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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 VA板带正电,若它的电容为3 μF,且P到A板距离为1 mm求:

1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136×10-4 C 2-90 eV 3120 eV 43×104 N/C

解析1由Q=UC得Q=120×3×10-6 C=36×10-4 C

2EP=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Ek=qUAB,得Ek=120 eV

4E=N/C =3×104 N/C

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