How to Calculate Plate Area and Maximum Voltage of a Capacitor
What plate area is required if an air-filled, parallel-plate capacitor with a plate separation of 2.4 mm is to have a capacitance of 18 pF?
A. 48.3 square millimeters
What is the maximum voltage that can be applied to this capacitor without causing dielectric breakdown?
A. Approximately 7.2 kV
Answer:
The plate area required for an air-filled, parallel-plate capacitor with a plate separation of 2.4 mm and a capacitance of 18 pF is approximately 48.3 square millimeters. The maximum voltage that can be applied to this capacitor without causing dielectric breakdown is approximately 7.2 kV.
To calculate the plate area required for the given capacitor, we can use the formula for capacitance of an air-filled parallel-plate capacitor:
C = ε₀*A/d
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of each plate, and d is the separation between the plates.
By rearranging the formula to solve for A, we get:
A = Cd/ε₀
Plugging in the values of capacitance (18 pF), separation distance (2.4 mm), and permittivity of free space, we find that the plate area required is approximately 48.3 square millimeters.
To determine the maximum voltage that can be applied without causing dielectric breakdown, we can use the breakdown voltage formula for air:
Vb = d/t
where Vb is the breakdown voltage, d is the distance between the plates, and t is the breakdown strength of air.
Plugging in the values of separation distance (2.4 mm) and breakdown strength of air, we find that the maximum voltage is approximately 7.2 kV.