Electric Field of a charged rod

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SUMMARY

The discussion focuses on calculating the electric field generated by a charged conducting rod and its surrounding coaxial cylindrical shell. The rod has a radius of 1.30 mm and a length of 11.00 m, with a net charge of Q1 = +1.50 x 10-12C, while the shell carries a charge of Q2 = -2.00Q1. The electric field is evaluated at two radial distances: r = 1.50R2 and r = 4.50R1, utilizing Gauss' theorem and the formula for the electric field, E = k(Q/(r√(r² + (L/2)²))).

PREREQUISITES
  • Understanding of Gauss' theorem in electrostatics
  • Familiarity with electric field calculations
  • Knowledge of cylindrical coordinates
  • Basic concepts of charge distribution in conductors
NEXT STEPS
  • Study the application of Gauss' law for different geometries
  • Learn about electric field calculations for cylindrical conductors
  • Explore the effects of charge distribution on electric fields
  • Investigate the relationship between electric field strength and distance from charged objects
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone interested in electrostatics and electric field calculations related to charged conductors.

22steve
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Homework Statement



The figure below is a section of a conducting rod of radius R1 = 1.30 mm and length L = 11.00 m inside a thick-walled coaxial conducting cylindrical shell of radius R2 = 10.0R1 and the (same) length L. The net charge on the rod is Q1 = +1.50 10-12C that on the shell is Q2 = −2.00Q1.
Question: What is the magnitude E of the electric field at a radial distance of r = 1.50R2? and
What is the magnitude E of the electric field at a radial distance of r = 4.50R1

Homework Equations



Electric Field = k(Q/(r (sq rt (r^2 + (L/2)^2)))
(I believe that's the only relevant equation)

The Attempt at a Solution



8.99^9 (1.5e-12/(.000585 x sqrt(.000585^2 + (.0013/2)^2)) = 26359
 
Last edited:
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Use Gauss' theorem to find the electric field.
 
but all I know is the charge, how can I go from flux to electric field?
 
Flux per unit area is the electric field.
 

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