Q. 23.8( 17 Votes )

A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1.

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.


Answer :

Given: Radius of circular plates R = 6.0 cm = 0.06 m

Capacitance C = 100 pF = 100 × 10-12 F


Voltage V = 230 V


Angular frequency, ω = 300 rad s-1


The figure is:



(a) The root mean square value of the conduction current is given by the following relation:


Irms = V/Xc


Where Irms is the root mean square value of the conduction current


V is the voltage


Xc is the capacitive reactance and is given by Xc = 1/ωc


Ir. m. s = V × ωC


Substituting the values we get


Irms = 230 V×300 rad s-1×100×10-12F


Irms = 6.9 × 10-6A= 6.9 μ A


(b) Yes, the conduction current is equal to the displacement current.


(c) Amplitude of magnetic field B at a point 3.0 cm from the axis between the plates is given by the following expression:



Distance between the two plates = 3.0 cm = 0.03 m


μ0 is the free space permeability and is equal to 4π×10-7NA-2


I0 is equal to the maximum value of current i.e. √2I




On calculating, we get 162.63


B = 166 × 10-11 T


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