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Three identical large parallel conducting plates are held fixed with separation d between adjacent plates. Each plate has area A. Plate 1 carries charge +Q; plates 2 and 3 are uncharged and are connected to each other through an inductor of inductance L and a switch S. Wire resistance is negligible. Define C = epsilon0 * A / d and ignore fringe effects. After the switch S is closed, what is the maximum current that flows through the inductor?

1. Q / sqrt(L * C)
2. Q / sqrt(2 * L * C)
3. 3 * Q / (2 * sqrt(L * C))
4. Q / (2 * sqrt(L * C))

Correct answer: Q / sqrt(2 * L * C)

Solution

When plate 1 carries +Q and plates 2-3 are neutral, charge distributes so inner face of plate 2 holds -Q/2 and inner face of plate 3 holds -Q/2 (by symmetry and charge neutrality). This means each of the two capacitors (1-2 and 2-3) holds charge Q/2. Total stored energy = (Q/2)² / (2C) + (Q/2)² / (2C) = Q² / (4C). At maximum current all energy is in the inductor: (1/2) * L * I² = Q² / (4C), giving I = Q / sqrt(2 * L * C).

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