题目内容
(15分)如图所示,两个板长均为L的平板电极,平行正对放置,两极板相距为d,极板之间的电势差为U,板间电场可以认为是匀强电场。一个带电粒子(质量为m,电荷量为+q)从正极板边缘以某一初速度垂直于电场方向射入两极板之间,到达负极板时恰好落在极板边缘。忽略重力和空气阻力的影响。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027185626921.jpg)
(1)极板间的电场强度E的大小;
(2)该粒子的初速度v0的大小;
(3)该粒子落到下极板时的末动能Ek的大小。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027185626921.jpg)
(1)极板间的电场强度E的大小;
(2)该粒子的初速度v0的大小;
(3)该粒子落到下极板时的末动能Ek的大小。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027187491030.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718578592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718593985.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027187491030.png)
试题分析:(1)两极板间的电压为U,两极板的距离为d,所以电场强度大小为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718578592.png)
(2)带电粒子在极板间做类平抛运动,在水平方向上有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718780465.png)
在竖直方向上有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718796668.png)
根据牛顿第二定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718812566.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718827504.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718843730.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718593985.png)
(3)根据动能定理可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002718905944.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027187491030.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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