Capacitance of Coaxial Cable Configuration with Multiple Conductors

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SUMMARY

The discussion focuses on calculating the capacitance per unit length of a coaxial cable configuration consisting of a solid inner cylinder and two cylindrical shields. The inner conductor has a diameter of 0.55 mm, the central shield has a diameter of 3.43 mm, and the outer shield has a diameter of 5.88 mm. The capacitance is derived using the equations ΔV = ∫E ds and C = Q/(ΔV), with the electric field calculated using Gauss's Law. It is clarified that the inner cylinder and outer shield do not share the same charge, highlighting the importance of understanding charge distribution in coaxial configurations.

PREREQUISITES
  • Understanding of Gauss's Law and its application to cylindrical charge distributions.
  • Familiarity with the concepts of electric fields and potential difference.
  • Knowledge of capacitance and its calculation methods.
  • Basic principles of electrostatics and charge distribution in conductors.
NEXT STEPS
  • Study the derivation of capacitance for coaxial cables in detail.
  • Learn about the implications of charge distribution in multi-conductor systems.
  • Explore the use of numerical methods for calculating electric fields in complex geometries.
  • Investigate the effects of dielectric materials on capacitance in coaxial configurations.
USEFUL FOR

Electrical engineers, physics students, and professionals involved in cable design and analysis will benefit from this discussion, particularly those focusing on capacitance and electrostatics in coaxial systems.

DRussian
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Homework Statement



I can't seem to understand this problem (I do see that the inner cylinder and outer shell will have the same charge, but I can't see what I'm supposed to assume about the middle shell):

A cable consists of three conductors, a solid inner cylinder and two thin cylindrical shields. All three conductors are coaxial, with the solid inner conductor, central shield, and outer shield having diameters 0.55 mm, 3.43 mm, and 5.88 mm, respectively. Assume the space between the conductors is filled with air. The inner cylinder and outer shield are connected at one end of the cable.

What is the capacitance per unit length of this configuration?

Homework Equations



ΔV = ∫E ds
C = Q/(ΔV)

The Attempt at a Solution



I tried finding the resulting change in voltage from the inner cylinder to the first inner shell.

Solving from Gauss's Law, I first got that the magnitude of the electric field of a cylindrical charge distribution having linear charge density λ is (k is Boltzmann's constant) E = 2kλ/r.
Integrating from the radius of the center to the first inner shell, I get:

ΔV = ∫E ds = -2kλ*ln(b/a) (where b = radius from center of capacitor to first inner shell and a = radius of inner solid conductor).

This is where I got confused; how would I deal with the second, outer shell into here? How would I account that the E-field is in the opposite direction in the second space (between the two hollow shells) as in the first space (between the solid conductor and first shell)?
 
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DRussian said:
I do see that the inner cylinder and outer shell will have the same charge,
No, the cylindrical core and the outermost shield will not have the same charge. What is it that will be the same for them?
 

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