题目内容
如图,为了确定电场中P点场强大小,用细线悬挂一个带负电的小球去探测,当球在P点静止后,测出悬线与竖直方向夹角为45º。已知P点的场强方向是水平方向,小球重力为4×10-3N,带电量为1×10-8C,试求P点场强的大小和方向? ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626400711708.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626400711708.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162640227793.png)
试题分析:带电小球受力分析如图:由受力分析可知,带电小球受到的电场力水平向左,小球带负电,
因而电场方向水平向右……(4分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626404142914.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162640632675.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162640788689.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162640227793.png)
点评:关键是对小球受力分析,根据平衡条件,列式求解
![](http://thumb2018.1010pic.com/images/loading.gif)
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