题目内容
一个质量m=40g,带电量q=-3×10-8C的半径极小的小球用丝线悬挂起来处在某匀强电场中,电场线水平.当小球静止时,测得悬线与竖直方向夹角为37o,如图所示,求该电场的场强大小及方向.(g=10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241516305752763.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241516305752763.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151630622653.png)
试题分析:如图对带电小球进行受力分析由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151630684845.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151630809479.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241516308711349.png)
电场力方向水平向左,电荷带负电,所以电场强度方向水平向右。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241516309802461.jpg)
点评:本题中把电场力看成一个普通的力,根据共点力平衡,利用三角函数关系即可求得未知量。
![](http://thumb2018.1010pic.com/images/loading.gif)
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