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Posted by on Tuesday, February 19, 2008 at 7:35pm.

A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it. View Figure A large potential difference is established between the wire and the outer cylinder, with the wire at a higher potential; this sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted into an audible click. Suppose the radius of the central wire is 145 micrometers and the radius of the hollow cylinder is 1.80 centimeters.

What potential difference V_wc between the wire and the cylinder produces an electric field of 2.00 \times 10^{4} volts per meter at a distance of 1.20 centimeters from the axis of the wire? (Assume that the wire and cylinder are both very long in comparison to their radii.)

  • long physics E&M - , Tuesday, February 19, 2008 at 11:15pm

    See my response to your second post of the same question.

  • long physics E&M - , Sunday, October 3, 2010 at 6:34pm

    i have the same problem.

  • long physics E&M - , Thursday, October 6, 2011 at 8:56pm

    IDENTIFY: The wire and hollow cylinder form coaxial cylinders. Problem 23.61 gives ( ) 1
    ln( / )
    ab E r V
    b a r
    = .
    SET UP: a =145×10−6 m , b = 0.0180 m .
    EXECUTE: 1
    ln( )
    ab E V
    b a r
    = and ln( (2.00 104 N C)(ln(0.018 m 145 10 6 m))0.012 m 1157 V. ab V = E b/a)r = × × − =
    EVALUATE: The electric field at any r is directly proportional to the potential difference between the wire and the
    cylinder.

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