题目内容
(18分)
如图15所示,PQ、MN两极板间存在匀强电场,两极板间电势差为U、间距为d,MN极板右侧虚线区域内有垂直纸面向内的匀强磁场。现有一初速度为零、带电量为q、质量为m的离子从PQ极板出发,经电场加速后,从MN上的小孔A垂直进入磁场区域,并从C点垂直于虚线边界射出。求:
(1)离子从小孔A射出时速度v0;
(2)离子带正电还是负电?C点离MN板的距离?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222140959598.jpg)
如图15所示,PQ、MN两极板间存在匀强电场,两极板间电势差为U、间距为d,MN极板右侧虚线区域内有垂直纸面向内的匀强磁场。现有一初速度为零、带电量为q、质量为m的离子从PQ极板出发,经电场加速后,从MN上的小孔A垂直进入磁场区域,并从C点垂直于虚线边界射出。求:
(1)离子从小孔A射出时速度v0;
(2)离子带正电还是负电?C点离MN板的距离?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222140959598.jpg)
(1)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142511.gif)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142609.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142511.gif)
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142609.gif)
(1)由动能定理得:qU=
(4分)
解得离子从小孔A射出时速度v0=
(4分)
注:用牛顿第二定律求解同样给分
(2)由左手定则可知离子带负电(3分)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222141886569.jpg)
离子进入磁场后,洛仑兹力提供向心力,由
牛顿第二定律:qv0B=
(3分)
解得:r=
=
(4分)
所以C点离MN板的距离为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142609.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214157441.gif)
解得离子从小孔A射出时速度v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142511.gif)
注:用牛顿第二定律求解同样给分
(2)由左手定则可知离子带负电(3分)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222141886569.jpg)
离子进入磁场后,洛仑兹力提供向心力,由
牛顿第二定律:qv0B=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214204420.gif)
解得:r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214204441.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142609.gif)
所以C点离MN板的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122214142609.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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