[题目]把一个3 pF的平行板电容器接在9 V的电池上.(1)保持与电池的连接.两极板的距离减半.极板上的电荷增加还是减少?电荷变化了多少?(2)移去电池后两极板的距离减半.极板间的电势差增大还是减小?电势差变化了多少? 题目和参考答案——青夏教育精英家教网—

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【题目】把一个3 pF的平行板电容器接在9 V的电池上。

（1）保持与电池的连接，两极板的距离减半，极板上的电荷增加还是减少？电荷变化了多少？

（2）移去电池后两极板的距离减半，极板间的电势差增大还是减小？电势差变化了多少？

【答案】（1）电荷增加 2．7×10－11 C （2）电势差减小 4．5 V

【解析】（1）当电源与电容器一直连接时，电容器两极间电压保持不变，电容器原来带电荷量：Q＝UC＝9×3×10－12 C＝2．7×10－11 C

当两极板的距离减半时，C′＝2C＝6×10－12 F

电容器后来带电荷量：Q′＝UC′＝5．4×10－11 C

极板上电荷增加了ΔQ＝Q′－Q＝2．7×10－11 C

（2）移去电池，电容器上电荷量Q不变，由C′＝得电势差U′＝＝V＝4．5 V，由此可知距离减半时，极板间电势差减小。

故电势差改变量为ΔU＝U－U′＝4．5 V

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