题目内容
如图所示,光滑水平桌面上有A、B两个带电小球(可以看成点电荷),A球带电量为+2q,B球带电量为-q,由静止开始释放后A球加速度大小为B球的两倍。现在AB中点固定一个带电C球(也可看作点电荷),再由静止释放A、B两球,结果两球加速度大小相等。则C球带电量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241325529042938.png)
A.
B.
C.
D.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553200376.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241325529042938.png)
A.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132552950398.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553138375.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553169383.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553200376.png)
AD
A、B两个带电小球带异号电荷,相互吸引,根据由静止开始释放后A球加速度大小为B球的两倍,由牛顿第二定律可知B球质量为A球的两倍。现在AB中点固定一个带电C球,设C球带电量为Q,A、B两个带电小球之间距离为2r,A球质量为m,B球质量为2m。若两球加速度方向相反,对A球,由牛顿第二定律,k
-k
=ma;对B球,由牛顿第二定律,k
+k
=2ma;联立解得:Q=
,选项A正确。若两球加速度方向相同,对A球,由牛顿第二定律,k
-k
=ma;对B球,由牛顿第二定律,k
+k
=2ma;联立解得:Q=
,选项D正确。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553247703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553278637.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553247703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553372557.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132552950398.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553278637.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553247703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553247703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553372557.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132553200376.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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