Electric Field and an Uniformly Charge Plastic Pipe

[Swagger]

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?

 

[siddharth]

Hint: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html"

Have you figured out what gaussian surface to use? Once you do that, you'll be able to calculate the charge enclosed.

 

[Swagger]

I don't understand

 

[siddharth]

What don't you understand?

 

[Swagger]

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.

 

[siddharth]

Yes, but more important than the formula is the underlying principle. Do you know how you got that formula?

 

[Vedj]

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.