题目内容
如图所示,真空中
点处固定一点电荷
,同时在
点通过绝缘细线悬挂一带电荷量为
质量为
的小球,开始时细线与小球处在水平位置且静止,释放后小球摆到最低点时,细线的拉力为
,则固定电荷
在最低点
处产生的场强大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241507202061686.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719192292.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719301333.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719192292.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719426310.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719519337.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719629483.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719301333.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150719941309.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241507202061686.jpg)
A.![]() | B.![]() | C.![]() | D.![]() |
C
试题分析:小球从A到B过程,受重力、拉力和静电力,只有重力做功,运用动能定理得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150720736772.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150720799588.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150720877805.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150721001518.png)
故B滴点的场强为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150721079771.png)
点评:本题关键是对小球从A到B过程运用动能定理列式求出最低点速度,然后对最低点位置运用牛顿第二定律和向心力公式求解静电力,最后根据场强定义得到电场强度.
![](http://thumb2018.1010pic.com/images/loading.gif)
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