题目内容
(12分)两根长均为L的绝缘细线下端各悬挂质量均为m的带电小球A和B,带电量分别为+q和―q。若加上水平向左的场强为E的匀强电场后,使连接AB的长也为L的绝缘细线绷紧,且两球均处于平衡状态,如图所示,则匀强电场的场强大小E应满足什么关系?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250018050113680.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250018050113680.jpg)
E≥
mg/3q+kq/L2。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001805026344.png)
试题分析:取A为研究对象,设OA线拉力为T1,AB线拉力为T2,
经受力分析得:水平方向 T1cos600 + T2+FBA="F" (3分)
竖直方向 T1sin600 ="mg" (3分)
由此可得Eq=mgtg300 + T2+kq2/L2,
故场强的大小E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001805026344.png)
因为绝缘细线绷紧,故T2≥0,所以E≥
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001805026344.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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