Electric Potential: Geiger Coutner

In summary: Thanks for catching that!In summary, the wire has a potential difference of 870 volts between it and the cylinder, and the electric field at the wire's surface is 8.149075e7 V/m.
  • #1
Bryon
99
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?
 
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  • #2
Check your expression used for the charge density.
 
  • #3
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
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
Check your value of a (the wire radius). Could be an order of magnitude issue.
 
  • #6
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!
 
  • #7
[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
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!
 

FAQ: Electric Potential: Geiger Coutner

1. What is an Electric Potential?

Electric potential, also known as voltage, is the amount of potential energy that a charged particle possesses at a specific point in an electric field. It is measured in volts (V) and is a fundamental concept in understanding the behavior of electricity.

2. How is Electric Potential related to Geiger Counters?

Geiger counters use the concept of electric potential to detect and measure radiation. The high voltage applied to the electrodes of a Geiger counter creates a strong electric field, which causes ionizing radiation to produce a current that can be measured and counted.

3. What factors affect Electric Potential in a Geiger Counter?

The electric potential in a Geiger counter can be affected by several factors, including the voltage applied to the electrodes, the type and intensity of radiation being detected, and the distance between the electrodes.

4. How does Electric Potential influence the accuracy of Geiger Counter readings?

The electric potential in a Geiger counter can greatly affect the accuracy of radiation readings. If the potential is too low, the sensitivity of the counter may be insufficient to detect low levels of radiation. On the other hand, if the potential is too high, it can cause the counter to saturate and give inaccurate readings.

5. How is Electric Potential measured in a Geiger Counter?

Electric potential in a Geiger counter is typically measured using a voltmeter, which is connected to the electrodes of the counter. The voltmeter displays the voltage and allows for adjustments to be made to ensure accurate readings.

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