题目内容
如图所示,带正电小球质量为
,带电量为
,置于光滑绝缘水平面上的A点,当空间存在着斜向上的匀强电场时,该小球从静止开始始终沿水平面作匀加速直线运动,当运动到B点时测得其速度
,此时小球的位移为
,求此匀强电场场强E的取值范围。(g=10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224505793639.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450532519.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450548361.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450548472.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450563446.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224505793639.jpg)
设电场方向与水平面夹角为θ,由动能定理![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450594740.gif)
得
V/m,由题意可知,θ>0,所以当
V/m
为使小球始终沿水平面运动,电场力在竖直方向的分力必须小于等于重力,即:qEsinθ≤mg
所以![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224506411229.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224506571619.gif)
即:7.5×104V/m<E≤1.25×105V/m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450594740.gif)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450610930.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122450626473.gif)
为使小球始终沿水平面运动,电场力在竖直方向的分力必须小于等于重力,即:qEsinθ≤mg
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224506411229.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241224506571619.gif)
即:7.5×104V/m<E≤1.25×105V/m
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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