[题目]如图所示.已知平行板电容器两极板间距离d＝4 mm.充电后两极板电势差为120 V.A板带正电.若它的电容为3 μF.且P到A板距离为1 mm.求:(1)每个极板的带电荷量,(2)一个电子在P点具有的电势能,(3)一个电子从B板出发到A板获得的动能,(4)两板间的电场强度. 题目和参考答案——青夏教育精英家教网—

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【题目】如图所示，已知平行板电容器两极板间距离d＝4 mm，充电后两极板电势差为120 V。A板带正电，若它的电容为3 μF，且P到A板距离为1 mm。求：

（1）每个极板的带电荷量；

（2）一个电子在P点具有的电势能；

（3）一个电子从B板出发到A板获得的动能；

（4）两板间的电场强度。

【答案】（1）3．6×10－4 C （2）－90 eV （3）120 eV （4）3×104 N/C

【解析】（1）由Q＝UC得Q＝120×3×10－6 C＝3．6×10－4 C

（2）EP＝eφP＝edPB＝－90 eV

（3）因为电子从B板出发到A板的过程中电场力做正功，电势能减小，动能增加，所以由动能定理有Ek＝qUAB，得Ek＝120 eV

（4）E＝＝N/C =3×104 N/C

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