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.

Question

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 2.00e^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.)

Answer 1

Treat the geiger counter as a cylindrical capacitor. The formulas you need to solve this problem can be found at

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html
The E field at radius r can be used to calculate the charge per unit length on the wire. The capacitance C per unit length is determined by the geometry. Once you have C and the charge per unit length, the Voltage is Q/C

Answer 2

thanks i normally look at that site but I wasn't thinking capacitors

Answer 3

To find the potential difference (V_wc) between the wire and the cylinder that produces an electric field of 2.00 × 10^4 volts per meter at a distance of 1.20 centimeters from the wire's axis, you can use the following formula:

Electric field (E) = V_wc / r

where:
- E is the electric field.
- V_wc is the potential difference.
- r is the distance from the wire's axis.

In this case, we know that the electric field (E) is 2.00 × 10^4 volts per meter, and the distance from the wire's axis (r) is 1.20 centimeters.

Converting the distance to meters:
r = 1.20 centimeters = 0.012 meters

Plugging these values into the formula, we can solve for V_wc:

2.00 × 10^4 = V_wc / 0.012

To isolate V_wc, we can multiply both sides of the equation by 0.012:

0.012 × 2.00 × 10^4 = V_wc

V_wc = 240 volts

Therefore, the potential difference V_wc between the wire and the cylinder that produces an electric field of 2.00 × 10^4 volts per meter at a distance of 1.20 centimeters from the axis of the wire is 240 volts.

Answer 4

To determine the potential difference (V_wc) between the wire and the cylinder that produces an electric field of 2.00e^4 volts per meter at a distance of 1.20 centimeters from the axis of the wire, you can use the formula for the electric field created by a cylindrical conductor:

E = (V_wc) / (ln(R2/R1))

Where:
- E is the electric field strength
- V_wc is the potential difference between the wire and the cylinder
- R1 is the radius of the wire
- R2 is the radius of the hollow cylinder

In this case, you are given that the electric field strength (E) is 2.00e^4 volts per meter and the distance from the axis of the wire to where the electric field is measured (R2) is 1.20 centimeters.

First, convert the distance from centimeters to meters:
R2 = 1.20 cm = 0.012 m

Next, substitute the given values into the formula and solve for V_wc:
2.00e^4 = V_wc / ln(0.018/0.000145)

Take the natural logarithm (ln) of the ratio of the cylinder radius to the wire radius:
ln(0.018/0.000145) ≈ ln(124.138)

Now, multiply both sides of the equation by ln(124.138) to isolate V_wc:
2.00e^4 * ln(124.138) = V_wc

Using a calculator, evaluate the right side of the equation:
V_wc ≈ 17130 volts

Therefore, the potential difference V_wc between the wire and the cylinder that produces an electric field of 2.00e^4 volts per meter at a distance of 1.20 centimeters from the axis of the wire is approximately 17130 volts.