A parallel-plate capacitor of capacitance C is charged to a potential …

Question

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

Accepted 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_

Answer

{'text': 'K² V', 'label': 'A'}

Answer

{'text': 'K V', 'label': 'C'}

Answer

{'text': 'V', 'label': 'D'}