把质量为2g的带电小球A用细绳吊起来，若将带电小球B靠近带电小球A，当B球带电荷量QB=4×10-6C时，两个带电体恰在同一高度，且相距l=30cm，绳与竖直方向的夹角为45°，如图，试求：
（1）A球所受的静电力．
（2）A球所带的电荷量．
（1）分析带电小球A受力，由力的平衡有
tanθ=F mg
∴F=mgtanθ=2×10-3×10×tan45°=0.02N
故A球所受的静电力为0.02N．
（2）根据F=k
得 F=kq1q2 r2 QAQB l2
∴QA=
=F•l2 kQB
C=5×10-8C0.02×0.32 9×109×4×10-6
故A球所带的电荷量为5×10-8C．
请仔细阅读小题，从A、B、C、D中找出其相对应的礼仪规则, 并回答第5小题。 小题1:The first and greatest rule. Talking is not allowed, even including talking quietly during the music. 小题2:The musicians do not need your help, and your neighbors need silence. Learn to tap (轻敲) your fingers quietly on the chair - it’s a good finger exercise. 小题3:Drinks and candies are not allowed. 小题4:Remember that you are representing your school, and you want to be on your best behavior. There are many eyes looking at you.
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