题目内容
如图所示,匀强电场分布在正方形ABCD区域内,M、N分别为AB边和BC边的中点。一个具有初动能E0的带电粒子射入电场(沿纸面运动)。如果带电粒子从M点垂直于电场方向进入电场后,恰好从D点离开电场。带电粒子从D点离开电场时的动能为________;如果带电粒子从N点垂直于BC边方向射入电场,它离开电场时的动能为__________。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028001836537.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028001836537.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002800214573.png)
试题分析:(1)设带电粒子的质量为m、电量为q、初速为v,则 E0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002800230334.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002800230334.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028002611060.png)
带电粒子从M点射入,从D点射出,电场力做功 W=qE?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002800292391.png)
(2)带电粒子从N点垂直于BC边方向射入电场,做匀加速直线运动,离开电场时的动能为E2,据动能定理有 qEL=E2-E0,解得 E2=E0+qEL=3E0.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目