题目内容
两平行金属板相距为d,电势差为U,一电子质量为m.电量为e,从O点沿垂直于极板的方向射出,最远到达A点,然后返回.如图所示,
=h,此电子具有的初动能是:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145545006800.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145544928378.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145545006800.png)
A.![]() | B.edUh; | C.![]() | D.![]() |
D
试题分析:电子受到的电场力做负功,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241455455991210.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145545537474.png)
点评:本题难度较小,粒子在只受电场力作用下做匀减速直线运动,电场力做功等于动能变化量,本题应该注意的是动能定理公式中电荷的正负、电势差下角标、电势差的正负的带入容易出错
![](http://thumb2018.1010pic.com/images/loading.gif)
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