Electric Potential: Geiger Coutner

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Homework Help Overview

The discussion revolves around calculating the electric field at the surface of a wire within a Geiger counter setup. The problem involves parameters such as tension, potential difference, and dimensions of the wire and cylinder, all of which are relevant to the electric field calculation.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of charge density and electric field, questioning the accuracy of their expressions and values used. There is an attempt to clarify the logarithmic relationship in the charge density formula and its implications on the electric field calculation.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and suggesting checks on specific values. There is recognition of potential errors in magnitude and unit conversions, indicating a productive exploration of the problem.

Contextual Notes

Participants note issues with unit conversions and the potential for order of magnitude errors in their calculations. The original poster expresses uncertainty about the correct approach to finding the electric field, indicating a need for further clarification on the setup and parameters involved.

Bryon
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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?
 
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Check your expression used for the charge density.
 
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!
 
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?
 
Check your value of a (the wire radius). Could be an order of magnitude issue.
 
I got 3070469.779V/m (still doesn't 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!
 
[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:
 
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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