Gauss' Law using linear charge density

In summary, a problem is given regarding the electric field produced by a charge distributed along a nonconducting rod and a conducting cylindrical shell. Using Gauss's Law, the magnitude of the electric field at a specific distance from the axis of the shell is determined to be 3.15e2 N/C. The surface charge density on the inner and outer surfaces of the shell can be calculated using the same formula, but the net charge on the shell being zero is also taken into consideration. Finally, Gauss's Law is used again to derive an equation for the electric field near the surface, which is then used to solve for the surface charge density on the inner and outer surfaces of the shell.
  • #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...)
 
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  • #2
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
does the fact that the net charge on the shell is zero have any real relevance to what I'm doing?
 
  • #4
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?
 
  • #5
they are opposites...
 
  • #6
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
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
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
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.
 

What is Gauss' Law using linear charge density?

Gauss' Law using linear charge density is a fundamental law in electromagnetism that relates the electric field at a point to the linear charge density (charge per unit length) along a line passing through that point. It is a special case of Gauss' Law, which is a more general law that relates the electric field to the total charge enclosed by a surface.

What is the formula for Gauss' Law using linear charge density?

The formula for Gauss' Law using linear charge density is E = λ/2πε0r, where E is the electric field, λ is the linear charge density, ε0 is the permittivity of free space, and r is the distance from the line charge to the point where the electric field is being calculated.

How is Gauss' Law using linear charge density applied in real life?

Gauss' Law using linear charge density is used in many real-life applications, such as calculating the electric field around power lines, determining the electric field inside a charged wire, and understanding the behavior of electric currents in wires. It is also important in the design of electrical devices and systems.

What are the units of measurement for linear charge density?

The units of measurement for linear charge density are coulombs per meter (C/m). This represents the amount of charge per unit length along a line.

Can Gauss' Law using linear charge density be used for all types of charge distributions?

No, Gauss' Law using linear charge density is only applicable for charge distributions that are symmetric about a line. For more complex charge distributions, Gauss' Law in its more general form must be used.

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