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