The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space between the c

Question

The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space between the conductors is filled with polyethylene, which has a dielectric constant of 2.30 and a dielectric strength of 18.0 3 106 V/m. What is the maximum potential difference this cable can withstand?

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Dâu 4 weeks 2021-08-17T16:25:48+00:00 1 Answers 0 views 0

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    2021-08-17T16:27:01+00:00

    Answer:

    The maximum potential difference is 186.02 x 10¹⁵ V

    Explanation:

    formula for calculating maximum potential difference

    V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

    where;

    Ke is coulomb’s constant = 8.99 x 10⁹ Nm²/c²

    k is the dielectric constant = 2.3

    b is the outer radius of the conductor = 3 mm

    a is the inner radius of the conductor = 0.8 mm

    λ is the linear charge density = 18 x 10⁶ V/m

    Substitute in these values in the above equation;

    V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

    Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

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