How an Electric Field is affected by an Insulator

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SUMMARY

The discussion centers on calculating the electric field and electric flux around a uniformly charged filament surrounded by an uncharged cardboard cylinder. The filament is 7.70 m long with a total positive charge of 2.00 µC, and the cylinder has a length of 1.50 cm and a radius of 10.0 cm. The electric field at the surface of the cylinder is determined using the formula E = λ/2πRε, where λ is the linear charge density. The participants clarify that the presence of the cardboard cylinder does not alter the electric field due to its status as an insulator, which does not create a significant dipole effect in this scenario.

PREREQUISITES
  • Understanding of electric fields and flux in electrostatics
  • Familiarity with the concept of linear charge density (λ)
  • Knowledge of the properties of insulators and dielectrics
  • Basic proficiency in applying Gauss's law
NEXT STEPS
  • Study the derivation and application of Gauss's law in electrostatics
  • Learn about the dielectric constant and its effect on electric fields
  • Explore the concept of electric flux and its calculation through different surfaces
  • Investigate the behavior of electric fields in the presence of various materials
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone looking to deepen their understanding of electric fields and the effects of insulators on these fields.

Iftekhar Uddin
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I looked through this forum post for an answer to my problem. It gave me enough insight to try and come to some conclusions but I'm still confused on some parts here and there about how I got my final answer.

1. Homework Statement

A uniformly charged, straight filament 7.70 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder 1.50 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder.
(a) Using reasonable approximations, find the electric field at the surface of the cylinder.

(b) Using reasonable approximations, find the total electric flux through the cylinder.


Homework Equations



The Electric Field of a filament is λ/2πRε
λ=Total charge/length

The Attempt at a Solution

(I got the solution but I have questions about part A)
[/B]
* I found that the answer to A was the electric field of a filament with r = .1m. I used this equation on an earlier problem that asked for the electric field at x distance away from the filament. But this is the same for this cardboard cylinder? I don't have to multiply it by the length of the cylinder? Why? The only reason I could think for this is that I'm finding a uniform E so I'm just finding E at a distance of R radially. Is that correct?

* Is cardboard not an insulator? If so, it doesn't affect the electric field at all? I thought that as an insulator it'd affect the electric field so that the outside surface of the insulator would have an E of 0.
 
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As far as I remember, The net electric field is less than the applied electric field because it will create dipoles and will have an electric field opposite to the applied so it becomes less. However it can never be zero inside or outside.

It should give you the dialectic constant of the cardboard.
Someone can confirm this.
 
Hmmm maybe I'm not far enough into physics for this because I got the right answer without adjusting for this dipole situation.
 

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