Electric Field and an Uniformly Charge Plastic Pipe

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  • #1
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A plastic pipe has an inner radius of a = 35.00 cm and an outer radius of b = 71.00 cm. Electric charge is uniformly distributed over the region a < r < b. The charge density in this region is 30.00 C/m3

A)Calculate the magnitude of the electric field at r = 0.44 m.

B)Calculate the magnitude of the electric field at r = 1.59 m.



I need some help on where to begin. I know the e-field for cylindrical sym. is E=2kQin/rL

I also know that Qin=charge density * Volume.

How do I find the volume so I can find the Qin in both A and B?
 

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  • #2
siddharth
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Hint: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html" [Broken]

Have you figured out what gaussian surface to use? Once you do that, you'll be able to calculate the charge enclosed.
 
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I dont understand
 
  • #4
siddharth
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What don't you understand?
 
  • #5
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I don't understand those formulas, we learned E=2kQin/rL. Was my approach the wrong way of doing it. Finding Qin and then using E=2kQin/rL to get the answer.
 
  • #6
siddharth
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Yes, but more important than the formula is the underlying principle. Do you know how you got that formula?
 
  • #7
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Sorry to revive this old thread, but it is the exact same problem I am working on.

What is the formula used to solve the equation? I thought all I would have to do is E = lambda/2pi*epsilon*r^2 but I was wrong.

In the previous posts, in the equation E=2kQin/rL, what is L? Also, how do I find the volume so I can find the Qin in both A and B? Thank you.
 

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