Question

A coaxial cable consists of an inner conductor or radius ri = 0.40 cm and an outer conductor of radius ro. When manufactoring such a cable, its size is determined by the voltage between the two conductors and by the insulating material used to insulate the two conductors from each other. We want to calculate an appropriate size for the following cable: Assume that the concentric insulating layers are rubber, with dielectric constant εr = 3.2, and polystyrene with dielectric constant εp = 2.6. If we want this cable to have a rating of 20 kV, what should the outer radius ro and the radius rp be? Assume that the maximum electric field in the insulating layers should not exceed 25% of their dielectric strength (that is a conservative estimate, in order to avoid breakdowns due to power surges etc.). Assume that the inner conductor carries linear charge density rho. Use the following data: dielectric strength of rubber = 25 × 106 V/m dielectric strength of polystyrene = 20 × 106 V/m

Answer 1

Answer: ro= 0.113

Step-by-step explanation:

Data Given: Where D= Outer diameter

d= Inner diameter

ri= 0.4cm

r°= ?

Charge 1 and Charge 2 (q) = K i.e constant of proportionality

Where K = 1/4πe°. er; 3.2, ep; 2.6 ( Relative permeability)

(a) Approximate value size for the coaxial cable

C= 7.354 x Er/ Log 10 (D÷d)

C = 7.354 x 3.2/ Log 2.6

= 23.5/0.419

= 56.0cm

(b) Outer radius r° when dealing with 20kv

r°= impedance value (pie ÷ 20) d^2

r° = 2( π÷20)0.6^2

= 2( 0.1571) 0.36

r° = 0.113

(c) Assuming The Maximum Electric Field; Fmax = 25%

Given Vin = 25x10^6v/m, Vout = 20x 10^6 v/m.

Impedance p= √Vin x vout/ Z° (i)

Where Z° = electric surge ~25%

Subtituting Value of Vin & Vout In Equation (i)

P = 2.05x10^-8ohms

Note : Length of a conducting material determines the strength of dielectric material between the coaxial cable