题目内容
(15分)如图所示,水平放置的平行板电容器,两板间距为d=8cm,板长为L=25cm,接在直流电源上,有一带电液滴以υ0=0.5m/s的初速度从板间的正中央水平射入,恰好做匀速直线运动,当它运动到P处时迅速将下板向上提起
cm,液滴刚好从金属板末端飞出,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445543276588.png)
求(1)将下板向上提起后,液滴的加速度大小和方向;
(2)液滴从金属板末端飞出时的速度大小;
(3)液滴从射入运动到P点所用时间。(g取10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144554218373.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445543276588.png)
求(1)将下板向上提起后,液滴的加速度大小和方向;
(2)液滴从金属板末端飞出时的速度大小;
(3)液滴从射入运动到P点所用时间。(g取10m/s2)
(1)
(2)
(3)0.3s
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144554374544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144554436683.png)
试题分析:(1)带电液滴在板间受重力和竖直向上的电场力,因为液滴匀速运动,所以有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445547321228.png)
当下板向上提后,d减小,E增大,电场力增大,故液滴向上偏转,在电场中做类平抛运动.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445548883203.png)
(2)因液滴刚好从金属末端飞出,所以液滴在竖直方向上的位移是d/2,
设液滴从P点开始在匀强电场中的飞行时间为,则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445549661668.png)
竖直方向的速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144555029798.png)
从金属板末端飞出的速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445550911128.png)
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241445553103763.png)
点评:带电粒子在电场中的偏转类似于类平抛运动,分清各个方向上的运动性质是关键,
![](http://thumb2018.1010pic.com/images/loading.gif)
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