题目内容
真空中水平放置的两金属板相距为d,两板电压是可以调节的,一个质量为m、带电量为+q的粒子,从负极板中央以速度vo垂直极板射入电场,当板间电压为U时,粒子经d/4的距离就要返回,若要使粒子经d/2才,返回,可采用的方法是( )
A.vo增大1倍 | B.使板间电压U减半 |
C.vo和U同时减半 | D.初速增为2vo,同时使板间距离增加d/2 |
B
试题分析:A、由题知粒子进入电场做匀减速直线运动,加速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154200454747.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154200844592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241542009531101.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201031323.png)
B、使板间电压U减半,位移增大为原来的2倍,即粒子经过
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201156433.png)
C、
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201031323.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201343338.png)
D、初速增为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201764408.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154201156433.png)
故选B
点评:关键是将电场力看成普通的力,由牛顿第二定律和运动学规律得出粒子位移的表达式。
![](http://thumb2018.1010pic.com/images/loading.gif)
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