题目内容
如图,带电量为+q的点电荷与均匀带电薄板相距为2d,点电荷到带电薄板的垂线通过板的几何中心.若图中a点处的电场强度为零,根据对称性,带电薄板在图中b点处产生的电场强度大小为________,方向________.(静电力恒量为k)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241648363233740.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241648363233740.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164836541572.png)
试题分析:q在a点形成的电场强度的大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164836791716.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164836978578.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164836978578.png)
点评:题目中要求的是薄板形成的场强,看似无法解决; 但注意a点的场强是由薄板及点电荷的电场叠加而成,故可求得薄板在a点的电场强度,而薄板两端的电场是对称的,故由对称性可解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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