题目内容
如图所示是匀强电场中的一组等势面.若A、B、C、D相邻两点间的距离都是2cm,则该电场的场强为 V/m,到A点距离为1.5cm的P点电势为 V.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241441428542218.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241441428542218.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144142870633.png)
A、B两点沿场强方向的距离
d=ABsin60°=0.02×
,
所以E=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241441430261555.png)
UPB=-EPBsin60°=-577.4×0.005×
V="-2.5" V
即φP-φB="-2.5" V,φP="-2.5" V
d=ABsin60°=0.02×
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144142917815.png)
所以E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241441430261555.png)
UPB=-EPBsin60°=-577.4×0.005×
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144143120447.png)
即φP-φB="-2.5" V,φP="-2.5" V
![](http://thumb2018.1010pic.com/images/loading.gif)
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