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3 years ago

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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 b

etween 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?

1 answer:

ZanzabumX [31]3 years ago

8 0

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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Explanation:

Part a)

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Part c)

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$v_f^2 - v_y^2 = 2a \Delta y$

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