Electric field of coax cable problem

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SUMMARY

The discussion focuses on calculating the electric field of a coaxial cable with a charged inner conductor of 8.3 µC and a surrounding oppositely charged conductor of -8.3 µC. The radius of the inner conductor is 2.917 mm, while the outer conductor has a radius of 5.997 mm. The length of the cable is 60 m, and the electric field is to be determined halfway between the two conductors using Gauss's Law. The incorrect calculation of the electric field yielded a value of 5.58 x 10^11 V/m, indicating a need for reevaluation of the applied formulas and assumptions regarding charge distribution.

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  • Understanding of Gauss's Law in electrostatics
  • Familiarity with cylindrical coordinates and geometry
  • Knowledge of electric field calculations for charged conductors
  • Basic proficiency in physics equations related to electric fields
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  • Review the application of Gauss's Law for non-uniform charge distributions
  • Learn about the derivation of the electric field for coaxial cables
  • Study the impact of conductor length on electric field calculations
  • Examine the use of numerical methods for complex charge distributions
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Physics students, electrical engineers, and anyone involved in the study of electrostatics and electric field calculations in coaxial cable systems.

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A coaxial cable has a charged inner conductor, with charge 8.3 \mu C and radius 2.917 mm, and a surrounding oppositely charged conductor, with charge -8.3 \mu C and radius 5.997 mm. Assume the region between the conductors is air and neglect end effects. The length of the cable is 60 m. What is the magnitude of the electric field halfway between the two cylindrical conductors. Answer in units of V/m.
I got help with this problem, but it still isn't working out. We used Gauss's Law. To find the radius of the Gaussian surface, we used R-r/2 + r.
So .005997 - .002917 /2 + .002917 = .004457
Then to get the E, you do q_e_n_c / E_o *A
With the area of a cylinder= 2\pi r^2 + 2\pi *r*h
So 8.3 x 10^-6 / 8.85 x 10^-12 *(2\pi (.004457)^2 + 2 \pi * (.004457)(60))<br /> Which gave me 5.58 x 10^11, which isn&#039;t right.. can someone please help me?
 
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Ok so the equation that this page gives for the E-field is
E= \lambda / 2\pi * E_o * r, but since the charge isn't uniform, I don't think I can use it...
 

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