A parallel plate capacitor of area A=16 cm² and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A₀=3.2 cm² inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through A₀ is __________ mA .

Solution

Physics sir · 3 steps

The displacement current spreads evenly over the plate area, so a smaller patch gets its area's share of 6 A.

1. Total displacement current. Between the plates, the displacement current I_d = ε₀ dΦ_E/dt equals the conduction current in the wires. Across the full plate area A it is therefore 6 A.
2. Uniform field. The field between the plates is uniform, so the rate of change of flux, and with it the displacement current, is shared in proportion to area. A surface of area A₀ picks up the fraction A₀/A of the total.

   I_(d,0) = I A₀/A = 6 A × (3.2 cm²)/(16 cm²) = 1.2 A

3. So. The displacement current through A₀ is 1.2 A, which is 1200 mA. The 10 cm plate separation plays no part.

The common slip: 6000. Taking the whole 6 A as the displacement current through A₀ gives 6000 mA. Only the fraction A₀/A = 0.2 of the flux passes through the smaller surface.

The idea to keep. Inside a capacitor the displacement current equals the circuit current over the full plate area. For any smaller parallel patch, scale it by the area ratio.

Answer from the NTA final answer key. Solution by PrepWiser.