题目内容
如图所示,光滑绝缘细杆竖直放置,它与以正电荷Q为圆心的某圆交于B、C两点,质量为m、带电荷量-q的有孔小球从杆上A点无初速度下滑,已知q
Q,AB=h,小球滑到B点时的速度大小为
求:
(1)小球由A到B的过程中电场力做的功;
(2)A、C两点的电势差
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241146373592296.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637343211.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637343290.gif)
(1)小球由A到B的过程中电场力做的功;
(2)A、C两点的电势差
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241146373592296.jpg)
(1)小球由A到B的过程中电场力做功![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637375389.gif)
(2)A、C两点电势差为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637375389.gif)
(2)A、C两点电势差为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
因为Q是点电荷,所以以Q为圆心的圆面是一个等势面,这是一个重要的隐含条件。由A到B过程中电场力是变力,所以不能直接用W=Fs来解,只能考虑应用功能关系。
(1)因为杆是光滑的,所以小球从A到B过程中只有两个力做功:电场力的功WAB和重力的功mgh,由动能定理得:
WAB+mgh=
,
代入已知条件vB=
得电场力做功
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637437603.gif)
(2)因为B、C在同一个等势面上
所以φB=φc,即UAB=UAC
由 WAB=qUAB
得 UAB=UAC=
= ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
故A、C两点电势差为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
(1)因为杆是光滑的,所以小球从A到B过程中只有两个力做功:电场力的功WAB和重力的功mgh,由动能定理得:
WAB+mgh=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637406310.gif)
代入已知条件vB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637343290.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637437603.gif)
(2)因为B、C在同一个等势面上
所以φB=φc,即UAB=UAC
由 WAB=qUAB
得 UAB=UAC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637468298.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
故A、C两点电势差为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114637390323.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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