Electric Potential: Geiger Coutner

AI Thread Summary
The discussion revolves around calculating the electric field at the surface of a wire in a Geiger counter setup. The initial attempt involved finding the charge density using the potential difference and the geometry of the wire and cylinder, but the calculations yielded incorrect results. It was identified that the logarithmic expression for charge density was misapplied, leading to confusion over the values used. A significant error was also noted regarding the conversion of the wire's radius from centimeters to meters, which impacted the final calculations. The conversation emphasizes the importance of careful unit conversion and correct application of formulas in electrostatics.
Bryon
Messages
98
Reaction score
0

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?
 
Physics news on Phys.org
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!
 
\frac{1.1 \times 10^{-4}cm}{2} \times \frac{1 m}{1 \times 10^2 cm} = 5.50 \times 10^{-7} m

: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!
 
Back
Top