题目内容
如图所示,A、B、C三点为一直角三角形的三个顶点,∠B = 30°,现在A、B两点放置两点电荷
、
,测得C点场强的方向与AB平行,则电荷量大小之比
为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241458209601738.jpg)
A、1:1 B、1:2 C、1:4 D、1:8
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145820804374.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145820850371.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145820913474.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241458209601738.jpg)
A、1:1 B、1:2 C、1:4 D、1:8
D
试题分析:由A、B在C点产生的场强EC,可作出矢量图,可知A一定带负电,且
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145820975693.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241458210691163.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241458212401015.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241458214591009.png)
点评:电场是矢量,叠加遵循平行四边形定则,然后结合几何关系解题
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目
题目内容