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Coaxial cable: Magnitude of the electric field; capacitance of the cable

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Neglect end effects. The region between the conductors is air. Ke =1/4*pi*epsilon= 8.98755e9 N m^2/C^2.

A coaxial cable has a charged inner conductor (with charge +4.8 microcoulombs and radius 1.199 mm) and a surrounding oppositely charged conductor (with charge -4.8 microcoulombs and radius 7.405 mm).

a) What is the magnitude of the electric field halfway between the two cylindrical conductors in V/m?

b) What is the capacitance of this cable in nF?

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Solution Summary

With formulas and calculations, the problems are solved.

Solution Preview

See attached graph.

According to Gauss theorem:
Within the Guassian surface:
Flux = charge inside the surface/epsilon
E*2pi*r*l = Q/epsilon
=> E(electric field strength) = Q/(2pi*r*e*l)
here, e =epsilon
=> E = ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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