题目内容
长为L的平行金属板水平放置,两极板带等量的异种电荷,板间形成匀强电场,一个带电量为+q、质量为m的带电粒子,以初速度v0紧贴上极板垂直于电场线方向进入该电场,刚好从下极板边缘射出,射出时速度恰与下极板成30º角,如图所示,不计粒子重力,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016537194021.jpg)
求:(1)粒子末速度的大小 (2)匀强电场的场强 (3)两板间的距离
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016537194021.jpg)
求:(1)粒子末速度的大小 (2)匀强电场的场强 (3)两板间的距离
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653765688.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653734614.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653750917.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653765688.png)
试题分析:(1)粒子垂直进入匀强电场,做类平抛运动
由速度关系得粒子末速度:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016537811001.png)
(2)竖直方向的速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016537811007.png)
由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653797591.png)
由平抛规律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653812465.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653828500.png)
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653750917.png)
(3)由平抛规律得竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653859645.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653875447.png)
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001653765688.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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