:
- \( a = 0.14 \, \text{mm} = 0.14 \times 10^{-3} \, \text{m} \)
- \( b = 0.54 \, \text{mm} = 0.54 \times 10^{-3} \, \text{m} \)
- \( \epsilon_r = 3.1 \)
First, we need to convert the given dimensions to meters and substitute them into the formula.
\[ C' = \frac{2 \pi \epsilon_0 \epsilon_r}{\ln(0.54 \times 10^{-3} / 0.14 \times 10^{-3})} \]
Next, we calculate the natural logarithm:
\[ \ln(0.54 \times 10^{-3} / 0.14 \times 10^{-3}) = \ln(0.54 / 0.14) \approx \ln(3.857) \approx 1.35 \]
Now we substitute the values:
\[ C' = \frac{2 \pi \times (8.854 \times 10^{-12} \,
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