题目内容
带电量为 1.0×10-2库的粒子,在电场中先后飞经A、B两点,飞经A点时的动能为10焦耳,飞经B点时的动能为4 0焦耳。已知A点的电势为一700伏,求:(l)电荷从A到B电场力做多少功?(2)带电粒子在A点的电势能是多少?(3) B点的电势是多少?
(l) 30J (2) -7J (3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336464673.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336464673.png)
带电粒子仅在电场力作用下,电场力做正功,电势能减小,动能增加,所以
(1) 电场力做功为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336495728.png)
(2) 在A点的电势能为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336526759.png)
(3) 动能的增加量等于电势能的减小量,故B点的电势能为
,
,联立解得:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336464673.png)
(1) 电场力做功为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336495728.png)
(2) 在A点的电势能为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336526759.png)
(3) 动能的增加量等于电势能的减小量,故B点的电势能为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336542774.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336573646.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124336464673.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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