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A parallel-plate capacitor has capacitance $C$. If BOTH the plate area and the plate separation are doubled, the new capacitance is
A$C$
B$4C$
C$2C$
D$C/2$
Answer & Solution
Correct answer: A. $C$
1. For a parallel-plate capacitor $C = \dfrac{\varepsilon_0 A}{d}$.
2. Doubling $A$ doubles the numerator; doubling $d$ doubles the denominator.
3. The two changes cancel: $C' = \dfrac{\varepsilon_0(2A)}{2d} = \dfrac{\varepsilon_0 A}{d} = C$ (unchanged).
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Long Answer Q.18 / Evaluation Q.13_
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