A 29.0-m length of coaxial cable has an inner conductor that has a diameter of 2

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? C = 20.5 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nF (b) What is the potential difference between the two conductors? V = kV

Explanation / Answer

(a) The capacitance of the cable is:

C = l / 2k*ln(b/a)

where l = 29 m is the length of the cable, a = 2.58 mm is the diameter of the inner conductor and b = 7.27 mm is the diameter of the outer conductor. The capacitance of the cable then becomes:

C = 29 / (2 x 9 x 10^9 x ln(7.27 / 2.58))

C = 1.555 x 10^-9 F

C = 1.555 nF

(b)

The potential difference can be obtain in two ways. One way is to use the potential difference between two charged concentric shells and the other from the capacitance of the cable. Using the first method gives:

V = 2ke ln(b/a)

where = q/l is the charge per unit length, so:

V = 2 × 9 × 10^9 × (8.10 × 10^6 / 29) × ln ( 7.27 / 2.58)

V = 5.2 x 10^3 = 5.2 kV

Using the second method we get:

V = Q / C = (8.10 × 10^6) / (1.555 × 10^9) = 5.2 kV

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