题目内容
(13分)如图所示,在两条平行的虚线内存在着宽度为L、场强为E的匀强电场,在与右侧虚线相距也为L处有一与电场平行的屏。现有一电荷量为+q、质量为m的带电粒子(重力不计),以垂直于电场线方向的初速度v0射入电场中,v0方向的延长线与屏的交点为O。试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250014369222122.png)
(1)粒子从射入到打到屏上所用的时间;
(2)粒子打到屏上的点P到O点的距离;
(3)粒子在整个运动过程中动能的变化量。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250014369222122.png)
(1)粒子从射入到打到屏上所用的时间;
(2)粒子打到屏上的点P到O点的距离;
(3)粒子在整个运动过程中动能的变化量。
(1)
(2)
(3) ![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436969846.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436938499.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436953860.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436969846.png)
试题分析:(1)根据题意,粒子在垂直于电场线的方向上做匀速直线运动,所以粒子从射入到打到屏上所用的时间t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436938499.png)
(2)设粒子射出电场时沿平行电场线方向的速度为vy,
根据牛顿第二定律,粒子在电场中的加速度为:a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437156547.png)
所以vy=a
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437172425.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437203681.png)
所以粒子刚射出电场时的速度方向与初速度方向间夹角的正切值为:
tan α=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437218451.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437234715.png)
设粒子在电场中的偏转距离为y,则 y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437250338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437265682.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001437281802.png)
又s=y+Ltanα,解得:s=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436953860.png)
(3)整个运动过程只有电场力做功,根据动能定理:△EK=qEy=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001436969846.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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