As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and the cloud to be the other plate. The altitude of the cloud is 600m and its area is 6.5 km2. How much charge can the cloud hold before the dielectric strength of air is exceeded and the lightning results?
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To determine the amount of charge a cloud can hold before lightning occurs, we need to use the formula for the capacitance of a parallel-plate capacitor:
C = (ε₀ * A) / d
Where:
C = Capacitance
ε₀ = Permittivity of free space (constant)
A = Area of the plates
d = Distance between the plates (in this case, the altitude of the cloud)
Given that the altitude of the cloud is 600m and the area is 6.5 km², we need to convert these values to the appropriate SI units before we can proceed with the calculation.
1 km² = 1,000,000 m²
1 m = 100 cm
So, the area of the cloud can be represented as:
A = 6.5 km² = (6.5 * 1,000,000) m² = 6,500,000 m²
The altitude of the cloud is already in meters, so we can use it as is:
d = 600 m
Now, we substitute these values into the formula to find the capacitance:
C = (ε₀ * A) / d
The permittivity of free space, ε₀, is approximately 8.854 x 10⁻¹² F/m.
C = (8.854 x 10⁻¹² F/m * 6,500,000 m²) / 600 m
Simplifying the equation:
C = 3.041 x 10⁻⁴ F
The capacitance of the cloud is approximately 3.041 x 10⁻⁴ Farads.
To find the amount of charge the cloud can hold before lightning occurs, we can use the equation:
Q = C * V
Where:
Q = Charge
C = Capacitance
V = Voltage
The dielectric strength of air is approximately 3 x 10⁶ V/m. Therefore, the voltage required to exceed this threshold is the product of the dielectric strength and the distance between the plates:
V = d * E
V = 600 m * 3 x 10⁶ V/m
V = 1.8 x 10⁹ V
Now, we can calculate the maximum charge the cloud can hold:
Q = (3.041 x 10⁻⁴ F) * (1.8 x 10⁹ V)
Q ≈ 5.4748 x 10⁵ C
Therefore, the cloud can hold approximately 54,748 Coulombs of charge before the dielectric strength of air is exceeded, leading to a lightning discharge.