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Gauss' Law

  • Thread starter suspenc3
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  • #1
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Homework Statement


A long, straight wire has a linear charge density of magnitude 3.6nC/m. The wire is to be enclosed by a thin, no-conducting cylinder of ouside radius 1.5cm, coaxil witht he wire. The cylinder is to have positive charge on its outside surface with a surface charge density [tex]\sigma[/tex] such that the net external electric field is zero. Calculate the required [tex]\sigma[/tex].


Homework Equations


[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]


The Attempt at a Solution



The electric field can be found, but then how do I go about finding the required [tex]\sigma[/tex] value?
 
Last edited:

Answers and Replies

  • #2
Gyroscope
Sorry, misread.
 
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  • #3
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Thats what it sayss....
 
  • #4
Gyroscope
The wire has obviously negative charge. Hasn't it?
Apply Gauss's Law, to a gaussian surface, I recommend you to use another cylinder, and put the condition that:

[tex]\oint E\cdot dS=0[/tex]
 
  • #5
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Im still confused
 
  • #6
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The total field is zero right? It is due to a superposition of the field from the wire and the field from the cylindrical shell.
 
  • #7
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So Do I just find an electric field for the gaussian surface that when summed with the electric field already found wll equal zero?
 
  • #8
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not sure what exactly you mean there
 
  • #9
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Since the total electrical field is 0, and I can find [tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex], cant I just find [tex]E=\frac{\sigma}{\epsilon_0}[/tex]

and then [tex]\frac{\lambda}{2\pi \epsilon_0 r} = -\frac{\sigma}{\epsilon_0}[/tex] and then solve for [tex]\sigma[/tex]

Sorry if this make no sense, I dont really understand this stuff.
 

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