Question

A 29.0-m length of coaxial cable has an inner conductor that has a diameter of 2.58 mm and carries a charge of 8.10 µC. The surrounding conductor has an inner diameter of 7.27 mm and a charge of −8.10 µC. Assume the region between the conductors is air.

(a) What is the capacitance of this cable?
(b) What is the potential difference between the twoconductors?

Answer 1

(a) 1.556 x 10^-9 F

(b) 5204 V

Step-by-step explanation:

Length, L = 29 m

inner diameter = 2.58 mm

outer diameter = 7.27 mm

q = 8.10 micro coulomb = 8.10 x 10^-6 C

inner radius, r1 = 2.58 / 2 = 1.29 x 10^-3 m

outer radius, r2 = 7.27 / 2 = 3.635 x 10^-3 m

(a) The formula for the cylindrical capacitor is given by

`C =(2\pi \epsilon _(0)L)/(ln\left ( (r_(2))/(r_(1)) \right ))`

where, L is the length of the conductor, r2 be the outer radius and r1 be the inner radius.

`C =(2* 3.14* 8.854* 10^(-12)* 29)/(ln\left ( (3.635)/(1.29)\right ))`

C = 1.556 x 10^-9 F

(b) V = q / C

`V = (8.10* 10^(-6))/(1.556* 10^(-9))`

V = 5204 V