题目内容
如图所示,电量Q为固定的正点电荷,A、B两点在Q的正上方和Q相距分别为 h和0.25 h,将另一点电荷从 A点由静止释放,运动到B点时速度正好又变为零。若此电荷在A点处的加速度大小为
,试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154300810367.png)
(1)此电荷在B点处的加速度。
(2)A、B两点间的电势差(用Q和h表示)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154300717477.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154300810367.png)
(1)此电荷在B点处的加速度。
(2)A、B两点间的电势差(用Q和h表示)
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154301060860.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154300935924.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154301060860.png)
试题分析:(1)这一电荷必为正电荷,设其电荷量为q,由牛顿第二定律,
在A点时
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241543012321076.png)
在B点时
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241543013411259.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154301466510.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154300935924.png)
(2)从A到B过程,由动能定理
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241543017151096.png)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824154301060860.png)
点评:难度中等,注意电场力与重力做功的特点,只与初末位置有关,再利用W=qU处理问题时,要注意W和U的下脚标
![](http://thumb2018.1010pic.com/images/loading.gif)
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