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When the separation between the plates of an air parallel-plate capacitor is doubled (keeping plate area constant), the capacitance
A{'text': 'remains unchanged', 'label': 'A'}
B{'text': 'becomes one-fourth', 'label': 'B'}
C{'text': 'doubles', 'label': 'C'}
D{'text': 'halves', 'label': 'D'}
Answer & Solution
Correct answer: D. {'text': 'halves', 'label': 'D'}
1. C = ε₀ A / d.
2. Doubling d gives C' = ε₀ A / (2d) = C/2.
3. So capacitance halves; the field also halves at fixed Q.
_Source: NCERT Class 12 Physics Part I, Ch 2 §2.12_
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