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Asked in JEE Main 2003 · Conducting slab inserted
A conducting slab of thickness t inserted parallel to the plates leaves an effective air gap of d-t, so C=(ε₀A)/(d-t).
The field inside the conductor is zero, which is why only the remaining gap counts.
With foil of negligible thickness, t→0.
C→(ε₀A)/d, the original value, so the capacitance remains unchanged.
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