题目内容
(7分)如图所示,用30cm的细线将质量为4×10-3㎏的带电小球P悬挂在O点下,当空中有方向为水平向右,大小为1×104N/C的匀强电场时,小球偏转37°后处在静止状态。(
)(取g=10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447464914004.png)
(1)分析小球的带电性质?
(2)求小球的带电量?
(3)如果剪断细绳小球的加速度?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144746397913.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447464914004.png)
(1)分析小球的带电性质?
(2)求小球的带电量?
(3)如果剪断细绳小球的加速度?
(1)正电 (2)
(3)12.5 m/s2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144746600561.png)
试题分析:(1)小球受力如图,故带正电。(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447466634437.png)
(2)由图知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144746725773.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241447473801233.png)
(3)绳剪断后,合力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144747458861.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824144747521533.png)
点评:本题学生要作出小球的受力分析图,据图去分析各力之间的关系。
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目