How Does Electric Field Vary with Distance in a Cylindrical Setup?

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SUMMARY

The electric field E in a cylindrical setup varies with distance r based on the position relative to the line charge and the cylindrical shell. For r < a, E is calculated as E = λ / (2πεr) using Gauss's Law. Between the inner radius a and outer radius b, the electric field is zero due to the conducting nature of the shell. For r > b, the electric field is given by E = (λs + λ) / (2πεr), accounting for the total charge enclosed by the Gaussian surface.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and charge distributions
  • Knowledge of cylindrical coordinates
  • Basic concepts of conductors in electrostatics
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Explore the concept of electric fields in conductors
  • Learn about charge density and its effects on electric fields
  • Investigate the implications of cylindrical symmetry in electrostatics
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields in cylindrical geometries.

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Homework Statement


based on the diagram attached, find the electric field E that varies with r.
the line and the metallic cylindrical shell are infinitely long.
the outer radius of the cylindrical shell is b and the inner one is a
and the line charge density for the line is [tex]\lambda[/tex]
and the line charge density for the cylindrical shell is [tex]\lambda[/tex]s

Homework Equations


Gauss's Law ( [tex]\oint[/tex]E dA = Qin / [tex]\epsilon[/tex]

The Attempt at a Solution


by using gauss's law with cylindrical gaussian surface coaxial with the line,
when r < a, i have E = [tex]\lambda[/tex] / (2*pi*[tex]\epsilon[/tex]*r);
for a< r < b, i have E = 0 since the shell is a conductor;

but for r > b, is it E = ([tex]\lambda[/tex]s + [tex]\lambda[/tex]) / (2*pi*[tex]\epsilon[/tex]*r) since the total charge enclosed is the sum of charges in the two bodies?

Pls help me. thanks
 

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Your calculations are correct.
 

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