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

Answer:

(a) 1.556 x 10^-9 F

(b) 5204 V

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

$C =\frac{2\pi \epsilon _{0}L}{ln\left ( \frac{r_{2}}{r_{1}} \right )}$

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

$C =\frac{2\times 3.14\times 8.854\times 10^{-12}\times 29}{ln\left ( \frac{3.635}{1.29}\right )}$

C = 1.556 x 10^-9 F

(b) V = q / C

$V = \frac{8.10\times 10^{-6}}{1.556\times 10^{-9}}$

V = 5204 V