Question

A parallel plate capacitor (Fig. 8.7) made of circular plates each of radiusR = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected toa 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 axisbetween the plates.

Answer 1

(a)

It is given that the radius of each circular plate, R=6⋅0 cm, the capacitance, C=100 pF, the capacitor is connected to a supply of V=230Vac supply and the angular frequency, ω=300 rads −1 .

The formula of rms value of conduction current is,

I= V X C

Substitute the values.

I=230×300×100× 10 −12 =6⋅9× 10 −6  A =6⋅9 μA

Thus, the rms value of conduction current is 6⋅9 μA.

(b)

In this parallel plate capacitor, the conduction current will be equal to the displacement current.

Therefore, the conduction current is equal to the displacement current.

(c)

It is given that the distance between the plates from the axis, r=3⋅0cm.

The formula of magnetic field is,

B= μ 0 r 2π R 2 I 0

The maximum value of current is,

I 0 = 2 I

Substitute the values.

B= 4π× 10 −7 ×0⋅03× 2 ×6⋅9× 10 −6 2π× ( 0⋅06 ) 2 =1⋅63× 10 −11 T

Thus, the value of the amplitude of magnetic field is 1⋅63× 10 −11  T.