Coaxial cable: Magnitude of the electric field; capacitance of the cable
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?
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 Summary
With formulas and calculations, the problems are solved.
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See attached file for clarity.
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