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If the distance between the parallel plates of a capacitor is halved and the dielectric constant is doubled, then the capacitance

1. Decreases two times
2. Increases two times
3. Increases four times
4. Remains the same

Correct answer: Increases four times

Solution

For a parallel-plate capacitor, $C=\kappa\varepsilon_0 A/d$. If the plate separation is halved, capacitance doubles; if the dielectric constant is doubled, capacitance doubles again. Therefore the total capacitance becomes four times.

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