题目内容
(本题12分)如图所示的装置中,平行板电场中有一质量为m,带电量为q的小球,用长L的细线拴住后在电场中处于平衡状态(即平衡位置),此时线与竖直方向的夹角为θ,两板间的距离为d,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032364752042.png)
(1)小球带何种电荷?
(2)两板间的电势差是多少?
(3)把线拉成竖直向下的位置,放手后小球到达平衡位置时的速度为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032364752042.png)
(1)小球带何种电荷?
(2)两板间的电势差是多少?
(3)把线拉成竖直向下的位置,放手后小球到达平衡位置时的速度为多大?
(1)负电 (2)
(3) ![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032365221127.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003236506954.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032365221127.png)
试题分析:(1)由小球处于平衡状态,根据受力分析可知小球受电场力方向水平向左,小球带负电 (2分)
(2)设此时绳子拉力为T,板间电场强度为E,由平衡条件得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003236537660.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003236553682.png)
又由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003236584534.png)
由上述①②③式可解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003236506954.png)
(3)从放手到平衡位置,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032366151362.png)
由上述①②③⑤可解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250032365221127.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目