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Electric Field problem

  1. Oct 10, 2005 #1
    I've been working on a problem but I dont understand how to solve it.

    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. A large potential difference is established between the wire and the outer cylinder, with the wire at 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 to an audible "click". The tube of a Geiger counter has a long, hollow, metal cylinder 2.00 cm in diameter. Along the axis of the tube is a wire 0.127 mm in diameter running its full length. When the tube is operating, a voltage of 850 V is applied between the two conductors.

    Find the electric field strength at the outer surface of the wire & the electric field strength at the inner surface of the cylinder.

    I was trying to use this eqn: E(r)=V/ln(b/a)*(1/r) where b=radius of the cylinder and a=radius of wire.

    Any help would be appreciated. Thank you.
  2. jcsd
  3. Oct 10, 2005 #2


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    The form of the E-field is correct (I didn't check the ln(b/a) factor ; it must be something of the kind ; you can verify it by integrating the E-field along a radial line from the surface of the wire to the radius of the tube). So your formula gives you the E-field for an arbitrary distance r from the center of the wire. They ask you to give the value of the E-field for two particular places ; why can't you just put in the numbers for those places (particular values of r) ?

    EDIT: a remark to the problem of general nature: you need to use special gas mixtures. A geiger counter doesn't work in AIR :smile: The reason is that oxygen is way too electronegative, and has "eaten" all electrons drifting before they reach the anode wire.
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