题目内容
如图,在空间固定有两个等量正点电荷,MN为它们连线的中垂线,O点为连线与中垂线的交点。另一带负电的点电荷,仅在电场力作用下由MN上A处静止释放,在运动过程中负点电荷漏失了一部分电荷,但质量保持不变,负点电荷达B处(图中未画出)时速度变为零,则下列说法正确的是
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250050566051681.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250050566051681.png)
A.OA的距离大于OB的距离 | B.A处场强可能大于B处场强 |
C.A处电势必高于B处电势 | D.负电荷在A处电势能大于B处电势能 |
BC
试题分析:如果负电荷的电量不变,则由对称关系,负电荷应运动到O点以下关于O点对称的位置B1,即满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005056605762.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005056620798.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005056636651.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005056651554.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005056667502.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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