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Capacitance of a parallel-plate capacitor (area A, separation d) when a dielectric slab of thickness t (t < d) and dielectric constant K is inserted between the plates is
A{'text': 'ε₀ A / (d − t + t/K)', 'label': 'A'}
B{'text': 'ε₀ A K / d', 'label': 'B'}
C{'text': 'ε₀ A / d', 'label': 'C'}
D{'text': 'ε₀ A (d − t) / (K t)', 'label': 'D'}
Answer & Solution
Correct answer: A. {'text': 'ε₀ A / (d − t + t/K)', 'label': 'A'}
1. Treat the gap as two capacitors in series: an air gap of (d − t) and a dielectric slab of t.
2. C_air = ε₀ A / (d − t), C_die = K ε₀ A / t.
3. 1/C = (d − t)/(ε₀ A) + t/(K ε₀ A) ⇒ C = ε₀ A / [(d − t) + t/K].
_Source: NCERT Class 12 Physics Part I, Ch 2 §2.13 derivation_
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