【題目】如圖所示，已知平行板電容器兩極板間距離d＝4 mm，充電后兩極板電勢差為120 V。A板帶正電，若它的電容為3 μF，且P到A板距離為1 mm。求：

（1）每個極板的帶電荷量；

（2）一個電子在P點具有的電勢能；

（3）一個電子從B板出發(fā)到A板獲得的動能；

（4）兩板間的電場強度。

【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）因為電子從B板出發(fā)到A板的過程中電場力做正功，電勢能減小，動能增加，所以由動能定理有Ek＝qUAB，得Ek＝120 eV

（4）E＝＝N/C =3×104 N/C

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