题目内容
长为l的平行金属板,板间形成匀强电场,一个带电为+q、质量为m的带电粒子,以初速v0紧贴上板垂直于电场线方向射入该电场,刚好从下板边缘射出,末速度恰与下板成30°,如图所示.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241251393324802.jpg)
小题1:粒子末速度的大小;
小题2:匀强电场的场强;
小题3:两板间的距离d.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241251393324802.jpg)
小题1:粒子末速度的大小;
小题2:匀强电场的场强;
小题3:两板间的距离d.
小题1:V =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139363425.gif)
小题2:E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139441508.gif)
小题3:d =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139597253.gif)
由运动学公式和牛顿定律求解。
(1)由图可得
V= V0/cos30°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241251396284652.jpg)
(2)∵ V⊥= V0tan30°=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139846294.gif)
t=L/V0∴V⊥=at=
,解得:E=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139441508.gif)
(3)d =
·t=
L.
(1)由图可得
V= V0/cos30°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139363425.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241251396284652.jpg)
(2)∵ V⊥= V0tan30°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139846294.gif)
|
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139862436.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139441508.gif)
(3)d =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139893232.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125139597253.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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