Question

A 1.0 m long piece of coaxial cable has a wire with a radius of 1.1 mm and a concentric conductor with inner radius 1.3 mm. The area between the cable and the conductor is filled with a dielectric. If the voltage drop across the capacitor is 6000 V when the line charge density is 8.8 μC/m, find the value of the dielectric constant. (k = 1/4πε₀ = 8.99 × 109 N · m²/C²)A) 4.8
B) 5.3
C) 4.4
D) 5.7

Answer 1

Answer:

C) 4.4

Explanation:

The potential of a cylindrical capacitor is given by the formula:

$V=\frac{2kq}{L\epsilon}ln(\frac{a}{b})\\\\\epsilon=\frac{2kq}{LV}ln(\frac{a}{b})$

where:

k : Coulomb Constant

L : length of the capacitor

a : outer radius

b : inner radius

V : potential

By replacing we obtain:

$\epsilon=\frac{2(8.89*10^{9}N/m^2C^2)(8.8*10^{-6}C)}{(1m)(6000V)}ln(\frac{1.3mm}{1.1mm})=4.35$

Hence, the answer is C) 4.4 (4.35 is approximately 4.4)

hope this helps!!