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A parallel-plate capacitor filled with air is first charged so that the potential difference across it is V volts. The charging battery is then removed, and the plate separation is increased using an insulating handle. What happens to the potential difference between the plates?

1. It remains unchanged
2. It becomes zero
3. It increases
4. It decreases

Correct answer: It increases

Solution

With the battery removed the charge Q is constant. Increasing separation reduces C = e0*A/d, so V = Q/C increases.

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