A parallel plate capacitor with plate area A and plate separation d is charged by constant current I. A plane surface of area A/2, parallel to the plates is drawn simultaneously between the plates. The displacement current through this area is (a) I (b) I/2 (c) I/4 (d) 1/8

Answer: (b). Between the plates the field is uniform, so the displacement current is spread evenly over the plate area A. Total displacement current through area A equals the conduction current I. Through area A/2: Id = I/2/A = I/2 So the answer is (b) I/2.

A parallel plate capacitor with plate area A and plate separation d… | MHT CET 2025 (20 Apr, Shift 2)

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Capacitors in DC Circuits & RC

A parallel plate capacitor with plate area A and plate separation $d$ is charged by constant current $I$ . A plane surface of area $\frac{A}{2}$ , parallel to the plates is drawn simultaneously between the plates. The displacement current through this area is

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Answer: (b)

Between the plates the field is uniform, so the displacement current is spread evenly over the plate area $A$ .
Total displacement current through area $A$ equals the conduction current $I$ .
Through area $\dfrac{A}{2}$ : $I_d = I\cdot\dfrac{A/2}{A} = \dfrac{I}{2}$
So the answer is (b) $\dfrac{I}{2}$ .

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