题目内容
如图所示,A、B、C为匀强电场中的三点,构成边长为a的等边三角形,场强为E,方向平行于ABC平面,已知电子从A运动到B时,动能增加
;质子从A运动到C时动能减少
,则该匀强电场的场强E为____________,方向___________。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241751078957482.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175107427392.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175107646458.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241751078957482.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175108129770.png)
试题分析:根据动能定理可得:
A→B过程:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175108410725.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175108660788.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175108894515.png)
A→C过程:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175109097799.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175109331929.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175109549526.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175109814629.png)
由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175110064818.png)
根据电场线与等势线垂直,由高电势处指向低电势处,则得知,电场强度E的方向垂直BD并由C指向A.
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175110313846.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824175108129770.png)
点评:本题根据动能定理求出电势差,找出等势点,根据电场线与等势面之间的关系判断电场强度的方向,是经常采用的思路.
![](http://thumb2018.1010pic.com/images/loading.gif)
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