What Is the Electric Field Strength in a Geiger Counter?

[Kudo Shinichi]

HELP!A problem on electricity?

Homework Statement

A Geiger counter consists of a metal cylinder 2.0 cm in diameter along whose axis is stetched a wire 0.013 cm in diameter. If 850 volts are supplied between them, what is the electric field strength at the surface of a) the wire and b) the cylinder?

The Attempt at a Solution

I know the formula v=-E*d
v=voltage
E=electric field
d= distance between the charges where we are looking for potential difference
since I am given v
so, 850=-E*d
the problem also provides the diameter for both cylinder and wire, I can use them to get the length if I know either the area or volume for both of them. However, I don't, so is there any other way to get the length for both objects to solve for electric field?

Thank you very much

 

[mgb_phys]

The field is just between the inside wire and the outside cylinder - the length doesn't matter, it shoudl be the same field at any point along their length.

Whats the distance between the wires?

 

[Kudo Shinichi]

The field is just between the inside wire and the outside cylinder - the length doesn't matter, it shoudl be the same field at any point along their length.

Whats the distance between the wires?

Sorry, I don't know, I have typed everything on the question
That is why I want to solve for the length of the wire and the cylinder so i can use that as distance(I am not really sure whether it's right or not...)
Then if I don't know the distance between the wires is there any other method to solve the problem?

 

[mgb_phys]

A Geiger counter is a tube (a hollow cylinder) with a wire down the middle along the axis.
So it's just the distance from the surface of the wire own the middle to the outside tube.

 

[Kudo Shinichi]

A Geiger counter is a tube (a hollow cylinder) with a wire down the middle along the axis.
So it's just the distance from the surface of the wire own the middle to the outside tube.

Then the distance is just the diameter of the cylinder minus the diameter of the wire 2-0.013=1.987
and divide the distance with voltage to get electric field

 

[mgb_phys]

yes,
Roughly it's just the radius of the outer cylinder

 

[Kudo Shinichi]

yes,
Roughly it's just the radius of the outer cylinder

Therefore, the distance is 1-0.013
Sorry, I don't quite get it, why is it the distance we are looking for? Can you expain it briefly. Thank you

 

[mgb_phys]

Electric field is voltage/distance = V/m