题目内容
(16分)如图所示,一质量为
的带电小球,用长为
的绝缘细线悬挂在场强大小为E、方向水平向右的匀强电场中,静止时悬线与竖直方向成
角,重力加速度大小为g.
(1)指出小球所带电荷的电性;
(2)求小球所带电荷量的大小;
(3)若将细线突然剪断,求小球运动时的加速度大小.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241229285713840.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928478346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928493285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928524302.png)
(1)指出小球所带电荷的电性;
(2)求小球所带电荷量的大小;
(3)若将细线突然剪断,求小球运动时的加速度大小.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241229285713840.png)
(1)小球带负电 (2)
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928587850.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928618657.png)
:(1)小球受到的电场力向左,与场强方向相反;故小球带负电荷.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241229286654613.png)
(2)对小球受力分析,受重力、电场力和拉力,如图,根据共点力平衡条件,有:
qE="mgtanθ"
则所带电量为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928587850.png)
(3)剪短细线后,小球受到重力和电场力,合力恒定,如图,故做初速度为零的匀加速直线运动;则:
F=
=ma则小球的加速度为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928618657.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241229286654613.png)
(2)对小球受力分析,受重力、电场力和拉力,如图,根据共点力平衡条件,有:
qE="mgtanθ"
则所带电量为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928587850.png)
(3)剪短细线后,小球受到重力和电场力,合力恒定,如图,故做初速度为零的匀加速直线运动;则:
F=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928758592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122928618657.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目