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A parallel-plate capacitor with plate area A and plate separation d is charged by a constant current i. Consider a plane surface of area A/4 placed parallel to the plates and located symmetrically between them. What is the displacement current through this surface?

1. i
2. i/2
3. i/4
4. zero

Correct answer: i/4

Solution

The displacement current density Jd = epsilon₀ * dE/dt is uniform between the plates and equals i/A (from total current i through area A). The displacement current through area A/4 is Jd * (A/4) = (i/A) * (A/4) = i/4.

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