A 75.0-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 aninner 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=nF(b) What is the potential difference between the two conductors?AV =kV

Given The length of the cylinder is L=75 m. The inner diameter of the coaxial cable is din=2.58…

Answered: A 75.0-m length of coaxial cable has an…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 39P

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Question

A 75.0-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=
nF
(b) What is the potential difference between the two conductors?
AV =
kV

Expert Solution

Part (a)

Given

The length of the cylinder is $L = 75 m$ .
The inner diameter of the coaxial cable is $d_{i n} = 2.58 mm = 2.58 \times 10^{- 3} m$ .
The outer diameter of the coaxial cable is $d_{o} = 7.27 mm = 7.27 \times 10^{- 3} m$ .
The magnitude of the change on the conductor is $Q = 8.10 μC = 8.10 \times 10^{- 6} C$ .

The capacitance of the coaxial or cylindrical cable is calculated as,

$\begin{array}{rcl} C & = & \frac{2 π ε_{0} L}{\ln (\frac{r_{0}}{r_{i n}})} \\ = & \frac{2 π \times \frac{1}{4 π k} \times L}{\ln (\frac{\frac{d_{o}}{2}}{\frac{d_{i n}}{2}})} \\ = & \frac{L}{2 k \ln (\frac{d_{o}}{d_{i n}})} \end{array}$

Here, k is the coulomb constant whose value is $9 \times 10^{9} {Nm}^{2} / C^{2}$ .

Substitute the known values.

$\begin{array}{rcl} C & = & \frac{75 m}{2 \times 9 \times 10^{9} {Nm}^{2} / C^{2} \times \ln (\frac{7.27 \times 10^{- 3} m}{2.58 \times 10^{- 3} m})} \\ = & \frac{75 m}{2 \times 9 \times 10^{9} {Nm}^{2} / C^{2} \times \ln (2.81782)} \\ = & \frac{75 m}{2 \times 9 \times 10^{9} {Nm}^{2} / C^{2} \times 1.0359} \\ = & 4.02 \times 10^{- 9} F \times \frac{1 nF}{10^{- 9} F} \\ = & 4.02 nF \end{array}$

Thus, the capacitance of the cable is 4.02 nF.

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