A coaxial cable has shielding with inner diameter 8.5 mm and conductorwith outer diameter 2.5 mm. Assume that the permittivity of the insulatoris 2.10.(a) (1 marks) If that capacitance of the cable must not be greater than120 pF, then what is the longest that we can make the cable?
Question
A coaxial cable has shielding with inner diameter 8.5 mm and conductorwith outer diameter 2.5 mm. Assume that the permittivity of the insulatoris 2.10.(a) (1 marks) If that capacitance of the cable must not be greater than120 pF, then what is the longest that we can make the cable?
Solution
The capacitance per unit length for a coaxial cable can be calculated using the formula:
C = 2πε / ln(b/a)
where:
- ε is the permittivity of the insulator,
- a is the radius of the inner conductor,
- b is the radius of the outer conductor,
- ln is the natural logarithm.
Given that ε = 2.1 * ε0, a = 1.25 mm, and b = 4.25 mm, we can substitute these values into the formula:
C = 2π * 2.1ε0 / ln(4.25/1.25)
The permittivity of free space ε0 is approximately 8.85 * 10^-12 F/m, so we can substitute this value in:
C = 2π * 2.1 * 8.85 * 10^-12 F/m / ln(4.25/1.25)
This will give us the capacitance per unit length in Farads per meter (F/m).
The total capacitance of the cable is given as 120 pF, or 120 * 10^-12 F. To find the maximum length of the cable, we can divide the total capacitance by the capacitance per unit length:
L = 120 * 10^-12 F / (2π * 2.1 * 8.85 * 10^-12 F/m / ln(4.25/1.25))
This will give us the length in meters.
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