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Gauss' Law using linear charge density

  • Thread starter catie1981
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  • #1
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Homework Statement



A charge of uniform linear density 2.80 nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius = 5.20 cm, outer radius = 10.8 cm). The net charge on the shell is zero. (a) What is the magnitude (in N/C) of the electric field at distance r = 16.0 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?


Homework Equations



E= lambda / (2pi epsilon r) I think....that's all

The Attempt at a Solution



I have the solution for part a) which I used the above formula to determine as 3.15e2 N/C. However, for parts b) and c), I am unsure as to whether I am to use that same formula, or if I'm supposed to use integral E*dA*cos(theta), and if so, would the value I determined from the first part be the E for the second formula? thanks! (I'm actually kinda lost on how to start the b & c parts....)
 

Answers and Replies

  • #2
G01
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For b), try using Gauss's Law for the space inside the cylindrical shell.

HINT: What must the E-Field be equal to in that region?

Once you have done that, can you find the charge for c)?
 
  • #3
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does the fact that the net charge on the shell is zero have any real relevance to what I'm doing?
 
  • #4
G01
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does the fact that the net charge on the shell is zero have any real relevance to what I'm doing?
Yes it does. What does this tell you about the charge on the outside of the shell if you know the charge on the inner wall?
 
  • #5
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they are opposites....
 
  • #6
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but I still don't know what I'm supposed to do with Gauss' Law here....I can't figure out what values are supposed to correspond to what, etc. *arrrghhh*
 
  • #7
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alright for the cylindrical shell, what form of Gauss' Law should I be using? is it

E = (rho*R^2)/(2*epsilon*r) and if it is, what is the difference between R and r?

And if not, arrrgh again! cause I don't know which one to use!!
 
  • #8
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so this is apparently not the equation because no matter how I manipulate it, I'm not getting the answer....someone, please help....this is my last homework problem for this section......
 
  • #9
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oooo, I figured it out....how silly of me :)~
 
  • #10
G01
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oooo, I figured it out....how silly of me :)~
Sry, I got caught up with work today and didn't have time to check the forum. Otherwise I would have answered your questions.

Good Job!:approve:
 
  • #11
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There is not net electric field in within the cylindrical shell and since it doesn't have a net charge either, you can simply use Gauss's Law in respect to the axis for the first one. For the 2nd and 3rd part...using Gauss's Law you can derive an equation for the electric field near the surface and you can use that to get your answer...what you'd be missing is the value for the electric field. At each point in the conductor, net E is 0N/C, so as you approach the surface, net E is still relatively 0N/C...but there is still a collective charge on the surface, so for net E to be 0, it has to produce an E equal to the external electric field. In this case, you dont have to worry whether you look at it inside or outside the conductor because you'll be dealing with distances supposedly infinitely close the surface.
 

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