题目内容
如图所示,在水平面上固定一个半圆形细管,在直径两端A、B分别放置一个正点电荷Q1、Q2,且Q2=8Q1.现将另一正点电荷q从管内靠近A处由静止释放,设该点电荷沿细管运动过程中最小电势能的位置为P,并设PA与AB夹角为
,则以下关系正确的是
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035121794890.png)
A. tan
=4 B. tan
=8 C. tan
=2 D. tan
=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512272386.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512163297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035121794890.png)
A. tan
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512163297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512163297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512163297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512163297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512272386.png)
C
试题分析:A点释放后,在AB端点电荷的合电场作用下,电场力做正功,电势能减少,当合场强方向与运动路径垂直,即与半径共线时,电场力不再做正功,此后开始做负功,所以合场强与半径共线时即是电势能最小的P点。电场合成如下图,根据几何关系有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035122881917.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003512304521.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035123353296.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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