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.)
See my response to your second post of the same question.
i have the same problem.
To determine the potential difference (V_wc) between the wire and the cylinder, we can use the formula for electric field (E) produced by a uniformly charged cylinder:
E = (λ / (2 * ε₀)) * (1 - (r² / R²))
Where λ is the charge density, ε₀ is the permittivity of free space, r is the distance from the axis of the wire, and R is the radius of the cylinder.
In this case, we are given the electric field (2.00 * 10^4 V/m), the distance from the axis (1.20 cm or 0.012 m), and the radius of the cylinder (1.80 cm or 0.018 m). We need to solve for the potential difference (V_wc).
First, let's convert the radius and distance from centimeters to meters:
r = 0.012 m
R = 0.018 m
Next, we rearrange the formula for electric field to solve for λ:
λ = E * (2 * ε₀) / (1 - (r² / R²))
Now we can calculate λ:
λ = (2.00 * 10^4 V/m) * (2 * ε₀) / (1 - (0.012² m² / 0.018² m²))
Next, we need to calculate the charge density (λ). The wire is insulated from the hollow cylinder, so it will assume a uniform charge density to achieve the desired electric field.
The charge density is given by:
λ = Q / L
Where Q is the total charge on the wire and L is the length of the wire.
Since the wire and cylinder are both very long, we can assume the wire extends to infinity in both directions. Therefore, the length of the wire is not a factor here.
Now we equate the two forms of λ:
λ = (2 * ε₀ * E) / (1 - (r² / R²)) = Q / L
Since Q / L is constant, let's call it k:
k = (2 * ε₀ * E) / (1 - (r² / R²))
Now we can solve for Q, the charge on the wire:
Q = k * L
We can also find the potential difference (V_wc) between the wire and the cylinder using:
V_wc = Q / C
Where C is the capacitance between the wire and the cylinder.
Since the wire and cylinder form a parallel plate capacitor, the capacitance can be calculated using:
C = (2πε₀) / ln(R / r)
Now we have all the information to calculate V_wc:
V_wc = Q / C = (k * L) / ((2πε₀) / ln(R / r))
Substituting the given values:
V_wc = k * (2πε₀) * ln(R / r) / L
Therefore, to find the potential difference (V_wc) between the wire and the cylinder, you need to calculate the charge density (λ) using the electric field (E), the distance from the axis (r), and the radius of the cylinder (R). Then, calculate the potential difference using the equations described above.
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.