题目内容
如图所示,A、B、C、D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=15V,?B =3V,?C =-3V,由此可知D点电势?D=______V;若该正方形的边长为2cm,且电场方向与正方形所在平面平行,则场强为E=________V/m。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020059112194.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020059112194.jpg)
9V;450
V/m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002005911322.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020059423910.jpg)
试题分析:(1)根据匀强电场分布电势特点可知AC连线中点电势与BD连线中点电势相等,且电势大小为两端点电势和的一半,即?A+?C = ?B +?D,则?D=9V;
(2)连接A、C,用E、F将AC三等分,可知F点电势为3V,连接BF,则BF为等势线。作AG┴BF,则AG为电场线。现求BF长度。由余弦定理得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020059581778.png)
又由于△ABF面积为△ABC的
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002005958383.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250020059891129.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002005989666.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006020740.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002006036606.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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