题目内容
(10分)如图所示,一条长为L的绝缘细线一端固定在O点,另一端拴有一个质量为m的带电小球,将它置于场强大小为E的水平方向的匀强电场中,当小球处在细线与竖直方向的夹角为α的A点时处于平衡状态,问:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017514545175.png)
(1)小球带何种电荷?所带电荷量为多少?
(2)在平衡位置以多大的速度vA释出小球,才能使之在电场中作竖直平面内的完整圆周运动?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017514545175.png)
(1)小球带何种电荷?所带电荷量为多少?
(2)在平衡位置以多大的速度vA释出小球,才能使之在电场中作竖直平面内的完整圆周运动?
(1)正电;
;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751485880.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751470765.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751485880.png)
试题分析:(1)小球受重力mg、电场力Eq、线的拉力T作用,如图可知小球受电场力与电场方向平行,故小球带正电。 (1分)
Eq=mgtanα,所以:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751501852.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017515176487.png)
(2)简化处理,将复合场(重力场和电场)等效为重力场,小球在等效重力场中所受重力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751532466.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751548976.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017515631099.png)
小球在A点处于平衡状态,若小球在A点以速度VA开始绕O点在竖直平面内作圆周运动,若能通过延长线上的B点(等效最高点)就能做完整的圆周运动,在B点根据向心力公式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751579915.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751595408.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001751610635.png)
又因仅重力、电场力对小球做功,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017516411097.png)
由以上二式解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017516412171.png)
(其它解法也可以,各种表达自行酌情给分)
![](http://thumb2018.1010pic.com/images/loading.gif)
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