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

Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10^-12 F Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s^–1
Ans (a).

Rms value of conduction current, I = v/ x_c

Where, Xc = Capacitive reactance =1/ωC

Therefore ,I = V x ωC = 230 x 300 x 100 x 10^–12 = 6.9 x 10^-6 A = 6.9 μA

Hence, the rms value of conduction current is 6.9μA.

Ans (b).

Yes, conduction current is equal to displacement current.

Ans (c).

Magnetic field is given as: B = μ₀rI₀/2πR²

Where,

μ₀ = Free space permeability = 4π x 10^-7 NA^–2

Io = Maximum value of current = √2I

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

Therefore, B = (4π x 10^-7  x 0.03 x √2 x 6.9 x 10^-6) /(2π x 0.06²)

=1.63  x 10^-11 T

Hence, the magnetic field at that point is 1.63 x 10^-11 T.