Q. 23.8( 17 Votes )

# 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.

Answer :

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:

I_{rms} = V/X_{c}

Where I_{rms} is the root mean square value of the conduction current

V is the voltage

X_{c} is the capacitive reactance and is given by X_{c} = 1/ω_{c}

⇒ I_{r. m. s} = V × ω_{C}

Substituting the values we get

I_{rms} = 230 V×300 rad s^{-1}×100×10^{-12}F

I_{rms} = 6.9 × 10^{-6}A= 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

μ_{0} is the free space permeability and is equal to 4π×10^{-7}NA^{-2}

I_{0} is equal to the maximum value of current i.e. √2I

On calculating, we get 162.63

⇒ B = 166 × 10^{-11} T

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