题目内容
如图,悬挂在O点的一根不可伸长的绝缘细线下端有一个带电量不变的小球甲。在两次实验中,均缓慢移动另一带同种电荷的小球乙。当乙到达悬点O的正下方并与甲在同一水平线上,甲处于受力平衡时,悬线偏离竖直方向的角度为θ,若两次实验中乙的电量分别为q1和q2, θ分别为30°和45°。求:q2 / q1的值( 要求在右图中对甲进行受力分析)。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241221206806474.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241221206806474.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120696494.gif)
设甲球带电量为Q ,甲球受重力、静电斥力和线的拉力作用,它的受力分析如右所示图。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241221207122645.jpg)
设线长为l , 甲乙两球相距为r ,而由图中几何关系可知:
r = l sin θ ①
由题意,甲球处于平衡状态,甲球受静电
斥力:
F = mg tan θ ②
又由库仑定律有:
F =
③
经①②③整理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120743830.gif)
进一步整理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120883800.gif)
由于两次实验中乙的电量分别为q1和q2,使得悬线偏离竖直方向的角度为θ分别为θ1=30°和
θ2=45°,其余量相同,所以有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120946990.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412212094673.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241221207122645.jpg)
设线长为l , 甲乙两球相距为r ,而由图中几何关系可知:
r = l sin θ ①
由题意,甲球处于平衡状态,甲球受静电
斥力:
F = mg tan θ ②
又由库仑定律有:
F =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120727448.gif)
经①②③整理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120743830.gif)
进一步整理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120883800.gif)
由于两次实验中乙的电量分别为q1和q2,使得悬线偏离竖直方向的角度为θ分别为θ1=30°和
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412212093072.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122120946990.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412212094673.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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