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

AI Thread Summary
To calculate the electric field inside and outside a charged metal tube, use the formula E = λ/(2πεr), where λ is the charge per unit length and ε is the permittivity of free space. For a radial distance of 3.15 cm (inside the tube), the electric field is 0 N/C, as the electric field inside a conductor is zero. For a radial distance of 10.5 cm (outside the tube), the electric field calculates to 12,569 N/C. The radius of the tube (5.56 cm) is only relevant for determining whether the point of interest is inside or outside the tube. Understanding Gauss' Law is crucial for solving such problems effectively.
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=\lambda/2\pi\epsilonr


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 \epsilon_{O} 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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