# 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. 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 = 1 66 × 10-11 T

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