A pair of parallel plates are use to create a capacitor. The gap between the pla
ID: 1581968 • Letter: A
Question
A pair of parallel plates are use to create a capacitor. The gap between the plates is filled with air. The area of each plate is 5.20 x 10^-3 m^2. If the electric field between the plates exceeds a value of 1.70 x 10^6 V/m, the air in the capacitor will breakdown and a spark will jump from one plate to another. The capacitor needs to be able to withstand a voltage of 3600 V.
What is the minimum distance that can separate the two plates?
What would be the capacitance of this capacitor if the plates are separated by the minimum distance?
What is the maximum charge that could be held on one plate of this capacitor (while the opposite plate holds the opposite charge)?
Explanation / Answer
seperation = V/E = 3600 / (1.7*10^6)
= 0.002 m
Capacitance = Aeo / d
= 5.2*10^-3*8.85*!0^-12 / 0.002
= 2.17*10^-11 F = 21.7*10^-12 F = 21.7 pF
maximum charge = CV = 21.7 * 3600 = 78234 pF = 78.23 nC
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