Finding the surface charge density

  • Thread starter jheld
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  • #1
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Homework Statement


A 1.0-mm-diameter wire has 1000 excess electrons per centimeter of length. What is the surface charge density?


Homework Equations


[tex]\eta[/tex] = Q/A


The Attempt at a Solution


[tex]\eta[/tex] = (100000 * 1.6*10^-19)/([tex]\pi[/tex] * (5*10^-4)2)
But that gets me 2.037 * 10^-8 C/m^2

The answer is 5.1 * 10^-12 C/m^2

Any ideas?
 

Answers and Replies

  • #2
Doc Al
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What's the surface area (not the cross-sectional area) of that section of wire?
 
  • #3
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Okay, well I'm guessing that this wire is of cylindrical form...
SA = 2pir^2 + 2pirh

the problem is with the height. you might suppose part of that is the diameter...but it's really not. so, i'm still unsure.
 
  • #4
Doc Al
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Okay, well I'm guessing that this wire is of cylindrical form...
SA = 2pir^2 + 2pirh
Don't include the circular end pieces, just the outer surface.

the problem is with the height. you might suppose part of that is the diameter...but it's really not. so, i'm still unsure.
The height corresponds to the length.
 
  • #5
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Okay.
eta = 10 * 1.6*10^-19 C/(2pi * (5 * 10^-4 m) * (10^-3 m)) = 5.1 * 10^-12 C/m^2
That's the right answer :)
But, I'm a little confused as to why I only take the area of the outer surface and not the end pieces as well? Is that because I'm only concerned with one "portion" of the wire, and I'm therefore only calculating a piece and not the whole thing?
 
  • #6
Doc Al
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Note that you're given the charge per centimeter, which only sits on the outside of the wire. Think of the wire as being very long and that you are just looking at a typical one cm section somewhere in the middle. The "ends" of that section have no charge--only the outside counts. Make sense?
 
  • #7
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Yeah, thanks that really helped. Thanks for the help :)
 

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