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  • A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R is equal to 6.0 cm has a capacitance C is equal to 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1.

2.(b) 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 an (angular) frequency of 300 rad s ^{-1}. Is the conduction current equal to the displacement current?

 

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According to questions, it is given that the radius of each circular plate,\ R = 6.0 \, \text{cm} \, the capacitance,\ C = 100 \, \text{pF} \, the capacitor is connected to a supply of \ V = 230 \, \text{V (AC supply)} \,, and the angular frequency, \ \omega = 300 \, \text{rad/s}

The formula for the RMS value of conduction current is given by -

\ I = V \cdot \omega \cdot C \

Substitute the values:

\\ I = 230 \cdot 300 \cdot 100 \cdot 10^{-12} \ \\I = 6.9 \cdot 10^{-6} \, \text{A} = 6.9 \, \mu\text{A} \

Thus, the RMS value of the conduction current is \( 6.9 \, \mu\text{A} \).

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-

\ I_{\text{displacement}} = I_{\text{conduction}} = 6.9 \, \mu\text{A}. \

Thus, the conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying the Ampere – Maxwell Law.

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