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【題目】把一個3 pF的平行板電容器接在9 V的電池上。

（1）保持與電池的連接，兩極板的距離減半，極板上的電荷增加還是減少？電荷變化了多少？

（2）移去電池后兩極板的距離減半，極板間的電勢差增大還是減小？電勢差變化了多少？

【答案】（1）電荷增加 2．7×10－11 C （2）電勢差減小 4．5 V

【解析】（1）當電源與電容器一直連接時，電容器兩極間電壓保持不變，電容器原來帶電荷量：Q＝UC＝9×3×10－12 C＝2．7×10－11 C

當兩極板的距離減半時，C′＝2C＝6×10－12 F

電容器后來帶電荷量：Q′＝UC′＝5．4×10－11 C

極板上電荷增加了ΔQ＝Q′－Q＝2．7×10－11 C

（2）移去電池，電容器上電荷量Q不變，由C′＝得電勢差U′＝＝V＝4．5 V，由此可知距離減半時，極板間電勢差減小。

故電勢差改變量為ΔU＝U－U′＝4．5 V

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