题目内容
(8分)如图所示,绝缘细线长为L,最大张力为T,一端固定于O点,另一端连接一带电荷量为q,质量为m的带正电小球,若T>mg,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447236532131.jpg)
(1)若要小球保持静止,可在O点放置一带电量为多少的点电荷;
(2)若要小球静止时细线与竖直方向成α角,需叠加匀强电场的最小电场强度E为多大.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447236532131.jpg)
(1)若要小球保持静止,可在O点放置一带电量为多少的点电荷;
(2)若要小球静止时细线与竖直方向成α角,需叠加匀强电场的最小电场强度E为多大.
(1)Q2≤
(2)E=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723746760.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723699873.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723746760.png)
试题分析:(1)若放置负电荷,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723902659.png)
可得Q1≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723965706.png)
若放置正电荷,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144724027654.png)
可得Q2≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723699873.png)
(2)由受力分析,有 qE=mgsinα
可得E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144723746760.png)
点评:当处于平衡状态的物体受到三个力,一个力是恒力,第二个力方向不变,第三个力大小与方向都发生变化时可以利用矢量三角形法即图解法求解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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