# A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF.

Question:

A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius = 6.0 cm has a capacitance = 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 at a point 3.0 cm from the axis between the plates. Solution:

Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F

Supply voltage, V = 230 V

Angular frequency, $\omega=300 \mathrm{rad} \mathrm{s}^{-1}$

(a) Rms value of conduction current, $I=\frac{V}{X_{C}}$

Where,

XC = Capacitive reactance

$=\frac{1}{\omega C}$

∴ I = V × ωC

= 230 × 300 × 100 × 10−12

= 6.9 × 10−6 A

= 6.9 μA

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

(b) Yes, conduction current is equal to displacement current.

(c) Magnetic field is given as:

$B=\frac{\mu_{0} r}{2 \pi R^{2}} I_{0}$

Where,

$\mu_{0}=$ Free space permeability $=4 \pi \times 10^{-7} \mathrm{~N} \mathrm{~A}^{-2}$

$I_{0}=$ Maximum value of current $=\sqrt{2} I$

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

$\therefore B=\frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi \times 0(06)^{2}}$

= 1.63 × 10−11 T

Hence, the magnetic field at that point is 1.63 × 10−11 T.