题目内容
(12分)如图所示,有一初速可忽略的电子经电压U1=500V加速后,进入两块水平放置、间距为d=2cm、电压为U2=10V的平行金属板A、B间。若电子从板正中央水平射入,且恰好能从B板的右端射出.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250006341634579.jpg)
(1)金属板的长度L;
(2)电子离开电场的偏转的角度正切值tanθ;
(3)电子从B板右端射出电场时的动能Ek为多少电子伏特。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250006341634579.jpg)
(1)金属板的长度L;
(2)电子离开电场的偏转的角度正切值tanθ;
(3)电子从B板右端射出电场时的动能Ek为多少电子伏特。
试题分析:(1)由动能定理和平抛知识可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634179836.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634195563.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634210645.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634241765.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634257541.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250006342881073.png)
(2)偏转的角度正切值:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250006343041043.png)
(3)由动能定理得:Ek=eU1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000634335338.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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