Question

A parallel plate capacitor is constructed of two circular disks, radius 5.26 cm, sharing a common axis. If the air-gap separation distance between them is 2.39 mm, what is the effective capacitance, in pF?

Answer 1

Here ,

distance between plates , d = 2.39 mm = 0.00239 m

r = 5.26 cm = 0.0526 m

effective capacitance = epsilon * Area/distance

effective capacitance = 8.854 *10^-12 * pi * 0.0526^2/0.00239

effective capacitance = 3.22 *10^-11 F

effective capacitance = 32.2 pF

the effective capacitance is 32.2 pF