题目内容
如图所示,已充电的平行板电容器两极板水平放置,开关S断开时,电容器内有一带电尘埃恰好处于静止状态.已知C="4" μF,L="0.1" mH.现突然将S闭合并开始计时,求t=2π×10-5 s时,尘埃运动的加速度.(设尘埃未与极板相碰,g取10 m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241207016651748.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241207016651748.jpg)
20 m/s2,方向竖直向下
设电容器上极板带正电,下极板带负电,则带电尘埃带负电,由平衡条件知
mg=qE
电路振荡周期T=2π![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120701680433.png)
=2π
=4π×10-5 s
因为t=2π×10-5 s=
,所以t=2π×10-5 s时,电容器恰好反向充电完毕,此时尘埃受向下的重力和电场力作用,则qE=mg
据牛顿第二定律有mg+qE=ma
所以a="2g=20" m/s2,方向竖直向下.
mg=qE
电路振荡周期T=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120701680433.png)
=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120701727761.png)
因为t=2π×10-5 s=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120701758406.png)
据牛顿第二定律有mg+qE=ma
所以a="2g=20" m/s2,方向竖直向下.
![](http://thumb2018.1010pic.com/images/loading.gif)
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