A capacitor is built from two circular plates of radius R = 2.1 cm separated (in

Question

A capacitor is built from two circular plates of radius R = 2.1 cm separated (in air) by a distance d = 2.4 mm . An oscillating voltage, V=V0cos(t), with a frequency of 77.5 Hz is placed across the capacitor. The peak displacement current in the capacitor is 35 A .

Part A: What is the peak "conduction" currentI in the wires connecting the voltage source to the capacitor?

I = ____3.5*10^-5____A (already found)

Part B: Find the value of V0.

V0 = ________ _____

Part C: Find the maximum rate of change of the electric flux, dE / dt between the capacitor plates. Ignore any fringing of the fields near the edges of the plates.

(dE / dt)max   = _____________ V * m/s

Explanation / Answer

part B:

capacitance of a parallel plate capacitor is given by

C=epsilon*area/distance between plates

where epsilon=electrical permitivity of free space=8.85*10^(-12)

area=pi*R^2=1.3854*10^(-3) m^2

then C=8.85*10^(-12)*1.3854*10^(-3)/(2.4*10^(-3))

=5.1087*10^(-12) F

then caapcitive reactance magnitude=1/(2*pi*frequency*C)

=1/(2*pi*77.5*5.1087*10^(-12))

=4.0198*10^8 ohms

then as given that current is 35 uA

==>V0/capacitive reactance=35 *10^(-6)

==>V0=4.0198*10^8*35*10^(-6)

==>V0=1.407*10^4 volts

part C:

as electric flux =integration of epsilon*E*dS

where E=electric field

dS=incremental area element

here E=voltage/distance between plates

=V0*cos(w*t)/(2.4*10^(-3))

=5.8622*10^6*cos(w*t)

then flux=integration of 8.85*10^(-12)*5.8622*10^6*cos(w*t)*dS

=5.188*10^(-5)*cos(w*t)*total area

=5.188*10^(-5)*cos(w*t)*1.3854*10^(-3)

=7.1875*10^(-8)*cos(w*t)

then time derivative of flux=-7.1875*10^(-8)*w*sin(w*t)

then maximum value of time derivative of flux=7.1875*10^(-8)*w

using the value of w=2*pi*frequency=2*pi*77.5=486.95 rad/sec

we get maximum value of time derivative of flux=7.1875*10^(-8)*486.95=3.5*10^(-5) V.m/s

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