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?

Answer 1

The maximum potential difference is 186.02 x 10¹⁵ V

Step-by-step explanation:

formula for calculating maximum potential difference

`V = (2K_e \lambda)/(k)ln((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 = (2K_e \lambda)/(k)ln((b)/(a)) = (2*8.99*10^9*18*10^6 )/(2.3)ln((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