题目内容
两板水平放置,相距为d,电压为U1;C、D两板竖直放置,相距也是d,电压为U2。
今有一电子经电压U0加速后,平行于金属板进入电场,当电子离开电场时,偏离子入
射方向多少距离?这时它的动能多大?(设极板间的电场是匀强电场,并设电子未与极
板相碰)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082411561590210088.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241156159172107.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615933973.gif)
粒子在X、y方向均作类平抛运动,由mVo2/2=eU0(1分),
Y方向偏距△y=
(3分),
同理,X方向△x=
(3分),
合偏距为
(2分),(2)
总动能为EK=eU0+e△y
+e△x
(3),
得EK=
.
Y方向偏距△y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615948931.gif)
同理,X方向△x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615964499.gif)
合偏距为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615980970.gif)
总动能为EK=eU0+e△y
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615995365.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115616011375.gif)
得EK=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115615933973.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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