题目内容
(8分)如图所示,一质量为m、带电量为q的金属小球,用绝缘细线悬挂在水平向右的匀强电场中。静止时悬线向右与竖直方向成θ角,已知重力加速为g。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418158103952.jpg)
(1)判断小球带何种电荷
(2)求电场强度E
(3)若把绝缘体细线剪断,则剪断细线瞬间小球的加速度是多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418158103952.jpg)
(1)判断小球带何种电荷
(2)求电场强度E
(3)若把绝缘体细线剪断,则剪断细线瞬间小球的加速度是多大?
(8分)解:(1)向右偏说明电场力向右,所以小球带正电.……(2分)
(2)受力分析:重力,拉力,电场力
……(3分)
(3) 细线剪断之前:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141815919790.png)
细线剪断,F消失,小球受到的合力大小等于F
……(3分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418160442129.jpg)
(2)受力分析:重力,拉力,电场力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141815810678.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141815857661.png)
(3) 细线剪断之前:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141815919790.png)
细线剪断,F消失,小球受到的合力大小等于F
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141815981754.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418160442129.jpg)
本题考查的是匀强电场对电荷的作用问题,由于场强方向向右,带电体所受电场力向右,可知带正电;根据受力平衡可计算出电场力进而求出场强;根据牛顿定律可计算出加速度;
![](http://thumb2018.1010pic.com/images/loading.gif)
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