Guass Law involving conducting cylinders

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SUMMARY

The discussion centers on applying Gauss's Law to coaxial cylindrical conductors. The inner cylinder has a radius of 2.0 cm and a charge of +8.0 nC, while the outer cylinder has an inner radius of 6.0 cm, outer radius of 7.0 cm, and a charge of -16.0 nC. The x-component of the electric field at a point 4.0 cm from the origin is determined using the enclosed charge and linear charge density. The conclusion emphasizes that the length of the cylindrical section does not affect the final electric field calculation due to cylindrical symmetry.

PREREQUISITES
  • Understanding of Gauss's Law and its application
  • Knowledge of electric fields generated by charged conductors
  • Familiarity with cylindrical symmetry in electrostatics
  • Concept of linear charge density (λ) and its relation to total charge
NEXT STEPS
  • Study the derivation and applications of Gauss's Law in different geometries
  • Learn how to calculate electric fields using linear charge density
  • Explore the concept of cylindrical symmetry in electrostatics
  • Investigate the effects of varying charge distributions on electric fields
USEFUL FOR

Students of electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields in coaxial cylindrical systems.

maiad
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Homework Statement


Two coaxial cylindrical conductors are shown in perspective and cross-section above. The inner cylinder has radius a = 2.0 cm, length L = 10.0 m and carries a total charge of Qinner = + 8.0 nC (1 nC = 10-9 C). The outer cylinder has an inner radius b = 6.0 cm, outer radius c = 7.0 cm, length L = 10.0 m and carries a total charge of Qouter = - 16.0 nC (1 nC = 10-9 C). What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r = 4.0 cm from the origin and makes an angle of 30.0o with the x-axis?

https://www.smartphysics.com/Content/Media/Images/EM/IE/cylinderx/coaxial.gif

Homework Equations


E A= q(enclosed)/\epsilon


The Attempt at a Solution



I simply treated the enclosed charge as +8nC, but I've seen ppl using linear charge density and yet they still got the answer. Is just just fluke luck that i got the answer? This is an online hmwk grading system btw.
 
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maiad said:
I simply treated the enclosed charge as +8nC, but I've seen ppl using linear charge density and yet they still got the answer. Is just just fluke luck that i got the answer?
No, it's not a fluke. You are to assume that this section of length L is just a piece out of a much longer cable (or you are looking only at the field near the center of the piece). The field is cylindrically symmetric, so the length you choose to analyze doesn't matter.

Regardless of how you choose to solve it, if you use Gauss's law you'll need the charge enclosed in whatever Gaussian surface you use. That charge can be thought of as λL, where λ is the linear charge density. Note that L drops out of your final answer.
 

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