题目内容
如图所示,ABCD是匀强电场中一正方形的四个顶点,已知A、B、C三点的电势分别为
A="15" V,
B ="3" V,
C=" -3" V, 由此可得D点电势为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020277981881.jpg)
A.6V B.9 V C.12 V D.15 V
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002027767336.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002027767316.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002027782210.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020277981881.jpg)
A.6V B.9 V C.12 V D.15 V
B
试题分析:连接AC,将AC三等分,标上三等分点E、F,则根据匀强电场中沿电场线方向相等距离,电势差相等可知,E点的电势为3V,F点的电势为9V.连接BE,则BE为一条等势线,根据几何知识可知,DF∥BE,则DF也是一条等势线,所以D点电势UD=9V,故选项B正确.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020278132958.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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