How Do You Calculate the Electric Field Inside and Outside a Charged Metal Tube?

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Homework Help Overview

The discussion revolves around calculating the electric field inside and outside a charged metal tube, specifically focusing on a long, thin-walled metal tube with a given charge per unit length. The problem involves determining the electric field at two radial distances from the tube.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss' Law and the significance of the radius of the tube in relation to the radial distances provided. There is uncertainty about how to apply the given equation for electric field strength and how to incorporate the different radii into the calculations.

Discussion Status

Some participants have provided guidance regarding the relevance of the charge density and the role of the tube's radius in determining whether a point is inside or outside the cylinder. There is acknowledgment of the permittivity of free space, and while some numerical values are mentioned, the focus remains on understanding the underlying concepts rather than reaching a definitive solution.

Contextual Notes

The problem is constrained by the need to apply specific equations related to electric fields and the conditions of being inside or outside the charged tube. The original poster expresses confusion regarding how to use the given radius values in the context of the problem.

ghetto_bird25
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hey guys i was wondering if anyone could help me with this question, kinda got stuck

Homework Statement


Figure 24-31 shows a section of a long, thin-walled metal tube of radius R = 5.56 cm, with a charge per unit length λ = 7.34 x 10-8 C/m. What is the magnitude E of the electric field at radial distance (a)r = 3.15 cm and (b)r = 10.5 cm.
http://edugen.wiley.com/edugen/courses/crs1141/art/qb/qu/c24/Fig23_34.gif

Homework Equations


E=[tex]\lambda[/tex]/2[tex]\pi[/tex][tex]\epsilon[/tex]r


The Attempt at a Solution


well i tried using that equation but got stuck with the radius and where to plug it in since you get 5.56 cm, but wasn't too sure how to exactly plug in the other radius. also i was wondering if that was how you do the question and if you had to plug in the radius in that equation and not R=5.56cm
 
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You are given the charge density per meter. Think about Gauss' Law. The only role that R = 5.56 cm plays in the problem is to determine whether r is inside or outside of the cylinder.
 
ic so would [tex]\epsilon[/tex][tex]_{O}[/tex] be 8.85e-12 C^2/N m^2
 
If that's the usual value for the permittivity of free space, yes.
 
ok thanks that realli helped, btw the answer for this is a) 0 N/C b)12569 N/C
thanks again
 

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