The radiation detector known as a Geiger counter uses a...

The radiation detector known as a Geiger counter uses a closed, hollow, cylindrical tube with an insulated wire along its axis. Suppose a Geiger tube, as it's called, has a 1.0 -mm-diameter wire in a tube with a 25 mm inner diameter. The tube is filled with a low-pressure gas whose dielectric strength is $1.0 \times$ $10^{6} \mathrm{V} / \mathrm{m} .$ What is the maximum potential difference between the wire and the tube?

Key Concepts

Dielectric Strength

Dielectric strength is the maximum electric field magnitude that an insulating material can withstand without experiencing electrical breakdown. It is a critical parameter in designing devices that rely on gas or solid insulators, as it determines the maximum potential difference that can be applied before the material begins to conduct due to ionization.

Cylindrical Geometries

Cylindrical geometries, such as those found in coaxial structures, are frequently used in various electrical components including capacitors and detectors. In these setups, the electric field distribution is governed by the cylindrical symmetry of the arrangement, meaning that the field strength changes with radial distance from the central axis, and this must be taken into account when calculating potential differences.

Electric Field Distribution in Coaxial Systems

In coaxial systems, the electric field is determined using Gauss’s Law, with the field typically being inversely proportional to the radial distance from the center conductor. This relationship is used to derive expressions for both the electric field and the resulting potential difference between the inner and outer conductors, which is crucial for determining operational voltage limits.

Potential Difference and Breakdown Voltage

The maximum potential difference, or breakdown voltage, in a system is determined by connecting the calculated electric field distribution to the dielectric strength of the insulating medium. By ensuring that the electric field anywhere in the device does not exceed the material's dielectric strength, one can compute the maximum voltage that the system can safely handle without initiating the breakdown process.

Transcript

00:01 So the electric field on the infinitely long line with the charge should be equal to landa over 2 pi, epsilon 0 times r.

00:08 Landa is the linear charge density, epsilon zero is the permittivity of free space.

00:12 R is the radius of the tube.

00:17 And we also know that electric view can be equal to negative divv over dr, which means that electric view is negative derivative of the potential.

00:27 So as you can tell, based on the equation, the electric view is related to 1 over r and we know potential is related to e because v is equal to er okay so therefore we know that the minimum radius in this case should be the radius from outside which is the outer radius because the outer diameter is smaller than the inner diameter so therefore we know that if the radius is at maximum level i mean i'm sorry at minimum level so that means the electric field should be at maximum level.

01:05 Okay? and then if electrofews at maximum level, that means the potential is at maximum level as well.

01:14 So therefore, we have emax is equal to landa over 2 pi times epsilon 0, 10 rl, okay? because rll is equal to r minimum, since the outer radius is the smallest radius in this case.

01:27 So if we do some arrangement here, we have the linear charge density landa is equal to 2 pi times epsilon 0 times rr, then times maximum electric fuel, e max.

01:37 So therefore, have e is equal to land out over 2 pi epsilon 0 times r, which is 2 pi epsilon 0 x x x x x x x x x x x x x x x x x x x will give us e x x over r.

01:55 Remember potential is equal to negative integral of the e and the radius is ranging from the inner radius to the outer radius which means that it's from r into r out.

02:10 So therefore, eventually we'll have v is equal to negative rr times e max and then times the integral.

02:18 And what's inside integral is d .r who are.

02:21 So now that's plug in the values to determine the integral...

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