A 5.00 pF parallel-plate air-filled capacitor with circular plates is to be used

Question

A 5.00 pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 * 10^2 V. The electric field between the plates is to be no greater than 1.00 * 10^4 N/C. As a budding electrical engineer for Live-Wire Electronics, your task is to design the capacitor by finding what its physical dimensions and separation must be.
a) What must the radius in cm of the plates in the designed capacitor be?
b) What must the separation of the plates in cm in the designed capacitor be?

Explanation / Answer

capacitance C = 5 pF = 5 * 10 ^-12 F potential difference V = 100 V maximum electric field E = 1* 10 ^ 4 N / C we know E = V / d from this separtion of plates d = V / E                                                = 10^-2 m we know C = A / d from this Area of cross section of the plates A = Cd / where = permitivity of free space = 8.85 * 10 ^-12 C^ 2/ Nm^ 2 plug the values we get A = 5.649 * 10 ^-3 m^ 2 we know A = r^ 2 from this radius r = [ A/ ]                             = 0.0424 m                             = 0.0424 * 100 cm                             = 4.24 cm (b). sepration of the plates d = 10^-2 m = 10^ -2 * 100 cm = 1 cm

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