题目内容
如图所示,有一电子(电荷量为e)经电压U0加速后,进入两块间距为d、电压为U的平行金属板间.若电子从两板正中间垂直电场方向射入,且正好能穿过电场,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242046222051374.png)
(1)金属板AB的长度;
(2)电子穿出电场时的动能.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242046222051374.png)
(1)金属板AB的长度;
(2)电子穿出电场时的动能.
(1)L=d
.(2)e(U0+
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204622922609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204623749441.png)
试题分析:(1)设电子飞离加速电场时的速度为v0,由动能定理得eU0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204624482338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204625216376.png)
设金属板AB的长度为L,电子偏转时间t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204625871428.png)
电子在偏转电场中产生偏转加速度a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204626542624.png)
电子在电场中偏转y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204624482338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204624482338.png)
由①②③④得:L=d
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204622922609.png)
(2)设电子穿过电场时的动能为Ek,根据动能定理Ek=eU0+e
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204623749441.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824204623749441.png)
点评:电子先在加速电场中做匀加速直线运动,后在偏转电场中做类平抛运动,根据电子的运动的规律逐个分析即可
![](http://thumb2018.1010pic.com/images/loading.gif)
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