题目内容
如图所示,光滑绝缘竖直细杆与以正点电荷O为圆心的圆周交于B、C两点。一质量m、带电量为-q的空心小球从杆上A点无初速下落。设AB =" BC" = h,小球滑到B点的速度为
试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241808486042083.jpg)
(1)小球滑至C点的速度;
(2)A、C两点的电势差。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848276523.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241808486042083.jpg)
(1)小球滑至C点的速度;
(2)A、C两点的电势差。
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848931914.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848666638.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848931914.png)
试题分析:(1)BC两点是等式点,从B到C过程中,电场力做功为零,由B至C过程,应用动能定理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241808492751170.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848666638.png)
(2)由A→B应用动能定理
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241808498671274.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824180848931914.png)
点评:关键是知道BC两点处于同一个等势面上
![](http://thumb2018.1010pic.com/images/loading.gif)
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