题目内容
如图所示,在水平向右场强为E的匀强电场中,有一质量为m、电荷量为q的点电荷从A点由静止释放,仅在由场力的作用下经时间t运动到B点.求.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250009240951491.jpg)
(1)点电荷从A点运动到B点过程中电场力对点电荷做的功;
(2)A、B两点间的电势差。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250009240951491.jpg)
(1)点电荷从A点运动到B点过程中电场力对点电荷做的功;
(2)A、B两点间的电势差。
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924126741.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924110746.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924126741.png)
试题分析:(1)设电荷受到的电场力为F,运动的加速度为a,在t时间内运动的距离为s,电场力对电荷做的功为W,则
F=Eq F=ma a= Eq/m s=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924141544.png)
联立解得W=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924110746.png)
说明:由动量定理,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924219638.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924453780.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924485941.png)
(2)设A、B两点间的电势差为UAB,则UAB=W/q;所以UAB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000924126741.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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