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?
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?