题目内容
两个大小相同的小球带有同种电荷(可看作点电荷),质量分别为m1和m2,带电量分别是q1和q2,用两等长的绝缘线悬挂后,因静电力而使两悬线张开,分别与竖直方向成夹角α1和α2,如图9-36-6所示,若α1=α2,则下述结论正确的是( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162623925629.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162623925629.png)
A.q1一定等于q2 | B.一定满足![]() |
C.m1一定等于m2 | D.必定同时满足q1=q2,m1=m2 |
C
试题分析:可任选
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162624237362.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162624393386.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162624237362.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162624798885.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162624939457.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626252822305.png)
点评:做本题的关键是对两小球受力分析,结合平衡条件分析解题,比较简单
![](http://thumb2018.1010pic.com/images/loading.gif)
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