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A parallel-plate capacitor of capacitance C is charged to a potential V by a battery, then disconnected. A dielectric slab of dielectric constant K is fully inserted between the plates. The new potential across the capacitor is
A{'text': 'K² V', 'label': 'A'}
B{'text': 'V / K', 'label': 'B'}
C{'text': 'K V', 'label': 'C'}
D{'text': 'V', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': 'V / K', 'label': 'B'}
1. Battery is disconnected, so Q is constant.
2. Inserting a dielectric raises capacitance: C' = K C.
3. V' = Q / C' = (CV) / (KC) = V / K.
4. Energy stored falls to U/K (the rest goes into pulling the slab in).
_Source: NCERT Class 12 Physics Part I, Ch 2 §2.13_
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