Gauss' Law using linear charge density

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Homework Help Overview

The problem involves a long, thin, nonconducting rod with a uniform linear charge density and a coaxial conducting cylindrical shell. Participants are tasked with determining the electric field at a specific distance from the shell and the surface charge densities on the shell's inner and outer surfaces.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Gauss's Law and whether to use different formulas for different parts of the problem. There is uncertainty about how the net charge on the shell affects the calculations and the relevance of the electric field inside the shell.

Discussion Status

Some participants have offered hints about using Gauss's Law for the cylindrical shell and have pointed out the implications of the net charge being zero. Others express confusion about which form of Gauss's Law to apply and how to relate the variables involved.

Contextual Notes

There is mention of the net charge on the shell being zero, which raises questions about the charge distribution on the shell's surfaces. Participants are also navigating the constraints of homework rules and the need to derive values from given information.

catie1981
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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...)
 
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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)?
 
does the fact that the net charge on the shell is zero have any real relevance to what I'm doing?
 
catie1981 said:
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?
 
they are opposites...
 
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*
 
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!
 
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...
 
oooo, I figured it out...how silly of me :)~
 
  • #10
catie1981 said:
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
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 don't 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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