Find charge field within coaxial cables

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SUMMARY

The discussion focuses on calculating the electric field E(r) within coaxial cylinders, specifically an inner solid cylinder with radius a and a charge per unit length of λ, and an outer hollow conductor with inner radius b, outer radius c, and charge per unit length of -2λ. The key equation derived is E(r) = λ/(2πrε₀), confirmed through the application of Gauss's law, which states that the electric field is uniform and perpendicular to the area element dA. The solution correctly integrates the charge density and applies the appropriate limits for the coaxial geometry.

PREREQUISITES
  • Understanding of Gauss's law and its application in electrostatics
  • Familiarity with coaxial cable geometry and charge distribution
  • Knowledge of electric field calculations in cylindrical coordinates
  • Basic calculus for integration and manipulation of equations
NEXT STEPS
  • Study the derivation of electric fields using Gauss's law in different geometries
  • Learn about charge density and its implications in electrostatics
  • Explore the effects of varying charge distributions on electric fields
  • Investigate the properties of conductors in electrostatic equilibrium
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Students and professionals in physics and electrical engineering, particularly those studying electrostatics and the behavior of electric fields in coaxial cable systems.

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Homework Statement


two very long coaxial cylinders. Inner has radius a and is solid with charge per unit length of λ. Volume is also uniform but not defined by a parameter. Outer is hollow with inner radius b and outer radius c. Outer cylinder is a conductor with charge per unit length of -2λ. Find E(r) for all r in terms of given parameters. you can also view attached picture.


Homework Equations


∫E.dA=qenclosed0


The Attempt at a Solution


I'm hung up right away on what to do with the inner radius from 0->a I've got the left side of the equation I think, but I don't know how to get qenclosed in terms of r, if it needs to be. I expect to integrate on that part. Anywho here's where I'm at.


E2∏rh=(1/ε0)∫dq

I've got λ=dq/dh where h represents the length of the cable being considered. All the examples seem to use charge density rather than charge per unit length, using this gives me:

E2∏rh=(1/ε0)∫λdh which is not in terms of r. I suppose I could still integrate it and cancel h on both sides, let me know if that's wrong.

E2∏rh=(λh/ε0)→E2∏r=(λ/ε0)

E(r)=λ/(2∏r*ε0)

Is this wrong for any reason? It doesn't feel right to not use the radius on the right side.
 

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Let's consider the cable's length finite, L. The total charge of the inner cylinder would be Q = λL. Applying Gauss' theorem ∫E*dA = λL/ε°, since E is uniform and perpendicular to dA: E∫dA = λL/ε° <=> E(2πrL) = λL/ε° <=> E = λ/2πε°r.
The units on both sides are also ok, so your formula is right.
 
Last edited:

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