题目内容
如图所示,电量为Q1、Q2的两个正点电荷分别置于A点和B点,两点相距L.在以L为直径的光滑绝缘的半圆环上,穿有负点电荷q(不计重力)且在P点平衡,PA与AB夹角为α,则
应为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241228500701885.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122849946520.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241228500701885.png)
A.![]() | B.![]() | C.![]() | D.![]() |
C
对小球进行受力分析如图所示:根据库仑定律有:F1=k
,r1=Lcosα…①
F2=k
,r2=Lsinα…②
根据平衡条件有:F1sinα=F2cosα…③
联立①②③解得:tan3α=
,故ABD错误,C正确.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122850710526.png)
F2=k
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122850741531.png)
根据平衡条件有:F1sinα=F2cosα…③
联立①②③解得:tan3α=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122850788453.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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