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Electric Potential: Geiger Coutner

  • Thread starter Bryon
  • Start date
  • #1
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Homework Statement


A particular Gieger counter has a metal cylinder with an inner diameter of 1.9 cm along whose axis is stretched a wire with 24 N of tension. The potential difference between the wire and the cylinder is 870 volts. The wire has a length of 8 cm and an outer diameter of 1.1 X 10-4 cm.

What is the electric field at the surface of the wire?


Homework Equations



Vb - Va = ∫E∙dl = λ/(2*pi*ε*r)*(lna/b)

λ = (V*2*PI*ε*ln(a/b))

E = λ/2*PI*ε*R

The Attempt at a Solution



First I have to find the charge density of the wire:
λ = 870*2*PI*(8.85e-12)*ln(0.0095/(5.5e-5)) = 2.49226e-7

Then it should have been pretty easy to find the electric field from that:
E = (2.49226e-7)/(2*PI*(8.85e-12)*(5.5e-5) = 8.149075e7 V/m

But it did not like my answer. I am wondering that to find the electric field i have to use the entire radius of the cylinder and then set that equal to the voltage potential. Which then the charge density can be found? Maybe I worked this out backwards. Any ideas?
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
Check your expression used for the charge density.
 
  • #3
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Oh...I think I see what it is, ln(a/b) should be under it all: λ = (V*2*PI*ε/(ln(a/b)).
And hopefully that should get me to what I want to find! Thanks!
 
  • #4
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I still have it wrong:

λ = 870*2*PI*(8.85e-12)/(ln(0.0095/(5.5e-5))) = 9.370541921e-9C

Using
E = λ /(2*PI*R*(8.85e-12)) and putting in the different radii I get incorrect answers. What did I do wrong here?
 
  • #5
gneill
Mentor
20,793
2,773
Check your value of a (the wire radius). Could be an order of magnitude issue.
 
  • #6
99
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I got 3070469.779V/m (still doesnt like the answer) for the wire of radius 0.000055m. Which is ten times greater than my previous answer. It does look like its just a magnitude issue, problably from hitting the wrong button! I will come back to this later on today....and as always thanks for the help!
 
  • #7
gneill
Mentor
20,793
2,773
[tex] \frac{1.1 \times 10^{-4}cm}{2} \times \frac{1 m}{1 \times 10^2 cm} = 5.50 \times 10^{-7} m [/tex]

:smile:
 
  • #8
99
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Oh! Interesting I never thought to convert it. I had assumed it was already in meters. Wow that was a big goof on my part!
 

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