题目内容
图甲所示的平行板电容器板间距离为d,两板所加电压随时间变化图线如图乙所示,t=0时刻,质量为m、带电量为q的粒子以平行于极板的速度V0射入电容器,t1=3T时刻恰好从下极板边缘射出电容器,带电粒子的重力不计,求:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634840871.gif)
(1)平行板电容器板长L;
(2)子射出电容器时偏转的角度φ;
(3)子射出电容器时竖直偏转的位移y。
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634840871.gif)
(1)平行板电容器板长L;
(2)子射出电容器时偏转的角度φ;
(3)子射出电容器时竖直偏转的位移y。
(1)L =3v0T (2)φ=arctan
(3)![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871734.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871517.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871734.gif)
(1)t=3T,○○方向匀速直线运动,L=v0t=3v0T
(2)射出电容器时,偏转角度为φ, tgφ=vy/v0, vy=2at=
,
∴φ=arctan![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871517.gif)
(3)电场方向粒子先匀加速再匀减速然后又匀加速至出电容器,
0~T,
T~2T,![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634934754.gif)
2T~3T,![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/2014082411563494973.gif)
∴ ![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871734.gif)
(2)射出电容器时,偏转角度为φ, tgφ=vy/v0, vy=2at=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634887488.gif)
∴φ=arctan
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871517.gif)
(3)电场方向粒子先匀加速再匀减速然后又匀加速至出电容器,
0~T,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634918760.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634934754.gif)
2T~3T,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/2014082411563494973.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634965692.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824115634871734.gif)
![](http://thumb.zyjl.cn/images/loading.gif)
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