Question

A parallel plate capacitor (Figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 Y ac supply with an (angular) frequency of 300 rad s¹.
(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 0 cm from the axis between the plates.

Answer 1

Here,

C = 100 pF = 100 x 10^-12 F

E_v = 230V

ω = 300 rads^-1 and R = 6 cm = 6 x 10^-2 m

(a) The r.m.s value of the conduction current is given by

= 230 x 300 x 100 x 10^-12

= 6.9 x 10^-6 A

(b) Yes the conduction current and the displacement current will be equal. It is true, even if the conduction current is oscillating.

(c) Let us first derive expression for the amplitude of magnetic field at a distance x from the axis of the capacitor. To do so, place a circular loop of radius r between the two plates of the capacitor with its face parallel to the plates and its centre on the axis of the plates.

According to modified Ampere’s circuital law,

The amplitude of magnetic field at a distance r from the axis of the plates is then, given by.